This question asks to prove the following:

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.

The assumption that $Y$ is complete seems redundant, and the assumption that $X$ is complete is invoked only when applying the uniform boundedness principle (if $X$ is complete, it must be nonmeager by the Baire category theorem). So I'm trying to either prove that $T$ is still bounded if $X$ and $Y$ are arbitrary normed vector spaces (over $\mathbb{R}$ or $\mathbb{C}$), or else to find a counterexample that illustrates that $X$ must be complete. Any suggestions would be greatly appreciated!


How about this for a proof that $T$ is bounded?

Consider the map $U:Y^*\to X^*$ given by $Uf = f \circ T$. We first show that $U$ has a closed graph: if $f_n \to f$ in $Y^*$ and $Uf_n = f_n\circ T \to g$ in $X^*$, then for all $x \in X$, it follows that $f_n\circ Tx \to f\circ Tx$ and $f_n\circ Tx \to g(x)$. Hence $g = f \circ T$.

Since $X^*$ and $Y^*$ are complete, by the closed graph theorem, it follows that $U$ is continuous.

Now consider the dual map $U^*:X^{**} \to Y^{**}$. See that $U^* \big |_X = T$: $$ \text{for $x \in X$, $g \in Y^*$}, \qquad (U^*x)g = (x\circ U)(g) = x(Ug) = x(g \circ T) = (g\circ T)(x) = g(T(x)), $$ where we identify $x \in X$ with the element in $X^{**}$ via $x(f) = f(x)$ for $f \in X^*$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.