$g(T)$ bounded implies $T$ bounded, if $T$ is linear and $g$ is bounded linear functional

Question

Suppose $X$ is a Banach space and $y$ is a normed linear space and $T :X\to Y$ is linear map such that for every bounded linear functional $g\in Y^*$ we have $g( T)$ is bounded . Show that $T$ is bounded.

I am thinking of using closed graph theorem, $(x_n,Tx_n)\to(x,y)$. Since $T$ is lineaer,and somehow if i could show $T(x)=y$ then i would be done. But I am not able , if you have any idea then please.....

Answer 1

You can use the canonical isometric injection $\iota : Y \hookrightarrow Y''$, where $\iota y$ for $y \in Y$ is defined by $(\iota y)(\varphi) = \varphi(y)$ for all $\varphi \in Y'$. Isometry of this map follows essentially by Hahn-Banach.

Then use the uniform boundedness principle on the family $(\iota(Tx))_{x \in B_1(0)}$.

This family is pointwise bounded, because for $\varphi \in Y'$, you have

$$ |(\iota(Tx))(\varphi)| = |\varphi(Tx)| \leq \Vert \varphi \circ T\Vert \cdot \Vert x \Vert \leq \Vert \varphi \circ T \Vert $$

for all $x \in B_1 (0) \subset X$.