Let $E$, $F$ be Banach vector spaces and $T:E\to F$ a linear continuous operator such that its image $\text{Im }T$ is a closed subspace. Prove that $\text{Im }T^{\ast }$ is also closed and $\text{Im }T^{\ast }=\text{ann }\left ( \ker T \right )$, where for every subset $X\subseteq E$ we denote $\text{ann }(X):=\left \{\varphi \in E^{\ast}:\varphi \left (X\right )=\left \{0\right \}\right \}$.
Conclude that $\text{Im }T$ is closed if and only if $\text{Im }T^{\ast }$ is closed.
This was my attempt, a not concluding one: Suppose that $\text{Im }T$ is closed. We consider the co-restriction $T^{\ast}:F^{\ast}\to \overline{\text{Im }T^{\ast}}$ and the canonical map $Q:F^{\ast}\to F^{\ast}/{\ker T^{\ast}}$. By the first isomorphism theorem for Banach Spaces, this induces a continuous linear operator $\overline{T^{\ast}}:F^{\ast}/{\ker T^{\ast}}\to \overline{\text{Im }T^{\ast}}$. So, it suffices to show that $\overline{T^{\ast}}:F^{\ast}/{\ker T^{\ast}}\to \overline{\text{Im }T^{\ast}}$ is surjective.
Since $\overline{\text{Im }T^{\ast}}$ is a closed subspace of $E^{\ast}$, it is a Banach Space. Since $F^{\ast}$ is a Banach Space, $F^{\ast}/{\ker T^{\ast}}$ is also a Banach Space. Also, it is straightforward to prove that $\text{ann }(\text{Im }T)=\ker T^{\ast}$ and $\text{Im }T^{\ast}\subseteq \text{ann }(\ker T)$, so we want $\overline{T^{\ast}}:F^{\ast}/\text{ann }(\text{Im }T)\to \overline{\text{Im }T^{\ast}}$ to be surjective.
I do not know whether this is the correct way to solve the problem. Maybe it is possible to show that $\overline{T^{\ast}}$ is surjective, but I am still unable to show the equality $\text{Im }T^{\ast }=\text{ann }\left ( \ker T \right )$ and the final equivalence. How would you solve this exercise?
