\section{Introduction}
	\subsection{Statement of Results for Bernoulli Convolutions}

	The main result of this is paper is to give a sufficient condition for a self similar measure to be absolutely continuous. For simplicity we will first state this result in the case of Bernoulli convolutions. First we need to define Bernoulli convolutions.

	\begin{definition}[Bernoulli Convolution]
		Given some $\lambda \in (0,1)$ we define the Bernoulli convolution with parameter $\lambda$ to be the law of the random variable $Y$ given by,
		\begin{equation*}
		Y = \sum_{i=0}^{\infty} X_n \lambda^n.
		\end{equation*}
		Where each of the $X_n$ are independent i.i.d. random variables which have probability $\frac{1}{2}$ of being $1$ and probability $\frac{1}{2}$ of being $-1$. We will denote this measure by $\mu_{\lambda}$.
	\end{definition}

	Bernoulli convolutions are the most well studied examples of self-similar measures which are an important object in fractal geometry. We will dsicuss these further in Section \ref{litreview}. Despite much effort it is still not known for which $\lambda$ the Bernoulli convolution with parameter $\lambda$ is absolutely continuous. The results of this paper contribute towards answering this question.

	\begin{definition}[Mahler Measure]
		Given some algebraic number $\alpha_1$ with conjugates $\alpha_2, \alpha_3, \dots, \alpha_n$ whose minimal polynomial (over $\mathbb{Z}$) has leading coefficient $C$ we define the Mahler measure of $\alpha_1$ to be,
		\begin{equation*}
		M_{\alpha_1} = |C| \prod_{i=1}^{n}  \max \{|\alpha_i|, 1\}
		\end{equation*}
	\end{definition}

	\begin{theorem} \label{mainbernoulli}
		Suppose that $\lambda \in \left( \frac{1}{2}, 1 \right)$ is algebraic and has Mahler measure $M_{\lambda}$. Suppose also that $\lambda$ is not the root of a non zero polynomial with coefficients $0, \pm 1$ and,
		\begin{equation*}
		(\log M_{\lambda} - \log 2) (\log M_{\lambda})^2 < \frac{1}{27} (\log M_{\lambda} - \log \lambda^{-1})^3 \lambda^4 \label{mainbernoullieqn}
		\end{equation*}
		then the Bernoulli convolution with parameter $\lambda$ is absolutely continuous.
	\end{theorem}

	This is a corollary of a more general statement about a more general class of self similar measures which we will dicuss in Section \ref{intro:generalIFS}. The requirement \eqref{mainbernoullieqn} is equivalent to $M_{\lambda} < F(\lambda)$ where $F:(\frac{1}{2}, 1) \to \mathbb{R}$ is some strictly increasing continuous function satisfying $F(\lambda) > 2$. Figure \ref{fgraph} displays the graph of $F$.

	\begin{figure}[h]
			\centering
			\includegraphics[scale=0.8]{fgraph}
			\caption{The graph of $F$} \label{fgraph}
	\end{figure}

	It is worth noting that $F(\lambda) \to 2^{\frac{27}{26}} \approx 2.054$ as $\lambda \to 1$. The fact that $F(\lambda) > 2$ is important because the box packing principal means that the requirement that $\lambda$ is not the root of a polynomial with coefficients $\pm 1$ forces $M_{\lambda} \geq 2$. It is also worth noting that this result is stronger than the result obtained by Garsia in \cite{garsiasep} where he showed that if $M_{\lambda} = 2$ and $\lambda \in (\frac{1}{2}, 1)$ then the Bernoulli convolution with parameter $\lambda$ is absolutely continuous.

	In Figure \ref{numexamples} there are some examples of $\lambda$ satisfying the conditions of Theorem \ref{mainbernoulli} found by a simple computer search.
	\begin{figure}[h]
	\begin{center}
		\begin{tabular}{ | m{19em} | c| c| }
			\hline
			Minimal Polynomial & Mahler Measure & $\lambda$\\
			\hline
			$X^{7}-X^{6}-2X^{5}-X^{2}+X+1$ & 2.010432 & 0.879161\\
			$X^{7}+2X^{6}-X-1$ & 2.015159 & 0.932864\\
			$X^{8}+2X^{7}-1$ & 2.007608 & 0.860582\\
			$X^{9}-2X^{8}-X+1$ & 2.003861 & 0.799533\\
			$X^{9}+2X^{8}-X-1$ & 2.003861 & 0.949560\\
			$X^{10}-2X^{9}-X^{2}+1$ & 2.005754 & 0.852579\\
			$X^{10}-2X^{9}-X+1$ & 2.001940 & 0.813972\\
			$X^{10}-X^{9}-2X^{8}-X^{5}+X^{4}+X^{3}-X^{2}+1$ & 2.014180 & 0.911021\\
			$X^{10}-X^{9}-X^{8}-2X^{7}-X^{5}+X^{4}+X^{2}+1$ & 2.012241 & 0.939212\\
			$X^{10}-X^{9}-X^{8}-X^{7}-2X^{6}-X^{5}+X^{3}+X^{2}+X+1$ & 2.017567 & 0.953949\\
			$X^{10}-2X^{8}-3X^{7}-2X^{6}-X^{5}+X^{3}+2X^{2}+2X+1$ & 2.008264 & 0.968459\\
			$X^{10}+X^{9}-2X^{8}+X^{7}+X^{6}-X^{5}+X^{4}-X^{3}+X-1$ & 2.016061 & 0.875809\\
			$X^{10}+2X^{9}-X^{4}-1$ & 2.030664 & 0.946934\\
			$X^{10}+2X^{9}-1$ & 2.001936 & 0.888810\\
			$X^{10}+3X^{9}+3X^{8}+3X^{7}+2X^{6}-2X^{4}-3X^{3}-3X^{2}-2X-1$ & 2.047156 & 0.984474\\
			\hline
		\end{tabular}
	\end{center}
	\caption{Examples of $\lambda$ for which the Bernoulli convolution with parameter $\lambda$ can be shown to be absolutely continuous using the methods of this paper} \label{numexamples}
	\end{figure}

	The smallest value of $\lambda$ which we were able to find for which the Bernoulli convolution with parameter $\lambda$ can be shown to be absolutely continuous using these method is  $\lambda \approx 0.799533$ with minimal polynomial $X^9 - 2 X^8 - X + 1$. This is much smaller than the examples given in  \cite{varjupaper} the smallest of which was $\lambda = 1 - 10^{-50 }$. We also show that for all $n \geq 13$ the Bernoulli convolution with parameter given by the root of the polynomial $X^n - 2X^{n-1} - X + 1$ in $\left( \frac{1}{2}, 1 \right)$ is absolutely continuous. Finally we give some explicit examples of absolutely continuous self similar measures in $\mathbb{R}^2$ which cannot be expressed as the product of absolutely continuous measures of $\mathbb{R}$.

	\subsection{Review of Existing Literature} \label{litreview}
	For a thorough survey on Bernoulli convolutions see \cite{firstrev} or \cite{secondrev}. For a review of the most recent developments see \cite{varjulitrev}. We will only discuss the case of unbiased Bernoulli convolutions in this review. Bernoulli convolutions were first introduced by Jessen and Winter in \cite{intobaper}. They have also been studied by Erd\H{o}s. In \cite{erdosnotcont} Erd\H{o}s showed that the Bernoulli convolution with parameter $\lambda$ is not absolutely continuous whenever $\lambda^{-1} \in (1,2)$ is a Pisot number. In his proof he exploited the property of Pisot numbers that powers of Pisot numbers approximate integers exponentially well. These are currently the only values of $\lambda \in \left( \frac{1}{2}, 1 \right)$ for which the Bernoulli convolution with parameter $\lambda$ is known not to be absolutely continuous.

	The typical behaviour for Bernoulli convolutions with parameters in $\left( \frac{1}{2}, 1 \right)$ is absolute continuity. In \cite{erdosabscon} by a beautiful combinatorial argument Erd\H{o}s showed that there is some $c < 1$ such that for almost all $\lambda \in (c,1)$ the Bernoulli convolution with parameter $\lambda$ is absolutely continuous. This was extended by Solomyak in \cite{solomyak} to show that we may take $c=\frac{1}{2}$. This was later extended by Smerkin in \cite{firstsmherkin} where he showed that the set of exceptional parameters has Hausdorff dimension $0$. These results have been further extended by for example Shmerkin in \cite{shmerkin} who showed that for every $\lambda \in \left( \frac{1}{2}, 1 \right)$ apart from an exceptional set of zero Hausdorff dimension the Bernoulli convolution with parameter $\lambda$ is absolutely continuous with density in $L^q$ for all finite $q>1$.

	In a ground breaking paper \cite{hochman} Hochman made progress on a related problem by showing that if $\lambda \in (\frac{1}{2}, 1)$ is algebraic and not the root of a polynomial with coefficients $-1, 0, +1$ then the Bernoulli convolution with parameter $\lambda$ has dimension $1$. Much of the progress in the last decade builds on the results of Hochman.

	There are relatively few known explicit examples of $\lambda$ for which the Bernoulli convolution with parameter $\lambda$ is absolutely continuous. It can easily be shown that for example the Bernoulli convolution with parameter $2^{-\frac{1}{k}}$ is absolutely continuous when $k$ is a positive integer. This is because it may be written as the convolution of the Bernoulli convolution with parameter $\frac{1}{2}$ with another measure. In \cite{garsiasep} Garsia showed that if $\lambda \in (\frac{1}{2}, 1)$ has Mahler measure $2$  then the Bernoulli convolution with parameter $\lambda$ is absolutely continuous. It is worth noting that the condition that $\lambda$ has Mahler measure $2$ implies that $\lambda$ is not the root of a polynomial with coefficients $0 \pm 1$. This means that my result is very similar to the result of Garsia. There has also been recent progress in this area by Varj\'u in \cite{varjupaper}. In his paper he showed that providing $\lambda$ is sufficiently close to $1$ depending on the Mahler measure of  $\lambda$ the Bernoulli convolution with parameter $\lambda$ is absolutely continuous. The techniques we will use in this paper are similar in many ways to those used by Varj\'u however, we introduce several crucial new
	ingredients. Perhaps the most important innovation of this paper is the quantity,
	which we call the ``detail of a measure around a scale'', which we will use in place
	of entropy. We will discuss this further in Section 1.4.

	\subsection{Results for More General IFS} \label{intro:generalIFS}

	In this Section we will discuss how the results of this paper apply to a more general class of iterated function system.

	\begin{definition}[Iterated Function System]
		Given some $n, d \in \mathbb{N}$, some continuous bijective functions $S_1, S_2, ..., S_n : \mathbb{R}^d \to \mathbb{R}^d$ and a probability vector $\mathbf{p} = (p_1, p_2, \dots, p_n)$ we say that $F = (n, d, (S_i)_{i=1}^n, \mathbf{p})$  is an iterated function system.
	\end{definition}

	\begin{definition}[Self similar measure]
		Given some iterated function system $F = (n, d, (S_i)_{i=1}^n, (p_i)_{i=1}^n)$ in which all of the $S_i$ are similarities with Lipsitz constant less than $1$ we say that a probability measure $\mu$ is a self-similar measure generated by $F$ if,
		\begin{equation*}
		\mu = \sum_{i=1}^n p_i \mu \circ S_i ^{-1}.
		\end{equation*}
	\end{definition}

	It is a result of J. Hutchinson in \cite{hutchinsonpaper} section 3.1 part 5 that under these conditions there is a unique self similar measure given by some iterated function system.

	Given an iterated function system satisfying these conditions let $\mu_F$ denote the unique self similar measure generated by $F$.

	This paper will only deal with a very specific class of iterated function systems. Specifically we care about the case where there is some orthogonal transformation $U$, some constant $\lambda \in (0, 1)$ and vectors $a_1, \dots, a_n \in \mathbb{R}^d$ such that for all $i = 1, \dots, n$ we have,
	\begin{equation*}
	S_i: x \mapsto \lambda U x + a_i.
	\end{equation*}

	I will describe such iterated function systems and the self similar measures generated by them as having a uniform contraction ratio with uniform rotation. It is easy to see from the definitions that self-similar measures with uniform rotation and uniform contraction ratio can be expressed in the following way.

	\begin{lemma}
		Let $F = (n, d, (S_i)_{i=1}^n, (p_i)_{i=1}^n)$ be an iterated function system with uniform contraction ratio and uniform rotation. Let $\lambda \in (0,1)$, let $U$ be an orthonormal transformation and let $x_1, \dots, x_n \in \mathbb{R}^d$ be vectors such that,
		\begin{equation*}
		S_i : x \mapsto\lambda U x +a_i.
		\end{equation*}
		Let $X_0, X_1, X_2, \dots$ be i.i.d. random variables such that $\mathbb{P}[X_0 = a_i] = p_i$ for $i=1,\dots,n$ and let,
		\begin{equation*}
		Y = \sum_{i=0}^{\infty} \lambda^i U^i X_i.
		\end{equation*}
		Then the law of $Y$ is $\mu_F$.
	\end{lemma}

	\begin{definition}
		We define the $k$-step support of an iterated function system $F$ to be given by,
		\begin{equation*}
		V_{F,k} := \left\{ S_{j_1} \circ S_{j_1}  \circ \dots \circ S_{j_k} (0) : j_1, j_2, \dots, j_k \in \{ 1, 2, ..., n\}\right\}.
		\end{equation*}
	\end{definition}
	\begin{definition}
		Let $F$ be an iterated function system. We define the seperation of $F$ after $k$ steps to be,
		\begin{equation*}
		\Delta_{F, k} := \inf \{ |x-y| : x, y \in V_{F,k}, x \neq y \}.
		\end{equation*}
	\end{definition}

	\begin{definition}
		Given an iterated function system $F$ let the splitting rate of $F$, which we will denote by $M_F$, be defined by,
		\begin{equation*}
		M_F := \limsup \limits_{k \to \infty} \left(\Delta_{F, k} \right) ^{-\frac{1}{k}}.
		\end{equation*}
	\end{definition}

	\begin{definition}
		With $F$ and $X_i$ defined as above let $h_{F,k}$ be defined by,
		\begin{equation*}
		h_{F,k} := H\left( \sum_{i=0}^{k-1} X_i \right).
		\end{equation*}
		Here $H(\cdot)$ denotes the Shannon entropy.
	\end{definition}

	\begin{definition}[Garsia Entropy]
		Given an iterated function system $F$ with uniform contraction ratio and uniform rotation define the \emph{Garsia entropy} of $F$ by,
		\begin{equation*}
		h_F := \liminf_{k \to \infty} \frac{1}{k} h_{F, k}.
		\end{equation*}
	\end{definition}

	I now have the notation necessary to describe the main result.

	\begin{theorem} \label{mainresult}
		Let $F$ be an iterated function system on $\mathbb{R}^d$ with uniform contraction ratio and uniform rotation. Suppose that $F$ has Garsia entropy $h_F$, splitting rate $M_F$, and uniform contraction ratio $\lambda$. Suppose further that,
		\begin{equation*}
		(d \log M_F - h_F)(\log M_F)^2 < \frac{1}{27} (\log M_F - \log \lambda^{-1})^3 \lambda^4.
		\end{equation*}
		Then the self similar measure $\mu_F$ is absolutely continuous.
	\end{theorem}

	We will give examples of self similar measures which can be shown to be absolutely continuous using these methods in Section \ref{examples}.

