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If I know $\int_{[0,1]} f_{n}(x) g(x) dx \rightarrow \int_{[0,1]} f(x) g(x) dx$ as $n \rightarrow \infty$ for all $g \in L^2([0,1])$ (weak convergence in $L^2$)

and $|f_n(x)|_{L^2} <C$ (uniformly bounded in $L^2$-Norm),

does then exist a subsequence $f_{n_k}$ such that $\lim_{n \rightarrow \infty}f_{n_k}(x) = f(x)$ for almost every $x \in [0,1]$?

Edit: If uniform boundedness in $L^2$ norm is not enough would $f_n$ being uniformly bounded in $L^\infty$ norm help to get a subsequence?

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    $\begingroup$ Taking $f_n(x)=\sin nx$ gives a counterexample. $\endgroup$ – David Mitra Jul 10 '14 at 13:03
  • $\begingroup$ A weakly convergent sequence is automatically norm bounded. $\endgroup$ – user940 Jul 10 '14 at 13:06
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    $\begingroup$ As for the edit: the counterexample $f_n(x) = \sin(nx)$ of David Mitra is uniformly bounded in $L^\infty$ so it still gives a negative answer to your question. $\endgroup$ – user38355 Jul 10 '14 at 16:45
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The example by David Mitra is perfectly valid, but it may be a little hard to see exactly what happens to $\sin nx$ as $n\to\infty$. Here's a slightly modified example: Rademacher functions $$r_n(x) = \operatorname{sign}\sin (2^n \pi x), \quad x\in [0,1]$$ These converge to $0$ weakly in $L^2$, but $|r_n|=1$ a.e.

To prove weak convergence, first consider $\int r_n g$ when $g$ is continuous, putting the contributions of adjacent $\pm $ intervals against each other. Then conclude by density (using the fact that $\|r_n\|_2=1$ and Hölder's inequality).


For complex-valued functions, $\exp(i nx)$ is a good counterexample.

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