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?

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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.


1 Answer 1


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 $z=P_\infty x_0$.

  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Nov 7, 2021 at 19:02
  • $\begingroup$ $H_{n+1}$ is a closed subspace of $H_n$. That holds in the application. $\endgroup$ Nov 8, 2021 at 1:02
  • $\begingroup$ Thanks for clarifying. How do you conclude that $\{x_n\}$ is Cauchy? Wouldn't we need $\sum_n\|h_n\|<\infty$ for that?. $\endgroup$
    – 0xbadf00d
    Nov 10, 2021 at 19:29
  • 1
    $\begingroup$ @0xbadf00d For $m<n$ we have $$\|x_m-x_n\|^2=\sum_{k=m+1}^n \|y_k\|^2$$ by Pythagoras. $\endgroup$ Nov 11, 2021 at 16:26
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    $\begingroup$ Yes, you're right. In general, square-summability wouldn't be enough, but here orthogonality saves the day. $\endgroup$
    – 0xbadf00d
    Nov 11, 2021 at 16:52

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