(It turns out that the claim is incorrect as pointed out in the comment. Will get back soon)
Let $A, B$ be normed linear spaces and let $S$ be a bounded linear map from $A$ to $B$. Define the map $S^* : B^* \rightarrow A^*$ by $S^*(g)(v) = g(S(v))$ for $g \in B^*, v \in A$. It’s easy to see that $S^*$ is also a bounded linear map.
Prove that the kernel of $S$ is isomorphic to the annihiltor of the range of $S^*$, where the annihilator of a subspace $Y$ of a normed linear space $X$ is defined as $\{f \in X^* \mid f(a) = 0, \forall a \in Y \}.$
I tried search online but most of the results are for Hilbert spaces. I tried to build an isomorphism from the kernel to the annihilator of range of adjoint (which is a subspace of $A^{**}$) by sending $v$ to the evaluation at $v$. However, I am not able to show that this map is surjective, namely for any function $\in A^{**}$ that vanishes at the range of the adjoint, it must be the evaluation at some $v \in$ the kernel of $S$.
Thank you!