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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!

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3 Answers 3

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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.

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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 $C$).

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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 $||T(x)|| \leq C||x||$.

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