Closed Graph Theorem Application Exercise

Question

Let E be a Banach space and $T:E\to E'$ a linear operator such that $\langle Tx,y\rangle=\langle Ty,x\rangle$ for all $x,y\in E$. Here $E'$ is the dual space of $E$. Show that $T$ is a bounded operator. I recognize that this is an answer on the Stack: Prove that a linear operator $T:E \rightarrow E'$ such that $\langle Tx,y \rangle=\langle Ty,x\rangle$ is bounded. However, I am trying to understand why exactly the first equality in the equality works. When we say $Tx_n \to z$ in $E'$, don't we mean that $\| Tx_n- z \|_{E'} \to 0$? If this was the case, is the first equality assuming convergence in dual norm implies point wise convergent? I tried to prove this, but I can't seem to prove it. In general, is my claim even true?

Answer 1

Let me write an answer for this question from scratch:

Since we are working with Banach spaces, in order to show that $T$ is bounded, we can employ the closed graph theorem. So, we begin by a sequence $(x_n)\subset E$ such that $x_n\to x\in E$ and we also assume that $Tx_n\to f\in E'$ and we want to show that $f=Tx$. This is exactly how one applies the closed graph theorem. So, let's do that.

Note that, in $E'$, if $f_n\to f$ in norm, then for $z\in E$ we have $0\le |f_n(z)-f(z)|=|(f_n-f)(z)|\le\|f_n-f\|\cdot\|z\|$ and since $\|f_n-f\|\to0$, by the sandwich theorem we also have $|f_n(z)-f(z)|\to0$, so $f_n(z)\to f(z)$.

Now let $y\in E$. We have $f(y)=\lim_{n\to\infty}Tx_n(y)=\lim_{n\to\infty}Ty(x_n)$ (here we used the hypothesis on $T$) and $\lim_{n\to\infty}Ty(x_n)=Ty(x)$ (here we used continuity of $Ty$) and $Ty(x)=Tx(y)$ (here we used the hypothesis on $T$ again). So $f(y)=Tx(y)$ for all $y\in E$, i.e. $f=Tx$ as we wanted.