	\subsection{Outline of Proof}
	The main innovation of  this paper is to use a different quantity for measuring the smoothness of a measure at a given scale. Before defining this quantity we will need the following definition.
	\newtheorem*{definition:spherical_normal}{Definition \ref{spherical_normal}}
	\begin{definition:spherical_normal}
		Given an integer $d \in \mathbb{N}$ and some $y>0$ let $n_y^{(d)}$ be the density function of the multivariate normal distributon with covariance matrix $yI$ and mean $0$. Specifically let,
		\begin{equation*}
		n_y^{(d)}(x) = (2 \pi y)^{-\frac{d}{2}} e^{-\frac{1}{2y} \sum_{i=1}^d x_i^2},
		\end{equation*}
		where the value of $d$ is clear from context we will usually just write $n_y$.
	\end{definition:spherical_normal}

	The quantity we use to measure the smoothness at a given scale is the following.
	\newtheorem*{definition:detail}{Definition \ref{detail}}
	\begin{definition:detail}[Detail Around a Scale]
		Given a compactly supported probability measure $\mu$ on $\mathbb{R}^d$ and some $r>0$ we define the \emph{detail of $\mu$ around scale $r$} by,
		\begin{equation*}
		s_r(\mu) :=  r^2 \frac{ \Gamma \left( \frac{d}{2} \right)}{2} \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \left| \left| \left. \mu * \frac{\partial}{\partial y} n_y \right|_{y=r^2} \right| \right|_1.
		\end{equation*}
	\end{definition:detail}

	Here the constant $r^2 \frac{ \Gamma \left( \frac{d}{2} \right)}{2} \left( \frac{d}{2e} \right) ^{-\frac{d}{2}}$ is choosen so that $s_r(\mu) \in (0,1]$ for all probability measures $\mu$ on $\mathbb{R}^d$. The main advantage of this quantity is that it allows much better quantative estimates for the increase in smoothness under convolution than those given in \cite{varjupaper}. Specifically we prove the following;

	\begin{theorem} \label{manyconv}
		Let $n \in \mathbb{N}$, $K>1$ and suppose that there are measures $\mu_1, \mu_2, \dots,$ $ \mu_n$ on $\mathbb{R}^d$ and postive real numbers $0 < \alpha_1 \leq \alpha_2 \leq \dots \leq \alpha_n < C_{K, d}^{-1}$. Let $m = \frac{\log n}{\log \frac{3}{2}}$ and let $r>0$. Suppose that for all $t \in \left[ 2^{-\frac{m}{2}}r, K^m \alpha_1^{-m 2^m }r \right]$ and $i \in \{1, 2, ..., n\}$ we have $s_t(\mu_i) \leq \alpha_i$. Then we have,
		\begin{equation*}
		s_r(\mu_1 * \mu_2 * \dots * \mu_n) \leq C_{K, d} ^ {n-1} \alpha_1 \alpha_2 \dots \alpha_n .
		\end{equation*}
	\end{theorem}

	Where $C_{K, d}$ is some constant depending only on $K$ and $d$ which is defined in Section 2. This is much stronger than the equivalent theorem given in \cite{varjupaper} in which P. Varj\'u proves that,

	\begin{equation*}
	(1 - H(\mu * \tilde{\mu} ; r |2r) ) \leq C \alpha^2 \left( \log \alpha^{-1} \right)^3
	\end{equation*}

	where $C = 10^8$,  $\alpha \in (0, \frac{1}{2})$ satisfies,

	\begin{equation*}
	\alpha \geq \sup_{\nu \in \{\mu, \tilde{\mu}\}, t \in [\alpha^3r, \alpha^{-3}r]}1- H(\nu : t | 2t)
	\end{equation*}

	and $1 - H(\mu ; r |2r) $ is a quantity defined in \cite{varjupaper} which measures how smooth a measure is at scale $r$. We will typically apply Theorem \ref{manyconv} in the case $K \to \infty$. It is worth noting that $C_{K, 1} \to \frac{8}{\Gamma \left( \frac{1}{2} \right)} \left( \frac{1}{2 e} \right) ^ {\frac{1}{2}} \approx 1.93577$. This is a much better constant than the one obtained in \cite{varjupaper}.

	Later in Section 2 we will show that if $s_r(\mu) \to 0$ sufficiently quickly then $\mu$ is absolutely continuous. Specifically we will show,

	\newtheorem*{lemma:isabscont}{Lemma \ref{isabscont}}
	\begin{lemma:isabscont}
		Suppose that $\mu$ is a measure on $\mathbb{R}^d$ and that there exists some constant $\beta>1$ such that for all sufficiently small $r>0$ we have,
		\begin{equation*}
		s_r(\mu) < \left( \log r^{-1} \right) ^{-\beta},
		\end{equation*}
		then $\mu$ is absolutely continuous.
	\end{lemma:isabscont}

	In order to show that a self similar measure $\mu_F$ is absolutely continuous we will express it as a convolution of  many measures and using Theorem \ref{manyconv} show that detail around a scale decreases under convolution. In the case of Bernoulli convolutions this is done by noting that the Bernoulli convolution with parameter $\lambda$ is the law of the random varaible $Y$ given by,

	\begin{equation*}
	Y = \sum_{i=0}^{\infty} \pm \lambda^i.
	\end{equation*}

	With each $\pm$ being indpendent with probability $\frac{1}{2}$ of being positive or negative. We then let,
	\begin{equation*}
	Y_j = \sum_{i=m_j}^{m_{j+1}-1} \pm \lambda^i
	\end{equation*}
	for $j = 0, \dots, n$ for some $n$. We also take $m_{n+1} = \infty$ and $m_0 = 0$. We then let $\nu_j$ be the law of $Y_j$ and note that the Bernoulli convolution with parameter $\lambda$ which I will denote by $\mu_{\lambda}$ is given by,
	\begin{equation*}
	\mu_{\lambda} = \nu_0 * \dots * \nu_n
	\end{equation*}
	To make this argument work we need to find some upped bound for the detail of the measures $\nu_j$. In the case of Bernoulli convolutions this is done by using entropy and a result dating back to at least Garsia in Lemma 1.51 in \cite{garsiasep}. This can be generalized to fit the requirements of the more general case. Specifically the two results we will need are the following,
	\newtheorem*{theorem:entropytodetail}{Theorem \ref{entropytodetail}}
	\begin{theorem:entropytodetail}
		Let $\mu$ and $\nu$ be compactly supported probability measures on $\mathbb{R}^d$ let $r, u$ and $v$ be positive real numbers such that $r^2=u+v$. Then,
		\begin{equation*}
		s_r(\mu * \nu)\leq \frac{r^{-2}}{2} \Gamma \left( \frac{d}{2} \right)  \left( \frac{d}{2e} \right) ^{-\frac{d}{2}}
		\sqrt{ \frac{\partial }{\partial u} H(\mu * n_u) \frac{\partial }{\partial v} H(\nu * n_v) }.
		\end{equation*}
	\end{theorem:entropytodetail}
	And,
	\newtheorem*{lemma:garsia}{Lemma \ref{garsia}}
	\begin{lemma:garsia}
		Let $\lambda$ be an algebraic number,  let $M_{\lambda}$ be it's Mahler measure and let $l$ denote the number of conjugates of $\lambda$ which have modulus $1$. Then there is some $c_{\lambda} > 0$ such that the following holds. Suppose that $p$ is a polynomial of degree $n$ with coefficients $-1, 0, 1$ and that $p(\lambda) \neq 0$. Then,
		\begin{equation*}
		\left| p(\lambda) \right| > c_{\lambda} n^l M_{\lambda}^{-n}.
		\end{equation*}
	\end{lemma:garsia}
	We use these to lemmas to get an upper bound on the detail of $\nu_j$ around some given scales and then using Theorem \ref{manyconv} we can get a bound on the detail of $\mu_{\lambda}$ around some given scale.

	One of the ways in which our method differs from that used by Varju in \cite{varjupaper} is that we  do not have a way of dealing with the ``low entropy" regime and instead use this result straight after finding the initial measures with small entropy given by Theorem \ref{entropytodetail}. This means that our method is currently unable to deal with parameters with large Mahler measure.

	\subsection{Structure of the Paper}
	In Section 2 of the paper we introduce a new quantity for measuring the smoothness of a measure at a given scale which we call the detail around a scale and give some of its properties. In Section 3 we give some basic properties of entropy which we will use later in the paper. In section 4 we prove an upper bound for the detail around a scale in terms of some quantity involving entropy.

	In Section 5 we give a bound for the gap in entropy using the Garsia entropy and the initial seperation. In Section 6 we prove the main results of the paper using a combinatorial argument and the results of the previous sections to express the self-similar measure as the convolution of many measures each of which have at most some detail around a given scale. Finally in Section 7 we give some explicit examples of self-similar measures which may be shown to be absolutely continuous using the methods of this paper.

	\subsection{Notation}
	Here we will summarize some of the commonly used notation in the paper.

	\newcommand{0.5cm}{0.5cm}
	\begin{center}
		\begin{tabular}{ m{4em}  |m{28em} }

			\hline
			$\mathbb{N}$ & Natural numbers $\{1, 2, \dots \}$.\\
			& \\
			$M_{\lambda}$ & The Mahler measure of the algebraic number $\lambda$.\\
			& \\
			$F$ & An iterated function system often given as $F=(n, d, (S_i)_{i=1}^n,  (p_i)_{i=1}^n)$ with the iterated function system being an iterated function system on $\mathbb{R}^d$, $S_i$ being similarities and $(p_i)_{i=1}^n$ being a probability vector.\\
			& \\
			$V_{F, k}$ & The $k$-step support of an iterated function system $F$.\\
			& \\
			$\Delta_{F,k}$ & $\inf \{|x-y| : x, y \in V_{F,k}, x \neq y \}$.\\
			& \\
			$M_F$  & $\limsup \limits_{k \to \infty} (\Delta_{F, k}) ^ {- \frac{1}{k}}$.\\
			& \\
			$\left| \left| \cdot \right| \right|_1$ & The $L^1$ norm of a signed measure defined by $\left| \left| \mu\right| \right|_1 = \sup \limits_{A \subset \mathbb{R}^d} \left(|\mu(A)| +|\mu(A^C)|\right) $.\\
			& \\
			$n_y^{(d)}$ & The density function of the normal distribution on $\mathbb{R}^d$ with mean zero and covariance matrix $yI$. The $(d)$ will often be ommitted where the value of $d$ is clear from context.\\
			& \\
			$N_{\Sigma}$ & The density function of the multivariate normal distribution with mean $0$ and covariance matrix $\Sigma$.\\
			& \\
			$s_r(\cdot)$ & The detail of a measure around a scale. See Definition \ref{detail} in Section 2.\\
			& \\
			$\mathcal{B}(\mathbb{R}^d)$ & The Borel $\sigma$-algebra on $\mathbb{R}^d$.\\
			& \\
			$H(\cdot)$ & The Shannon entropy. This can represent both the differential Shannon entropy and the discrete Shannon entropy. Which one will be clear from context.\\

			\hline
		\end{tabular}
	\end{center}

	\section{Detail around a Scale}
	In this section we will discuss some properties of detail around a scale as defined in the previous section. The main results will be proving Lemmas \ref{isabscont} and \ref{t2} as well as Theorem \ref{manyconv}.

	\begin{definition} \label{spherical_normal}
		Given an integer $d \in \mathbb{N}$ and some $y>0$ let $n_y^{(d)}$ be the density function of the multivariate normal distributon with covariance matrix $yI$ and mean $0$. Specifically let,
		\begin{equation*}
		n_y^{(d)}(x) = (2 \pi y)^{-\frac{d}{2}} e^{-\frac{1}{2y} \sum_{i=1}^d x_i^2}
		\end{equation*}
		where the value of $d$ is clear from context we will usually just write $n_y$.
	\end{definition}

	\begin{lemma}
		Let $\mu$ and $\nu$ be probability measures. Then we have,
		\begin{equation*}
		\left| \left| \mu * \nu * \frac{\partial}{\partial y} n_y \right| \right|_1 \leq \left| \left| \nu * \frac{\partial}{\partial y} n_y \right| \right|_1
		\end{equation*}
		in particular,
		\begin{equation}
		\left| \left| \mu *  \frac{\partial}{\partial y} n_y \right| \right|_1 \leq \left| \left|  \frac{\partial}{\partial y} n_y \right| \right|_1 = \frac{1}{y} \frac{2}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} . \label{maxderivative}
		\end{equation}
	\end{lemma}

	\begin{proof}
		For the first part simply write the measure $\nu * \frac{\partial}{\partial y} n_y $ as $\nu * \frac{\partial}{\partial y} n_y = \tilde{\nu}_+ - \tilde{\nu}_-$ where $\tilde{\nu}_+$ and $\tilde{\nu}_-$ are (non-negative) measures with disjoint support. Note that this means,
		\begin{equation*}
		\left| \left| \nu * \frac{\partial}{\partial y} n_y \right| \right|_1 = \left| \left| \tilde{\nu}_+ \right| \right|_1 + \left| \left| \tilde{\nu}_- \right| \right|_1
		\end{equation*}
		and so,
		\begin{eqnarray}
		\left| \left| \mu * \nu * \frac{\partial}{\partial y} n_y \right| \right|_1 \nonumber
		& = & \left| \left| \mu * \tilde{\nu}_+ - \mu * \tilde{\nu}_-\right| \right|_1 \nonumber\\
		& \leq & \left| \left| \mu \right| \right|_1 \left| \left| \tilde{\nu}_+ \right| \right|_1 + \left| \left| \mu \right| \right|_1 \left| \left| \tilde{\nu}_- \right| \right|_1\nonumber \\
		& = & \left| \left| \nu * \frac{\partial}{\partial y} n_y \right| \right|_1.\nonumber
		\end{eqnarray}

		 For the second part we need to compute,
		\begin{equation*}
		\int_{\mathbf{x} \in \mathbb{R}^d} \left| \frac{\partial}{\partial y} n_y \right| \, d \mathbf{x} .
		\end{equation*}
		To do this we work in polar coordinates. Let $r = \sqrt{ \sum_{i=1}^d x_i^2}$. Then we have,
		\begin{equation*}
		\frac{\partial}{\partial y} n_y = \left( \frac{r^2}{2y^2} - \frac{d}{2y} \right) (2 \pi y) ^ {-\frac{d}{2}} \exp(- \frac{r^2}{2y}) .
		\end{equation*}
		Noting that the $(d-1)$- dimensional Lesbegue measure of $S^{(d-1)}$ is $\frac{2 \pi^{\frac{d}{2}}}{\Gamma \left( \frac{d}{2} \right) }$ we get,
		\begin{eqnarray} \nonumber
		\int_{\mathbf{x} \in \mathbb{R}^d} \left| \frac{\partial}{\partial y} n_y \right| \, d \mathbf{x} &=& \frac{2 \pi^{\frac{d}{2}}}{\Gamma \left( \frac{d}{2} \right) } \left( -\int_{r=0}^{\sqrt{dy}} \left( \frac{r^2}{2y^2} - \frac{d}{2y} \right) (2 \pi y) ^ {-\frac{d}{2}} r^{d-1}\exp(- \frac{r^2}{2y}) \,dr  \right.\\
		&+& \left. \int_{\sqrt{dy}}^{\infty} \left( \frac{r^2}{2y^2} - \frac{d}{2y} \right) (2 \pi y) ^ {-\frac{d}{2}} r^{d-1}\exp(- \frac{r^2}{2y}) \,dr
		\right) .\nonumber
		\end{eqnarray}
		By differentiation it is easy to check that,
		\begin{equation*}
		\int \left( \frac{r^2}{y} - d \right) r^{d-1} e^{- \frac{r^2}{2y}} \, dr = - r^d e^{- \frac{r^2}{2y}} .
		\end{equation*}
		Hence,
		\begin{eqnarray} \nonumber
		\lefteqn{  -\int_{r=0}^{\sqrt{dy}} \left( \frac{r^2}{2y^2} - \frac{d}{2y} \right) (2 \pi y) ^ {-\frac{d}{2}} r^{d-1}\exp(- \frac{r^2}{2y}) \,dr} \\ \nonumber
		& + & \int_{\sqrt{dy}}^{\infty} \left( \frac{r^2}{2y^2} - \frac{d}{2y} \right) (2 \pi y) ^ {-\frac{d}{2}} r^{d-1}\exp(- \frac{r^2}{2y}) \,dr \\ \nonumber
		& = &  \frac{1}{2y} (2 \pi y) ^ {-\frac{d}{2}} 2 (dy) ^ {\frac{d}{2}} e^{-\frac{d}{2}}\\
		& = & \frac{1}{y} (2 \pi)^{-\frac{d}{2}} d^{\frac{d}{2}} e^{-\frac{d}{2}} .\nonumber
		\end{eqnarray}
		Meaning that,
		\begin{equation*}
		\int_{\mathbf{x} \in \mathbb{R}^d} \left| \frac{\partial}{\partial y} n_y \right| \, d \mathbf{x} = \frac{1}{y} \frac{2}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} .
		\end{equation*}
	\end{proof}

	\begin{definition}[Detail Around a Scale] \label{detail}
		Given a compactly supported probability measure $\mu$ on $\mathbb{R}^d$ and some $r>0$ we define the \emph{detail of $\mu$ around scale $r$} by,
		\begin{equation*}
		s_r(\mu) :=  r^2 \frac{ \Gamma \left( \frac{d}{2} \right)}{2} \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \left| \left| \left. \mu * \frac{\partial}{\partial y} n_y \right|_{y=r^2} \right| \right|_1 .
		\end{equation*}
		Note that this means $s_r(\mu) \in (0,1]$
	\end{definition}

	We am now almost ready to prove Lemma \ref{t2} and Theorem \ref{manyconv}. First we will need the following result.

