4
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.