Weak convergence and $\|f_{n}\|\rightarrow\|f\|$ implies strong convergence

Question

Let $\left\{ f_{n}\right\} _{n=1}^{\infty}$ be a sequence in $L^{p}\left([0,1]\right)$ for $p\geq1$. Suppose that there exists $f\in L^{p}\left([0,1]\right)$ satisfying $\lim_{n\rightarrow\infty}\int_{0}^{1}f_{n}\left(x\right)g(x)\ dx=\int_{0}^{1}f(x)g(x)\ dx$ for any $g\in L^{2}\left([0,1]\right)$. Prove that if $\lim_{n\rightarrow\infty}\|f_{n}\|_{p}=\|f\|_{p}$, then $\lim_{n\rightarrow\infty}\|f_{n}-f\|_{p}=0$.

I don't know how to approach to this question. Weak convergence does NOT imply the subsequence convergence. And I expanded $|f_{n}-f|^{p}$ but nothing I can obtain. Can anyone help me?

Try looking up here: http://math.stackexchange.com/questions/163209/weak-convergence-in-lp-plus-convergence-of-norm-implies-strong-convergence — Harnak, Dec 26 '16 at 14:31
@Harnak I already looked it up. But it's a little different. — kayak, Dec 26 '16 at 14:49
In what is it different? I think $g \in L^q$ by the way, cause that's the weak convergence for $L^p$ — Harnak, Dec 26 '16 at 14:54
If you want an answer that follows the formulation in Stein and Shakarchi, you can find it here. — Petra Axolotl, Oct 18 '22 at 02:59

score 0 · Answer 1 · answered Dec 26 '16 at 14:25

0

Partial answer, for $p=2$.

$$ \|\,f_n-f\|^2=(\,f_n-f,f_n-f)=(\,f_n,f_n-f)-(\,f,f_n-f) $$ and clearly $(\,f,f_n-f)\to 0$. Also $$ (\,f_n,f_n-f)=(\,f_n,f_n)-(f_n,f)\to \|\,f\|^2-\|\,f\|^2=0, $$ and finally, $\|\,f_n-f\|\to 0$.

answered Dec 26 '16 at 14:25

Yiorgos S. Smyrlis

83,933

It's only partial. I know it already. – kayak Dec 26 '16 at 14:51
While a partial answer (and identified as such), this clears away a good bit of the easily answered parts. – hardmath Dec 26 '16 at 15:08

Weak convergence and $\|f_{n}\|\rightarrow\|f\|$ implies strong convergence

1 Answers1