Can't figure out "standard application" of the Garsia-Rodemich-Rumsey Lemma

Question

I'm currently reading the paper http://arxiv.org/abs/0908.2473 and can't figure out what they call a "standard application" of the Garsia-Rodemich-Rumsey lemma (see p.8). Summed up, they have a sequence $(X_t^h)_{t \geq 0}$ of processes such that $X^h_0=0$ and show that for $0<s<t$ and $p \geq 1$ $$ \operatorname{E} [ | X_t^h |^p ]\leq C h^{2p} |t-s|^p.$$ From this estimate, they conclude by the "standard application" of the GRR lemma that $X_t^h$ converges to zero in probability, uniformly on compact sets, i.e. for any $T>0$, $$ \sup_{0 \leq t \leq T} |X_t^h| \to 0$$ in probability.

The lemma is as follows (taken from the book "Multidimensional Stochastic Processes and Rough Paths" by Friz and Victoir, p.573, the original paper is from the seventies):

Consider $f \in \mathcal{C}([0,T],E)$ where $(E,d)$ is a metric space. Let $\Psi$ and $p$ be continuous, strictly increasing functions on $[0,\infty)$ with $p(0)=\Psi(0)=0$ and $\Psi(x) \to \infty$ for $x \to \infty$. Then, $$\int_0^T \int_0^T \Psi \left( \frac{d(f_s,f_t)}{p(|t-s|)} \right) \, \mathrm{d}s \, \mathrm{d}t \leq F$$ implies, for $0 \leq s \leq t \leq T$, $$d(f_s,f_t) \leq 8 \int_0^{t-s} \Psi^{-1} \left( \frac{4F}{u^2} \right) \, \mathrm{d} p(u).$$

I naively tried taking $d(f_s,f_t)= \left(\operatorname{E} [|f_s-f_t|^p]\right)^{1/p}$ but then can't get the supremum inside the expectation. Can somebody help me out here?

Answer 1

Let $p \ge 1$ and $\alpha >p^{-1}$. From the GRR inequality, there exists a constant $C_{\alpha,p} >0$ such that for any continuous function $f$ on $[0,T]$, and for all $t,s \in [0,T]$ one has:

$ |f(t)-f(s)|^p \le C_{\alpha,p} |t-s|^{\alpha p-1} \int_0^T \int_0^T \frac{ |f(x)-f(y)|^p}{ |x-y|^{\alpha p+1}} dx dy. $

Using the inequality with $s=0$, $f(t)=X_t^h$ and taking the sup over $t \in [0,T]$ gives

$ \sup_{t \in [0,T]} | X_t^h |^p \le C_{\alpha,p} T^{\alpha p-1}\int_0^T \int_0^T \frac{ |X^h_x-X^h_y|^p}{ |x-y|^{\alpha p+1}} dx dy $

From your assumption, the right hand side converges to 0 in probability.