From Functional analysis, by Conway. I try to prove this exercise.
If $ X $ and $ Y $ are Banach spaces and $ B \in \mathscr B(Y^*, X^*) $, then there is an operator $ A $ in $ \mathscr B(X,Y) $ such that $ B=A^* $ if and only if $ B $ is weak$ ^* $-continuous.
Note that
- $X^* = $ the set of all continuous linear functional on $X$.
- $Y^* = $ the set of all continuous linear functional on $Y$.
- $\mathscr B(X,Y) = \{ f \colon X \to Y \mid f \text{ is continuous} \}$.
- $\mathscr B(Y^*,X^*) = \{ f \colon Y^* \to X^* \mid f \text{ is continuous} \}$.
Please help me.
If $A \in B(X,Y)$ then $A^{\ast} \in B(Y^{\ast},X^{\ast})$ is clearly weak-star continuous. Conversely, suppose $B \in B(Y^{\ast},X^{\ast})$ is weak-star continuous, we want to define $A : X\to Y$ such that $$ B(f)(x) = f(Ax) \quad\forall f \in Y^{\ast}, x \in X $$ Since $B$ is weak-star continuous, for any $x\in X$ fixed $$ \hat{x} \circ B : Y^{\ast} \to \mathbb{C} \text{ given by } f \mapsto B(f)(x) $$ is continuous. Hence, $\exists ! y\in Y$ such that $$ B(f)(x) = f(y) \quad\forall f\in Y^{\ast} $$ (This is a short lemma, that perhaps has been proved before in the textbook?). So we define $$ A : X\to Y \text{ by } x \mapsto y $$ where $y$ satisfies the above criterion. It remains to show that $A$ is continuous. This follows from the closed-graph theorem: If $x_n \to x$ and $A(x_n) \to y$, then for any $f\in Y^{\ast}$, $$ B(f)(x_n) \to B(f)(x) = f(Ax) $$ since $B(f) \in X^{\ast}$. Also, $$ B(f)(x_n) = f(A(x_n)) \to f(y) $$ Hence, $f(y) = f(Ax)$ for all $f\in Y^{\ast}$, so $y=Ax$, whence $A$ is bounded.