Bounded Linear Mappings of Banach Spaces This problem has been giving me some troubles.  Does anyone have any ideas on how to go about proving this?
Let $X$ and $Y$ be Banach spaces.  If $T: X \to Y$ is a linear map such that $f \circ T \in X^*$ for every $f \in Y^*$, then $T$ is bounded.
Thanks in advanced!
 A: By the uniform boundedness principle applied to the set of all $f \circ T$ with $|f| = 1$ (operator norm from $Y$ to ${\bf C}$), there are two possibilities: 1) there is a constant $C$ such that $|f(T(x))| \leq C||x||$ for all $x \in X$ and all $f$ with $|f| = 1$, or 2) there exists at least one $x$ such that ${\displaystyle \sup_{|f| = 1} |f(T(x))| = \infty}$.
The second option cannot hold since $|f(T(x))| \leq  ||T(x)||$ whenever $|f| = 1$. So the first option must occur; there is a constant $C$ such that $|f(T(x))| \leq C||x||$ for all $x \in X$ and all $f$ with $|f| = 1$. By the Hahn-Banach theorem, for any $y = T(x) \in Y$ one can create an $f$ with $|f| = 1$ such that $|f(T(x))| = ||T(x)||$. So we have $||T(x)|| \leq C||x||$. 
A: Let $A=\{f\circ T \mid f\in Y^* \land \Vert f \Vert = 1\}$ (so $A\subset X^*$). Let $x\in X$. Then ${|(f\circ T)x|}\leq {\Vert f \Vert} {\Vert Tx \Vert}$, so
$$
\sup_{\phi\in A} |\phi(x)| \leq \Vert Tx\Vert
$$
which is finite. By the uniform boundedness principle, we have $$\sup_{\phi\in A} \Vert \phi \Vert<\infty
$$
and
$$
\begin{align*}
\sup_{\phi\in A} \Vert \phi \Vert &= \sup_{\substack{f\in Y^* \\ \Vert f \Vert = 1}} \Vert f\circ T\Vert\\
&=  \sup_{\substack{f\in Y^* \\ \Vert f \Vert = 1}} \left(\sup_{\substack{x\in X \\ \Vert x \Vert = 1}} |(f\circ T)x|\right)\\
&=  \sup_{\substack{x\in X \\ \Vert x \Vert = 1}}\left(\sup_{\substack{f\in Y^* \\ \Vert f \Vert = 1}}  |(f\circ T)x|\right)
\end{align*}
$$
But
$$
\forall x\in X,\; \sup_{\substack{f\in Y^* \\ \Vert f \Vert = 1}}  |(f\circ T)x| \ge \Vert Tx \Vert,
$$
therefore
$$
\Vert T \Vert = \sup_{\substack{x\in X \\ \Vert x \Vert = 1}} \Vert Tx \Vert < \infty.
$$
So $T$ is bounded.
A: First, I claim that the unit ball of $Y^*$ is mapped into a bounded subset of $X^*$. This follows from the Banach-Steinhaus theorem. If $x \in X$, then $(f \circ T)(x) = f(T(x))$ is bounded as $f$ ranges over the elements of $Y^*$ of norm at most one. So we have the collection $\mathcal{C}$ of functionals $f \circ T$ on $X$, such that for each $x \in X$, $\sup_{r \in \mathcal{C}} ||r(x)|| < \infty$. This implies that $\mathcal{C}$ is a bounded set and that the transpose of $T$ is bounded. 
Now, if the transpose of a linear transformation $T$ is bounded by some $C$, then $T$ is itself bounded by $C$ (to see this, suppose $x \in X$ is of norm at most one; then the claim is that $|\ell(T(x))| \leq C$ for $\ell$ a functional on $Y$ of norm at most one, which is equivalent by Hahn-Banach. But this is $
|T^*(\ell)(x)|$, which by assumption is of norm at most $C$).
