This question arises as a variation of [this question](https://mathoverflow.net/questions/434183/quantitative-analytic-continuation-estimate-for-a-function-small-on-a-set-of-pos), which was helpfully answered in the negative. It turns out that for my application, a substantially weaker conjecture suffices, which fails to be counterexampled by the answers given in the previous question.


Define
$$A^\gamma(K,\delta) := \sup \left\{c\_0 : \exists (c\_j)\_{j=1}^\infty, |c\_j| \leq K\,, \lambda\bigg(\Big\{ x \in (0,1/2) : \Big|\textstyle\sum\_{j=0}^\infty c\_j x^j\Big| > \delta \Big\}\bigg) \leq \gamma\right\},$$
where by $\lambda(S)$ for a set $S \subseteq \mathbb{R}$ we mean the Lebesgue measure of $S$.


Question.
---------


Does there exists $\gamma>0,K>1$ such that
$$\limsup\_n A^\gamma(K^n,e^{-n}) =0?$$


Relation to Previous Question.
------------------------------


Iosif Pinelis' answer to the above linked question shows that for all $K>1$ there exists some $\gamma<1/2$ such that
$$\limsup\_n A^\gamma(K^n,e^{-n}) \geq 1.$$
One can verify that the $\gamma$ given by his example goes rapidly to $1/2$ as $K \downarrow 1$. It seems that by taking $\gamma$ small, we should be able to avoid examples such as Iosif Pinelis'.


Thoughts on Possible Paths to a Proof.
--------------------------------------


First, let's note that $A^\gamma$ is $1$-homogeneous, so that $A^\gamma(CK,C\delta) = C A^\gamma(K,\delta)$. Thus we can rewrite
$$A^\gamma(K^n,e^{-n}) = K^n A^\gamma(1, (Ke)^{-n}).$$
Thus it suffices to show that
$$A^\gamma(1,\delta) \leq C\delta^{1/N}$$
for some $N$ and $\delta$ small enough, since then, taking $1 < K \leq e^{1/N}$
$$A^\gamma(K^n,e^{-n}) = K^n A^\gamma(1, (Ke)^{-n}) \leq CK^n (Ke)^{-n/N} \leq CK^{-n/N} \to 0.$$


There is a fairly straightforward proof that, for any $\gamma$, $A^\gamma(1,\delta) \to 0$ as $\delta \downarrow 0$, e.g. using weak-\* compactness of the unit ball in $\ell^\infty$ applied to the $c\_j$ sequences. The proof is entirely non-quantitative, so the above remark reduces the question to giving a good rate for this convergence.


One thought on how one could try to show some quantitative control of $A^\gamma(1,\delta)$ as $\delta \to 0$ is by some sort of interpolation inequality. For any $\gamma, \delta$, we have the $\sup$ almost realized by some function $f\_{\gamma,\delta}(x) = \sum\_j c\_j x^j$, where
$$f(0) \geq \frac{1}{2}A^\gamma(1,\delta);\ \ |c\_j| \leq 1;\ \ \lambda(\{x : |f(x)| > \delta\})\leq \gamma.$$
Note that the control on the $c\_j$ allow us control infinite derivative order norms, in particular we can control $W^{k,\infty}$ norms for every $k$. The fact the $f\_{\delta,\gamma}$ is small on a large set allows us to show that "norms" like $\log L$ are small.
$$\|f\_{\delta,\gamma}\|\_{\log L} = \exp\left(\int\_0^{1/2} \log |f\_{\delta,\gamma}|\right) \leq \exp\left(\gamma\log 2 + (1/2-\gamma) \log \delta\right) \leq 2^\gamma \delta^{1/2-\gamma}.$$
One can similarly show that $L^\epsilon$ "norms" are small for $0<\epsilon <1$.


Thus if we can get an interpolation inequality controlling $L^\infty$, say between an appropriately defined $W^{\infty,\infty}$ norm and a $\log L$ "norm", we'd get the desired bound. Note this looks something like a [Gagliardo-Nirenberg interpolation](https://en.wikipedia.org/wiki/Gagliardo%E2%80%93Nirenberg_interpolation_inequality).



Quantitative analytic continuation estimate for functions small except on a small set
This conjecture is correct. Take $K=e$, and let $\gamma\leq 1/4$; we will fix $\gamma$ later.


First we give a crude estimate of $c\_0$.
Let $g(z)=\sum\_{1}^\infty c\_nz^n.$ Since $|c\_n|\leq e^n$,
we obtain $|g(z)|\leq e^n$ for $|z|\leq 1/2$ by the trivial estimate. Then by Cauchy, $|f'(z)|=|g'(z)|\leq 4\cdot e^n,\; |z|\leq 1/4$.
Then $$|c\_0-e^{-n}|\leq (1/4)\max\_{[0,1/4]}|f'|\leq (1/4)4\cdot e^n=e^n,$$
so $|c\_0|\leq 1.5\cdot e^n$. Therefore
$$|f(z)|\leq |c\_0|+|g(z)|\leq 3\cdot e^n,\quad |z|\leq 1/2.$$


Now consider the subharmonic function
$$u(z)=\log|f(z)|$$
in the region $D=\{z:|z|<1/2,z\not\in E\},$
where $E=\{ z\in(0,1/2): |f(z)|\leq e^{-n}\}.$
By the Two constants Theorem,
$$u(0)\leq -n\omega(0,E,D)+n\omega(0,C,D)+\log 3,\quad\quad\quad (1)$$
where $C=\{ z:|z|=1/2\}$ and $\omega$ is the harmonic measure. Now, according to a theorem of Beurling
(Nevanlinna, Analytic functions, Chap. IV, section 84),
$\omega(0,E,D)\geq \omega(0,E\_0,D\_0),$ where
$E\_0$ is the segment $[\gamma,1/2]$ and $D\_0$ is the complement of this segment to the disk $|z|<1/2$.


It is easy to obtain an explicit formula for this $\omega(0,E\_0,D\_0)$, (see for example Nevanlinna's book),
but we only need the fact that it tends to $1$ when $\gamma\to 0$, which is evident. Since
$\omega(0,C,D\_0)=1-\omega(0,E\_0,D\_0)$, we can fix $\gamma$ such that
$\omega(0,E\_0,D\_0)>\omega(0,C,D\_0)$. Then
from the inequality (1) we conclude that
$$\log|c\_0|=u(0)\to-\infty,\; n\to\infty.$$
This proves the result.


From the explicit expression of $\omega(0,E\_0,D\_0)$ we can obtain an explicit value of $\gamma$.


Remark. Beurling theorem is not necessary for mere existence of $\gamma$; it is only needed to obtain an explicit value. It is clear without Beurling that
$\omega(0,E,D)\to 1$ when $\gamma\to 0$, by an elementary "compactness argument".


