Let $(u_n)$ a bounded sequence of $L^p(\Omega )$, $p\in [1,\infty )$ where $\Omega\subset \mathbb R^d $ is a bounded domain. Then there is a subsequence that converge weakly to $u$.

The proof is as following, but I don't really understand it.

Let $q=\frac{p}{p-1}$. Since $L^q$ is separable, there is a dense $(\varphi_m)$. For each $m$, using Holder, $$\left|\int_\Omega u_n \varphi_m\right|\leq C\|\varphi_m\|_{L^q(\Omega )}.$$ Therefore, there is a subsequence (still denoted $u_n$) s.t. $$\int_\Omega u_n \varphi_m\to A(\varphi_m).$$

Q1) I don't understand what is $A(\varphi_m)$ and why we have this convergence.

Let assume $$\lim_{n\to \infty }\|u_n\|=\liminf_{n\to \infty }\|u_n\|.$$

Q2) Why $\lim_{n\to \infty }\|u_n\|$ exists ? I thought we have to prove the weak convergence and here we have the strong convergence ?

WLOG assume $$\lim_{n\to \infty }\int_\Omega u_n\varphi_m=A(\varphi_m),$$ for all $m$.

This implies that $$\lim_{n\to \infty }\int u_n\varphi=A(\varphi)$$ for all $\varphi\in L^q$.

Q3) Why do we have that ? I think it's by density, but why this limit exist ?

Since we have $$|A(\varphi)|\leq \lim_{n\to \infty }\|u_n\|_{L^p}\|\varphi\|_{L^q},$$ there $u\in L^p$ s.t. $$A(\varphi)=\int u\varphi$$ for all $\varphi\in L^p$, and the claim follow.

Q4)Here, I don't understand why such an $u$ exist.


If $p=1$ or $p=\infty$, the result is not true. So assume $1<p<\infty.$

$1).$ Define $A_n\in (L_q)^*$ by $A_n(\varphi)=\int u_n\varphi.$ Minkowski's Inequality implies that $|A_n(\varphi)|\le C\cdot \|\varphi\|_q$, where $C>0$ is a bound on $(u_n).$ Helley's Theorem now implies that there is an $A\in (L_q)^*$ and a subsequence $(A_{n_k})$, which for convenience, we still write $A_n$ such that $A_n(\varphi)\to A(\varphi).$ In particular, $A_n(\varphi_m)=\int u_n\varphi_m\to A(\varphi).$

$2).\ (\|u_n\|)$ is a sequence of positive numbers, so $\liminf \|u_n\|$ exists.

$3).$ Let $\epsilon>0$ and $\varphi \in L_q.$ Choose an integer $m$ such that $\|\varphi-\varphi_m\|<\epsilon$ and, for this $m$, an integer $N$ such that $n>N\Rightarrow |A_n(\varphi_m)-A(\varphi_m)|<\epsilon.$

Then, if $n>N,$ we have

$|\int u_n\varphi-A(\varphi)|=|A_n(\varphi)-A(\varphi)|\le $

$|A_n(\varphi)-A_n(\varphi_m)|+|A_n(\varphi_m)-A(\varphi_m)|+|A(\varphi_m)-A(\varphi)|\le $

$C\|\varphi-\varphi_m\|_q+\epsilon+\|A\|\cdot \|\varphi-\varphi_m\|_q=(C+1+\|A\|)\epsilon.$

$4).$ By the Riesz Representation Theorem, there is a $u\in L_p$ such that $A(\varphi)=\int u\varphi.$

I think there is a much easier way to do this. All you need to know is that $L_p$ is separable, $1<p<\infty.$ Then, I think we can argue this way:

$1).$ Define $A_n(\varphi)=\int u_n\varphi$.

$2).$ Apply Helly's Theorem directly to produce a subsequence, still called $A_n$ and an $A\in (L_q)^*$such that $A_n(\varphi)\to A(\varphi).$

$3).$ Now apply the Riesz Representation Theorem, to find a $u\in L_p$ such that $A(\varphi) =\int u\varphi.$

Combing $1).,\ 2).$ and $3).$ we have $\int u_n\varphi\to \int u\varphi.$

  • $\begingroup$ By Helly's theorem, you mean this one ? If yes, I don't understand the way you use it... Indeed, why is there a subsequence $A_n$ and $A\in (L^q)^*$ s.t. $A_n(\varphi)\to A(\varphi)$ ? en.wikipedia.org/wiki/Helly%27s_theorem $\endgroup$ – MSE Nov 25 '17 at 10:18
  • $\begingroup$ Let $X$ be a separable Banach space and $M\subseteq X^*$ a bounded set. Then every sequence of elements of $M$ contains a subsequence which is weak* convergent to an element of $X^*$. The proof is basically a diagonalization argument. $\endgroup$ – Matematleta Nov 25 '17 at 12:35

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.