Let $B\_t$ be the classic Brownian motion. I understand that, if $s>1/2$, almost surely $B\_t$ is nowhere $s$-Hölder continuous i.e. almost surely for no point $x$ it happens that $B\_t\in C^s(x)$.


Moreover I read a claim that said the same about any translation by a continuous function: given $s>1/2$ and $f$ continuous then almost surely $f+B\_t$ is nowhere $s$-Hölder continuous. Is this true and if so why?



Regularity of translations for Brownian motion
The result holds for any bounded function $f$, in the following sense: for any real $s>1/2$,
\begin{equation}
 P^\*(A)=0,
\end{equation}
where
\begin{equation}
A:=\Big\{\exists t\_0\in[0,1]\ \limsup\_{t\to t\_0}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}<\infty\Big\},
\end{equation}
$P^\*$ is the outer probability, $\limsup\_{t\to t\_0}:=\limsup\_{t\to t\_0,t\in[0,1]}$,
\begin{equation}
 W\_f:=W+f, 
\end{equation}
and $W$ is a standard Wiener process.


The proof is obtained by a straightforward adaptation of the proof of the Paley--Wiener--Zygmund theorem on the almost sure nowhere differentiability of the Brownian motion.


Indeed, since $W$ and $f$ are bounded,
\begin{equation}
 A\subseteq B=\bigcup\_{M=1}^\infty B\_M, 
\end{equation}
where
\begin{equation}
 B:=\Big\{\exists t\_0\in[0,1]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}<\infty\Big\},
\end{equation}
\begin{equation}
 B\_M:=\Big\{\exists t\_0\in[0,1]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}\le M\Big\}. 
\end{equation}
Next,
\begin{equation}
 B\_M:=B\_{M,1}\cup B\_{M,2},
\end{equation}
where
\begin{equation}
 B\_{M,1}:=\Big\{\exists t\_0\in[0,1/2]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}\le M\Big\}, 
\end{equation}
\begin{equation}
 B\_{M,2}:=\Big\{\exists t\_0\in[1/2,1]\ \sup\_{t\in[0,1]}\frac{|W\_f(t)-W\_f(t\_0)|}{|t-t\_0|^s}\le M\Big\}. 
\end{equation}


Let $r$ be any integer such that
\begin{equation}
r>1/(s-1/2). 
\end{equation}
Let then $n$ be any integer $\ge n\_r$, where $n\_r$ is the smaller integer $q$ such that $2^{q-1}>r$.


Assuming the event $B\_{M,1}$ occurs, let $K$ be a random integer in the set $\{1,\dots,2^{n-1}\}$ such that $t\_0\in[\frac{K-1}{2^n},\frac K{2^n}]$. Then for $j=1,\dots,r$
\begin{equation}
\begin{aligned}
 &\Big|W\_f\Big(\frac{K+j}{2^n}\Big)-W\_f\Big(\frac{K+j-1}{2^n}\Big)\Big| \\ 
 &\le\Big|W\_f\Big(\frac{K+j}{2^n}\Big)-W\_f(t\_0)\Big|
 +\Big|W\_f\Big(\frac{K+j-1}{2^n}\Big)-W\_f(t\_0)\Big| \\ 
 &\le M\Big|\frac{K+j}{2^n}-t\_0\Big|^s
 +M\Big|\frac{K+j-1}{2^n}-t\_0\Big|^s\le\frac{2Mj^s}{2^{sn}} \le\frac{2Mr^s}{2^{sn}} 
\end{aligned} 
\end{equation}
So,
\begin{equation}
B\_{M,1}\subseteq\bigcap\_{n\ge n\_r}\bigcup\_{k=1}^{2^{n-1}}C\_{n,k},
\end{equation}
where
\begin{equation}
 C\_{n,k}:=\bigcap\_{j=1}^r \Big\{\Big|W\_f\Big(\frac{k+j}{2^n}\Big)-W\_f\Big(\frac{k+j-1}{2^n}\Big)\Big|
 \le\frac{2Mr^s}{2^{sn}}\Big\}. 
\end{equation}
By the independence of the increments of the Wiener process and because the pdf of the normally distributed random variable $W\_f\big(\frac{k+j}{2^n}\big)-W\_f\big(\frac{k+j-1}{2^n}\big)$ is $\le2^{n/2-1}$, we have
\begin{equation}
\begin{aligned}
 P(C\_{n,k})&=\prod\_{j=1}^r P\Big(\Big|W\_f\Big(\frac{k+j}{2^n}\Big)-W\_f\Big(\frac{k+j-1}{2^n}\Big)\Big|
 \le\frac{2Mr^s}{2^{sn}}\Big) \\ 
& \le\Big(2^{n/2}\frac{2Mr^s}{2^{sn}}\Big)^r=\frac C{2^{(1+a)n}}, 
\end{aligned}
\end{equation}
where $C:=(2Mr^s)^r$ and $a:=r(s-1/2)-1>0$. So,
\begin{equation}
 P\Big(\bigcup\_{k=1}^{2^{n-1}}C\_{n,k}\Big)\le2^{n-1}\frac C{2^{(1+a)n}}\le\frac C{2^{an}}
\end{equation}
and
\begin{equation}
 P^\*(B\_{M,1})\le\lim\_{n\to\infty}\frac C{2^{an}}=0. 
\end{equation}
So, $P^\*(B\_{M,1})=0$. Similarly, $P^\*(B\_{M,2})=0$. So, $P^\*(B\_M)=0$, for all $M$.


Thus,
\begin{equation}
 P^\*(A)\le P^\*(B)\le\sum\_{M=1}^\infty P^\*(B\_M)=0, 
\end{equation}
as claimed. $\quad\Box$


