Let $H$ he a Hilbert space. Let. ${f_{n}}$ be a sequence in $H$.
a) Show that if $e_{n}$ is any orthonormal sequence in H, then $e_{n}$ converges weakly to $0$.
Sol:
For any $g \in H$, $\langle e_{n},g\rangle=0$ and $\langle 0,g\rangle =0$. Thus $\langle e_{n}, g \rangle \xrightarrow{w} \langle 0, g \rangle$, thus $e_{n}$ converges weakly to $0$.
b) If $f_{n}$ converges weakly to $f$ and $\|f\|= \lim_{n \to \infty} \|f_{n}\|$, then $f_{n}$ converges to $f$ in $H$.
Sol:
$$\begin{align*}\|f-f_{n}\|^{2}&=\langle f-f_n, f- f_n\rangle \\ &= \langle f,f\rangle-\langle f,f_n\rangle- \langle f_n,f\rangle+\langle f_n,f_n\rangle\end{align*}$$
Then, because of weak convergence $\langle f,f_n\rangle \to \langle f,f\rangle$ and $\langle f_n,f\rangle \to \langle f,f\rangle$. Then, using the result that $\|f_n\|^{2} \to \|f\|^{2}$.
I get that
$$\|f-f_{n}\|^{2} \to 0$$
c) Show that if $f_{n}$ converges weakly to $f$, then $f$ is in the closure of the subspace spanned by $f_{n}$.
Sol:
I can show that $f$ is in the closure, if there exists a sequence $(g_n)$ in $M=\text{span}\{f_n\}$, such that $g_{n} \to f$ in H. I know that $f_{n}$ converges weakly to $f$, so I thought of taking the sequence $g_n=f_n$, but it is not helping me make progress.
Can someone help? I think the rest of the proof is logical.
Best Answer
As pointed out in the comments, a) follows from Bessel's inequality, which gives the convergence of $\sum_n \lvert e_n,g\rvert^2$.
b) is fine.
For c), let $V:=\operatorname{Span}\{f_n,n\geqslant 1\}$. We know that $H=\bar{V}\oplus V^{\perp}$, where $\bar{V}$ denotes the closure of $V$ and $V^{\perp}$ the ortho-complement of $\bar{V}$. If $v\in V^{\perp}$, then $$ \langle f,v\rangle=\lim_{n\to\infty} \langle f_n,v\rangle=0$$ because $v$ is orthogonal to each $f_n$. Since $\left(V^{\perp}\right)^{\perp}=\bar{V}$, we get the wanted result.