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?
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. $\quad\Box$
If $f$ is absolutely continuous with $\int_0^1 f'(t)^2\,dt<\infty$, then, by Girsanov's formula (see e.g. Theorem 5.1. in [1]), the process $(B_t+f(t))_{t\in[0,1]}$ is a standard Brownian motion with respect to the measure $Q$ whose density $\frac{dQ}{dP}$ with respect to the distribution, say $P$, of the original standard Brownian motion $(B_t)_{t\in[0,1]}$ is given by the formula $$\frac{dQ}{dP}(x)=\exp\left(-\int_0^1 f'(t)dx_t - \frac{1}{2}\int_0^1 f'(t)^2\,dt\right).$$ It also follows that $P$ is absolutely continuous with respect to $Q$.
So, in this case, for any real $s>1/2$, the process $(B_t+f(t))_{t\in[0,1]}$ is $Q$-almost surely nowhere $s$-Hölder continuous, and hence $P$-almost surely nowhere $s$-Hölder continuous, as desired.
The case of general bounded functions $f$ is now considered in the other answer.
[1] Ioannis Karatzas and Steven E. Shreve, Brownian Motion and Stochastic Calculus, Second Edition, 1991.
