For some $\alpha>0$, $ e^L=P\left(\exp(\alpha\sup_{|s-t|\le\delta}\frac{|B_s-B_t|^2}{|s-t|})<\infty\right) $?

Question

I am reading one lecture note Dynamics for Spherical Models of Spin-Glass and Aging by Alice Guionnet. On page 124, it says that for some $\alpha>0$, $$ e^L=P\left(\exp(\alpha\sup_{|s-t|\le\delta}\frac{|B_s-B_t|^2}{|s-t|})<\infty\right) $$ Moreover, how to show that $$ E\left[\exp(\alpha\sup_{|s-t|<\delta}\frac{|B_s-B_t|^2}{|s-t|})\right]\le e^L $$

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I do not know what is $L$ here. I check the the reference and found that I just found a useful result: for $\alpha-$Holder continuity Brownian motion, 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

First, there is a typo there - should be E instead of P, as in the second display in your question. Second, the argument as written is not quite right, but it can be rescued. Indeed, $\sup_{t<\delta} (B_t/\sqrt{t})=\infty$, and therefore that expectation is $\infty$. The solution is not to divide by $|t-s|$ but rather by $|t-s|^{\alpha}$ for some $\alpha$ close to $1$- then the expectation will be finite (using arguments from [28] to control the mean of the sup, and Borel's lemma to control the deviations). The only difference is that instead of $1/M\delta$ you will have $1/M\delta^\alpha$, which is still fine for the rest of the argument.