	\begin{lemma} \label{xtoy}
		Let $y>0$. Then we have,
		\begin{equation*}
		\frac{1}{2} \bigtriangleup n_y = \frac{\partial}{\partial y} n_y
		\end{equation*}
		where $\bigtriangleup$ denotes the Laplacian,
		\begin{equation*}
		\bigtriangleup = \sum_{i=1}^d \frac{\partial^2}{\partial x_i^2} .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		This is just a simple computation. Simply note that,
		\begin{equation*}
		\frac{\partial}{\partial x_i} n_y = -\frac{x_i}{y} n_y
		\end{equation*}
		and so,
		\begin{equation*}
		\frac{1}{2} \sum_{i=1}^d \frac{\partial^2}{\partial x_i^2} n_y = \frac{1}{2}\left( \frac{|x|^2}{y^2} - \frac{d}{y} \right) n_y = \frac{\partial}{\partial y} n_y .
		\end{equation*}
	\end{proof}

	\begin{lemma}
		Let $\mu$ and $\nu$ be probability measures and let $y>0$. Then,
		\begin{equation*}
		\left| \left| \mu * \nu * \frac{\partial}{\partial y} n_y \right| \right|_1 \leq 2 \int_{\frac{y}{2}}^{\infty} \left| \left| \mu * \frac{\partial}{\partial v} n_v \right| \right|_1 \left| \left| \nu * \frac{\partial}{\partial v} n_v \right| \right|_1 \, dv .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		First note that,
		\begin{equation*}
		\left| \left| \mu * \nu * \frac{\partial}{\partial y} n_y \right| \right|_1  \leq \int_y^{w} \left| \left| \frac{\partial}{\partial u} \left( \mu * \nu * \frac{\partial}{\partial u} n_u\right) \right| \right|_1  \, du + \left| \left| \mu * \nu * \frac{\partial}{\partial w} n_w \right| \right|_1 .
		\end{equation*}
		Taking $w \to \infty$ and using \eqref{maxderivative} this gives,
		\begin{equation*}
		\left| \left| \mu * \nu * \frac{\partial}{\partial y} n_y \right| \right|_1
		 \leq  \int_y^{\infty} \left| \left| \frac{\partial}{\partial u} \left( \mu * \nu * \frac{\partial}{\partial u} n_u\right) \right| \right|_1  \, du .
		\end{equation*}
		We can then use Lemma \ref{xtoy} and standard properties of the convolution of distributions to move the derivatives around as follows,
		\begin{eqnarray}
		\frac{\partial}{\partial y} \left( \mu * \nu * \frac{\partial}{\partial y} n_y \right) & = & \frac{1}{2} \frac{\partial}{\partial y} \left( \mu * \nu * \bigtriangleup n_y \right) \nonumber\\
		& = & \frac{1}{2} \frac{\partial}{\partial y} \left( \mu * \nu * n_{y-a} *\bigtriangleup n_a \right) \nonumber\\
		& = & \frac{1}{2} \left(\mu * \frac{\partial}{\partial y} n_{y-a} \right) * \left( \nu * \bigtriangleup n_a \right) .\nonumber
		\end{eqnarray}
		Letting $a= \frac{1}{2}y$ and applying Lemma \ref{xtoy} again this gives,
		\begin{equation*}
		\frac{\partial}{\partial y} \left( \mu * \nu * \frac{\partial}{\partial y} n_y \right)  =
		\left(\mu * \left. \frac{\partial}{\partial u} n_{u} \right|_{u=\frac{y}{2}} \right) *
		\left(\nu * \left. \frac{\partial}{\partial u} n_{u} \right|_{u=\frac{y}{2}} \right) .
		\end{equation*}
		Where this is understood to mean that they are representitives of the same distribution and so in particular they have the same $L^1$ norm. This means,
		\begin{eqnarray}
		\lefteqn{\left| \left| \mu * \nu * \frac{\partial}{\partial y} n_y \right| \right|_1 } \nonumber\\
		& \leq & \int_y^{\infty} \left| \left| \frac{\partial}{\partial u} \left( \mu * \nu * \frac{\partial}{\partial u} n_u\right) \right| \right|_1  \, du\nonumber \\
		& = & \int_y^{\infty} \left| \left| \left(\mu * \left. \frac{\partial}{\partial v} n_{v} \right|_{v=\frac{u}{2}} \right) *
		\left(\nu * \left. \frac{\partial}{\partial v} n_{v} \right|_{v=\frac{u}{2}} \right) \right| \right|_1 \, du \label{intv1} \nonumber\\
		& \leq & \int_y^{\infty} \left| \left| \mu * \left. \frac{\partial}{\partial v} n_{v} \right|_{v=\frac{u}{2}} \right| \right|_1 \left| \left|
		\nu * \left. \frac{\partial}{\partial v} n_{v} \right|_{v=\frac{u}{2}} \right| \right|_1 \, du\nonumber\\
		& = & 2 \int_{\frac{y}{2}}^{\infty} \left| \left| \mu * \frac{\partial}{\partial v} n_v \right| \right|_1 \left| \left| \nu * \frac{\partial}{\partial v} n_v \right| \right|_1 \, dv \nonumber
		\end{eqnarray}
		as required.
	\end{proof}

	\begin{lemma} \label{t2}
		Let $\mu_1$ and $\mu_2$ be probability measures on $\mathbb{R}^d$, let $r>0$, $\alpha_1,\alpha_2>0$ and let $K>1$. Suppose that for all $t \in [r/\sqrt{2}, K\alpha_1^{-\frac{1}{2}} \alpha_2^{-\frac{1}{2}}r]$ and for all $i\in \{1, 2\}$ we have,
		\begin{equation*}
		s_r(\mu_i) \leq \alpha_i
		\end{equation*}
		then,
		\begin{equation*}
		s_r(\mu_1*\mu_2) \leq C_{K.d} \alpha_1 \alpha_2
		\end{equation*}
		where,
		\begin{equation*}
		C_{K,d} = \frac{8}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} \left( 1 + \frac{1}{2K^2} \right) .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		We have,
		\begin{eqnarray}
		s_r(\mu_1*\mu_2)  & = & r^2 \frac{ \Gamma \left( \frac{d}{2} \right)}{2} \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \left| \left| \left. \mu *\nu *  \frac{\partial}{\partial y} n_y \right|_{y=r^2} \right| \right|_1\nonumber\\
		& \leq & r^2 \Gamma \left( \frac{d}{2} \right) \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \int_{\frac{r^2}{2}}^{\infty} \left| \left| \mu * \frac{\partial}{\partial v} n_v \right| \right|_1 \left| \left| \nu * \frac{\partial}{\partial v} n_v \right| \right|_1 \, dv\nonumber\\
		& \leq & r^2 \frac{4}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} \int_{\frac{r^2}{2}}^{\infty} v^{-2} \alpha_1 \alpha_2 \, dv\nonumber\\
		& + & r^2 \frac{4}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} \int_{K^2\alpha_1^{-1}\alpha_2^{-1}r^2}^{\infty} v^{-2} \, dv\nonumber\\
		& = & \frac{8}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} \left( 1 + \frac{1}{2K^2} \right) \alpha_1 \alpha_2.\nonumber
		\end{eqnarray}
	\end{proof}

	When we apply this lemma we will do it in the case where $K$ is large and the only important property of $C_{K,d}$ is that $C_{K,d} \to \frac{8}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} $ as $K \to \infty$. In the case $d=1$ we get $\frac{8}{ \Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} \approx 1.93577$. This result is in many ways similar to Theorem 2 in \cite{varjupaper} though is much stronger as it does not include any logorithm factors and has a much smaller constant (the constant given in \cite{varjupaper} is $10^8$). We will now prove Theorem \ref{manyconv}

	\newtheorem*{theorem:manyconv}{Theorem \ref{manyconv}}
	\begin{theorem:manyconv}
		Let $n \in \mathbb{N}$, $K>1$ and suppose that there are measures $\mu_1, \mu_2, \dots,$ $ \mu_n$ on $\mathbb{R}^d$ and postive real numbers $0 < \alpha_1 \leq \alpha_2 \leq \dots \leq \alpha_n < C_{K, d}^{-1}$. Let $m = \frac{\log n}{\log \frac{3}{2}}$ and let $r>0$. Suppose that for all $t \in \left[ 2^{-\frac{m}{2}}r, K^m \alpha_1^{-m 2^m }r \right]$ and $i \in \{1, 2, ..., n\}$ we have $s_t(\mu_i) \leq \alpha_i$. Then we have,
		\begin{equation*}
		s_r(\mu_1 * \mu_2 * \dots * \mu_n) \leq C_{K, d} ^ {-n+1} \alpha_1 \alpha_2 \dots \alpha_n .
		\end{equation*}
	\end{theorem:manyconv}

	\begin{proof}
		I will prove this by induction. The case $n=1$ is trivial. Suppose that $n>1$. Let $n' = \left\lceil \frac{n}{2} \right\rceil$ and let $m' = \frac{\log n'}{\log \frac{3}{2}}$. Define $\nu_1, \nu_2, ..., \nu_{n'}$ and $\beta_1, \beta_2, ..., \beta_{n'}$ as follows. For $i=1, 2, ..., \left \lfloor \frac{n'}{2} \right \rfloor$ let $\nu_i = \mu_{2i-1} * \mu_{2i}$ and $\beta_i = C_{K, d} \alpha_{2i-1} \alpha_{2i}$ and if $n$ is odd let $\nu_{n'} = \mu_n$ and $\beta_{n'} = \alpha_n$. Note that $C_{K,d} ^ {n'-1} \beta_1 \beta_2 \dots \beta_{n'} = C_{K,d}^{n-1} \alpha_1 \alpha_2 \dots \alpha_n$ so we need to show that $n'$, $\left(\nu_i \right)_{i=1}^{n'}$ and $\left(\beta_i \right)_{i=1}^{n'}$ satisfy the inductive hypothesis.

		I want to use Lemma \ref{t2}. This means that we need to show that if $t \in \left[ 2^{-\frac{m'}{2}}r, K^{m'} \beta_1^{-m'2^{m'} }r \right]$ and $q \in \left[ 2^{-\frac{1}{2}} t, K \alpha_{2i-1} ^{-\frac{1}{2}}\alpha_{2i}^{-\frac{1}{2}} t \right]$ then $q \in  \left[ 2^{-\frac{m}{2}}r, K^m \alpha_1^{-m 2^m }r \right]$. It is sufficient to show that,
		\begin{equation*}
		\left[ 2^{-\frac{m'+1}{2}} r, K^{m'+1} \beta_1^{-m'2^{m'} } \alpha_1^{-1} r\right] \subset \left[ 2^{-\frac{m}{2}}r, K^m \alpha_1^{-m 2^m }r \right] .
		\end{equation*}
		Note that $m' +1 \leq m$ so $2^{-\frac{m'+1}{2}}r \geq 2^{-\frac{m}{2}}r$. Also we have,
		\begin{eqnarray}
		K^{m'+1} \beta_1^{-m'2^{m'} } \alpha_1^{-1} r & \leq & K^{m} \left( \alpha_1^{2} \right) ^{-m' 2^{m'}} \alpha_1^{-1}r\nonumber\\
		& = & K^m \alpha_1^{-1 - m' 2^{m'+1}}r\nonumber\\
		& \leq & K^m \alpha_1^{-m 2^{m}}r\nonumber
		\end{eqnarray}
		as required.
	\end{proof}
	It is worth noting that the only properties of $m$ we have used are that $m\geq1$ when $n>1$ and that $m\geq m'+1$. This means that it is possible to choose $m$ such that $m\sim \log _2 n$. It turns out that this doesn't make any difference to the bound in Theorem \ref{mainresult}.

	\begin{lemma} \label{isacont}
		Let $\mu$ be a probaility measure on $\mathbb{R}^d$ and let $y>0$. Suppose that,
		\begin{equation} \label{normdecay}
		\int_{0^+}^y \left| \left| \mu * \frac{\partial}{\partial u} n_u \right| \right|_1 \, du < \infty
		\end{equation}
		then $\mu$ is absolutely continuous.
	\end{lemma}

	\begin{proof}
		The condition \eqref{normdecay} implies that the sequence $\mu * n_u$ is Cauchy as $u \to 0$ wrt $\left| \left| \cdot \right| \right|_1$. Since the space $L^1$ is complete this means that there is some absolutely continuous measure $\tilde{\mu}$ such that $\mu * n_u \to \tilde{\mu}$ with respect to $L^1$ as $u \to 0$.

		Suppose for contradiction that $\tilde{\mu} \neq \mu$. The set of open cuboids is a $\pi$-system generating $\mathcal{B}(\mathbb{R}^d)$ so this means that there exists real numbers $a_1, \dots, a_d$ and $b_1, \dots, b_d$ such that,
		\begin{equation*}
		\mu((a_1, b_1) \times \dots \times (a_d, b_d)) \neq \tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d)) .
		\end{equation*}

		There are now two cases to deal with. First we will deal with the case where,
		\begin{equation*}
		\mu((a_1, b_1) \times \dots \times (a_d, b_d)) > \tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d)) .
		\end{equation*}
		This means that there is some $\varepsilon > 0 $ such that,
		\begin{equation*}
		\mu((a_1+\varepsilon , b_1-\varepsilon ) \times \dots \times (a_d+\varepsilon , b_d-\varepsilon )) >\tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d)) +\varepsilon .
		\end{equation*}
		We now consider $\mu*(n_u|_{B_{\varepsilon}})$ where $B_r$ is the ball center $0$ with radius $r$. We have,
		\begin{eqnarray}
		\lefteqn{\left(\mu* n_u \right) ((a_1, b_1) \times \dots \times (a_d, b_d))}\nonumber\\
		& \geq & \left(\mu*(n_u|_{B_{\varepsilon}})\right) ((a_1, b_1) \times \dots \times (a_d, b_d))\nonumber\\
		& \geq & \left| \left| n_u|_{B_{\varepsilon}} \right| \right|_1 \mu((a_1+\varepsilon , b_1-\varepsilon ) \times \dots \times (a_d+\varepsilon , b_d-\varepsilon )) \nonumber\\
		& \geq & \left| \left| n_u|_{B_{\varepsilon}} \right| \right|_1(\tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d)) +\varepsilon)\nonumber\\
		& \to & \tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d)) +\varepsilon\nonumber
		\end{eqnarray}

		as $u \to 0$. This contradicts the requirement,
		\begin{equation*}
		\left(\mu* n_u \right) ((a_1, b_1) \times \dots \times (a_d, b_d)) \to \tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d))
		\end{equation*}
		as $u \to 0$.

		The case,
		\begin{equation*}
		\mu((a_1, b_1) \times \dots \times (a_d, b_d)) < \tilde{\mu}((a_1, b_1) \times \dots \times (a_d, b_d))
		\end{equation*}
		is almost identical.

		Together these cases show that $\mu = \tilde{\mu}$ and so in particular $\mu$ is absolutely continuous.
	\end{proof}

	\begin{lemma} \label{isabscont}
		Suppose that $\mu$ is a measure on $\mathbb{R}^d$ and that there exists some constant $\beta > 1$ such that for all sufficiently small $r>0$ we have,
		\begin{equation*}
		s_r(\mu) < \left( \log r^{-1} \right) ^{-\beta}
		\end{equation*}
		Then $\mu$ is absolutely continuous.
	\end{lemma}
	\begin{proof}
		This follows immediately from Lemma \ref{isacont}.
	\end{proof}

	\section{Proporties of Entropy}
	Here we will discuss some basic properties of entropy which we will later use to bound $s_r(\cdot)$. We will be studying the differential entropy of quantities like $\mu * n_y$ for some compactly supported measure $\mu$. None of the results in this section are particularly significant and the purpose of this section is more to lay the ground work for the next section.
	\begin{definition}
		Let $\mu$ be a probability measure which is absolutely continuous with respect to the Lesbegue measure and has density function $f$. Then we define the entropy of $\mu$ to be,
		\begin{equation*}
		H(\mu) := \int_{\mathbb{R}^d} - f(x) \log f(x) \, dx .
		\end{equation*}
		Here we define $0 \log 0 = 0$
	\end{definition}
	\begin{definition}
		Let $\mu$ be an atomic measure. Then $H(\mu)$ will denote the Shannon entropy of $\mu$.
	\end{definition}
	Using $H$ to denote two different quantities shouldn't cause confusion as we will only be using measures which are either purely atomic or absolutely continuous.
	\begin{definition}[Finite Entropy]
		Given some probability measure $\mu$ on $\mathbb{R}^d$ we say that $\mu$ has \emph{finite entropy} if $\mu$ is absolutely continuous and has a density function $f$ such that the function $f \, \log \, f$ is in $L^1$.
	\end{definition}
	\begin{lemma} \label{convincrease}
		Let $\mu$ and $\nu$ be probability measures on $\mathbb{R}^d$ such $\mu$ is absolutely continous with respect to the Lesbegue measure and has finite entropy. Then,
		\begin{equation*}
		H(\mu * \nu) \geq H(\mu) .
		\end{equation*}
	\end{lemma}
	\begin{proof}
		This follows from Jensen's inequatity. Define the function,
		\begin{align*}
		h  :  [0, \infty) &\to  \mathbb{R} \\
		x & \mapsto x \log x\\
		\end{align*}
		where as above we take $0 \log 0 = 0$. Note that $h$ is concave. Let $f$ be a probability density function for $\mu$. We then have,
		\begin{eqnarray}
		H(\mu * \nu) &=& -\int_{- \infty}^{\infty} h \left( \int_{-\infty}^{\infty} f(y-x) \,\nu(dx) \right) \, dy\nonumber\\
		& \geq & -\int_{- \infty}^{\infty} \int_{-\infty}^{\infty} h(f(y-x)) \, \nu(dx) \, dy \nonumber\\
		& =& -\int_{- \infty}^{\infty} \int_{-\infty}^{\infty} h(f(y-x)) \, dy \, \nu(dx) \nonumber\\
		& = & \int_{- \infty}^{\infty} H(\mu) \, \nu(dx)\nonumber\\
		& = & H(\mu) .\nonumber
		\end{eqnarray}
	\end{proof}

	\begin{definition}[Rapidly Decaying Measure]
		I will decribe an absolutely continuous measure $\mu$ as being a \emph{Rapidly Decaying Measure} if it has a density function $f$ such that  there are some constants $C, \varepsilon > 0$ such that for all $x \in \mathbb{R}^d$ we have,
		\begin{equation*}
		f(x) < C \exp \left( -\varepsilon |x|^2 \right).
		\end{equation*}
	\end{definition}

