show that operator $A$ that $(A x)(y)=(A y)(x)$ is continuous

Question

Let $X$ be a Banach space and let $A: X \rightarrow X^{*}$ be a linear operator satisfying $$(A x)(y)=(A y)(x)$$ Show that $A$ is a continuous operator, i.e. $A \in L\left(X, X^{*}\right)$.

I know that if a operator is linear and bounded then it is continuous, so i just have to show that this is bounded, right? I was thinking that $A$ is identity, can you give me a hint to show that is bounded. Or any idea to solve the problem

Answer 1

Suppose that $x_n \to x \in X$ and $Ax_n \to \psi \in X^{*}$. Then for all $y \in X$, $$\psi(y) =\lim_n (Ax_n)(y)= \lim_n (Ay)(x_n)= (Ay)(x)=(Ax)(y)\,.$$ Thus $\psi=Ax$, and we have verified that the graph of $A$ is closed. Therefore, by the closed graph theorem, $A$ is continuous.

https://en.wikipedia.org/wiki/Closed_graph_theorem_(functional_analysis)

Edit: After (slowly) typing this proof, I saw that the same suggestion was made in the comments by Ryszard Szwarc.