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.
Best Answer
In the case that $X$ is reflexive, I believe this is true, but I haven't checked it out; but I'm pretty sure that the map $B(X^*)\to B(X^{**})$ composed with the canonical isometric isomorphism $B(X^{**})\cong B(X)$ will give you the desired surjectivity.
However, the well-defined linear isometry $B(X)\to B(X^*)$, $T\mapsto T^*$ is not surjective in general. Take a non-reflexive Banach space $X$ (like $c_0$, or $\ell^1$) and let $j:X\to X^{**}$ be the canonical inclusion. Since $j$ is not surjective, let $\chi\in X^{**}$ be an element outside of $j(X)$. Fix a non-zero functional $\psi_0\in X^*$ and define an operator $P:X^*\to X^*$ by $P(\phi):=\chi(\phi)\cdot\psi_0$. This is a well-defined, bounded operator (and actually its range is one dimensional, but we dont care about it).
Now assume that $P$ is in the range of $\alpha:B(X)\to B(X^*)$, so there exists a bounded operator $T\in B(X)$ such that $P=T^*$, i.e. $P(\phi)=T^*\phi=\phi\circ T$ for all $\phi\in X^*$, so $\phi(Tx)=\chi(\phi)\cdot\psi_0(x)$ for all $x\in X$ and all $\phi\in X^*$. Since $\psi_0$ is non-zero, find $x_0\in X$ such that $\psi_0(x_0)\ne0$. We then have $$\phi(Tx_0)=\chi(\phi)\cdot\psi_0(x_0)$$ and this is true for all $\phi\in X^*$; set $\psi_0(x_0):=\lambda\in\mathbb{C}\setminus\{0\}$. We have just shown that $\chi=\frac{1}{\lambda}j_{Tx_0}=j_{\frac{1}{\lambda}Tx_0}\in j(X)$, which is a contradiction.
Note that this answer works for any non-reflexive space; in other words, if the claim in my first paragraph is true (which i think it is) we get the following:
Comment: I've spent a few hours thinking about this, so I have to give some credit to GEdgar. I knew the counter-example would come from the non-reflexive world from the first moment I started on this, but I was trying to give an answer specifically for $X=c_0$; after seeing GEdgar's comment I realized there is no need to go in a specific space, the abstraction actually helps here.