# Prove that a linear operator $T:E \rightarrow E'$ such that $\langle Tx,y \rangle=\langle Ty,x\rangle$ is bounded

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 $E$. I have to prove that $T$ is a bounded operator. I tried to use the closed graph theorem, but I can't prove that the graph of T is close. I would appreciate it if anyone could help me. Thank you.

## 2 Answers

Let $x_n\to x$ in $E$ and $Tx_n\to z$ in $E'$. Note that for all $y\in E$ we have $$\langle z,y\rangle =\lim\limits_{n\to\infty}\langle Tx_n, y\rangle =\lim\limits_{n\to\infty}\langle Ty, x_n\rangle =\langle Ty, \lim\limits_{n\to\infty}x_n\rangle =\langle Ty, x\rangle =\langle Tx, y\rangle$$ Since $y\in E$ is arbitrary $z=Tx$. By closed graph theorem $T$ is bounded.

It is also possible to apply Banach-Steinhaus to obtain the continuity.

• could you if you don't mind give me a sketch of the proof ? – rapidracim Jan 12 '18 at 20:35
• Let $B \subset X$ be the unit ball and consider the family of functionals $\{ Tx \mid x \in B \} \subset X'$. To check that it is pointwise bounded, take a point $y \in X$ and $(Tx)(y) = \langle Tx , y \rangle = \langle Ty, x \rangle \le \|Ty\|$. – gerw Jan 12 '18 at 21:21