Kolmogorov’s Continuity Theorem

The aim is to give a proof of the Kolmogorov’s Continuity Theorem, also known as Kolmogorov-Chentsov Theorem. The main reference we are following is Probability Theory, An Analytic View by Daniel W. Stroock.

Theorem : Suppose a stochastic process $X=\{X(t):0\le t\le T\}$ takes values in a Banach space $B$ so that for some $p>1$, $C>0$, $r>0$,

$$\mathbb{E}{\left[{\Vert X(t)-X(s)\Vert }_B^p\right]}^{\frac{1}{p}}\le C|t-s{|}^{\frac{1}{p}+r}$$

for all $s,t\in [0,T]$. Then there exists a family $\{\tilde{X}(t):t\in [0,T]\}$ of random variables such that $X(t)=\tilde{X}(t)$ almost surely for all $t\in [0,T]$ and $t\in [0,T]\mapsto \tilde{X}(t,\omega )\in B$ is continuous for all $\omega \in \mathrm{\Omega }$. In fact, for each $\alpha \in (0,r)$, we have

$$\mathbb{E}{\left[\underset{0\le s<t\le T}{sup}{\left(\frac{{\Vert \tilde{X}(t)-\tilde{X}(s)\Vert }_B}{(t-s{)}^{\alpha }}\right)}^p\right]}^{\frac{1}{p}}\le K{T}^{\frac{1}{p}+r-\alpha }$$

for a suitable constant $K=K(\alpha ,r,C)$.

Proof : By rescaling time, we can assume without loss of generality, that $T=1$.

Now, for $n\ge 0$, we define

$$M_n:=\underset{1\le m\le 2^n}{max}{\Vert X\left(m{2}^{-n}\right)-X((m-1)2^{-n})\Vert }_B$$

and observe that

$$\mathbb{E}{\left[M_n^p\right]}^{\frac{1}{p}}\le \mathbb{E}\left[{\left(\sum _{m=1}^{2^n}{\Vert X\left(m{2}^{-n}\right)-X((m-1)2^{-n})\Vert }_B^p\right)}^{\frac{1}{p}}\right]$$

which is bounded by $C{2}^{-rn}$.

Next, consider the path $t⇝X(t)$ on each interval $[(m-1)2^{-n},m{2}^{-n}]$ and linearize it to $t⇝X_n(t)$. It is easy to see that

$$\underset{t\in [0,1]}{max}{\Vert X_{n+1}(t)-X_n(t)\Vert }_B=\underset{1\le m\le 2^n}{max}\Vert X((2m-1)2^{-n-1})-\frac{X((m-1)2^{-n})-X(m{2}^{-n})}{2}\Vert$$

which is bounded by $M_{n+1}$.

So,

$$\mathbb{E}{\left[\underset{t\in [0,1]}{sup}{\Vert X_{n+1}(t)-X_n(t)\Vert }_B^p\right]}^{\frac{1}{p}}\le C{2}^{-rn}$$

and so there exists a measurable $\tilde{X}:[0,1]\times \mathrm{\Omega }\to B$ satisfying

$$\mathbb{E}{\left[\underset{t\in [0,1]}{sup}{\Vert \tilde{X}(t)-X_n(t)\Vert }_B^p\right]}^{\frac{1}{p}}\le \frac{C{2}^{-rn}}{1-2^{-r}}$$

and $t⇝\tilde{X}(t,\omega )$ is continuous for all $\omega \in \mathrm{\Omega }$.

Next, notice that for all $t\in [0,1]$,

$${\Vert \tilde{X}(\tau )-X(t)\Vert }_B\underset{\tau \to t}{\overset{p}{\to }}0$$

and hence $\tilde{X}(\tau )=X(t)$ almost surely.

Finally, note that for $2^{-n-1}\le t-s\le 2^{-n}$, we have

$$\begin{array}{rl}{\Vert \tilde{X}(t)-\tilde{X}(s)\Vert }_B& \le {\Vert \tilde{X}(t)-X_n(t)\Vert }_B+{\Vert X_n(t)-X_n(s)\Vert }_B+{\Vert X_n(s)-\tilde{X}(s)\Vert }_B\\ & \le 2\underset{\tau \in [0,1]}{sup}{\Vert \tilde{X}(\tau )-X_n(\tau )\Vert }_B+2^n(t-s)M_n\end{array}$$

and hence

$$\frac{{\Vert \tilde{X}(t)-\tilde{X}(s)\Vert }_B}{(t-s{)}^{\alpha }}\le 2^{\alpha n+\alpha +1}\underset{\tau \in [0,1]}{sup}{\Vert \tilde{X}(\tau )-X_n(\tau )\Vert }_B+2^{n+\alpha n}{M}_n$$

which implies

$$\begin{array}{rl}\mathbb{E}{\left[\underset{0\le s<t\le 1}{sup}{\left(\frac{{\Vert \tilde{X}(t)-\tilde{X}(s)\Vert }_B}{(t-s{)}^{\alpha }}\right)}^p\right]}^{\frac{1}{p}}& \le C\sum _{n=0}^{\mathrm{\infty }}(\frac{2^{\alpha n+\alpha +1-rn}}{1-2^{-r}}+2^{\alpha n-rn})\\ & \le \frac{5C}{(1-2^{-r})(1-2^{\alpha -r})}\end{array}$$

hence completing the proof.