# Let (X, ||.||) be a Banach space and T: X $\rightarrow$ $X^*$ a linear operator with the property (Tx)(y) = (Ty)(x) for every x, y $\in$ X.

Let (X, ||.||) be a Banach space and T: X $$\rightarrow$$ $$X^*$$ a linear operator with the property (Tx)(y) = (Ty)(x) for every x, y $$\in$$ X.

Let $$(X, ||\cdot||)$$ be a Banach space and $$T: X \rightarrow X^*$$ a linear operator with the property $$(Tx)(y) = (Ty)(x)$$ for every $$x, y \in X$$. To prove that $$T$$ is a bounded operator, we need to show that $$T$$ maps bounded sets in $$X$$ to bounded sets in $$X^*$$.

Let $$B$$ be a bounded set in $$X$$, which means there exists a constant $$M > 0$$ such that $$||x|| \leq M$$ for all $$x \in B$$.

Now, consider the set $$\{Tx : x \in B\}$$ in $$X^*$$. We want to show that this set is bounded in $$X^*$$.

For any $$x \in B$$ and $$y \in X$$, by the property of $$T$$ given, we have:

$$|(Tx)(y)| = |(Ty)(x)| \leq ||Ty|| \cdot ||x||$$

Have I started correct? How to continue?

Since $$X$$ is Baanch we have that $$X^{*}$$ is too. Now, let $$(x_n,Tx_n) \longrightarrow (x,\varphi) \in X \times X^{*}$$. It suffices to prove that $$Tx = \varphi$$ to prove that the graph of $$T$$ is closed.
Indeed, observe for $$y \in X$$, we have: $$(Tx)y = (Ty)x = \lim_n (Ty)(x_n) = \lim_n (Tx_n)y = \varphi(y)$$ where the second equality follows from $$Ty$$ being a continuous linear functional on $$X$$. Hence since $$y \in X$$ was arbitrary, we have $$Tx=\varphi$$ so that $$T$$ is bounded by the Closed Graph theorem.
We want to show that $$\sup_{\|x\|= 1} \|T x\| < \infty$$. By the uniform boundedness principle applied to the set of functionals $$\{T x \in X^* : \|x\| = 1\}$$, it suffices to show that for each fixed $$y \in X$$, $$\sup_{\|x\| \le 1} |T(x)(y)| < \infty$$.
But this supremum can be written as $$\sup_{\|x\|\le 1} |T(y)(x)| = \|Ty\| < \infty$$ since $$y$$ is fixed $$Ty \in X^*$$