TITLE: If the adjoint of an operator is bounded is the operator as well?
QUESTION [5 upvotes]: Let $X$ and $Y$ be normed vector spaces. Let $T: X \to Y$ be a linear operator. Let $T^* : Y^* \to X^*$ be the adjoint of $T$ defined by $T^*(f) =f \cdot T$. Show that if $T^*$ is bounded then $T$ is bounded.
I know that the converse of this statement is true but I'm not sure about this direction. If it is not true, I am wondering if there is some extra hypothesis which makes it true.

REPLY [3 votes]: Let $x \in X$. As a consequence of the Hahn-Banach Theorem we have:
$(*)$ $||Tx||= \sup\{|f(Tx)|: f \in Y^*, ||f||=1\}$.
For $f \in Y^* with ||f||=1$ we have, since $T^*$ is bounded:
$|f(Tx)|=|(T^*f)(x)| \le ||T^*f||*||x|| \le ||T^*||*||f||*||x||=||T^*||*||x||$. 
This and $  (*)$ shows now that
$||Tx|| \le ||T^*||*||x||$.