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?

  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.