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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!

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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 $f \in X^*$.

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