I tried to prove the following:
Let $X,Y$ be normed spaces and $\Phi:\mathcal{L}(X,Y)\to\mathcal{L}(Y^*,X^*)$, $\Phi(A)=A^*$ is surjective. Then $Y$ is reflexive. Here $X^*$ $(A^*)$ denotes the dual space of $X$ (the dual operator of $A$).
My try is: Let $F\in Y^{**}\setminus\{0\}$ and $x\in X\setminus\{0\}$. By Hahn-Banach theorem, there exists $\varphi\in X^*$ such that $\varphi(x)=1$. Denote operator $A:Y^*\to X^*$, $A(\phi)y=\varphi(y)F(\phi)$; this operator is bounded and linear. Due to our assumption, there exists $B\in\mathcal{L}(X,Y)$ such that $B^*=A$. Now we choose $\phi\in Y^*$ and compute: $$F(\phi)=\varphi(x)F(\phi)=A(\phi)x=B^*(\phi)x=\phi(Bx).$$ Hence $Y$ is reflexive due to the definition of reflexivity.
Is this proof correct ? If yes, why $Bx$ does not depend on choice of $x$, only on choice of $F$ ?
Thank you for any comments.