Weak convergence in $L^2$ implies strong convergence?

Question

Consider a sequence of (real-valued) functions $f_n$ in $L^2(\Omega)$ where $\Omega$ is a smooth, bounded domain in $\mathbb{R}^d$. If $\lVert f_n \rVert_{L^2} < M$ (uniformly bounded in $n$), then we know that there exists a weakly convergent subsequence, and a limit function $f \in L^2(\Omega)$:

$$\int_{\Omega} (f_{n_k} - f)g \ dx = 0 \quad \forall \ g \in L^2(\Omega)$$

So, if we use $f_{n_k}$ and $f$ as the test functions (in place of $g$), we can show the following:

$$\int_{\Omega} \lvert f_{n_k} \rvert^2 - \int_{\Omega} \lvert f \rvert^2 = \int_{\Omega} (f_{n_k} - f)f_{n_k} + \int_{\Omega} (f_{n_k} - f)f \xrightarrow[k\rightarrow\infty]{} 0$$

due to the weak convergence. This establishes convergence of the norms of the sequence of functions, thus implying the strong convergence $f_{n_k} \rightarrow f$ in $L^2(\Omega)$. Can someone please explain what is wrong with this argument, since we know that weak convergence doesn’t imply strong convergence?

Answer 1

Weak convergence implies as you said that for each fixed $g$ you have $$ \int_{\Omega} (f_{n_k} - f)g \to 0 $$

You wrongly concluded that this implies that $$ \int_{\Omega} (f_{n_k} - f)f_{n_k} \to 0 $$

Answer 2

Firstly, $lim_k\int_Ω|_{_}|^2−\int_Ω||^2=0$ means sequence $|_{_}|$ converges to $|f|$, which is the sequence of real(complex) number converges to a real(complex) number.(norm converge)

Strongly converge in $L^2$ space is: $$lim_k ||f_{n_k}-f||_{L^2}=0$$.

And due to the weakly convergence, we cannot let $g$ in $\int_Ω(_{_}-)gdx$ change at the same time as $k$ limit to $\infty$, which means the for every fixed $g$. Actually, we could choose the $g_{n_k}$ such that

$$\int_Ω(_{_}-)g_{n_k}dx=c \quad (constant) \quad \forall k$$

So, weakly convergence also cannot imply norm convergence.