The map $B(X,Y) \to B(Y^*, X^*), T \mapsto T^*$ is not necessarily onto.

Question

There is the following result:

Let $X,Y$ be normed spaces. Then the map $T \mapsto T^*$ is an isometric isomorphism from $B(X,Y)$ into $B(Y^*, X^*)$.

I want to show the map $T \mapsto T^*$ is not necessarily onto. In order to do so, I consider the Banach spaces $B(\mathbb{F}, c_0)$ and $B(c_0^*, \mathbb{F}^*)$ and want to show that there is no continuous map $T \mapsto T^*$ from $B(\mathbb{F}, c_0)$ onto $B(c_0^*, \mathbb{F}^*)$. That is, there is an operator in $B(c_0^*, \mathbb{F}^*)$ that is not the adjoint of an operator in $B(\mathbb{F}, c_0)$.

Answer 1

Hint/check:

• $B(\mathbb F, c_0)$ is isometric to $c_0$,

• $B(c_0^* ,\mathbb F^*)$ is isometric to $c_0^{**}$,

• Under this identification, the mapping $T\mapsto T^*$ can be treated as the canonical embedding $c_0 \to c_0^{**}$.

Key word: Reflexive Banach space