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.


1 Answer 1


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 $L^2(\mathbb{R}^3,L^2_{loc}(\mathbb{R}^3))$.

  • $\begingroup$ Hello @Mindlack, thanks for your answer which i'm reading at the moment. A quick remark to make sure I understand : isn't it $\psi_k^{-1}(\theta(n)) = \phi_{k+1} \circ \phi_{k+2} \circ \dots \circ \phi_n (n) \geq n$ ? $\endgroup$
    – Velobos
    Jun 1, 2021 at 16:05
  • $\begingroup$ 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 $y|_{B(0,k)}=y_k$. $\endgroup$
    – Velobos
    Jun 1, 2021 at 16:19
  • $\begingroup$ @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 $L^2(\mathbb{R}^3,L^2_{loc}(\mathbb{R}^3))$. $\endgroup$
    – Aphelli
    Jun 1, 2021 at 17:15
  • $\begingroup$ 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 ? $\endgroup$
    – Velobos
    Jun 2, 2021 at 12:31
  • 1
    $\begingroup$ Oops. You’re right. $\endgroup$
    – Aphelli
    Jun 2, 2021 at 14:48

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