	\begin{proposition}
		Let $X_1$, $X_2$ and $X_3$ be independent absolutely continuous random variables with finite entropy. Then,
		\begin{equation*}
		H(X_1 + X_2 + X_3) + H(X_1) \leq H(X_1 + X_2) + H(X_1 + X_3) .
		\end{equation*}
	\end{proposition}
	\begin{proof}
		This is proven in for example \cite{entropysubmodality} Theorem 1. Note that the author of \cite{entropysubmodality} doesn't explicitly state that the random variables need to have finite entropy but this is implicitly used in the proof.
	\end{proof}
	\begin{corollary} \label{2ndordent}
		Suppose the $\mu_1, \mu_2, \mu_3$ are absolutely continuous, rapidly decaying probability measures which have finite entropy. Then,
		\begin{equation*}
		H(\mu_1 * \mu_2 * \mu_3) + H(\mu_1) \leq H(\mu_1 * \mu_2) + H(\mu_1 * \mu_3) .
		\end{equation*}
	\end{corollary}

	\begin{definition}
		Given some function $f : \mathbb{R} \to \mathbb{R}$ we define the \emph{right hand directional derivative} of $f(x)$ with respect to $x$ by,
		\begin{equation*}
		\frac{\partial}{\partial x^+} f(x) := \lim_{h \to 0^+} \frac{f(x+h) - f(x)}{h} .
		\end{equation*}
		Similarly we define the  \emph{left hand directional derivative} of $f(x)$ with respect to $x$ by,
		\begin{equation*}
		\frac{\partial}{\partial x^-} f(x) := \lim_{h \to 0^-} \frac{f(x+h) - f(x)}{h} .
		\end{equation*}
	\end{definition}

	\begin{lemma} \label{entroprop}
		Let $\mu$ be a rapidly decaying or compactly supported probability measure on $\mathbb{R}^d$. Then the function $\mathbb{R}_{>0} \to \mathbb{R}$, $y \mapsto H(\mu * n_y)$;
		\begin{enumerate}
			\item is increasing \label{inc}
			\item has both directional derivatives everywhere \label{diff}
			\item is continuous \label{cont}
			\item both directional derivatives $\frac{\partial}{\partial y^+} H(\mu * n_y)$  and $\frac{\partial}{\partial y^-} H(\mu * n_y)$ are decreasing as a function of $y$ \label{deder}
		\end{enumerate}
	\end{lemma}
	\begin{proof}
		Part \ref{inc} follows immediately from Lemma \ref{convincrease}
		\paragraph{}
		To prove part \ref{diff} we will show that if $y>0$ and $\epsilon_1, \epsilon_2$ are non-zero real numbers such that $-y < \epsilon_1 < \epsilon_2$ then,
		\begin{equation*}
		\frac{H(\mu * n_{y+\epsilon_1}) - H(\mu*n_y)}{\epsilon_1} \geq \frac{H(\mu * n_{y+\epsilon_2}) - H(\mu*n_y)}{\epsilon_2} .
		\end{equation*}
		To prove this we will start with the case where $\frac{\epsilon_1}{\epsilon_2} \in  \mathbb{Q}$. In this case there are integers $a, b \in \mathbb{Z}$ and some $\epsilon_3 >0$ such that $\epsilon_1 = a \epsilon_3$ and $\epsilon_2 = b \epsilon_3$. We know by substiuting $\mu_1 = \mu * n_{y+n\epsilon_3}, \mu_2 = n_{\epsilon_3}$ and $\mu_3 = n_{(m-n)\epsilon_3}$ into Proposition \ref{2ndordent} that for any $n, m \in \mathbb{Z}$ with $n < m$ and $y + n \epsilon_3 > 0$ we have,
		\begin{equation*}
		H(\mu * n_{y + (n+1) \epsilon_3}) -  H(\mu * n_{y+n\epsilon_3}) \geq H(\mu * n_{y+(m+1)\epsilon_3}) - H(\mu * n_{y+m\epsilon_3})
		\end{equation*}
		summing this means that,
		\begin{equation*}
		|b| \frac{\epsilon_1}{|\epsilon_1|} [H(\mu * n_{y+\epsilon_1}) - H(\mu * n_y)] \geq |a| \frac{\epsilon_2}{|\epsilon_2|} [H(\mu * n_{y+\epsilon_2}) - H(\mu * n_y)]
		\end{equation*}
		as both sides can be as the sum of terms of the form $H(\mu * n_{y + (n+1) \epsilon_3}) -  H(\mu * n_{y+n\epsilon_3})$. Dividing through gives the required result. The case where $\frac{\epsilon_1}{\epsilon_2} \notin \mathbb{Q}$ simply follows from continuity.

		The existence of partial derivates now follows from the fact that the limit required is the limit of a bounded monotonic function and so exists.
		\paragraph{}
		Part \ref{cont} follows immediately from Part \ref{diff}.
		\paragraph{}
		Part \ref{deder} follows from the fact that by Proposition \ref{2ndordent} we have if $0 < y_1 < y_2$ and $\delta > -y_1$ then,

		\begin{equation*}
		\frac{H(\mu * n_{y_1+\delta}) - H(\mu * n_{y_1})}{\delta} \geq \frac{H(\mu * n_{y_2+\delta}) - H(\mu * n_{y_2})}{\delta}
		\end{equation*}

		and so we can take $\delta \to 0$ from either side.

	\end{proof}

	\begin{lemma}
		Given any $0 < a < b$ we have,
		\begin{equation*}
		H(\mu * n_b) - H(\mu * n_a)  = \int_{a}^{b} \frac{\partial}{\partial y^+} H(\mu * n_y) \, dy .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		Define the function,
		\begin{eqnarray}
		t : (a, b) \times (0, \infty) & \to & \mathbb{R}\nonumber\\
		(y, h) & \mapsto & \frac{H(\mu * n_{y+h}) - H(\mu * n_y)}{h}\nonumber
		\end{eqnarray}
		clearly by the arguments of the previous section $t$ is decreasing in $h$. We also have that $t(y , h) \to \frac{\partial}{\partial y^+} H(\mu * n_y) $ as $h \to 0$. This means that by the monotone convergence theorem we have,
		\begin{equation*}
		\int_{a}^{b} t(y, h) \, dy \to \int_{a}^{b} \frac{\partial}{\partial y^+} H(\mu * n_y) \, dy
		\end{equation*}
		as $h \to 0$.

		We also have,
		\begin{equation*}
		\int_{a}^{b} t(y, h) \, dy = \frac{1}{h} \int_a^{a+h} H(\mu * n_y) \, dy - \frac{1}{h} \int_b^{b+h} H(\mu * n_y) \, dy
		\end{equation*}
		by the continuity of $H(\mu * n_y)$ this tends to $ H(\mu * n_b) - H(\mu * n_a) $ as $h \to 0$.
	\end{proof}

	It is worth noting that there is no reason to suspect that $H(\mu * n_y)$ is not for example $C^1$. Indeed it is not too difficult to show that it is continuous diferentiable at all but at most countably many points using arguments similar to those earlier in the section. Proving that $H(\mu * n_y)$ is $C^1$ is not needed for the arguments of this paper and so has not been done.

	\begin{lemma}
		Let $\alpha >0$ and let $T_{\alpha}: \mathbb{R}^d \to \mathbb{R}^d$ be defined by $T_{\alpha}:x \to \alpha x$. Let $\mu$ be an absolutely continuous probability measure with finite entropy. Then we have,
		\begin{equation*}
		H(\mu \circ T_{\alpha}^{-1}) = H(\mu) + d \log \alpha .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		Note that if $f$ is a density function of $\mu$ then then function $\tilde{f}$ defined by,
		\begin{equation*}
		\tilde{f}: x \mapsto \frac{1}{\alpha^d} f \left(\frac{x}{\alpha}\right)
		\end{equation*}
		is a density function for $\mu \circ T_{\alpha}^{-1}$. Therefore we have,
		\begin{eqnarray}
		H(\mu \circ T_{\alpha}^{-1}) & = & -\int_{x \in \mathbb{R}^d} \frac{1}{\alpha^d} f \left(\frac{x}{\alpha}\right)  \log \left( \frac{1}{\alpha^d} f \left(\frac{x}{\alpha}\right)  \right) \, dx\nonumber\\
		& = & -\int_{y \in \mathbb{R}^d} f \left(y\right)  \log \left( \frac{1}{\alpha^d} f \left(y\right)  \right) \, dy\nonumber\\
		& = & H(\mu) + d \log \alpha .\nonumber
		\end{eqnarray}
	\end{proof}

	\begin{corollary}
		Let $\alpha >0$ and let $T_{\alpha}:x \to \alpha x$. Let $\mu$ be a compactly supported probability measure and let $y>0$. Then we have,
		\begin{equation*}
		\left. \frac{\partial}{\partial u} H(( \mu \circ T_{\alpha}^{-1}) * n_{u}) \right|_{u=\alpha ^2y} = \alpha^{-2} \frac{\partial}{\partial y} H(\mu * n_y) .
		\end{equation*}
	\end{corollary}

	\begin{proof}
		Note that we have,
		\begin{equation*}
		H(( \mu \circ T_{\alpha}^{-1}) * n_{\alpha ^2y}) = H(\mu * n_y) +\log \alpha
		\end{equation*}
		and so,
		\begin{equation*}
		\frac{\partial}{\partial y} H(( \mu \circ T_{\alpha}^{-1}) * n_{\alpha ^2 y})  =  \frac{\partial}{\partial y} H(\mu * n_y)
		\end{equation*}
		the result follows from a change of variables.
	\end{proof}

	\begin{definition} \label{nonprobent}
		Given a (not necessarily probability) measure $\mu$ it is convenient to define,
		\begin{equation*}
		H(\mu) := \left| \left| \mu \right| \right|_1 H\left(\frac{\mu}{\left| \left| \mu \right| \right|_1}\right) .
		\end{equation*}
	\end{definition}

	\begin{lemma} \label{concave}
		Let $\mu$ and $\nu$ be measures with finite entropy either both atomic or both absolutely continuous. Then,
		\begin{equation*}
		H( \mu + \nu) \geq H(\mu) + H(\nu) .
		\end{equation*}
	\end{lemma}
	\begin{proof}
		This follows easily from an application of Jensen's inequality to the function $-x \log x$ and Definition \ref{nonprobent}.
	\end{proof}

	\begin{lemma} \label{maxinc}
		Let $k \in \mathbb{N}$, let $\mathbf{p} = (p_1, \dots, p_k)$ be a probability vector and let $\mu_1, \dots, \mu_k$ be absolutely continuous probabiliy measures with finite entropy such that $\left| \left| \mu_i \right| \right|_1 = p_i$. Then,
		\begin{equation*}
		H(\sum_{i=1}^k \mu_i) \leq H(\mathbf{p}) + \sum_{i=1}^k H(\mu_i)
		\end{equation*}
		in particular,
		\begin{equation*}
		H\left(\sum_{i=1}^k \mu_i\right) \leq \sum_{i=1}^k H(\mu_i) + \log k .
		\end{equation*}
	\end{lemma}
	\begin{proof}
		Let the density function of $\mu_i$ be $f_i$ for $i=1, \dots,k$. Let $h:[0, \infty) \to \mathbb{R}$ be defined by $h : x \to - x \log x$ for $x > 0$ and $h(0) = 0$. Note that $h$ satisfies $h(a+b) \leq h(a) + h(b)$. We have,
		\begin{eqnarray}
		H\left(\sum_{i=1}^k \mu_i\right) & = & \int_{- \infty}^{\infty} h(f_1 + \dots + f_k)\nonumber\\
		& \leq & \sum_{i=1}^k\int_{- \infty}^{\infty} h(f_i)\nonumber\\
		&=& H(\mathbf{p}) + \sum_{i=1}^k H(\mu_i) .\nonumber
		\end{eqnarray}
	\end{proof}

	\begin{lemma} \label{sepinc}
		Let $\mu$ and $\nu$ be probability measures of $\mathbb{R}^d$. Suppose that $\mu$ is a discrete measure supposrted on finitely many points with seperation at least $2R$ and that $\nu$ is an absolutely continuous measure with finite entropy whose support is contained in a ball of radius $R$.  Then,
		\begin{equation*}
		H(\mu * \nu) = H(\mu) + H(\nu) .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		Let $n \in \mathbb{N}$, $p_1, p_2, \dots, p_n \in (0, 1)$ and $x_1, x_2, \dots, x_n \in \mathbb{R}^d$ be choosen such that,
		\begin{equation*}
		\mu = \sum_{i = 1}^n p_i \delta_{x_i}
		\end{equation*}

		And let $f$ be the density function of $\nu$. Note that this means that the density function of $\mu * \nu$, which we will denote by $g$,  can be expressed as,

		\begin{equation*}
		g(x)  =
		\begin{cases}
		p_i f(x - x_i)  & |x_i - x| < R \text{ for some } i\\
		0 & \text{Otherwise}\\
		\end{cases}
		\end{equation*}

		This means that,

		\begin{eqnarray}
		H(\mu * \nu) & = & \sum_{i=1}^{n} \int_{B_R(x_i)} - g(x) \log \, g(x) \, dx\nonumber\\
		& = & \sum_{i=1}^{n} \int_{B_R(0)} - p_i f(x) \log \left( p_i f(x) \right) \, dx\nonumber\\
		& = & \sum_{i=1}^{n} \int_{B_R(0)} -  p_i f(x) \log \left( f(x) \right) \, dx\nonumber\\
		& + & \sum_{i=1}^{n} \int_{B_R(0)} -  p_i f(x) \log \left( p_i\right) \, dx\nonumber\\
		& = & H(\mu) + H(\nu)\nonumber
		\end{eqnarray}

	\end{proof}

	\section{Entropy Bound for $s_r(\cdot)$} \label{entropybound2}
	In this section we wish to come up with a bound on the amount of detail around a scale for $\mu$ in terms of directional derivatives of $H(\mu * n_y)$.

	\begin{definition}
		Let $f:[0,1] \to [0,1]$ be defined by $f(\delta) = \frac{1}{2} (1 + \delta) \log (1 + \delta) + \frac{1}{2} (1-\delta) \log (1-\delta)$.
	\end{definition}

	Note that $f$ is convex.

	\begin{lemma} \label{jensondiscrete}
		Let $\mu$ and $\nu$ be two absolutely continuous probability measures on $\mathbb{R}^d$ with density functions $a$ and $b$. Then,
		\begin{equation*}
		f\left(\frac{1}{2}||\mu - \nu ||_1\right) \leq H\left(\frac{\mu + \nu}{2}\right) - \frac{1}{2} H(\mu) - \frac{1}{2} H(\nu) \label{2mb}
		\end{equation*}
	\end{lemma}
	\begin{proof}
		We simply note that,
		\begin{eqnarray}
		\lefteqn{f\left(\frac{1}{2}||\mu - \nu ||_1\right)}\nonumber\\
		&=& f \left( \int \left| \frac{a(x) - b(x)}{a(x) + b(x)} \right| \frac{a(x)+b(x)}{2} \, dx\right)\nonumber\\
		&\leq &\int \frac{a(x) + b(x)}{2} f \left(\left|\frac{a(x) - b(x)}{a(x)+b(x)}\right|\right)\,dx \label{bj}\\
		&=& \int \frac{a+b}{2} \left( \frac{1}{2} \left(1+\frac{a-b}{a+b} \right) \log \left(1+\frac{a-b}{a+b} \right) \right. \nonumber\\
		&& \left.+ \frac{1}{2} \left(1-\frac{a-b}{a+b} \right) \log \left(1-\frac{a-b}{a+b} \right) \right) \, dx \nonumber \\
		& = & \int \frac{a}{2} \log \left( \frac{2a}{a+b} \right) + \frac{b}{2} \log \left( \frac{2b}{a+b} \right) \, dx \nonumber\\
		&=& H\left(\frac{\mu + \nu}{2}\right) - \frac{1}{2} H(\mu) - \frac{1}{2} H(\nu)\nonumber
		\end{eqnarray}
		with \eqref{bj} following from Jensen's inequaility.
	\end{proof}

	\begin{definition}
		Let $t\in \mathbb{R}^d$. Then $B_k^{(t)}$ is the convolution of $k$ copies of $\frac{1}{2} (\delta_t + \delta_{-t})$.
	\end{definition}

	\begin{lemma} \label{binenb}
		Let $n \in \mathbb{N}$, let $\mu$ be a probability measure on $\mathbb{R}$ and let $t \in \mathbb{R}^d$. Then,

		\begin{equation}
		\sum_{k=0}^{n-1} f\left( \frac{1}{2} \left| \left| \mu * B_k^{(t)} * (\delta_t - \delta_{-t}) \right| \right|_1\right) \leq H\left( \mu * B_n^{(t)}\right) - H(\mu)  . \label{eb}
		\end{equation}
	\end{lemma}

	\begin{proof}
		I will prove this by induction. The case $n=1$ follows from putting $\mu * \delta_t$ and $\mu* \delta_{-t}$ into Lemma \ref{jensondiscrete}. For the case $n>1$ we simply note that by putting $\mu * B_{n-1}^{(t)}*\delta_t$ and $\mu * B_{n-1}^{(t)}*\delta_{-t}$ into Lemma \ref{jensondiscrete} we have,
		\begin{equation} \
		f \left( \frac{1}{2}\left| \left| \mu * B_{n-1}^{(t)} * (\delta_t - \delta_{-t}) \right| \right|_1\right)  \leq H(\mu*B_n^{(t)}) - H(\mu*B_{n-1}^{(t)})  .
		\end{equation}
		The result then follows from the inductive hypothesis.
	\end{proof}

