I want to prove that if $T:X \to Y$ is a continuous linear operator and $X,Y$ are infinite dimensional Banach spaces, with $TX$ closed, then
$$ T^*Y^* = (\text{ker}T)^{\perp} $$
In one direction:
Let $x^* \in T^*Y^*$, then there exists a $y^* \in Y^*$ such that $x^* = T^*y^* \implies x^* = y^* T$ by definition of the adjoint. Now, let $m \in \text{ker}T$, then $Tm=0 \implies y^*Tm = 0 $ and so $x^*m=0$, which implies that $x^* \in (\text{ker}T)^{\perp}$.
I am confused about the other direction. Let $x^* \in (\text{ker}T)^{\perp}$, then it must be that $\text{ker}T \subset \text{ker} x^*$. Since $TX$ is closed, we can use the first isomorphism theorem to say that $X/ \text{ker} T \cong TX$ ... but i'm not sure how this helps or how to proceed from here..any hints?
Suppose $\{T(x_n)\} \to 0$. By the isomorphism you have obtained in your answer it follows that $x_n+ker(T) \to 0$. This means $x_n+y_n \to 0$ for some sequence $\{y_n\}\subset ker (T)$. But then $x^{*} (x_n+y_n)\to 0$. Since $x^{*}(y_n)=0$ for all $n$ we get $x^{*} (x_n)\to 0$. Using this fact it follows that the linear functional $y^{*}$ on $T(X)$ defined by $y^{*}(Tx)=x^{*}(x)$ is well defined and continuous. By Hahn - Banach Theorem this extends to an element of $Y^{*}$. Now chek that $x^{*}=T^{*}y^{*}$.