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?


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.