	The idea now is to let $t= r/\sqrt{n}$ for some where $r= (r_1, 0 , \dots, 0) \in \mathbb{R}^d$ and consider the limit $n \to \infty$. This will give us an expression in terms of convolutions with normal distributions. The next lemma will be used to deal with the right hand side of \eqref{eb} in this limit.

	\begin{lemma} \label{hconv1}
		Let $\lambda_n$ be a sequence of probability measures on $\mathbb{R}^d$ such that $\lambda_n \to \lambda$ in distribution and let $\mu$ be a probability measure with finite entropy and with Lipshitz density function $g$. Suppose further that there exists constants $C>0$ and $\varepsilon>0$ such that for all $n \in \mathbb{N}$ and for all $x \in \mathbb{R}^d$ we have,

		\begin{equation}
		(\lambda_n * g) (x), g(x), (\lambda * g) (x) \leq C e^{-\varepsilon |x|^2} \label{ub}
		\end{equation}
		then $H(\mu * \lambda_n) \to H(\mu * \lambda)$ as $n \to \infty$.
	\end{lemma}

	\begin{proof}
		First note that,
		\begin{equation*}
		(\lambda_n * g) (x) = \int_{-\infty}^{\infty} g(x-y) \, \lambda_n(dy) .
		\end{equation*}
		Since $g(x-\cdot)$ is Lipshitz, by the Portmanteau Lemma we have that this converges to $\int_{-\infty}^{\infty} g(x-y) \, \lambda(dy)$. This means that for all $x$ we have,
		\begin{equation*}
		(\lambda_n * g) (x) \to (\lambda * g) (x)
		\end{equation*}
		and consequently,
		\begin{equation*}
		- (\lambda_n * g) (x) \log (\lambda_n * g) (x) \to - (\lambda * g) (x) \log ( \lambda * g) (x) .
		\end{equation*}
		We wish to apply the dominated converge theorem. For this we will use \eqref{ub}. First note that if $0 \leq x \leq y$ then,
		\begin{equation*}
		\sup \limits_{x \in [0,y]} |x \log x | =
		\begin{cases}
		\max\{e^{-1}, y \log y\}, & \text{if } y \geq e^{-1} \\
		-y \log y, & \text{otherwise}
		\end{cases}
		.
		\end{equation*}
		This means that we can find $C', R$ such that the function $a: \mathbb{R} \to \mathbb{R}$ defined by,
		\begin{equation*}
		a: x \mapsto
		\begin{cases}
		C', & \text{if } |x| \leq R\\
		C'x^2e^{-\varepsilon x^2} & \text{otherwise}
		\end{cases}
		\end{equation*}
		is a dominating function and so the lemma holds.
	\end{proof}
	We now need to find the limit of the left hand side of \eqref{eb}.

	\begin{definition}[Wasserstein Distance] Let $\mu$ and $\nu$ be two probability  measures on $\mathbb{R}$. Then the first Wasserstein Distance between $\mu$ and $\nu$ is given by,
		\begin{equation*}
		W_1(\mu, \nu) = \inf\limits_{\gamma \in \Gamma} \int_{(x, y) \in \mathbb{R}^d \times \mathbb{R}^d} |x-y| \,d\gamma(x,y)
		\end{equation*}
		Where $\Gamma$ is the set of copplings between $\mu$ and $\nu$.
	\end{definition}

	\begin{lemma}
		Suppose that $f$ is $\alpha$- Lipshitz and bounded, and let $\mu$ and $\nu$ be probability measures. Then for all $x \in \mathbb{R}$ we have,
		\begin{equation*}
		|(\mu * f)(x) - (\nu * f)(x)| \leq \alpha W_1(\mu, \nu) .
		\end{equation*}
	\end{lemma}
	\begin{proof}
		Let $\gamma \in \Gamma$. Note that,
		\begin{eqnarray}
		\lefteqn{|(\mu * f)(x) - (\nu * f )(x)|}\nonumber\\
		&=& \left| \int_{(y,z) \in \mathbb{R}^d\times \mathbb{R}^d} f(x-y) -f(x-z) \, d \gamma(y,z) \right|\nonumber \\
		&\leq & \int_{(y,z) \in \mathbb{R}^d \times \mathbb{R}^d} |f(x-y) -f(x-z)| \, d \gamma(y,z) \nonumber\\
		&\leq & \alpha\int_{(y,z) \in \mathbb{R}^d \times \mathbb{R}^d} |y-z| \, d \gamma(y,z) .\nonumber
		\end{eqnarray}
		Taking the infium of the right hand side gives the required result.
	\end{proof}

	\begin{definition}
		Given some convariance matrix $\Sigma$ let $N_{\Sigma}$ denote the multivariate normal distribution with mean $0$ and covariance matrix $\Sigma$.
	\end{definition}

	\begin{lemma} \label{cor1}
		There is some constant $C$ such that,
		\begin{equation*}
		W_1\left(B_k^{(t)}, N_{kt t^{T}}\right) \leq C \left| \left| t \right| \right|^2
		\end{equation*}
		here $t t^{T}$ denotes the matrix $\Sigma$ with $\Sigma_{ij} = t_i t_j$ and $\left| \left| \cdot \right| \right|$ denotes the standard Euclidian norm.
	\end{lemma}
	\begin{proof}
		This follows trivially from many different non-uniform Berry-Essen bounds. For example it follows from \cite{nonuniformberryesseenerickson} Theorem 1. This Theorem states (amongst other things) that the Wasserstein distance between the sum of some independent mean $0$ random variables and a normal distribution with the same variance is at most a constant multiplied by the sum of the expectations of the absolute values of the cubes of the random variables divided by the variance of the sum. This gives the required bound.
	\end{proof}

	\begin{lemma}
		Let $\mu$ be a probability measure on $\mathbb{R}^d$ with density function $g$ such that $g$ is twice differentiable, the function $a: \mathbb{R} \to \mathbb{R}$ defined by $a(x) = \sup \limits_{y \in B_1(x)} \left| \frac{\partial^2}{\partial y_1^2} g(y)\right|$ is in $L^1$ and $g$ is bounded and Lipshitz. Then there is a constant $C >0$ depending only on $\mu$ such that for any $k \in \mathbb{N}$ and any $t_1 \in (0, 1)$ letting $t=(t_1, 0, \dots, 0)$ we have,
		\begin{equation*}
		\left| \,\, \left| \left| \mu * B_k^{(t)} * (\delta_t - \delta_{-t}) \right| \right|_1-2t _1\left| \left| \frac{\partial g}{\partial x_1} * N_{ktt^{T}} \right| \right|_1     \,\, \right| \leq C t_1^2 .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		First note that,
		\begin{eqnarray}
		\lefteqn{\left| \,\, \left| \left| \mu * B_k^{(t)} * (\delta_t - \delta_{-t}) \right| \right|_1-2t _1\left| \left| \frac{\partial g}{\partial x_1} * N_{ktt^{T}} \right| \right|_1     \,\, \right|}\nonumber\\
		&\leq& \left| \left| ( \mu * (\delta_t - \delta_{-t}) - 2t_1 \frac{\partial g}{\partial x_1} ) * B_k^{(t)} \right| \right| _ 1 + 2t_1 \left| \left| \frac{\partial g}{\partial x_1} * (B_k^{(t)}- N_{ktt^{T}}) \right| \right|_1 .\nonumber
		\end{eqnarray}

		By the mean value theorem we have,
		\begin{equation*}
		g* (\delta_t - \delta_{-t} ) (x) = 2 t_1 \frac{\partial g}{\partial y_1}(u)
		\end{equation*}
		for some $u$ in the line segment between $x-t$ and $x+t$.  This means,
		\begin{equation*}
		\left| \left( g * (\delta_t - \delta_{-t} ) - 2t_1 \frac{\partial g}{\partial x_1} \right) (x) \right| \leq 2t_1^2a(x)
		\end{equation*}

		and so,
		\begin{equation*}
		\left| \left| ( \mu * (\delta_t - \delta_{-t}) - 2t_1 \frac{\partial g}{\partial x_1}) * B_k^{(t)} \right| \right| _ 1 \leq \left| \left| ( \mu * (\delta_t - \delta_{-t}) - 2t_1 \frac{\partial g}{\partial x_1}) \right| \right| _ 1 \leq C t_1^2 .
		\end{equation*}

		We then note that by Lemma \ref{cor1} we have,
		\begin{equation*}
		W_1(B_k^{(t)}, N_{ktt^{T}}) \leq C t_1
		\end{equation*}

		giving,
		\begin{equation*}
		2t_1\left| \left| \frac{\partial g}{\partial x_1} * (B_k^{(t)}- N_{kt^2}) \right| \right|_1 \leq Ct_1^2 .
		\end{equation*}
		Adding these terms gives the required result.
	\end{proof}

	\begin{proposition} \label{longform}
		Let $r>0$ and let $\mu$ be a probability measure with density function $f$ such that,
		\begin{enumerate}
			\item There are $C, \varepsilon >0$ such that $f(x) < C e ^ {-\varepsilon |x| ^2}$
			\item $f$ is twice differentiable
			\item $f$ and $\frac{\partial f}{\partial x_1}f$ are bounded and Lipshitz
			\item The function $g(x) = \sup \limits_{y \in B_1(x)} \left| \frac{\partial^2}{\partial y_1^2} f(y)\right|$ is in $L^1$ .
		\end{enumerate}
		Then we have,
		\begin{equation*}
		\frac{1}{2} \int_0^{r_1^2} \left| \left| \frac{\partial}{\partial x_1} \mu * N_{ye_1} \right| \right|_1^2 \, dy \leq H(\mu * N_{r_1^2e_1}) - H(\mu) .
		\end{equation*}
		Here $e_1$ denota the matrix with $e_{11} = 1$ and all other entries equal to $0$.
	\end{proposition}
	\begin{proof}
		The idea is to fix $r >0$ and then let $t = \frac{r}{\sqrt{n}}$ where $r = (r_1,0,0\dots,0)$ and take the limit $n \to \infty$ in Lemma \ref{binenb}. From Lemma \ref{hconv1} we immediately have that,
		\begin{equation*}
		H(\mu * B_n^{(t)}) - H(\mu) \to H(\mu * n_r^2) - H(\mu) .
		\end{equation*}
		For the other side note that $f(\delta) = \frac{1}{2} \delta^2 + O(\delta^3)$ and so,
		\begin{eqnarray}
		f\left(\frac{1}{2} \left| \left| \mu * B_k^{(t)} * (\delta_{t} - \delta_{-t}) \right| \right|_1\right)\nonumber &=&\frac{1}{2} t_1^2 \left| \left| \frac{\partial}{\partial x_1} \mu * N_{kt_1^2e_1} \right| \right|_1^2 + O(t_1^3)\nonumber\\
		&=& \frac{1}{2} \frac{r_1^2}{n} \left| \left| \frac{\partial}{\partial x_1} \mu * N_{kt_1^2e_1} \right| \right|_1^2 + O(n^{-\frac{3}{2}})\nonumber
		\end{eqnarray}
		with the error term uniform in $k$. This means that,
		\begin{eqnarray}
		\lefteqn{\sum_{k=0}^{n-1} f(\frac{1}{2} \left| \left| \mu * B_k^{(t)} * (\delta_t - \delta_{-t}) \right| \right|_1)}\nonumber\\
		& = & \sum_{k=1}^{n-1} \frac{1}{2} \frac{r_1^2}{n} \left| \left| \frac{\partial}{\partial x_1} \mu * N_{kt_1^2e_1} \right| \right|_1^2 + O(n^{-\frac{1}{2}})\nonumber\\
		& \to & \frac{1}{2} \int_0^{r_1^2} \left| \left| \frac{\partial}{\partial x_1} \mu * N_{ye_1} \right| \right|_1^2 \, dy\nonumber
		\end{eqnarray}
		as required with the last line following from for example the dominated convergence theorem.
	\end{proof}

	\begin{definition}
		Given some $y = (y_1, y_2, ..., y_d) \in \mathbb{R}^d$ let $\tilde{n}_{y}$ be the multivariante normal with diagnol covariance matrix with the entries on the diagonal given by $y$ and zero mean.
	\end{definition}

	\begin{corollary} \label{tmpc}
		Let $y = (y_1, y_2, ..., y_d)$ such that $y_1, ..., y_d >0$ and let $\mu$ be a compactly supported probability measure on $\mathbb{R}^d$. Then,
		\begin{equation*}
		\frac{1}{2} \left| \left| \frac{\partial}{\partial x_i} \mu * \tilde{n}_y \right| \right|_1^2 \leq \frac{\partial}{\partial y_i ^{+}} H(\mu * \tilde{n}_y) .
		\end{equation*}
	\end{corollary}
	\begin{proof}
		This follows from Proposition \ref{longform}.
	\end{proof}
	\begin{corollary} \label{tmpc2}
		Suppose that $y>0$. Then,
		\begin{equation*}
		\frac{1}{2} \sum_{i=1}^d \left| \left| \frac{\partial}{\partial x_i} \mu * n_y \right| \right|_1^2 \leq \frac{\partial}{\partial y} H(\mu * n_y)
		\end{equation*}
	\end{corollary}
	\begin{proof}
		This follows from the chain rule.
	\end{proof}

	\begin{theorem} \label{entropytodetail}
		Let $\mu$ and $\nu$ be compactly supported probability measures on $\mathbb{R}^d$ let $r, u$ and $v$ be positive real numbers such that $r^2=u+v$. Then,
		\begin{equation*}
		s_r(\mu * \nu)\leq \frac{r^{2}}{2} \Gamma \left( \frac{d}{2} \right)  \left( \frac{d}{2e} \right) ^{-\frac{d}{2}}
		\sqrt{ \frac{\partial }{\partial u} H(\mu * n_u) \frac{\partial }{\partial v} H(\nu * n_v) }
		\end{equation*}
	\end{theorem}
	\begin{proof}
		First note that by Corollary \ref{tmpc2} we know that,
		\begin{equation}
		\frac{1}{2} \sum_{i=1}^d \left| \left| \frac{\partial}{\partial x_i} \mu * n_u \right| \right|_1^2 \leq \frac{\partial}{\partial u} H(\mu * n_u) \label{tmpe1}
		\end{equation}
		and,
		\begin{equation}
		\frac{1}{2} \sum_{i=1}^d \left| \left| \frac{\partial}{\partial x_i} \nu * n_v \right| \right|_1^2 \leq \frac{\partial}{\partial v} H(\nu * n_v) \label{tmpe2} .
		\end{equation}
		Letting $y=r^2$ we have that,
		\begin{equation*}
		\frac{\partial^2}{\partial x_i^2} \mu * \nu * n_y = \frac{\partial}{\partial x_i} \mu * n_u  * \frac{\partial}{\partial x_i} \nu * n_v
		\end{equation*}
		meaning,
		\begin{equation*}
		\left| \left| \frac{\partial^2}{\partial x_i^2} \mu * \nu * n_y \right| \right|_1 \leq \left| \left| \frac{\partial}{\partial x_i} \mu * n_u \right| \right|_1 \left| \left| \frac{\partial}{\partial x_i} \nu * n_v \right| \right|_1
		\end{equation*}
		and so by Cauchy-Swartz,
		\begin{equation*}
		\sum_{i=1}^{d} \left| \left| \frac{\partial^2}{\partial x_i^2} \mu * \nu * n_y \right| \right|_1 \leq \left( \sum_{i=1}^d \left| \left| \frac{\partial}{\partial x_i} \mu * n_u \right| \right|_1^2 \right) ^{\frac{1}{2}} \left( \sum_{i=1}^d \left| \left| \frac{\partial}{\partial x_i} \nu * n_v \right| \right|_1^2 \right) ^{\frac{1}{2}} .
		\end{equation*}
		Hence by Lemma \ref{xtoy} and equations \eqref{tmpe1} and \eqref{tmpe2} we have that,
		\begin{equation*}
		\left| \left| \frac{\partial}{\partial y} \mu * \nu * n_y \right| \right|_1 \leq \sqrt{ \frac{\partial }{\partial u} H(\mu * n_u) \frac{\partial }{\partial v} H(\nu * n_v) } .
		\end{equation*}
		The result follows by the definition of detail around a scale.
	\end{proof}

	\section{Bounds on Gap in Entropy} \label{initialgaps}

	In this section we will be finding estimates on the difference in entropy of the convolution of a discrete measure with two different sphereical normal distributions in terms of the variance of the normal distributions, the entropy of the measure, the radius of the support of the measure, and the minimum distance between two distinct points in the support of the measure.
	\begin{lemma} \label{lowerbound2}
		Let $n \in \mathbb{N}$, let $x_1, \dots, x_n \in \mathbb{R}^d$ be such that $|x_i-x_j| \geq 2R$ for $i \neq j$ and let $\mathbf{p} = (p_1, p_2 , \dots, p_n)$ be a probability vector. Let,
		\begin{equation*}
		\mu =  \sum_{i = 1}^n p_i \delta_{x_i}
		\end{equation*}
		then,
		\begin{equation*}
		H(\mu * n_{r^2} ) \geq 	d\log r +  H (\mathbf{p}) -c
		\end{equation*}
		for some constant $c$ depending only on $d$ and the ratio $R/r$.
	\end{lemma}

	\begin{proof}
		Given $k \in \mathbb{N}$ and define,
		\begin{equation*}
		\tilde{n}_k := n_{r^2} | _ {A_{(k-1)R, kR}}
		\end{equation*}
		where $A_{a,b} := \{x \in \mathbb{R}^d : |x| \in [a, b)\}$.
		Given $k \in \mathbb{N}$ and $i \in \mathbb{N}$ such that $i \leq n$ define,
		\begin{equation*}
		\mu_{i,k} := \delta_{x_i} * \tilde{n}_k .
		\end{equation*}
		Given $k \in \mathbb{N}$ and $j \in \mathbb{Z} / k \mathbb{Z}$ define,
		\begin{equation*}
		\nu_{j, k} := \sum_{i \equiv j \,\, ( \text{mod } k)} \delta_{x_i} .
		\end{equation*}
		We  have by Lemma \ref{maxinc},
		\begin{equation*}
		\sum_{j=1}^k H(\nu_{j, k}) \geq H(\mathbf{p}) - \log k .
		\end{equation*}
		Also by Lemma \ref{sepinc},
		\begin{equation*}
		H(\nu_{j,k} * \tilde{n}_k) = \left| \left| \tilde{n}_k \right| \right|_1 H(\nu_{j,k}) + \left| \left| \nu_{j,k} \right| \right|_1 H(\tilde{n}_k)
		\end{equation*}
		meaning that,
		\begin{eqnarray}
		H(\mu * \tilde{n}_k) & = & H \left( \sum_{j=1}^k \nu_{j,k} * \tilde{n}_k \right)\nonumber\\
		& \geq & \sum_{j=1}^k H (\nu_{j,k} * \tilde{n}_k )\nonumber\\
		& \geq & \left| \left| \tilde{n}_k \right| \right|_1 H ( \mathbf{p}) + H(\tilde{n}_k) -  \left| \left| \tilde{n}_k \right| \right|_1 \log k .\nonumber
		\end{eqnarray}
		We now simply wish to sum over $k$. It is not immediately clear that Lemma \ref{concave} holds for series as well as sums so to deal with this given $N \in \mathbb{N}$ we define,
		\begin{equation*}
		\tau_N := n_{r^2} | _{A_{NR, \infty}} .
		\end{equation*}
		This means that we have,
		\begin{equation*}
		H(\mu * n_{r^2}) \geq H(\tau_N) + \sum_{k=1}^N H(\mu*\tilde{n}_k) .
		\end{equation*}
		By Lemma \ref{maxinc} we know that,
		\begin{equation*}
		H(\tau_N) + \sum_{k=1}^N H(\tilde{n}_k) \geq H(n_{r^2}) - c_1
		\end{equation*}
		for some constant $c_1$ depending only on $R/r$.

