How to show that there exists $\beta$ s.t. $ E\left[\exp(\beta\sup_{|s-t|<\delta}\frac{|B_s-B_t|^2}{|s-t|})\right]\le e^L $?

Question

For a standard Brownian motion $B_t$, we know that $B_t$ is $\alpha-$Holder continuity for $\alpha<1/2$. That means there exists $\delta>0$ and $C>0$ s.t. $|s-t|<\delta$, $$ |B_s-B_t|<C|s-t|^{\alpha} $$ with $\alpha<1/2$.

Based on this result, I am reading a paper that the author says that for some $\beta>0$, $$ e^L=P\left(\exp(\beta\sup_{|s-t|<\delta}\frac{|B_s-B_t|^2}{|s-t|})<\infty\right) $$ He does not say what is $L$ here... I guess $L$ is the $L-$Holder continuity of $B_t$?

Question: How to show that there exists $\beta>0$ s.t. $$ E\left[\exp(\beta\sup_{|s-t|<\delta}\frac{|B_s-B_t|^2}{|s-t|})\right]\le e^L $$

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I just found a useful result: there exists $C=C(\alpha)>0$ s.t. $0<\epsilon\le 1$, $$ -C\epsilon^{-\frac{2}{1-2\alpha}}\le \log P(\sup_{|s-t|<\delta}\frac{|B_s-B_t|}{|s-t|^\alpha}\le \epsilon)\le -C^{-1}\epsilon^{-\frac{2}{1-2\alpha}} $$

Answer 1

Maybe I'm missing something here.

Let $$ X := \exp(\beta\sup_{|s-t|<\delta}\frac{|B_s-B_t|^2}{|s-t|})$$.

Then you claim there is $ \beta > 0 $ s.t. $EX \leq P(X<\infty)$. Note that $X \geq 1$ almost surely and $> 1 $ with positive probability, so that seems suspicious.Is some term missing? Is delta supposed to go to zero?