# Regularity of translations for Brownian motion

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?

• If $s=1/2$, then almost surely there will be points at which the Brownian motion is $s$-Hölder continuous -- see e.g. Remark 1.21 in Brownian Motion by Peter Mörters and Yuval Peres. I have edited your post accordingly. Feb 12, 2023 at 3:07

The result holds for any bounded function $$f$$, in the following sense: for any real $$s>1/2$$, $$$$P^*(A)=0,$$$$ where $$$$A:=\Big\{\exists t_0\in[0,1]\ \limsup_{t\to t_0}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}<\infty\Big\},$$$$ $$P^*$$ is the outer probability, $$\limsup_{t\to t_0}:=\limsup_{t\to t_0,t\in[0,1]}$$, $$$$W_f:=W+f,$$$$ and $$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, $$$$A\subseteq B=\bigcup_{M=1}^\infty B_M,$$$$ where $$$$B:=\Big\{\exists t_0\in[0,1]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}<\infty\Big\},$$$$ $$$$B_M:=\Big\{\exists t_0\in[0,1]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}\le M\Big\}.$$$$ Next, $$$$B_M:=B_{M,1}\cup B_{M,2},$$$$ where $$$$B_{M,1}:=\Big\{\exists t_0\in[0,1/2]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}\le M\Big\},$$$$ $$$$B_{M,2}:=\Big\{\exists t_0\in[1/2,1]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}\le M\Big\}.$$$$

Let $$r$$ be any integer such that $$$$r>1/(s-1/2).$$$$ Let then $$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{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} So, $$$$B_{M,1}\subseteq\bigcap_{n\ge n_r}\bigcup_{k=1}^{2^{n-1}}C_{n,k},$$$$ where $$$$C_{n,k}:=\bigcap_{j=1}^r \Big\{\Big|W_f\Big(\frac{k+j}{2^n}\Big)-W_f\Big(\frac{k+j-1}{2^n}\Big)\Big| \le\frac{2Mr^s}{2^{sn}}\Big\}.$$$$ By the independence of the increments of the Wiener process and because the pdf of the normally distributed random variable $$W_f\big(\frac{k+j}{2^n}\big)-W_f\big(\frac{k+j-1}{2^n}\big)$$ is $$\le2^{n/2-1}$$, we have \begin{aligned} P(C_{n,k})&=\prod_{j=1}^r P\Big(\Big|W_f\Big(\frac{k+j}{2^n}\Big)-W_f\Big(\frac{k+j-1}{2^n}\Big)\Big| \le\frac{2Mr^s}{2^{sn}}\Big) \\ & \le\Big(2^{n/2}\frac{2Mr^s}{2^{sn}}\Big)^r=\frac C{2^{(1+a)n}}, \end{aligned} where $$C:=(2Mr^s)^r$$ and $$a:=r(s-1/2)-1>0$$. So, $$$$P\Big(\bigcup_{k=1}^{2^{n-1}}C_{n,k}\Big)\le2^{n-1}\frac C{2^{(1+a)n}}\le\frac C{2^{an}}$$$$ and $$$$P^*(B_{M,1})\le\lim_{n\to\infty}\frac C{2^{an}}=0.$$$$ So, $$P^*(B_{M,1})=0$$. Similarly, $$P^*(B_{M,2})=0$$. So, $$P^*(B_M)=0$$, for all $$M$$.

Thus, $$$$P^*(A)\le P^*(B)\le\sum_{M=1}^\infty P^*(B_M)=0,$$$$ as claimed. $$\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.