		This means that,
		\begin{equation*}
		H(\mu * n_{r^2})  \geq \left( \sum_{k=1}^N \left| \left| \tilde{n}_k \right| \right|_1 \right) H(\mathbf{p}) + H(n_{r^2}) -c_1 -  \sum_{k=1}^N \left| \left| \tilde{n}_k \right| \right|_1 \log k .
		\end{equation*}
		The result follows by taking $N \to \infty$ and noting that $H(n_{r^2}) = d \log r + c$.
	\end{proof}

	\begin{definition}
		Given an interval $I \subset \mathbb{R}$ let,
		\begin{equation*}
		\nu_{F}^I := \Conv _{n \in \mathbb{Z} : \lambda^n \in I} \nu_F\circ U^{-n} \circ T_{\lambda^n}^{-1} .
		\end{equation*}
	\end{definition}

	\begin{lemma} \label{initialgap}
		Let $F$ be an iterated function system on $\mathbb{R}^d$ with uniform contraction ratio and uniform rotation. Let $h_F$ be its Garsia entropy, $M_F$ be its splitting rate, and $\lambda$ be its contraction ratio. Then for any $M>M_F$ there is some $c>0$ such that for all $n \in \mathbb{N}$ letting $\mu = \nu_F^{(\lambda^n, 1]}$ we have,
		\begin{equation*}
		H \left( \mu * n_{1}\right) - H(\left( \mu * n_{M^{-2n}} \right) < (d\log M- h_F)n + c .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		We simply bound $H \left( \mu * n_{1}\right)  \leq H(\mu_F*n_1)$ and $H(\mu * n_{M^{-2n}}) \geq -n d \log M +n h_F-c$. This means that,
		\begin{equation*}
		H \left( \mu * n_{1}\right) - H(\left( \mu * n_{M^{-2n}} \right) \leq (d\log M- h_F)n + c+H(\mu_F*n_1)
		\end{equation*}
		as required.
	\end{proof}
	This turns out to be the only place in the proof where we use the splitting rate and the Garsia entropy. This means that if it were possible to find a result like this for some iterated function system by some other means then this could also be used to show that an iterated function system is absolutely continuous.

	\section{Finding the Measures}
	In this section we will be showing that under certain conditions it is possible to express $\mu_F$ as the convolution of many measures each of which have at most a limited amount of detail at some given scale. After this we will use Theorem \ref{manyconv} to get an upper bound on the detail around a scale of $\mu_F$. Finally we will apply Lemma \ref{isabscont} to show that $\mu_F$ is absolutely continuous under these conditions.

	\begin{definition}
		Given some $r \in \left(0, \frac{1}{10} \right)$ and iterated function system $F$ on $\mathbb{R}^d$ we say that an interval $I$ is $\alpha$-admissible at scale $r$ if for all $t$ with,
		\begin{equation*}
		t \in \left[ \exp \left( -\left( \log \log r^{-1} \right) ^{10} \right)r,  \exp \left( \left( \log \log r^{-1} \right) ^{10} \right)r\right]
		\end{equation*}
		we have,
		\begin{equation*}
		\left. \frac{\partial}{\partial y^{+}} H(\nu_F^{I}*n_y) \right|_{y=t^2} \leq \alpha t^{-2} .
		\end{equation*}
	\end{definition}
	The precise values of the constants taken to be ten in the above definition makes no difference to the final result and so they are taken to be ten for simplicity.

	\begin{proposition} \label{manyadmissibletodetail}
		Let $\alpha, K >0$ and let $d \in \mathbb{N}$. Suppose that $\alpha < \frac{1}{8}\left(1 + \frac{1}{2K^2} \right)^{-1}$. Then there exists some constant $c$ such that the following is true.

		Let $F$ be an iterated function system on $\mathbb{R}^d$ with uniform contraction ratio and uniform rotation. Suppose that $r <c$ and  $n\in \mathbb{N}$ is even with,
		\begin{equation*}
		n \leq 10 \frac{\log \log r^{-1}}{\log \left( \frac{1}{8} \left(1 + \frac{1}{2K^2} \right)^{-1} \alpha^{-1} \right)}
		\end{equation*}
		and that $I_1, I_2, \dots, I_n$ are disjoint $\alpha$-admissible intervals at scale $r$ contained in $(0,1)$. Then we have,
		\begin{equation}
		s_r(\mu_F) \leq \Gamma \left( \frac{d}{2} \right) \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \frac{1}{8}  \left( 8\left(1 + \frac{1}{2K^2} \right) \alpha \right)^{\frac{n}{2}} . \label{admissibledetail}
		\end{equation}
	\end{proposition}
	We wish to show that $s_r(\mu_F) \leq \left( \log r^{-1} \right) ^{-\beta}$ for some $\beta >1$ for all sufficiently small $r$. For this it sufficies to show that we can find at least $\beta \frac{\log \log r^{-1}}{\frac{1}{2} \log \frac{1}{8 \alpha}}$ admissible intervals for some $\beta >1$. The important fact about equation \eqref{admissibledetail} is that for any $\lambda > \sqrt{8 \alpha}$ if $r$ is sufficienly small and $n$ satisfies the conditions of the proposition we have,
	\begin{equation*}
	s_r(\mu_F) \leq C \lambda^n .
	\end{equation*}
	\begin{proof}
		Throughout this proof let $c_1, c_2, ...$ denote constants depending only on $\alpha, K$ and $d$. The idea is to use Theorem \ref{manyconv} and Theorem \ref{entropytodetail}. First note that by Theorem \ref{entropytodetail} we know that for all
		\begin{equation*}
		t \in \left[ \frac{1}{\sqrt{2}}\exp \left( -\left( \log \log r^{-1} \right) ^{10} \right)r,  \frac{1}{\sqrt{2}} \exp \left( \left( \log \log r^{-1} \right) ^{10} \right)r\right]
		\end{equation*}
		and for $i = 1, 2, ... , \frac{n}{2}$ we have,
		\begin{equation*}
		s_r(\nu_F^{I_{2i-1}} * \nu_F^{I_{2i}}) \leq \Gamma \left( \frac{d}{2} \right) \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \alpha .
		\end{equation*}
		We now wish to apply Theorem \ref{manyconv}. To do this we simply need to check that,
		\begin{equation*}
		\left[ 2^{-\frac{m}{2}} r, K^m \alpha_1^{-m 2^m} r \right] \subset \left[ \exp \left( -\left( \log \log r^{-1} \right) ^{10} \right)r,  \exp \left( \left( \log \log r^{-1} \right) ^{10} \right)r\right]
		\end{equation*}
		where $m = \frac{\log \frac{n}{2}}{\log \frac{3}{2}}$ and $\alpha_1 = \Gamma \left( \frac{d}{2} \right) \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \alpha$. We note that,
		\begin{equation*}
		m \leq \frac{1}{\log \frac{3}{2}} \log \log \log r^{-1} + c_1
		\end{equation*}
		and so for all sufficiently small $r$ we have,
		\begin{equation*}
		2^{-\frac{m}{2}} r \geq \exp \left( -\left( \log \log r^{-1} \right) ^{10} \right)r .
		\end{equation*}
		For the other side note that,
		\begin{equation*}
		K^m \alpha_1^{-m 2^m}  \leq \exp \left( c_2 \left( \log \log \log r^{-1} \right) \left( \log \log r^{-1} \right)^{\frac{\log 2}{\log \frac{3}{2}}} +c_3 \right) .
		\end{equation*}
		Noting that $\frac{\log 2}{\log \frac{3}{2}}<10$ this means that for all sufficiently small $r$ we have,
		\begin{equation*}
		K^m \alpha_1^{-m 2^m} r \leq \exp \left( \left( \log \log r^{-1} \right) ^{10} \right)r .
		\end{equation*}
		This means that the conditions of Theorem \ref{manyconv} are satisfied and so,
		\begin{eqnarray}
		\lefteqn{s_r(\nu_F^{I_1} * \nu_F^{I_2} * \dots * \nu_F^{I_n}) }\nonumber\\
		& \leq & \left( \Gamma \left( \frac{d}{2} \right) \left( \frac{d}{2e} \right)^{-\frac{d}{2}} \alpha \right) ^{\frac{n}{2}} \left( \frac{8}{\Gamma \left( \frac{d}{2} \right)} \left( \frac{d}{2e} \right) ^{\frac{d}{2}} \left( 1 + \frac{1}{2K^2} \right) \right) ^{\frac{n}{2}-1}\nonumber\\
		& \leq & \Gamma \left( \frac{d}{2} \right) \left( \frac{d}{2e} \right) ^{-\frac{d}{2}} \frac{1}{8}  \left( 8\left(1 + \frac{1}{2K^2} \right) \alpha \right)^{\frac{n}{2}} .\nonumber
		\end{eqnarray}
		We conclude the proof by noting that,
		\begin{equation*}
		s_r(\mu_F) \leq s_r(\nu_F^{I_1} * \nu_F^{I_2} * \dots * \nu_F^{I_n}) .
		\end{equation*}
	\end{proof}

	\begin{lemma} \label{numadmissible}
		Suppose that $F$ is an iterated function system with uniform contraction ratio $\lambda$ and uniform rotation and that $M>M_F$. Let $\alpha \in (0,1)$. Suppose further that,
		\begin{equation*}
		h_F > (d-2\alpha \lambda^4) \log M
		\end{equation*}
		then there is some constant $c$ depending only on $M$, $F$ and $\alpha$ such that for all $n \in \mathbb{N}$ and all $r \in (0,\frac{1}{4})$ there are at least,
		\begin{equation*}
		\frac{1}{2 \log \lambda^{-1}} (h_F - (d-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} n - c
		\end{equation*}
		values of $k$ with,
		\begin{equation*}
		k \in [0, n \frac{\log M}{\log \lambda^{-1}}) \cap \mathbb{Z}
		\end{equation*}
		such that the interval,
		\begin{equation*}
		(\lambda^{n-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil+b},\lambda^{-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil-b}]
		\end{equation*}
		is $\alpha$-admissible at scale $r$. Here we take $b = \frac{1}{\log \lambda^{-1}} \left( \log \log r^{-1} \right) ^{10} +10$.
	\end{lemma}
	\begin{proof}
		Throughout this proof let $c_1, c_2, ...$ be constants depending only on $F$, $M$ and $\alpha$. Recall that by Lemma \ref{initialgap} we have,
		\begin{equation*}
		H(\nu_F ^{(\lambda^n,1]} *n_1) - H(\nu_F ^{(\lambda^n,1]} *n_{M^{-2n}}) < n (\log M - h_F) +c_1.
		\end{equation*}
		Define,
		\begin{equation*}
		S := \left\{ y \in (M^{-2n},1) : \frac{\partial}{\partial y^{+}} H(\nu_F^{(\lambda^n,1]} * n_y) \leq \frac{1}{y} \alpha \lambda^4\right\} .
		\end{equation*}
		Let $\mu$ be the measure on $(0,\infty)$ with density function $y \mapsto \frac{1}{y}$. Note that we have,
		\begin{eqnarray}
		\lefteqn{n (d\log M - h_F) +c_1}\nonumber\\
		& \geq & H(\nu_F ^{(\lambda^n,1]} *n_1) - H(\nu_F ^{(\lambda^n,1]} *n_{M^{-2n}})\nonumber\\
		& = & \int_{M^{-2n}}^1 \frac{\partial}{\partial y} H(\nu_F ^{(\lambda^n,1]} *n_{y}) \, dy\nonumber\\
		& \geq & \int_{y \in (M^{-2n},1) \backslash S} \frac{1}{y} \alpha \lambda^4 \, dy\nonumber\\
		& = & (2n \log M - \mu(S)) \alpha \lambda^4 .\nonumber
		\end{eqnarray}
		This means,
		\begin{equation*}
		\mu(S) \geq (h_F - (d-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} n - c_2 .
		\end{equation*}
		Now let,
		\begin{equation*}
		\tilde{S} = \left\{ k \in \mathbb{N}_0 : (\lambda^{2(k+1)}, \lambda^{2k}] \cap S \neq \emptyset \right\} .
		\end{equation*}
		Note that $\mu ((\lambda^{2(k+1)}, \lambda^{2k}]  ) = 2 \log \lambda^{-1}$ and so,
		\begin{equation*}
		|\tilde{S}| \geq \frac{1}{2 \log \lambda^{-1}} \mu(S) .
		\end{equation*}
		In particular,
		\begin{equation*}
		|\tilde{S}| \geq \frac{1}{2 \log \lambda^{-1}} (h_F - (d-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} n - c_3 .
		\end{equation*}
		Now we wish to show that if $k \in \tilde{S}$ then the interval,
		\begin{equation*}
		(\lambda^{n-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil+b},\lambda^{-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil-b}]
		\end{equation*}
		is $\alpha$-admissible at scale $r$. To do this suppose that,
		\begin{equation*}
		t \in \left[ \exp \left( -\left( \log \log r^{-1} \right) ^{10} \right)r,  \exp \left( \left( \log \log r^{-1} \right) ^{10} \right)r\right].
		\end{equation*}
		I wish to show that $\left. \frac{\partial}{\partial y^{+}} H(\nu_F^I* n_y) \right|_{y=t^2} \leq \alpha$ where
		\begin{equation*}
		I= {(\lambda^{n-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil+b},\lambda^{-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil-b}]} .
		\end{equation*}
		First choose $\tilde{k} \in \mathbb{Z}$ such that,
		\begin{equation*}
		\lambda^{\tilde{k} + 1} \leq t \leq \lambda ^ {\tilde{k}}
		\end{equation*}
		and choose $\tilde{t} \in (\lambda^{k+1}, \lambda^k]$ such that $\tilde{t}^2 \in S$. Note that we have,
		\begin{eqnarray}
		\left. \frac{\partial}{\partial y^{+}} H(\nu_F^{(\lambda^n,1]}*n_y) \right|_{y = \lambda^{2k}} &\leq &\left. \frac{\partial}{\partial y} H(\nu_F^{(\lambda^n,1]}*n_y) \right|_{y = \tilde{t}^2}\nonumber\\
		& \leq & \lambda^4 \alpha \tilde{t}^{-2}\nonumber\\
		& \leq & \lambda^2 \alpha \lambda^{-2k} .\nonumber
		\end{eqnarray}
		This means that,
		\begin{equation*}
		\left. \frac{\partial}{\partial y^{+}} H(\nu_F^{(\lambda^{n-k+\tilde{k}+1},\lambda^{-k+\tilde{k}+1}]}*n_y) \right|_{y = \lambda^{2(\tilde{k}+1)}} \leq \lambda^2 \alpha \lambda^{-2(\tilde{k}+1)}
		\end{equation*}
		and so,
		\begin{equation*}
		\left. \frac{\partial}{\partial y^{+}} H(\nu_F^{(\lambda^{n-k+\tilde{k}+1},\lambda^{-k+\tilde{k}+1}]}*n_y) \right|_{y = t^2} \leq \lambda^2 \alpha \lambda^{-2(\tilde{k}+1)}\leq \alpha t^{-2} .
		\end{equation*}
		It only remains to show that,
		\begin{equation*}
		(\lambda^{n-k+\tilde{k}+1},\lambda^{-k+\tilde{k}+1}] \subset (\lambda^{n-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil+b},\lambda^{-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil-b}] .
		\end{equation*}
		In other words it suffices to show that $\lambda^{\tilde{k} + 1} \geq \lambda ^{\lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil + b}$ and $\lambda^{\tilde{k} + 1} \leq \lambda ^{\lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil - b}$. Both of these statements follow immediately from the definition of $b$.
	\end{proof}

