# Weak convergence in $L^2(\mathbb{R}^3,L_{loc}^2(\mathbb{R}^3))$ thanks to diagonal extraction.

Let $$f_n$$ be a sequence bounded in $$L^2(\mathbb{R}^3,L_{loc}^2(\mathbb{R}^3))$$ which means that, for any bounded set in $$\mathbb{R}^3$$, one has for any $$n>0$$

$$\int_{\mathbb{R}^3} \int_{B} |f_n(x,y)|^2 \ dx dy \leq M$$ with a constant $$M$$ independent of $$n$$. I would like to prove that, up a to an extraction $$\sigma$$, $$(f_{\sigma(n)})_n$$ weakly converges toward $$f$$ in $$L^2(\mathbb{R}^3,L_{loc}^2(\mathbb{R}^3))$$.

Using Banach-Alaoglu, I know that there exists an extraction $$\sigma_B$$, depending on the set $$B$$, such that $$(f_{\sigma_B(n)})_n$$ weakly converges toward $$f_B$$ in $$L^2(\mathbb{R}^3,L^2(B))$$.

I know the classic procedure is to use diagonal extraction to get the result, but as I am unfamiliar with the method I don't know how to prove the result. Does anyone know how to justify properly this property or know a reference that explains how to deal with it ?

Any help is welcomed.

Diagonal extraction is the method of proof of the following (very general) property.

Let, for each $$n$$, $$X_n$$ be a sequentially compact space. Then $$P=\prod_{n \geq 1}{X_n}$$ is sequentially compact for pointwise convergence.

Let $$x_n = (x_n(i))_{i \geq 1}$$ be an element of $$P$$, for each $$n \geq 1$$.

Then the sequence $$(x_n(1))_n$$ is a sequence of $$X_1$$ so it has a convergent subsequence $$(x_{\phi_1(n)}(1))_n$$ (to some $$y_1 \in X_1$$). Let $$\psi_1 = \phi_1$$.

The sequence $$(x_{\psi_1(n)}(2))_n$$ is a sequence of $$X_2$$, so it has a convergent subsequence $$(x_{\psi_1(\phi_2(n))}(2))_n$$ (to some $$y_2 \in X_2$$). Let $$\psi_2 = \psi_1 \circ \phi_2$$.

The sequence $$(x_{\psi_2(n)}(3))_n$$ is a sequence of elements of $$X_3$$, so it has a convergent subsequence $$(x_{\psi_2(\phi_3(n))}(3))_n$$ (to some $$y_3 \in X_3$$). Let $$\psi_3 = \psi_2 \circ \phi_3$$.

By induction, you can construct increasing maps $$\phi_n: \mathbb{N}^* \rightarrow \mathbb{N}^*$$, such that, if $$\psi_k = \phi_1 \circ \ldots \phi_k$$, $$(x_{\psi_k(n)}(k))_n$$ is convergent to some $$y_k \in X_k$$.

In particular, if $$l \leq k$$, since $$\psi_l(\mathbb{N}^*) \supset \psi_k(\mathbb{N}^*)$$, $$(x_{\psi_k(n)}(l))$$ is convergent to $$y_l$$.

So you can extract subsequences of $$x_n$$ that will work for arbitrarily many coordinates, but how to get all of them? Enter the diagonal extraction: define $$\theta(n)=\psi_n(n)$$.

For $$n \geq k$$, $$\theta(n) \in \psi_k(\mathbb{N}^*)$$, and $$\psi_k^{-1}(\theta(n)) = \phi_{k+1} \circ \ldots \circ \phi_n(n) \geq n$$ so $$c_{k,n}:=\psi_k^{-1}(\theta(n)) \rightarrow \infty$$ (as $$k$$ is fixed but $$n$$ grows), thus $$x_{\theta(n)}(k) = x_{\psi_k(c_{k,n})}(k) \rightarrow y_k$$.

Of course, there's an abstract argument that is shorter and easier to understand (although it uses a stronger result). In your case, your $$X_n$$ are the $$\{\phi \in L^2(\mathbb{R}^3,L^2(B(0,n))) = L^2(\mathbb{R}^3 \times B(0,n)),\, \|phi\|^2 \leq M_{B(0,n)}\}$$ with its weak topology. Let $$C_n$$ be the closure of the image of $$f_n$$ in $$X_n$$, then $$C_n$$ is bounded (in norm) and compact (for the weak topology) hence metric compact.

Then $$\prod_n{C_n}$$ is metric compact for the product topology (by Tychonov), which is (elementarily) metric. Therefore there exists a subsequence $$f_{\theta(n)}$$ of the $$f_n$$ whose image in each $$C_k$$ is convergent, ie the images of the $$f_{\theta(n)}$$ in $$L^2(\mathbb{R}^3,L^2(B(0,k)))$$ converge.

Either way, I show that there is an extraction $$\sigma$$ such that the images of $$f_{\sigma(n)}$$ in any $$L^2(\mathbb{R}^3,L^2(B))$$ where $$B \subset \mathbb{R}^3$$ make a weakly convergent. But I'm not sure if this condition is sufficient enough to show weak convergence in $$L^2(\mathbb{R}^3,L^2_{loc}(\mathbb{R}^3))$$.

I'm not sure if that's enough to show that the $$f_{\theta(n)}$$ converge weakly in $$L^2(\mathbb{R}^3,L^2_{loc}(\mathbb{R}^3))$$.

• Hello @Mindlack, thanks for your answer which i'm reading at the moment. A quick remark to make sure I understand : isn't it $\psi_k^{-1}(\theta(n)) = \phi_{k+1} \circ \phi_{k+2} \circ \dots \circ \phi_n (n) \geq n$ ? Jun 1, 2021 at 16:05
• At the end, I think you proved that there is an extraction $\sigma$ such that $f_{\sigma_n}|_{B(0,k)}$ is weakly convergent toward a limit $y_k$ in $L^2(\mathbb{R}^3, L^2(B(0,k))$ for all $k$ am I right ? From there it is easy to see that there exists a unique $y \in L^2(\mathbb{R}^3, L^2_{loc}(\mathbb{R}^3))$ s.t $y|_{B(0,k)}=y_k$. Jun 1, 2021 at 16:19
• @Velobos . Re your first comment: yes, I corrected. For your second comment: it’s true, yes. But what I wasn’t sure about was whether that criterion was enough to ensure weak convergence to $y$ in $L^2(\mathbb{R}^3,L^2_{loc}(\mathbb{R}^3))$. Jun 1, 2021 at 17:15
• I'm not sure but shouldn't it be $X_n= \{ \varphi \in L^2(\mathbb{R}^3, L^2(B(0,n)), \ ||\varphi||^2_{L^2(\mathbb{R}^3, L^2(B(0,n))} \leq M\}$ where $M$ is my absolute constant ? Jun 2, 2021 at 12:31
• Oops. You’re right. Jun 2, 2021 at 14:48