Weak convergence in $L^p$ space.

Question

Let $p\in [1,\infty [$ and $\Omega \subset \mathbb R^n$ an open. Let $u_n,\in L^p$. We say that $u_n\rightharpoonup u$ (weak convergence) if $$\lim_{n\to \infty }\int_\Omega (u_n-u)\varphi=0$$ for all $\varphi\in L^{p'}$ (the dual of $L^p$).

Q1) First, what is a the intuition behind this definition ?

Q2) How can I bu sure that $(u_n-u)\varphi\in L^1$ ?In other word that $\int (u_n-u)\varphi$ exist for all $n$.

Answer 1

For $Q2$ use Hölder's inequality.

For $Q1$, I personally keep the intuition for the weak vs convergence in norm from the one of the pointwise vs uniform convergence on (say) bounded functions (which is usually taught earlier than $L^p$ spaces). I do not claim that this is the best or even the right way to think of it though.

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If you are in for a more advanced explanaition, look for the definition of the weak topology on a normed space. Since $L^{p'}$ is the dual of $L^{p}$ for $1 \le p < \infty$, the definition you wrote gives exactly the sequential characterisation of the weak topology.

Answer 2

To complete Marko Karbevski answer : for Q2), as you can remark, it's a convergence in $\mathbb R$. Here we consider $L^p$ for $p\in ]1,\infty [$. Let denote $$T(\varphi)=\int u\varphi\quad \text{and}\quad T_n(\varphi)=\int u_n\varphi.$$

So $T$ and $T_n$ are in $(L^{p'})^*$ (the dual of $L^{p'}$) You say that $(u_n)$ converge weakly if $$\lim_{n\to \infty }T_n(\varphi)=T(\varphi).$$ In other word, if the sequence $(T_n)$ converge pointwise to $T(\varphi)$.

Now, suppose $u_n\to u$ strongly. Then, $$|T_n(\varphi)-T(\varphi)|\leq \int |(u_n-u)\varphi|\leq \|u_n-u\|_{L^p}\|\varphi\|_{L^{p'}}\implies \|T_n-T\|=\sup_{\|\varphi\|_{L^{p'}\leq 1}}|T_n(\varphi)-T(\varphi)|\leq \|u_n-u_n\|_{L^p}\underset{n\to \infty }{\longrightarrow }0,$$ and thus the convergence of $(T_n)$ is uniform.