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If a sequence ${f_i},f \in {L^p}([0,1]){\kern 1pt} {\kern 1pt} (1 < p < \infty )$ such that ${f_i}$ converges weakly to $f$ and ${\left\| {{f_i}} \right\|_p} \to {\left\| f \right\|_p}$, then is ${\left\| {{f_i} - f} \right\|_p} \to 0$ right?

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Right. (more characters) –  Jonas Teuwen Oct 24 '11 at 16:12
    
Yes. This is called the Radon-Riesz property. –  Mark Oct 24 '11 at 16:41
    
I also heard the name "Kadec-Klee property" and another name which I forgot... –  Dirk Oct 24 '11 at 18:46

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up vote 5 down vote accepted

Assume WLOG that $\|f_i\| = \|f\| = 1$. Assume that we don't have convergence of $f_n$ to $f$ in norm. So by local uniform convexity we have a subsequence of $(f_n)$ still denoted by $f_n$ such that

$$\left \|\frac{f_n + f}{2}\right \| < C < 1 \text{ for all $n$}.\tag{1}$$

By Hahn-Banach there is a $\varphi \in L^q$ such that $\varphi(f) = 1$ and $\|\varphi\| = 1$.

For this $\varphi$ $(1)$ implies that $|\varphi(f_n) + 1| < 2C$, hence $f_n$ does not converge weakly to $f$.

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Very nice. This is an easily rememeberable proof. –  Olivier Bégassat Oct 24 '11 at 17:09
    
Very cool application of uniform convexity and it occurs to me that in Adam's book he has given an elementary proof of Riesz representation theorem by the same methods. –  Hezudao Oct 25 '11 at 9:39

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