	\begin{corollary} \label{oneadmissible}
		Suppose that $F$ is an iterated function system with uniform contraction ratio $\lambda$ and uniform rotation and that $M>M_F$. Suppose further that,
		\begin{equation*}
		h_F > (d-2\alpha \lambda^4) \log M
		\end{equation*}
		then there is some constant $c$ depending only on $M$, $F$ and $\alpha$ such that for all  $r \in (0,\frac{1}{4})$ and all $n \in \mathbb{N}$ with,
		\begin{equation}
		\frac{1}{2 \log \lambda^{-1}} (h_F - (d-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} n - c > 0 \label{atleastone}
		\end{equation}
		there is some $k$ with,
		\begin{equation*}
		k \in [\frac{1}{2 \log \lambda^{-1}} (h_F - (d-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} n - c , n \frac{\log M}{\log \lambda^{-1}}) \cap \mathbb{Z}
		\end{equation*}
		such that the interval,
		\begin{equation*}
		(\lambda^{n-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil+b},\lambda^{-k + \lceil \frac{\log r^{-1}}{\log \lambda^{-1}} \rceil-b}]
		\end{equation*}
		is $\alpha$-admissible at scale $r$. Here we take $b = \frac{1}{\log \lambda^{-1}} \left( \log \log r^{-1} \right) ^{10} +10$.
	\end{corollary}
	\begin{proof}
		Simply take the largest $k$ found by Lemma \ref{numadmissible}. Note that the condition \eqref{atleastone} is necessary to ensure that at least one such $k$ is found.
	\end{proof}

	\begin{lemma}
		Suppose that $c>0$, $A>1$ and the sequence $(x_n)_{n\in \mathbb{N}_0}$ of positive reals satisfies,
		\begin{equation*}
		x_{n+1} = A x_n +c
		\end{equation*}
		then,
		\begin{equation*}
		x_{n} = A^n (x_0+\frac{c}{A-1})-\frac{c}{A-1} .
		\end{equation*}
	\end{lemma}
	\begin{proof}
		Simply note that $x_{n+1} +\frac{c}{A-1} = A (x_n+\frac{c}{A-1} )$.
	\end{proof}

	\begin{corollary} \label{find2}
		Suppose that $c>0$, $A>1$ and the sequence $(x_n)_{n\in \mathbb{N}_0}$ of positive reals satisfies,
		\begin{equation*}
		x_{n+1} \leq A x_n +c
		\end{equation*}
		then,
		\begin{equation*}
		x_{n} \leq A^n (x_0+\frac{c}{A-1})-\frac{c}{A-1} .
		\end{equation*}
	\end{corollary}

	\begin{lemma} \label{find4}
		Suppose that $F$ is an iterated function system with uniform rotation and uniform contraction ratio $\lambda$. Let $M>M_F$, $\alpha \in (0, \frac{1}{8})$ and suppose that,
		\begin{equation*}
		h_F > (1-2 \alpha \lambda^4) \log M .
		\end{equation*}
		Then there exists some $c >0$ such that for every $r>0$ sufficiently small there are at least
		\begin{equation*}
		\frac{1}{\log A} \log \log r^{-1} - c \log \log \log r^{-1}
		\end{equation*}
		disjoint $\alpha$-admissable intervals at scale $r$ all of which are contained in $(0,1]$ . Here,
		\begin{equation*}
		A = \frac{2\alpha \lambda^4 \log M}{h_F - 2 \alpha \lambda^4 \log \lambda^{-1} - (d-2 \alpha \lambda^4) \log M} .
		\end{equation*}
	\end{lemma}
	\begin{proof}
		Throughout this proof $E_1, E_2, ...$ will denote error terms which may be bounded by $0 \leq E_i \leq c_i (\log \log r^{-1}) ^{c_i}$ for some positive constants $c_1, c_2, ...$ which depends only on $\alpha$, $F$ and $M$.

		Choose $N$ such that equation \eqref{atleastone} is satisfied for all $n \geq N$ in other words,
		\begin{equation*}
		\frac{1}{2 \log \lambda^{-1}} (h_F - (1-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} N- c > 0 .
		\end{equation*}
		The idea of the proof is to choose a sequence $\left( n_j \right)_{j=0}^{j_{\text{max}}}$ such that $N = n_0 < n_1 < n_2 < \dots <n_{j_{\text{max}}}$ such that by using Corollary \ref{oneadmissible} we are garanteed to find $j_{\text{max}}$ disjoint admissible intervals contained in $(0,1]$. Let $I_j$ denote the interval found using $n_j$ in Corollary \ref{oneadmissible}.

		Note that to be able to garantee that $I_j$ and $I_{j+1}$ are disjoint we need,
		\begin{equation*}
		(C -1)n_{j+1} \geq \frac{\log M}{\log \lambda^{-1}} n_j + E_1
		\end{equation*}
		where we take,
		\begin{equation*}
		C = \frac{1}{2 \log \lambda^{-1}} (h_F - (d-2\alpha \lambda^4) \log M) \alpha^{-1} \lambda^{-4} .
		\end{equation*}
		For this is it sufficient to take,
		\begin{equation*}
		n_{j+1} = \lceil A n_j + E_2 \rceil
		\end{equation*}
		where,
		\begin{equation*}
		A = \frac{2\alpha \lambda^4 \log M}{h_F - 2 \alpha \lambda^4 \log \lambda^{-1} - (d-2 \alpha \lambda^4) \log M}
		\end{equation*}
		meaning,
		\begin{equation*}
		n_{j+1} \leq  A n_j + E_3 .
		\end{equation*}
		This means we have,
		\begin{equation*}
		n_j \leq A^j \left( n_0 + E_4\right) .
		\end{equation*}
		Noting that $n_0 = N = E_5$ this means,
		\begin{equation*}
		n_j \leq A^j E_6 .
		\end{equation*}
		It must be the case that we are unable to continue this process after $j_{\text{max}}$. The only way that this can happen is if the interval given by continuing this process for one more step is not contained in $(0,1]$. This means that,
		\begin{equation*}
		\frac{\log M}{\log \lambda^{-1}} \left( A n_{j_{\text{max}}} + E_3\right) + E_7  > \frac{\log r^{-1}}{\log \lambda ^ {-1}}
		\end{equation*}
		and so,
		\begin{equation*}
		A^{j_{\text{max}}+1} E_8 > \frac{\log r^{-1}}{\log \lambda ^ {-1}} .
		\end{equation*}
		This means that,
		\begin{equation*}
		j_{\text{max}} > \frac{1}{\log A} \log \log r^{-1} - c \log \log \log r^{-1}
		\end{equation*}
		for some constant $c$ depending only on $\alpha, F$ and $M$ as required.
	\end{proof}

	I am now ready to prove the main result.
	\newtheorem*{theorem:mainresult}{Theorem \ref{mainresult}}
	\begin{theorem:mainresult}
		Let $F$ be an iterated function system on $\mathbb{R}^d$ with uniform contraction ratio and uniform rotation. Suppose that $F$ has Garsia entropy $h_F$, splitting rate $M_F$, and uniform contraction ratio $\lambda$. Suppose further that,
		\begin{equation*}
		(d \log M_F - h_F)(\log M_F)^2 < \frac{1}{27} (\log M_F - \log \lambda^{-1})^3 \lambda^4
		\end{equation*}
		then the self similar measure $\mu_F$ is absolutely continuous.
	\end{theorem:mainresult}
	\begin{proof}
		The idea is to use Proposition \ref{manyadmissibletodetail} and Lemma \ref{find4} to show that the detail around a scale decreases fast enough for us to be able to apply Lemma \ref{isabscont}.

		We are given that,
		\begin{equation*}
		(d \log M_F - h_F)(\log M_F)^2 < \frac{1}{27} \left( \log M_F - \log \lambda^{-1} \right) ^3 \lambda^4 .
		\end{equation*}

		Letting $\alpha = \frac{1}{18} \left( \frac{ \log M_F - \log \lambda^{-1}}{\log M_F} \right) ^2$ we get,

		\begin{equation*}
		d \log M_F - h_F < \frac{2}{3} (\log M_F - \log \lambda^{-1}) \alpha \lambda^4 .
		\end{equation*}

		Rearranging this gives,

		\begin{equation}
		\frac{4}{3} (\log M_F - \log \lambda^{-1}) \alpha \lambda^4 < h_F - 2 \alpha \lambda^4 \log \lambda^{-1} - (d-2 \alpha \lambda^4 ) \log M_F \label{getpos}
		\end{equation}

		which means that,

		\begin{equation*}
		\frac{2}{3} \left( \frac{\log M_F - \log \lambda^{-1}}{\log M_F} \right) \tilde{A} < 1
		\end{equation*}

		where,
		\begin{equation*}
		\tilde{A} = \frac{2 \alpha \lambda^4 \log M_F}{h_f-2\alpha \lambda^4 \log \lambda^{-1} - (d- 2 \alpha \lambda^4) \log M_F} .
		\end{equation*}
		This means that,
		\begin{equation*}
		\sqrt{8 \alpha} \tilde{A} < 1
		\end{equation*}
		and so,
		\begin{equation*}
		\frac{1}{\log \tilde{A}} > \frac{2}{\log \frac{1}{8 \alpha}} .
		\end{equation*}
		We now choose some $M>M_F$ such that,
		\begin{equation*}
		\frac{1}{\log A} > \frac{2}{\log \frac{1}{8 \alpha}}
		\end{equation*}
		where,
		\begin{equation*}
		\tilde{A} = \frac{2 \alpha \lambda^4 \log M_F}{h_f-2\alpha \lambda^4 \log \lambda^{-1} - (d- 2 \alpha \lambda^4) \log M_F} .
		\end{equation*}
		As mentioned previously this is enough to finish the proof. We wish to apply Lemma \ref{find4}. To do this we need to check that,
		\begin{equation*}
		h_F > (d - 2 \alpha \lambda^4) \log M .
		\end{equation*}
		To see this simply note that by \eqref{getpos} we have,
		\begin{eqnarray}
		h_F - (d - 2 \alpha \lambda^4) \log M_F &>& \frac{4}{3}(\log M_F - \log \lambda^{-1}) \alpha \lambda^4 +2\alpha \lambda^4 \log \, \lambda^{-1}\nonumber\\
		&>& 0 .\nonumber
		\end{eqnarray}
		This means that by Lemma \ref{find4} there is some $c_1>0$ such that for all sufficiently small $r>0$ we can find at least,
		\begin{equation*}
		\frac{1}{\log A} \log \,\log \, r^{-1} - c_1 \log \, \log \, \log \, r^{-1}
		\end{equation*}
		disjoint $\alpha$- admissible intervals at scale $r$. We now wish to apply Proposition \ref{manyadmissibletodetail}. We choose $K>1$ in this proposition such that,
		\begin{equation*}
		\frac{1}{\log A} > \frac{2}{\log \frac{1}{8 (1 + \frac{1}{2K^2})\alpha}} .
		\end{equation*}
		Applying Proposition \ref{manyadmissibletodetail} with $n$ the largest even number which is less than both $\frac{1}{\log A} \log \,\log \, r^{-1} - c_1 \log \, \log \, \log \, r^{-1}$ and $10 \frac{\log \log r^{-1}}{\log \left( \frac{1}{8} \left(1 + \frac{1}{2K^2} \right)^{-1} \alpha^{-1} \right)}$ we get that for all sufficiently small $r$,

		\begin{equation*}
		s_r(\mu_F) \leq c_2 (\log r^{-1})^{-\beta}(\log \log r ^{-1})^{c_3}
		\end{equation*}

		 or some $\beta>1$. By Lemma \ref{isabscont} this means that the measure $\mu_F$ is absolutely continuous.
	\end{proof}

	To prove Theorem \ref{mainbernoulli} we will first need the following result dating back at least to Garsia.

	\begin{lemma} \label{garsia} Let $\lambda$ be an algebraic number and denote by $d$ the number of its algebraic conjugates with modulos $1$. Then there is some constant $c_{\lambda}$ depending only on $\lambda$ such that whenever $p$ is a polynomial with degree $n$ and coefficients $-1, 0$ and $1$ such that $p(\lambda) \neq 0$ we have,
		\begin{equation*}
		|p(\lambda)| > c_{\lambda} n^{-d} M_{\lambda} ^ {-n} .
		\end{equation*}
	\end{lemma}

	\begin{proof}
		This is proven in \cite{garsiasep} Lemma 1.51
	\end{proof}

	\begin{corollary}
		Let $F$ be an iterated function system such that  $\mu_F$ is a (possibly biased) Bernoulli convolution with parameter $\lambda$. Then,
		\begin{equation*}
		M_F \leq M_{\lambda} .
		\end{equation*}
	\end{corollary}
	\newtheorem*{theorem:mainbernoulli}{Theorem \ref{mainbernoulli}}
	\begin{theorem:mainbernoulli}
		Suppose that $\lambda \in (\frac{1}{2}, 1)$ and has Mahler measure $M_\lambda$. Suppose also that $\lambda$ is not the root of a non zero polynomial with coefficients $0, \pm 1$ and,
		\begin{equation}
		(\log M_{\lambda} - \log 2) (\log M_{\lambda})^2 < \frac{1}{27} (\log M_{\lambda} - \log \lambda^{-1})^3 \lambda^4 \label{mainberneq}
		\end{equation}
		then the unbiased Bernoulli convolution with parameter $\lambda$ is absolutely continuous.
	\end{theorem:mainbernoulli}

	\begin{proof}
		To prove this simply note that letting $F$ be the iterated function system generating the Bernoulli convolution we have,
		\begin{equation*}
		M_F \leq M_{\lambda}
		\end{equation*}
		and,
		\begin{equation*}
		h_F = \log 2 .
		\end{equation*}
		We are now done by applying Theorem \ref{mainresult}.
	\end{proof}

	\section{Examples} \label{examples}
	In this section we will give some examples of self similar measures which can be shown to be absolutely continuous using the results of this paper.

	\subsection{Unbiased Bernoulli Convolutions} \label{findunbiased}
	In this subsection we will give explicit values of $\lambda$ for which the unbiased Bernoulli convolution with parameter $\lambda$ is absolutely continuous using Theorem \ref{mainbernoulli}.  we will do this by a simple computer search. We can ensure that $\lambda$ is not a root of a non-zero polynomial with coefficients $0, \pm 1$ by ensuring that it has a conjugate with absolute value greater that $2$.

	The computer search works by checking each integer polynomial with at most a given hieght and at most a given degree with leading coefficient $1$ and constant term $\pm 1$. The program then finds the roots of the polynomial. If there is one real root with modulos at least $2$ and at least one real root in $(\frac{1}{2}, 1)$ the program then checks that the polynomial is irreducible. If the polynomial is irreducible it then tests each real root in $(\frac{1}{2}, 1)$ to see if it satisfies equation \eqref{mainberneq}. In figure \ref{examples2}are the results for polynomials of degree at most $11$ and height at most $3$.

	The smallest value of $\lambda$ which we was able to find for which the Bernoulli convolution with parameter $\lambda$ can be shown to be absolutely continuous using this method is  $\lambda \approx 0.799533$ with minimal polynomial $X^9 - 2 X^8 - X + 1$.

