If the adjoint of an operator is bounded is the operator as well? 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.
 A: For any $x\in X$ and $\phi\in Y^*$, we have
$$|T^*\phi(x)|=|\phi(Tx)|\leq\|\phi\|\|T\|\|x\|.$$
It follows that
$$\|T^*\phi\|\leq\|\phi\|\|T\|,$$
thus by definition of the operator norm we have
$$\|T^*\|\leq \|T\|.$$
Similarly we have 
$$\|T^{**}\|\leq\|T^*\|,$$
where $T^{**}:X^{**}\to Y^{**}$ is the adjoint operator of $T^*$. Now let $J_X:X\to X^{**}$ be the canonical isometry given by
$$(J_Xx)(\psi)=\psi(x),\quad\forall x\in X,\psi\in X^*.$$
Then for any $x\in X, \phi\in Y^*$ we have
$$(T^{**}(J_Xx))(\phi)=(J_Xx)(T^*\phi)=(T^*\phi)(x)=\phi(Tx)=(J_Y(Tx))(\phi).$$
It follows that
$$|J_Y(Tx)(\phi)|\leq \|T^{**}\|\|J_Xx\|\|\phi\|=\|T^{**}\|\|x\|\|\phi\|,$$
thus
$$\|T^{**}\|\|x\|\geq \|J_Y(Tx)\|=\|Tx\|,$$
and hence
$$\|T^{**}\|\geq\|T\|.$$
Consequently we have
$$\|T\|\leq\|T^*\|,$$
which means $T$ is bounded. In fact we can see that $\|T\|=\|T^*\|$.
A: 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
$||Tx|| \le ||T^*||*||x||$.
