$L^2$ and $C^0$ norms combined

Question

We know that $L^2$ and $C^0$ are complete metric spaces.

Now consider for fixed $T$ the space of real-valued processes $X:\Omega\times [0,T] \rightarrow \mathbb R$ (with corresponding $\sigma$-algebra) such that

$$\Vert X \Vert = \mathbb E [\sup_{0 \leq s \leq T} |X_s| ^2] ^{1/2}< \infty$$

Is this a closed complete metric space?

My attempt: It must be. This space is a subspace of $L^2$ and therefore every Cauchy sequence in the above sense has $L^2$ limit. However, I don't see if this space is closed (i.e. $L^2$ limit also has finite norm in the above sense).

Thank you in advance!

Answer 1

For convenience let us denote this space $S$. I believe we can adapt the standard argument showing completion of the $L^p$ spaces. Let $(X^{(n)})\subset S$ be such that $\sum_{n=1}^\infty \|X^{(n)}\|<\infty$. To show completeness of $S$ it suffices to show that $(\sum_{n=1}^NX^{(n)})_N$ converges to some $X\in S$ in norm. We propose that this $X$ is actually $\sum_{n=1}^\infty X^{(n)}$.

We note that $\sup_{s\leq T}|X_s|^2=\left(\sup_{s\leq T}|X_s|\right)^2$. Thus, for any $N\in \mathbb N$, by the Minkowski inequality, $$\mathbb E\left[\sup_{s\leq T}|\sum_{n=1}^NX^{(n)}_s|^2\right]^{1/2}\leq\mathbb E\left[(\sum_{n=1}^N\sup_{s\leq T}|X^{(n)}_s|)^2\right]^{1/2}\leq\sum_{n=1}^N\mathbb E[\sup_{s\leq T}|X^{(n)}_s|^2]^{1/2}=\sum_{n=1}^N\|X^{(n)}\|.$$ Therefore, $$\lim_{N\to \infty}\mathbb E\left[\sup_{s\leq T}|\sum_{n=1}^NX^{(n)}_s|^2\right]^{1/2}\leq\sum_{n=1}^\infty\|X^{(n)}\|<\infty.$$ Now $\omega\mapsto (\sup_{s\leq T}\sum_{n=1}^N|X^{(n)}_s(\omega)|)^2$ converges to $\omega\mapsto (\sup_{s\leq T}\sum_{n=1}^\infty |X^{(n)}_s(\omega)|)^2$ pointwise monotonically, so the monotone convergence theorem thus gives us that $$\mathbb E\left[(\sum_{n=1}^\infty\sup_{s\leq T}|X^{(n)}_s|)^2\right]^{1/2}<\infty,$$ from which we deduce that $$\|X\|<\infty,$$ so $X\in S$, and $X$ is finite on a set of full measure. It also means for each $N\in\mathbb N$ that $\omega\mapsto \sup_{s\leq T}\sum_{n=1}^N|X_s^{(n)}(\omega)| $ is bounded by an $L^2$ function $\omega\mapsto \sup_{s\leq T}\sum_{n=1}^\infty|X_s^{(n)}(\omega)| $.

This implies that $\sup_{s\leq T}|X_s-\sum_{n=1}^NX^{(n)}_s|$ is also bounded by an $L^2$ function (via triangle inequality), so by the dominated convergence theorem $$\lim_{N\to \infty}\mathbb E\left[\sup_{s\leq T}|X_s-\sum_{n=1}^NX^{(n)}_s|^2\right]^{1/2}=0.$$ Thus the partial sums of $(X^{(n)})$ indeed converge to $X$ in $S$, which finishes the proof.