# If $(u_n)$ is bounded in $L^p$ there is a subsequence that converge weakly.

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 $\int u_n\varphi\to \int u\varphi.$

• 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 – MSE Nov 25 '17 at 10:18
• 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. – Matematleta Nov 25 '17 at 12:35