	I was also able to find an infinite family of $\lambda$ for which the results of this paper show that the Bernoulli convolution with parameter $\lambda$ is absolutely continuous. This family is found using the following lemma.
	\begin{figure}[H] \label{examples2}
	\begin{center}
	\begin{tabular}{ | m{20em} | c| c| }
		\hline
		Minimal Polynomial & Mahler Measure & $\lambda$\\
		\hline
		$X^{7}-X^{6}-2X^{5}-X^{2}+X+1$ & 2.010432 & 0.879161\\
		$X^{7}+2X^{6}-X-1$ & 2.015159 & 0.932864\\
		$X^{8}+2X^{7}-1$ & 2.007608 & 0.860582\\
		$X^{9}-2X^{8}-X+1$ & 2.003861 & 0.799533\\
		$X^{9}+2X^{8}-X-1$ & 2.003861 & 0.949560\\
		$X^{10}-2X^{9}-X^{2}+1$ & 2.005754 & 0.852579\\
		$X^{10}-2X^{9}-X+1$ & 2.001940 & 0.813972\\
		$X^{10}-X^{9}-2X^{8}-X^{5}+X^{4}+X^{3}-X^{2}+1$ & 2.014180 & 0.911021\\
		$X^{10}-X^{9}-X^{8}-2X^{7}-X^{5}+X^{4}+X^{2}+1$ & 2.012241 & 0.939212\\
		$X^{10}-X^{9}-X^{8}-X^{7}-2X^{6}-X^{5}+X^{3}+X^{2}+X+1$ & 2.017567 & 0.953949\\
		$X^{10}-2X^{8}-3X^{7}-2X^{6}-X^{5}+X^{3}+2X^{2}+2X+1$ & 2.008264 & 0.968459\\
		$X^{10}+X^{9}-2X^{8}+X^{7}+X^{6}-X^{5}+X^{4}-X^{3}+X-1$ & 2.016061 & 0.875809\\
		$X^{10}+2X^{9}-X^{4}-1$ & 2.030664 & 0.946934\\
		$X^{10}+2X^{9}-1$ & 2.001936 & 0.888810\\
		$X^{10}+3X^{9}+3X^{8}+3X^{7}+2X^{6}-2X^{4}-3X^{3}-3X^{2}-2X-1$ & 2.047156 & 0.984474\\
		$X^{11}-2X^{10}-X^{2}+1$ & 2.002899 & 0.861821\\
		$X^{11}-2X^{10}-X+1$ & 2.000973 & 0.826146\\
		$X^{11}-X^{10}-2X^{9}-X^{8}+X^{7}+2X^{6}+X^{5}-X^{4}-2X^{3}-X^{2}+X+1$ & 2.000730 & 0.876660\\
		$X^{11}-X^{10}-X^{9}-2X^{8}-X^{4}+X^{2}+X+1$ & 2.004980 & 0.952899\\
		$X^{11}+X^{10}-2X^{9}+X^{8}+X^{7}-2X^{6}+X^{5}+X^{4}-2X^{3}+X^{2}+X-1$ & 2.000730 & 0.831391\\
		$X^{11}+X^{10}-X^{9}+2X^{8}+X^{4}-X^{2}+X-1$ & 2.004980 & 0.805998\\
		$X^{11}+2X^{10}-X-1$ & 2.000973 & 0.959607\\
		$X^{11}+2X^{10}+X^{2}-1$ & 2.002899 & 0.810380\\
		\hline
	\end{tabular}
	\end{center}
	\caption{Examples of Parameters of Absolutely Continuous Bernoulli Convolutions}
	\end{figure}

	\begin{lemma} \label{infinitefamily}
		Suppose that $n \geq 5$ is an integer and let,
		\begin{equation*}
		p(X) = X^n - 2 X^{n-1} -X +1
		\end{equation*}
		Then $p$ has one root in the interval $(\left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}}, 1)$, one root in the interval $(2, 2+2^{2-n})$ and all of the remaining roots are contained in the interior of  the unit disk. Furthermore $p$ is irreducible.
	\end{lemma}

	Before proving this we need the following result.

	\begin{theorem}[Schur-Cohn test]
		Let $p$ be a polynomial of degree $n$ and let $p^*$ be its reciprocal adjoint polynomial defined by $p^*(z) := z^n \overline{p(\overline{z}^{-1})}$. Note that if,
		\begin{equation*}
		p(z) = a_n z^n + \dots + a_1 z + a_0
		\end{equation*}
		then,
		\begin{equation*}
		p^*(z) = \overline{a_0} z^n + \dots + \overline{a_{n-1}} z + \overline{a_n} .
		\end{equation*}
		Let $Tp := \overline{p(0)}p - \overline{p^*(0)}p^*$. Then the following results are true,
		\begin{enumerate}
			\item The roots of $p^*$ are the images of the roots of $p$ under inversion in the unit circle
			\item If $Tp(0) \neq 0$ then $p$, $p^*$ and $Tp$ share roots on the unit circle
			\item If $Tp(0) > 0$ then $p$ and $Tp$ have the same number of roots inside the unit disk
			\item If $Tp(0) < 0$ then $p^*$ and $Tp$ have the same number of roots outside the unit disk
		\end{enumerate}
	\end{theorem}
	\begin{proof}
		This was first proven in \cite{schuror}. For a reference in English see \cite{schureng}.
	\end{proof}

	We are now ready to prove Lemma \ref{infinitefamily}.

	\begin{proof}
		The idea is to use the Schur-Cohn test to show that all but one of the roots of $p$ are in the unit disk. We then identify the specified real roots using the intermediate value theorem. We cannot use the Schur Cohn test directly as we end up with $Tp(0) = 0$. We instead choose some $\alpha < 1$ and let,
		\begin{equation*}
		q = p(\alpha X) = \alpha^n X^n - 2 \alpha^{n-1} X^{n-1} - \alpha X + 1
		\end{equation*}
		meaning that we have,
		\begin{equation*}
		q* = X^n - \alpha X^{n-1} - 2 \alpha^{n-1} X +\alpha^{n}
		\end{equation*}
		and so,
		\begin{equation*}
		Tq = (-2\alpha^{n-1} + \alpha^{n+1}) X^{n-1} + (- \alpha + 2 \alpha^{n-1}) X + (1 - \alpha^{2n}) .
		\end{equation*}
		We can choose $\alpha$ such that the coefficient of $X$ disapears. To do this we require,
		\begin{equation*}
		\alpha = \left( \frac{1}{2} \right) ^ \frac{1}{2n-2} .
		\end{equation*}
		We then simply note that since,
		\begin{equation*}
		\left| -2\alpha^{n-1} + \alpha^{n+1} \right| = \sqrt{2}\left| 1 - \alpha^{2n} \right|
		\end{equation*}
		we must have that all of the roots of $Tq$ are in the unit disk. This means that all but one of the roots of $p$ are in the disk of radius $\alpha$ which in particular means that they are in the disk of radius $1$.
		\paragraph{}
		The other roots can be found be using the intermediate value theorem. First note that $p(2)=-1<0$ and that,
		\begin{eqnarray}
		p(2+2^{2-n}) &= &2^{2-n} (2 + 2^{2-n})^{n-1} -1 - 2^{2-n}\nonumber\\
		& > & 2-1 - 2^{2-n}\nonumber\\
		& > & 0\nonumber
		\end{eqnarray}
		so there is a root in $(2, 2+2^{2-n})$. We also have $p(1)=-1<0$ and,
		\begin{equation*}
		p\left( \left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}} \right)= \left( \frac{1}{2} \right) ^ {\frac{2n}{\sqrt{n-1}}} -2 \left( \frac{1}{2} \right) ^ {2\sqrt{n-1}} + 1 - \left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}} .
		\end{equation*}
		Note that,
		\begin{equation*}
		\frac{d}{dx} \left( \frac{1}{2} \right) ^ x = - \log 2 \left( \frac{1}{2} \right) ^ x
		\end{equation*}
		so noting that $n \geq 5$ we have,
		\begin{equation*}
		1-\left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}} >\frac{2}{\sqrt{n-1}} \left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}} \log 2 > \frac{1}{2 \sqrt{n-1}} .
		\end{equation*}
		Also note that for $x \geq 1$,
		\begin{equation*}
		\frac{d}{dx} 4^x = 4^x \log 4 > 4
		\end{equation*}
		so for $x \geq 1$ we have,
		\begin{equation*}
		4^x \geq 4x
		\end{equation*}
		meaning that,
		\begin{equation*}
		\frac{1}{2\sqrt{n-1}} \geq 2\left( \frac{1}{2} \right) ^ {2\sqrt{n-1}}
		\end{equation*}
		and so,
		\begin{equation*}
		p\left( \left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}} \right) > 0
		\end{equation*}
		so there is a root in the interval $(\left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}}, 1)$. Infact it must be in that interval $(\left( \frac{1}{2} \right) ^ {\frac{2}{\sqrt{n-1}}}, \left( \frac{1}{2} \right) ^ {\frac{1}{n-2}})$.

		The fact that $p$ is irreducible follows from the fact that it is an integer polynomial with non-zero constant coefficient and all but one of its factors contained in the interior of the unit disk. If $p$ where not irreducible then one of its factors would need to have all of its root contained in the interior of the unit disk. This would mean that the product of the roots of this factor would not be an integer which is a contradiction.
	\end{proof}

	We now simply let $\lambda_n$ be one of the roots (or more likely the unique root) of  $X^n - 2X^{n-1} - X + 1$ contained in the interval $\left( \left( \frac{1}{2} \right) ^{\frac{2}{\sqrt{n-1}}}, 1 \right)$. By Theorem \ref{mainbernoulli} to show that the Bernoulli convolution with parameter $\lambda_n$ is absolutely continuous it suffices to show that,

	\begin{equation*}
	(\log  (2 + 2^{2-n}) - \log 2) (\log  (2 + 2^{2-n}) )^2 <  \frac{1}{27} \left( \log (2 ) - \log 2 ^{\frac{2}{\sqrt{n-1}}}\right)^3 2^{-\frac{8}{\sqrt{n-1}}}
	\end{equation*}

	The left land side is decreasing in $n$ and the right hand side is increasing in $n$ and for $n=13$ the left hand side is less than the right hand side so for $n \geq 13$ we know that the Bernoulli convolution with parameter $\lambda_n$ is absolutely continuous. It is worth noting that we have $\lambda_n \to 1$ and $M_{\lambda_n} \to 2$ so all but finity many of these Bernoulli convolutions can be shown to be absolutely continuous by for example the results of  \cite{varjupaper}. Using the results of \cite{varjupaper} does however require a significantly higher value of $n$ to work. Indeed it requires $n \geq 10^{65}$.

	\subsection{Biased Bernoulli Convolutions} \label{findbiased}
	It is worth noting that all of the examples in the previous section also work with a small amount of bias as an arbitarily small amount of bias will reduce $h_F$ by an arbitarily small amount. In contrast to the results of \cite{varjupaper} it is not possible to find examples of highly biased Bernoulli convolutions that can be shown to be absolutely continuous using the results of this paper.

	\subsection{Other Self Similar Measures on $\mathbb{R}$}
	In this subsection we will briefly mention some  other examples of iterated function systems in one dimension which can be shown to be absolutely continuous by these methods.

	\begin{proposition}
		Let $q$ be a prime number for $i = 1, \dots, q-1$ let $S_i : x \mapsto \frac{q-1}{q} x + i$. Let $F$ be the iterated function system on $\mathbb{R}^1$ given by,
		\begin{equation*}
		 F =\left (q-1, 1, (S_i)_{i=1}^q, \left(\frac{1}{q-1}, \dots, \frac{1}{q-1}\right)\right)
		 \end{equation*}
		 Then we have $M_F \leq \log q$, $h_F = \log (q-1)$ and $\lambda = \frac{q-1}{q}$. Furthermore if $q \geq 17$ then $\mu_F$ is absolutely continuous.
	\end{proposition}
	\begin{proof}
		We note that any point in the $k$- step support of $F$ must be of the form $u = \sum_{i=0}^{k-1} x_i \left( \frac{q-1}{q} \right) ^i$ with $x_i \in \{1, \dots, q-1\}$. Suppose $u = \sum_{i=0}^{k-1} x_i \left( \frac{q-1}{q} \right) ^i$ and  $v = \sum_{i=0}^{k-1} y_i \left( \frac{q-1}{q} \right) ^i$ are two such points. We note that $q^{k-1} u, q^{k-1} v \in \mathbb{Z}$. This means that if $u \neq v$ then $|u-v| \geq q^{-k}$. This gives $M_F \leq \log q$.

		We can also note if $u = v$ then looking at $q^{k-1} u$ and $q^{k-1} v $ mod $q^i$ for $i = 1, \dots, k$ we see that we must have $(x_1, x_2, \dots, x_k) = (y_1, y_2, \dots, y_k) $. This means that $F$ has no exact overlaps and consequently $h_F = \log (q-1)$. We also note that $\lambda = \frac{q-1}{q}$ follows immediately from the definition of $F$.

		To show that $\mu_F$ is absolutely continuous by theorem \ref{mainresult} it is sufficient to check that,

		\begin{equation*}
		(\log q - \log (q-1)) (\log q)^2 < \frac{1}{27}\left(\log q - \log \left(\frac{q}{q-1}\right) \right)^3  \left( \frac{q-1}{q} \right)^4.
		\end{equation*}

		This is the same as showing that,

		\begin{equation}
		\left( \log \left( 1 + \frac{1}{q-1} \right) \right) < \frac{1}{27} \left( \frac{\log (q-1)}{\log q} \right)^2 (\log (q-1)) \left( \frac{q-1}{q} \right)^4. \label{qinequality}
		\end{equation}

		The left had side of \eqref{qinequality} is decreasing in $q$ and the left hand side is increasing in $q$. The inequality is satisfied for $q = 17$. This means that it is satisfied for $q \geq 17$.
	\end{proof}
	\subsection{Self Similar Measures on $\mathbb{R}^2$}
		In this section we will describe some examples of self similar measures on $\mathbb{R}^2$ which can be shown to be absolutely continuous using the methods of this paper and which cannot be expressed as the product of self similar measures on $\mathbb{R}$. This will be done by indenifying $\mathbb{R}^2$ with $\mathbb{C}$.

		\begin{proposition} \label{prop:2dexamples}
			Let $p$ be a prime number such that $p \equiv 3 \, (\text{mod } 4)$. Let $I_p$ denote the ideal $(p)$ in the ring $\mathbb{Z}[i]$. Note that  this is a prime ideal. Let $a_1, \dots, a_{m}$ be in different cosets of $I_p$. Choosen some $\alpha$ of the form $\alpha = \frac{a}{p}$ with $a \in \mathbb{Z}[i] \backslash I_p$ and $|\alpha| < 1$. Let $\lambda = |\alpha|$ and let $U : \mathbb{R}^2 \to \mathbb{R}^2$ be a rotation around the origin by $\arg \alpha$.

			For $i = 1, \dots, m$ let $S_i :\mathbb{R}^2 \to \mathbb{R}^2, \, x \mapsto \lambda U x+a_i$ and let $F$ be the iterated function system on $\mathbb{R}^2$ given by $F = \left( m, 2, \left( S_i \right)_{i=1}^{m}, \left( \frac{1}{m}, \dots, \frac{1}{m} \right) \right)$. Then we have $M_F \leq \log p$ and $h_F =  \log m$.
		\end{proposition}

		\begin{proof}
			To see that $M_F \leq \log p$ let $x = \sum_{i=0}^{k-1} x_i \alpha^i $ and $y = \sum_{i=0}^{k-1} y_i \alpha^i $ be two points in the $k$-step support of $F$. Note that $p^{k-1} (x-y) \in \mathbb{Z}[i]$ and so if $x \neq y$ then $|x-y| \geq p^{-k+1}$.

			To prove $h_F = \log m$ it suffices to show that $F$ has no exact overlaps. For this it suffices to show that if $x_1, \dots, x_k, y_1, \dots, y_k \in \{a_0, \dots, a_{m}\}$ and,
			\begin{equation}
			\sum_{i=0}^{k} x_i \alpha^i = \sum_{i=0}^{k} y_i \alpha^i \label{eq:primeidealthing}
			\end{equation}
			then $x_i = y_i$ for $i=1, \dots, k$. We will prove this by induction on $i$. For $i=k$ simply multiply both sides of \eqref{eq:primeidealthing} by $p^k$ and then work modulo the ideal $I_p$. This means that $x_k$ and $y_k$ must be in the same coset of $I_p$ which in particular means that they must be equal. The inductive step follows by the same argument.
		\end{proof}

		Note that the above proposition combined with Theorem \ref{mainresult} makes it very easy to give numerous examples of absolutely continuous iterated function systems in $\mathbb{R}^2$ which are not products of absolutely continuous iteratedfunction systems in $\mathbb{R}^1$. Some possible examples are given in the following Corollary.

	\begin{corollary}
			Let $p$ be a prime number such that $p \equiv 3 \, (\text{mod } 4)$. Let $I_p$ denote the ideal $(p)$ in the ring $\mathbb{Z}[i]$.  Let $a_1, \dots, a_{m}$ be in different cosets of $I_p$. Choosen some $\alpha$ of the form $\alpha = \frac{p-1 + i}{p}$. Let $\lambda = |\alpha|$ and let $U : \mathbb{R}^2 \to \mathbb{R}^2$ be a rotation around the origin by $\arg \alpha$.

			For $i = 1, \dots, m$ let $S_i :\mathbb{R}^2 \to \mathbb{R}^2, \, x \mapsto \lambda U x+a_i$ and let $F$ be the iterated function system on $\mathbb{R}^2$ given by $F = \left( m, 2, \left( S_i \right)_{i=1}^{m}, \left( \frac{1}{m}, \dots, \frac{1}{m} \right) \right)$. Suppose that,
			\begin{equation*}
			(2 \log p - \log m) (\log p)^2 < \frac{1}{27} \left(2 \log p - \log \frac{p}{p-1} \right)^3 \left( \frac{p-1}{p} \right)^4
			\end{equation*}
	\end{corollary}

	\begin{proof}
		This follows immedately from theorem \ref{mainresult} and proposition \ref{prop:2dexamples}.
	\end{proof}

	It is worth noting that the case $m=p^2$ follows from the results of \cite{garsiasep} so in this case the result of this paper can again be seen as a strengthening of the results of \cite{garsiasep}. It is also worth noting that in the case $m=p^2-1$ the conditions of this corollary are satisfies for all $p$ with $p \equiv 3 \, (\text{mod } 4)$.

	\section{Acknowlegdements}
	I would like to thank my supervisor Prof. P. Varj\'u for his helpful feedback and comments in the prepartion of this paper.
	\medspace
	\printbibliography
\end{document}