The question of interchanging inner product and limit has been posted here several times, especially in the case of Hilbert space. There are also several posts about the proof of the closeness of annihilator space. However, I have a more specific issue.
Let $(V, \|\ \cdot\ \|)$ be a Banach space on $\mathbb{C}$ and let $V^{*}$ be its dual, endowed with the dual norm $$\|f\|_{V^{*}}=\|f\|_{op}=\sup_{\|x\|_{V}\leq 1}|\langle x,f \rangle_{V, V^{*}}|.$$ For $M\subset V$, we define the annihilator of $M$ as $$M^{\bot}:=\{f\in V^{*}:\langle x,f\rangle_{V,V^{*}}=0\ \ \text{for all}\ \ x\in M\}\subset V^{*}.$$
I want to show:
$M^{\bot}$ is a closed linear subspace of $V$.
I have proved that $M^{\bot}$ is a linear subspace but I got stuck when I tried to show the closeness. There are several online notes that have proved this, but not in the language of inner product which is the one that brings the issue to us.
So, let $f\in \overline{M^{\bot}}$, then there exists a sequence of bounded linear functional, $(f_{n})_{n=1}^{\infty}\subset M^{\bot}$ such that $\lim_{n\rightarrow\infty}f_{n}=f.$ It then follows that, for any $x\in M,$ we have $$\langle x, f\rangle_{V,V^{*}}=\langle x, \lim_{n\rightarrow\infty}f_{n}\rangle_{V, V^{*}}=0,$$ since $\langle x, f_{n}\rangle_{V, V^{*}}=0$ for every $n$.
Is this correct? It looks really weird to me in the sense that: what does it mean by $\lim_{n\rightarrow\infty}f_{n}=f$ here? pointwise? in the norm?
If in the norm, do we actually have $$\langle f-f_{n}\rangle_{V^{*}}=\sup_{\|x\|_{V}\leq 1}|\langle x, f-f_{n}\rangle_{V,V^{*}}|\longrightarrow 0\ \ \text{as}\ \ n\rightarrow\infty?$$ If this is correct, then I do not know what to do next…
If pointwise, does it actually mean $$\lim_{n\rightarrow\infty}\langle x, f_{n}\rangle_{V,V^{*}}=\langle x, f\rangle_{V, V^{*}}?$$
Well then I do not even need to consider the interchange of limit and inner product.
I am really confused the sense of $\lim_{n\rightarrow\infty}f_{n}=f$ here…
Best Answer
You have to start with: let $(f_n)$ be a convergent sequence in $M^{\perp}$ with limit $f \in V^*$.
This means: $||f_n-f||_{op} \to 0.$
We have to show that $f \in M^{\perp}.$
Since
$|f_n(x)-f(x)| \le ||f_n-f||_{op} \cdot ||x||_V$ for all $n$ and all $x \in V,$ we get
$$f_n(x) \to f(x)$$
for all $x \in V$.
Since $f_n(x)=0$ for all $n$ and all $x \in M$, we conclude that $f(x)=0$ for all $x \in M.$
This gives $f \in M^{\perp},$ thus $M^{\perp}$ is closed.