# Reverse martingale convergence theorem in Banach spaces

In section 1.5 of a course given by Gilles Pisier, the author is claiming that in the excerpt below $$\operatorname E[\varphi_i\mid\mathcal A_{-n}]\to\operatorname E[\varphi_i\mid\mathcal A_{-\infty}]$$ in $$L^2(\operatorname P)$$ would follow from classical Hilbert space theory. What does he mean? How does he obtain this convergence? Clearly, we know that if $$\mathcal F\subseteq\mathcal A$$ is a $$\sigma$$-algebra on $$\Omega$$, then $$\pi_{\mathcal F}X:=\operatorname E\left[X\mid\mathcal F\right]\;\;\;\text{for }X\in\mathcal L^1(\operatorname P)$$ is a linear contraction from $$\mathcal L^p(\operatorname P)$$ to $$\mathcal L^p(\mathcal F,\operatorname P):=\{X\in\mathcal L^p(\operatorname P):X\text{ is }\mathcal F\text{-measurable}\}.$$ If $$p=2$$, then $$\pi_{\mathcal F}$$ is an orthogonal projection from $$\mathcal L^2(\operatorname P)$$ to $$\mathcal L^2(\mathcal F,\operatorname P)$$ and hence a self-adjoint operator on $$\mathcal L^2(\operatorname P)$$. By the tower property of conditional expectation, we obtain $$\pi_{\mathcal G}=\pi_{\mathcal F}\pi_{\mathcal G}=\pi_{\mathcal G}\pi_{\mathcal F}\tag2$$ for all $$\sigma$$-algebras $$\mathcal F,\mathcal G\subseteq\mathcal A$$ with $$\mathcal G\subseteq\mathcal F$$.

While all that is clear to me, I don't get which argument he is referring to.

Let $$\{H_n\}_{n \ge 0}$$ be a sequence of Hilbert spaces, with $$H_{n+1} \subset H_n$$ (Clarification: we assume that $$H_{n+1}$$ is a subspace of $$H_n$$) for all $$n\ge 0$$, and denote $$H_\infty=\cap_{n=1}^\infty H_n$$.
Claim: If $$P_n$$ is the orthogonal projection from $$H_0$$ to $$H_n$$, and $$P_\infty$$ is the orthogonal projection from $$H_0$$ to $$H_\infty$$, then for all $$x_0 \in H_0$$, we have $$P_n x\to P_\infty x$$ in norm as $$n \to \infty$$.
Proof: For all $$n \ge 1$$, write $$x_n:=P_n x_0$$ and $$y_n:=x_{n-1}-x_n$$. Then $$y_n$$ is orthogonal to $$H_n$$ for all $$n \ge 1$$, so the elements of the sequence $$\{y_n\}_{n \ge 1}$$ are pairwise orthogonal. Thus Bessel's inequality gives $$\sum_{n=1}^\infty \|y_n\|^2 \le \|x_0\|^2$$, whence $$\{x_n\}$$ is a Cauchy sequence in $$H_0$$, so it converges to some vector $$z$$. Since $$z \in H_n$$ for each $$n$$, we conclude that $$z \in H_\infty$$. Also, for all $$h \in H_\infty$$, we have $$0=\langle x_0-x_n,h\rangle \to \langle x_0-z,h\rangle$$ as $$n \to \infty$$, so $$x_0-z$$ is orthogonal to $$H_\infty$$. Therefore $$z=P_\infty x_0$$.
• Thank you for your answer. I need some clarification to understand your it: With "$H_{n+1}\subseteq H_n$" do you mean that $H_{n+1}$ is continuously embedded into $H_n$ or do you even require that $H_{n+1}$ is a closed subspace of $H_n$ (hence all Hilbert spaces are equipped with corresponding restrictions of the inner product on $H_0$)? In the former case, what do you mean by "orthogonal projection"? I only know this term in the sense of orthogonal projections to closed subspaces of Hilbert spaces. Nov 7, 2021 at 19:02
• $H_{n+1}$ is a closed subspace of $H_n$. That holds in the application. Nov 8, 2021 at 1:02
• Thanks for clarifying. How do you conclude that $\{x_n\}$ is Cauchy? Wouldn't we need $\sum_n\|h_n\|<\infty$ for that?. Nov 10, 2021 at 19:29
• @0xbadf00d For $m<n$ we have $$\|x_m-x_n\|^2=\sum_{k=m+1}^n \|y_k\|^2$$ by Pythagoras. Nov 11, 2021 at 16:26