Weighted global Holder property for Brownian motion paths 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?
 A: 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
