Let $\mathcal H^p$, with $p \in [1,\infty)$, be the space of all (continuous-time) martingales $M$ such that $$ \|M\|_{\mathcal H^p} := \mathbb E\left[\sup_t \left| M_t\right|^p\right]^{1/p} < \infty. $$

I want to show that (identifying indistinguishable martingales) $\mathcal H^p$ is complete. Could someone help me complete the following proof?

Suppose that $\left\{ X^n \right\}$ is a Cauchy sequence in $\mathcal H^p$, so that, for every $\varepsilon > 0$, there is an $N$ such that $$ \| X^n -X^m \|_{\mathcal H^p} < \varepsilon $$ for all $n,m \ge N$. Since $$ \sup_t \mathbb E\left[\left| X^n_t - X^m_t \right|^p\right] \le \| X^n -X^m \|_{\mathcal H^p}^p, $$ we have that $\left\{ X^n_t \right\}$ is uniformly (in $t$) Cauchy in $L^p$, and thus must converge uniformly in $L^p$ to some $X_t \in L^p$.

It is straightforward to verify that $X$ must be a martingale. How do you show that $X^n \to X$ in the $\mathcal H^p$-norm?

By passing to a subsequence if necessary, we can assume that $X^n_t \to X_t$ uniformly almost surely. That is, $$ \sup_t | X_t^n -X_t |^p \to 0 \quad \text{a.s.} $$ One might then be able to appeal to an appropriate convergence theorem to show that $$ \mathbb E \left[ \sup_t | X_t^n -X_t |^p \right] \to 0. $$ We can then use the fact that $\{X^n\}$ is Cauchy to recover convergence for the original sequence. Equivalently, one can show that $\sup_t | X_t^n -X_t |^p$ is uniformly integrable. Unfortunately, I don't know how to show this.

Another thought is to use Doob's $L^p$ inequality, but that only works for $p > 1$.

Any suggestions? I suspect this should be rather simple, but my analysis is quite rusty.


The proof is quite standard.

By Chebyshev and Borel-Cantelli one may find a subsequence $\{M^{(n)}\}$ such that $\sum_n(M^{(n)}_t-M^{(n-1)}_t)$ converges uniformly a.s.; denote its limits by $N_t$. It suffices to show $M^{(n)}\to N$ in $\mathcal H^p$, which amounts to prove $\sum_{n\ge n_0}\|M^{(n)}-M^{(n-1)}\|_{\mathcal H^p}$ converges to zero. By choosing the subsequence properly, e.g. in a way such that $\|M^{(n)}-M^{(n-1)}\|_{\mathcal H^p}<2^{-n}$, the conclusion is obvious.

  • $\begingroup$ Thanks a lot. I've forgotten the telescoping sum trick. Could you elaborate on the use of Chebyshev and Borel-Cantelli to find the subsequence? I know from other results that we can find this subsequence, but I don't believe they make use of the results you mention. $\endgroup$ – Theoretical Economist Sep 11 at 15:38
  • $\begingroup$ I think I'm missing something. Your answer argues that $$ \| M^{(n)} - N \| = \left\| \sum_{k\ge n} \left( M^{(k+1)} -M^{(k)} \right)\right\| \le \sum_{k \ge n} \left\| M^{(k+1)} - M^{(k)} \right\|. $$ However, the triangle inequality is valid only for finite sums. So actually we would need $$ \lim_{m\to\infty} \left\| \sum_{k = n}^m \left( M^{(k+1)} -M^{(k)} \right)\right\| = \left\| \sum_{k\ge n} \left( M^{(k+1)} -M^{(k)} \right)\right\|.$$ That is, we need to be able to take the limits inside the integral that defines the norm. How do you know this step is valid? $\endgroup$ – Theoretical Economist Sep 11 at 15:53
  • $\begingroup$ Apologies if I've misunderstood your answer. Would an appeal to Fatou work? $\endgroup$ – Theoretical Economist Sep 11 at 15:54
  • $\begingroup$ @TheoreticalEconomist The $L^p$ triangle inequality is valid for infinite sums: this follows from the triangle inequality for real numbers (which is valid for infinite sums) and monotone convergence theorem. $\endgroup$ – Cave Johnson Sep 12 at 1:27

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