# $E$ and $F$ Banach Spaces, $T:E\to F$ such that $\text{Im }T$ is closed. Then $\text{Im } T^{\ast }=\text{ann }\left ( \ker T \right )$

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?

Since $ImT$ is closed, $U:E/kerT\rightarrow ImT$ is bijective and invertible. Open mapping theorem implies that $U$ is invertible. You deduce that $U^*$ is invertible and its image is $(E/ker T)^*=ann(kerT)$. Let $f\in F^*, T^*(f)$ is a linear function which vanishes on $Ker T$ and is equal to $U^*(f_{\mid Im T})$. The image of $U^*$ is the image of $T^*$ and is $ann(Ker T)$ which is closed.