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It is well-known that the Brownian motion (Wiener process) is almost sure locally $\alpha$-Holder for any $\alpha<1/2$. That is, with probability 1 $$ \sup_{t,s\in[0,1]}\frac{|W_t-W_s|}{|t-s|^{\alpha}}<\infty. $$

On the other hand, Brownian motion is clearly not globally $\alpha$-Holder. Indeed, it follows from the law of iterated logarithm, that $$ \sup_{t\ge0}\frac{|W_t|}{|t|^{\alpha}}=\infty. $$

One can play with the first inequality and with the help of Kolmogorov theorem and Borell-Cantelli obtain something that looks like ``global-local'' weighted Holder $$ \sup_{\substack{t,s\ge0\\ |t-s|\in[0,1]}}\frac{|W_t-W_s|}{(t^\delta\vee s^\delta\vee1)|t-s|^{\alpha}}<\infty. $$ for any arbitrarily small $\delta>0$.

My question is whether a global weighted Holder of the form $$ \sup_{t,s\ge0}\frac{|W_t-W_s|}{(t^\gamma\vee s^\gamma\vee 1)|t-s|^{\alpha}}<\infty $$ holds? What is the optimal weights exponent $\gamma$ (can one say any better than $\gamma=1$)? Does there exist a global version of Kolmogorov continuity theorem that one can directly apply?

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    $\begingroup$ If $s=1$ and $t$ is large, the quantity in your last line is comparable to $|W_t|/t^{\alpha}$, which as you know has infinite supremum. So it can't hold in the form that you've stated. $\endgroup$ – Nate Eldredge Feb 8 '16 at 19:57
  • $\begingroup$ @NateEldredge Sorry I had a typo. One should read $t^\gamma\vee s^\gamma\vee1$. Then the quantity in my last line is comparable to $|W_t|/t^{\alpha+\gamma}$ which has a finite supremum. $\endgroup$ – Oleg Feb 9 '16 at 10:27
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Let $\alpha,\gamma >0$. As Nate notes, for the inequality $$ \sup_{t,s \ge 0}\frac{|W_t-W_s|}{(t^\gamma\vee s^\gamma\vee 1)|t-s|^{\alpha}}<\infty $$ to hold almost surely, it is necessary that $\gamma+ \alpha>1/2$. Of course $\alpha<1/2$ is also needed.

Claim: These conditions are also sufficient.

As observed by the original poster, the claim holds if we restrict attention to $|t-s| \le 1$. This is where the condition that $\alpha<1/2$ is needed. One proof for this case is quite similar to the argument below, considering separately $t \in [2^{n-1},2^n]$.

So it suffices to consider $1<|t-s|$. For integers $0 \le k \le n$ and a constant $R>1$, denote by $A_n(k,R)$ the event that $$ \frac{|W_t-W_s|}{(t^\gamma\vee s^\gamma\vee 1)|t-s|^{\alpha}}\ge R $$ holds for some $s<t$ in $[2^{n-1},2^n]$ that satisfy $2^{k-1} \le t-s \le 2^k$.

Since $\max_{t \le T} W_t$ has the same law as $|W_t|$ (see e.g., Theorem 2.21 in [1]), the standard bound for Gaussian tails (see, e.g., Lemma 12.9 in [1]) implies that $${\bf P}[A_n(k,R)] \le 2^n \cdot\exp(-cR^2 \cdot 2^{2n\gamma+2k\alpha-k}) \le 2^n\cdot\exp(-cR^2 \cdot 2^{2n(\gamma+\alpha-1/2)}) $$ where $c=c(\alpha,\gamma)$ is a constant. Summing this over all pairs of integers $0 \le k \le n$ and then taking $R \to \infty$ proves the claim.

[1] Peter Morters and Yuval Peres, Brownian Motion, Cambridge University Press (2010), available at http://research.microsoft.com/en-us/um/people/peres/brbook.pdf

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  • $\begingroup$ Yuval, thanks for your answer. Could you please clarify how does the event $A_n(k,R)$ depend on $R$? Did you mean something like $A_n(k,R):=\{ |W_t-W_s|/|t-s|^{\alpha}(t^\gamma\vee s^\gamma \vee 1)\le R\}$? $\endgroup$ – Oleg Feb 14 '16 at 13:09
  • $\begingroup$ I also did not find Lemma 12.1 in [1]. Did you mean Lemma 12.9? $\endgroup$ – Oleg Feb 14 '16 at 13:31
  • $\begingroup$ Hi Oleg, I corrected the definition of $A_n(k,R)$ and the citation. $\endgroup$ – Yuval Peres Mar 6 '16 at 8:24

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