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I'm interested in the behaviour of the stopped $\alpha$-Hölder norm of a one-dimensional real-valued Brownian motion $(B_t)_{t \geq 0}$ for $\alpha < 1/2$.

For fixed $T>0$, self similarity implies that $$\Bbb E\left[ \sup_{0 \leq s < t \leq T} \frac{|B_t-B_s|}{|t-s|^\alpha}\right] = T^{1/2 - \alpha}\Bbb E\left[ \sup_{0 \leq s < t \leq 1} \frac{|B_t-B_s|}{|t-s|^\alpha}\right].$$

Does the corresponding upper bound extend to the stopped Hölder norm? In other words, does there exist a random integrable constant $C$ such that for all bounded stopping times $\tau > 0$ (w.r.t. the sigma algebra generated by $B_t$), $$\Bbb E\left[ \sup_{0 \leq s < t \leq \tau} \frac{|B_t-B_s|}{|t-s|^\alpha}\right] \leq \Bbb E [C \tau^{1/2 - \alpha}],$$ where $C$ depends on $\alpha < 1/2$?

Thanks in advance.

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Yes, see in Revuz-Yor exercise 4.27 chapter IV. We will apply theorem 4.10 to the the process

$$A_{T}:=\sup_{0 \leq s < t \leq T} \frac{|B_t-B_s|}{|t-s|^\alpha}.$$

It is adapted, continuous and increasing. Using this Q&A, we indeed get by Markov inequality that

$$\lim_{b\to +\infty }P_{x}(A_{\lambda^{2}}>b\lambda)=0.$$

We also have by the Markov property that

$$ \begin{split} A_{T+S}-A_{S} & \leq\sup_{S \leq s < t \leq T+S} \frac{|B_t-B_s|}{|t-s|^\alpha}\\ & =\sup_{0 \leq s < t \leq T} \frac{|B_{t+S}-B_{s+S}|}{|t-s|^\alpha}\\ & =A_{T}\circ \theta_{S}, \end{split} $$ where $\theta_{S}$ is the shift of the path (here for the cross-term $$C_{S,T+S}:=\sup_{0 \leq s\leq S \leq t \leq T+S} \frac{|B_t-B_s|}{|t-s|^\alpha}$$ we used the bound $$ \frac{|B_t-B_s|}{|t-s|^\alpha}\leq \frac{|B_t-B_{S}|}{|t-S|^\alpha}+ \frac{|B_S-B_s|}{|S-s|^\alpha}$$

and so $C_{S,T+S}-A_{S}\leq A_{S,T+S}$.) Therefore, by theorem 4.10 we get for any moderate function F (i.e. increasing, continuous, vanishing at zero and $\sup_{x>0}\frac{F(ax)}{F(x)}=\gamma<\infty$ for some $a>1$)

$$\Bbb E\left[ F\left(\sup_{0 \leq s < t \leq \tau} \frac{|B_t-B_s|}{|t-s|^\alpha}\right)\right] \leq c_{F} \Bbb E [F(\tau^{1/2})].$$

So in particular we can take $F(x)=x^{p}$ for $p\geq 1$.

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    $\begingroup$ Could you supply a little more detail in the first step of your claim that $A_{T+S} -A_S =A_T\circ\theta_S$? Clearly $$ A_{T+S} = \max\left( A_S, \sup_{0\le s< S, S\le t\le T+S}{|B_t-B_s|\over |t-s|^\alpha}, A_{T}\circ\theta_S\right), $$ but after that ...? $\endgroup$
    – John Dawkins
    Commented Sep 21 at 17:09
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    $\begingroup$ @JohnDawkins I should have added an inequality, not equality. I fixed it, thank you. $\endgroup$
    – Thomas Kojar
    Commented Sep 21 at 17:50
  • $\begingroup$ Dear Thomas, thank you for the helpful reference and elucidating comments. I think you have a small typo here though! I think it should be $$A_T = \left(\sup_{0 \leq s < t \leq T} \frac{|B_t-B_s|}{|t-s|^\alpha}\right)^{\frac{1}{2(1/2-\alpha)}}?$$ For $F(x) = x^{2p(1/2-\alpha)}$, the desired inequality then follows (for each $p$) $\endgroup$
    – user2103480
    Commented Sep 22 at 21:02
  • $\begingroup$ @ThomasKojar I didn't mean to doubt the validity for general $p$. However, as it stands, Theorem 4.10 in Revuz-Yor would yield an upper bound of $\Bbb E[\tau^{p/2}]$ instead of $\Bbb E[\tau^{p(1/2-\alpha)}]$ for $A_T$ as defined in your answer $\endgroup$
    – user2103480
    Commented Sep 22 at 22:56
  • $\begingroup$ @user2103480 I see. Yes, for your particular bound you indeed need a slightly different $A_t$. So we need to replace the triangle inequality by the convexity inequality $$|a+b|^r<2^{r-1}(|a|^r+|b|^r), r\geq 1.$$ $\endgroup$
    – Thomas Kojar
    Commented Sep 22 at 23:07

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