Given $T:E^*\to E^*,$ does there exist $S:E\to E$ such that $\langle Te^*,e \rangle = \langle e^*,Se\rangle?$

Question

Let $E$ be a Banach space and $T:E\to E$ be a bounded linear operator. Denote $E^*$ a dual space of $E,$ that is, the space of all bounded linear functional $e^*:E\to\mathbb{R}.$

It is well known that if $T:E\to E$ is a bounded linear operator, then there exists a bounded linear operator $T^*:E^*\to E^*$ such that for all $e^*\in E^*$ and $e\in E,$ $$\langle Te^*,e \rangle = \langle e^*,Se\rangle.$$ In most literature, $T^*$ is called adjoint operator of $T.$

I would like to know its converse, that is,

Question: Given a bounded linear operator $T:E^*\to E^*,$ does there exist bounded linear operator $S:E\to E$ such that for all $e\in E$ and $e^*\in E^*,$ $$\langle Te^*,e \rangle = \langle e^*,Se\rangle?$$

If $T:E^*\to E^*$ is assumed to be weak$^*$-to-weak$^*$ continuous, then the question has positive answer.

What happens if we remove weak$^*$-to-weak$^*$ continuity? Is the answer still positive?

Answer 1

Let $T\in L(X,Y)$ (linear and bounded, $X,Y$ normed spaces). We can show that $$T^{**}\circ J_{X}=J_{Y}\circ T,$$ with $J_X$ the natural embedding from $X$ to $X^{**}$ via $(J_X x)(x^*)=x^*(x)$.

Back to your case: If $E$ is reflexive, then $S=J_{E}^{-1}\circ T^*\circ J_{E}$. With this you have that $T=S^*$ and so $$\langle Te^*,e\rangle=\langle S^*e^*,e\rangle=\langle e^*,Se\rangle.$$

Answer 2

You can easily prove the reverse of the sentence after your question: If an $S$ as in the question exists, then $T$ is weak*-weak* continuous.