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Let $\Omega \subset \mathbb{R}^{n}$ an open set and $(f_{n})_{n=1}^{\infty}$ a sequence in $L^{2}(\Omega)$ such that $||f_{n}||_{2} \to ||f||_{2}$, then there exists a subsequence $(f_{n_{k}})_{k}$ of $(f_{n})_{n=1}^{\infty}$ such that $\int_{\Omega} |f_{n_{k}} - f|^{2} \to 0$, when $k\to \infty$.

The only thing I thought: How the sequence $(||f_{n}||_2)_{n=1}^{\infty}$ is limited because is convergent and $L^{2}(\Omega)$ is reflexive, then there exists a subsequence $f_{n_{k}} \rightharpoonup g$, but i don't know how to continue.

Thanks

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    There seems to be a missing hypothesis in the statement. Take $f_n=h$ the characteristic function of [0,1], for all $n$, and $f$ that of $[-1,0]$. The norms are 1, but $f$ and $h$ are not equal a.e. – 2017-02-10
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    $f_{n}=\mathbb{1}_{[0,1]}$ and $f=\mathbb{1}_{[-1,0]}$? – 2017-02-10
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    Yes. But your question is probably related to this one: http://math.stackexchange.com/questions/163209/weak-convergence-in-lp-plus-convergence-of-norm-implies-strong-convergence In other words, you forgot to include that $f_n\rightharpoonup f$ in your hypothesis. – 2017-02-10
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    This exercise is the previous item to this question that is in the post. I imagine the teacher made some typos. – 2017-02-10
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    @ThibautDumont i can see that $||h||_{2} \to ||f||_{2}$, but , why the fact that $f$ e $h$ are not equal a.e implies that the statement of the question is wrong? I can't see this. – 2017-02-10
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    $f$ e $h$? $\,\,$ – 2017-02-10
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    @zhw. see the coment above – 2017-02-10
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    I see no comment about $f$ e $h$. – 2017-02-10
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    The comment of @ThibautDumont ..."Take $f_{n}=h$.." – 2017-02-10
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    In this example, the sequence $\int (f_n-f)^2 = \int (h-f)^2=2$ is constant and therefore cannot have a subsequence converging to $0$. – 2017-02-10

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