# weak convergence in $L^p$ plus convergence of norm implies strong convergence

Having trouble with this problem. Any ideas?

Let $\Omega$ be a measure space. Let $f_n$ be a sequence in $L^p(\Omega)$ with $1<p<\infty$ and let $f \in L^p(\Omega)$. Suppose that $$f_n \rightharpoonup f \text{ weakly in } \sigma(L^p,L^{p'})$$ and $$\|f_n\|_p \to \|f\|_p.$$

Prove that $\|f_n-f\|_p \to 0$.

Also, can you come up with a counter-example for the $L^1$ case?

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Do you know that $L^p$ is reflexive for $1<p < \infty$ while it is not for $p\in\{1,\infty\}$? –  user20266 Jun 26 '12 at 8:14

Since $1<p<\infty$, the space $L^p(\Omega)$ is uniformly convex. This follows from Clarkson's inequalities. Now we use the following theorem, which can be studied in Brezis' book on functional analysis (chapter III).
Theorem. Let $E$ be a uniformly convex Banach space, and let $\{x_n\}$ be a weakly convergent sequence in $E$, i.e. $x_n \rightharpoonup x$ for some $x \in E$. If $$\limsup_{n \to +\infty} \|x_n\| \leq \|x\|,$$ then $x_n \to x$ strongly in $E$.
Try to construct a counter-example in the $\sigma(L^1,L^\infty)$ topology.
I think I got one. Let $\Omega=(0,2\pi)$ and $f_n(x)=\sin(nx)+1$. Then $f_n \rightharpoonup 1 \text{ weakly in } \sigma(L^1,L^{\infty})$. Also, $\|f_n\|_1= 2\pi$ for all $n$. However, $\|f_n-1\|_1 = 4$ for all $n$ so we have NO strong convergence. –  chango Jun 26 '12 at 13:24