TITLE: Is the map sends $T$ to $T^*$ adjoint of $T$ surjective?
QUESTION [6 upvotes]: Let $B(X)$ denotes the set of all bounded linear operators from $X$ to $X$, where $X$ is a Banach space. Same is defined for the set $B(X^*)$, where $X^*$ denotes the set of all bounded linear functionals of $X$, that is all bounded linear functionals from $X$ to the field $\mathbb F$. In other words, $X^*$ is the dual space of $X$.
Let us now define a map $\alpha :B(X) \to B(X^*)$ by.
$$\alpha(T)=T^*, \quad T \in B(X)$$
where $T^*$ is the Banach space adjoint of the operator $T$.
That is, for Banach spaces $X,Y$ and $T\in B(X,Y)$ the adjoint operator $T^*: Y^* \to X^*$ is defined as
$$T^*(y^*)(x)=y^*(T(x)),~~~y^* \in Y^*,~~x \in X.$$
Now note that, since $\|T\|=\|T^*\|$, the map $\alpha$ is an isometry and also injective. Can I show that $\alpha$ is a linear isomorphism? Linearity of $\alpha$ comes from the linearity of adjoint operators but I am not able show that $\alpha$ is surjective. One of my friends told me $\alpha$ may not be surjective but can't give any argument.

REPLY [6 votes]: This is an addition to the excellent answer of @JustDroppedIn, where I provide full details of the claim he sketches.
Theorem: If $X$ is reflexive, the map $\varphi: B(X) \to B(X^*): T \mapsto T^*$
is surjective.
Proof: Consider the canonical embedding $j: X \to X^{**}: x \mapsto \operatorname{ev}_x$. Since $X$ is reflexive, $j$ is an isometric isomorphism. As mentioned by @JustDroppedIn, this implies that we have a natural map
$$\psi: B(X^*) \to B(X^{**}) \to B(X)$$
given by
$$\psi(T) = j^{-1} T^* j, \quad T \in B(X^*).$$
We claim that $\varphi \psi = \operatorname{id}_{B(X^*)}$ and surjectivity of $\varphi$ will follow.
For this, fix $T \in B(X^*)$. We have to show that
$$(j^{-1}T^* j)^* = T.$$
Fix $f\in X^*$. It suffices to show
$$(j^{-1}T^*j)^*(f) = T(f) \in X^*$$
so fix $x \in X$, and note that it suffices to show
$$((j^{-1} T j)^*(f))(x) = (T(f))(x).$$
From here, the proof is a simple calculation. Start by noting that $\operatorname{ev}_x \circ T = \operatorname{ev}_y$ for some $y \in X$. Hence, $(T(f))(x) = \operatorname{ev}_x(T(f)) = f(y)$ and thus
\begin{align*}((j^{-1}Tj)^*(f))(x)&=f(j^{-1}T^*j(x))= f(j^{-1}(\operatorname{ev_x}\circ T))= f(y) = (T(f))(x)\end{align*}
and we are done.