The Kolmogorov Continuity theorem (see for example the Wikipedia page) lets us prove that a stochastic process $X_t$ (on some complete metric space $(S,d)$) is Holder continuous almost surely provided we have a bound of the form $$\mathbb E\left[d(X_t,X_s)^\alpha\right]\le K |t-s|^{1+\beta}$$ where $\alpha, \beta>0$. More precisely, it shows that there exists a random variable $c>0$ such that $$d(X_t,X_s)\le c|t-s|^\gamma $$ where $\gamma\in (0,\frac\beta\alpha)$.

However, this doesn't seem to give any information on the Holder constant $c$. I am wondering if there is some relation between $K$ and $c$. For example, is it possible to estimate the moments $$\mathbb E[c^n]$$ in terms of $K$?

This seems to be related to this unanswered question many years ago.


2 Answers 2


One can apply a deterministic result, called Garsia--Rodemich--Rumsey inequality, to estimate $\mathrm{E}[||X||^\alpha_{\gamma;[0,T]}]$. Here is a particular form of this result, which is most convenient for us.

For any $\alpha >1$, $\delta> 1/\alpha$, there is a constant $C(\delta,\alpha)$ such that for any $f\in C([0,T],S)$ and $t,s\in[0,T]$ $$ d(f(t), f(s))^\alpha\le C(\delta,\alpha)|t-s|^{\delta\alpha - 1} \int_s^t \int_s^t \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv. $$ In particular, for $\gamma = \delta - 1/\alpha$ $$ ||f||_{\gamma;[0,T]}:= \sup_{0\le t<s\le T}\frac{d(f(t),f(s))}{|t-s|^\gamma}\le C(\delta,\alpha)^{1/\alpha}\bigg(\int_0^T\int_0^T \frac{d(f(u),f(v))^\alpha}{|u-v|^{\delta\alpha + 1}}du\,dv\bigg)^{1/\alpha}. $$

For any $\gamma<\beta/\alpha$ take $\delta\in (1/\alpha,\gamma+1/\alpha)$ to get $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\delta,\alpha) \int_0^T\int_0^T \frac{\mathrm{E} \left[d(X_u,X_v)^\alpha\right]}{|u-v|^{\delta\alpha + 1}}du\,dv \\ \le C(\delta,\alpha) K\int_0^T\int_0^T |u-v|^{\beta-\delta\alpha}du\,dv = C(\delta,\alpha,\beta,T)K. $$ As a result, $$ \mathrm E\left[||X||_{\gamma;[0,T]}^\alpha\right]\le C(\alpha,\beta,\gamma,T) K, $$ and the constant may be written explicitly.

This is slightly better than @Kostya_I's estimate $\mathrm{P}(||X||_{\gamma;[0,T]}^\alpha > x)\le CKx^{-1}$, which does not imply integrability.


Well, you can read a simple estimate off the proof of Kolmogorov criterion. If $(a,b)$ is a diadic interval, with $|b-a|=2^{-n}$, then, by Chebyshev's inequality, we have $$ \mathbb{P}(d(X_a,X_b)\geq M 2^{-\gamma n})\leq\frac{K2^{-n}2^{(\alpha\gamma-\beta)n}}{M^\alpha}. $$ Suppose we are interested in the Hölder norm on $[0,1]$. The union bound over diadic intervals gives $$ \mathbb{P}(\exists a,b\text{ diadic}:d(X_a,X_b)\geq M|a-b|^\gamma)\leq KM^{-\alpha}(1-2^{\alpha\gamma-\beta})^{-1}. $$ The $\gamma$-Hölder norm of $X_t$ and the norm of its restriction to diadic points differ by a factor of at most $2(1-2^{-\gamma})^{-1}$. Therefore, $$ \mathbb{P}(||X_t||_\gamma\geq M)\leq K\cdot C(\alpha,\beta,\gamma)\cdot M^{-\alpha}. $$ Taking $X_t=tX_1$ with a suitable $X_1$ shows that in general, this cannot be substantially improved (EDIT: in fact, it can be slightly improved, see the answer by zhoraster).

  • $\begingroup$ In fact $C = C(\alpha,\beta,\gamma,|b-a|)$. $\endgroup$
    – zhoraster
    Aug 19, 2017 at 5:39
  • $\begingroup$ you probably mean $C=C(\alpha,\beta,\gamma,T)$ in the notation of you answer? That's right, I'll edit the post accordingly. $\endgroup$
    – Kostya_I
    Aug 19, 2017 at 7:48
  • $\begingroup$ Yes, the constant does depend on the length of the interval. $\endgroup$
    – zhoraster
    Aug 19, 2017 at 8:33

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