# Weak convergence and weak star convergence.

If region $\Omega$ is bounded and $u_n$ has weak star convergence in $L^\infty ( \Omega)$ to some $u\in L^\infty(\Omega)$ , does it imply that $u_n$ converges weakly in any $L^p(\Omega)$ ?

I think i got it : If $sup$ of a function is finite then integral over a bounded region is finite with any $p$ norm . is it right ?

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$\{u_n\}\subset L^\infty(\Omega)$ converges in the weak star topology to $u\in L^\infty(\Omega)$ if $$\lim_{n\to\infty}\int_\Omega u_n\phi\,dx=\int_\Omega u\,\phi\,dx\quad\forall\phi\in L^1(\Omega).$$ Since $\Omega$ is bounded, $L^\infty(\Omega)\subset L^p(\Omega)\subset L^1(\Omega)$ for all $p\ge1$. It follows that $u_n$ converges weakly to $u$ in $L^p(\Omega)$ for all $p\in[1,\infty)$.
Did you use the reflexivity of $L^1(\Omega)$ here, to conclude weak and weak* convergent sequences are one and the same? – Host-website-on-iPage Jul 13 '12 at 17:14
$L^1$ is not reflexive. What I proved is that weak-$\star$ convergent sequences in $L^\infty$ are weak convergent in $L^p$, $1\le p<\infty$ (on a finite measure space.) – Julián Aguirre Jul 13 '12 at 17:22
I do not understand the logic here. Did you intend to write "if $u_n$ converges in the weak star topology then, in particular, the sequence of integrals (see formula) converges (because $L^1$ can be identified with a subspace of $(L^\infty)^*$)". Because the way you formulated it one does indeed get the impression you claim $L^1$ to be reflexive. – user20266 Jul 13 '12 at 17:36
@Thomas: this is perfectly correct and it's just plugging in the definition (I have no idea where reflexivity should be implied): the weak$^\ast$-topology is the topology on $L^\infty$ induced by the functionals $u \mapsto \int u \phi$ with $\phi \in L^1$. Now we also have $u,u_n \in L^p$ for all $1 \leq p \lt \infty$ by the first displayed inclusion, and we get from the hypothesis convergence of the integrals for all $\phi \in L^q \subset L^1$, which is weak convergence in $L^p$ by definition. – t.b. Jul 13 '12 at 18:51