It is well known that for a brownian process $B(t),t\geq 0$, it holds $$ \sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}<\infty $$ almost surely, for any $T>0$ and $\alpha<1/2$.
My question is: how about $$ \mathbb{E}\left[\left(\sup_{0\leq s<t\leq T}\frac{|B(t)-B(s)|}{|t-s|^\alpha}\right)^p\right] $$ for $p\geq 1$ ?
It looks like it is finite for all $p > 1$. Here is a neat "bootstrap" argument (I still cannot believe it works, but I fail to find an error).
Denote $$S_T = \sup_{0\le s<t\le T} \dfrac{|B(t) - B(s)|}{|t - s|^\alpha}.$$ By self-similarity of the Brownian motion, we have $$S_2 \stackrel{d}{=} 2^{1/2 - \alpha} S_1 .$$ On the other hand, considering the intervals $t, s \in [0, 1]$ and $t, s \in [1, 2]$ separately, we see that $$S_2 \stackrel{d}{\ge} \max(S_1, S_1'),$$ where $S_1'$ is an independent copy of $S_1$. It follows that $$\begin{aligned}\mathbb{P}(S_1 > x) & = \mathbb{P}(S_2 > 2^{1/2 - \alpha} x) \ge \mathbb{P}(\max(S_1, S_1') > 2^{1/2 - \alpha} x) \\ & = 2 \mathbb{P}(S_1 > 2^{1/2 - \alpha} x) - (\mathbb{P}(S_1 > 2^{1/2 - \alpha} x))^2 \\ & \ge (2 - \epsilon) \mathbb{P}(S_1 > 2^{1/2 - \alpha} x)\end{aligned}$$ for $x$ large enough (because we know that $S_1$ is finite a.s., so that $\mathbb{P}(S_1 > 2^{1/2 - \alpha} x)$ goes to zero as $x \to \infty$). By induction, $$\mathbb{P}(S_1 > 2^{n (1/2 - \alpha)} x) \le (2 - \epsilon)^{-n} \mathbb{P}(S_1 > x),$$ which means that $$\mathbb{P}(S_1 > y) \le C y^{-\frac{\log(2 - \epsilon)}{(1/2 - \alpha) \log 2}} .$$ In particular, $\mathbb{E}[S_1^p]$ is finite if $$p < \frac{\log(2 - \epsilon)}{(1/2 - \alpha) \log 2} $$ for some $\epsilon > 0$, that is, if $p < 2 / (1 - 2 \alpha)$.
On the other hand, $S_1$ is clearly an increasing function of $\alpha \in [0, 1/2)$, so that if $\mathbb{E}[S_1^p]$ is finite for some $\alpha$, it is also finite for all smaller values of $\alpha$. We conclude that $\mathbb{E}[S_1^p] < \infty$ for all $p > 1$ and $\alpha \in [0, 1/2)$.
Much more is true: we have $$\mathbb{E} [\exp(\epsilon \|B\|_{0,\alpha}^2)] < \infty$$ for some $\epsilon > 0$. Here $\|\omega\|_{0,\alpha} = \sup_{0 \le s < t \le T} \frac{|\omega(t)-\omega(s)|}{|t-s|^\alpha}$ is the Hölder norm you want.
This is basically Fernique's theorem but we need to be a little more careful because the Hölder space $C^{0,\alpha}([0,T])$ is non-separable. Instead, fix some $1/2 > \beta > \alpha$ and let $X$ be the closure of $C^{0,\beta}([0,T])$ in the $C^{0,\alpha}$ norm. Then $X$ is a separable Banach space (this follows for instance from the compact embedding).
We have $X \subset C([0,1])$ and $X$ has full Wiener measure, so we can view Wiener measure as a Borel probability measure on $X$. One can verify that it is in fact a Gaussian measure. (The issue is that $X$ has a larger dual than $C([0,1])$, and we have to check that these extra functionals still have a Gaussian distribution - but they're weak-* limits of functionals from $C([0,1])^*$.) So now Fernique's theorem applies on $X$ and gives the above result.
Some details can be found in
Baldi, P.; Ben Arous, G.; Kerkyacharian, G., Large deviations and the Strassen theorem in Hölder norm, Stochastic Processes Appl. 42, No. 1, 171-180 (1992). ZBL0757.60014.
in which they prove an even stronger statement, a large deviations principle.
First, by a self-similarity argument, it suffices to consider the case $T=1$. We can use the equivalence of the usual Hölder norm with the sequence norm, defined by $$ \lVert x\rVert_\alpha:=\sup_{j\geqslant 1}2^{j\alpha}\max_{1\leqslant k\leqslant 2^{j-1}}\left\lvert x\left(\left(2k\right)2^{-j}\right)-2x\left(\left(2k-1\right)2^{-j}\right) +x\left(\left(2k-2\right)2^{-j}\right)\right\rvert +\left\lvert x\left(0\right)\right\rvert+\left\lvert x\left(1\right)\right\rvert.$$ This was established in Ciesielski, Z. (1960). On the isomorphisms of the spaces Hα and m. Bull. Acad. Pol. Sci. Ser. Sci. Math. Phys. 8, 217–222.
Applying this to our problem, it suffices to consider the moments of the random variable $$ Y:=\sup_{j\geqslant 1}2^{j\alpha}\max_{1\leqslant k\leqslant 2^{j-1}}\left\lvert B\left(\left(2k\right)2^{-j}\right)-2B\left(\left(2k-1\right)2^{-j}\right) +B\left(\left(2k-2\right)2^{-j}\right)\right\rvert. $$ We control for a fixed $t$ the probability that $Y\gt 2t$ in the following way: $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty}\sum_{k=1}^{2^{j-1}} \Pr\left(\left\lvert B\left(\left(2k\right)2^{-j}\right)-2B\left(\left(2k-1\right)2^{-j}\right) +B\left(\left(2k-2\right)2^{-j}\right)\right\rvert\gt 2t2^{-j\alpha}\right). $$ Using stationarity of increments of standard Brownian motion, we derive the bound $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty} 2^{j } \Pr\left(\left\lvert B\left( 2^{-j}\right) \right\rvert\gt t2^{-j\alpha}\right). $$ By self-similarity, we derive that $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty} 2^{j } \Pr\left(\left\lvert N\right\rvert\gt t2^{j\left(1/2-\alpha\right)}\right), $$ where $N$ has a standard normal distribution. A comparison between series and integrals yields $$\tag{*} \Pr\left(Y\gt 2t\right)\leqslant \mathbb E\left[ \left\lvert N\right\rvert^{\frac 1{1/2-\alpha}}t^{-\frac 1{1/2-\alpha}} \mathbb 1\left\{\left\lvert N\right\rvert \gt t\right\} \right]. $$ Multiplying by $t^{p-1}$ and integrating over $(1,+\infty)$ shows that the Hölderian norm has moments of all orders. More can be deduced from the point of view of exponential moments. Indeed, for $\varepsilon\lt 1/2$, $$ \mathbb{E} \left[\exp\left(\varepsilon \|B\|_{\alpha}^2\right)\right] < \infty. $$ Multypling indeed in (*) by the derivative of $t\mapsto \exp\left(\varepsilon t^{2}\right)$ and integrate on $(1,+\infty)$.
