The image of dual operator $T^*$ is the annihilator of the kernel of $T$

Question

Let $T$ be an operator from $V\to W$, banach spaces, and we could assume that $\text{ker}(T)$ is closed. Then, is it true that $$ \text{Ker}(T)^\perp= \text{Im}(T^*) $$ where $\perp$ means the annihilator of $\ker{T}$, and $*$ the topological adjoint?

I have already searched in this forum for the answer; but I only found answers by adding also that $W$ be of finite dimension. Maybe through Hahn Banach theorem, the problem is solvable the same.

I am in particular thinking about adapting this solution: Image of dual map is annihilator of kernel.

Thanks.

Almost any book which discusses adjoint operators has a proof. In particular Rudins FA'has it. — Kavi Rama Murthy, Oct 08 '20 at 09:17
See https://math.stackexchange.com/questions/386395/annihilator-of-a-subspace — Kavi Rama Murthy, Oct 08 '20 at 09:31

score 0 · Answer 1 · answered Oct 08 '20 at 10:05

(My answer assumes that you work in a topological (and not algebraical) setting, e.g. $T$ is continuous and we also have topological dual spaces)

This cannot be true in general, since the left-hand side is always closed (even weak-* closed), but the right-hand side might not be closed.

In fact, this equality is (if $T$ is continuous) equivalent to $\operatorname{Im}(T^*)$ to be closed and to the closedness of $\operatorname{Im}(T)$. This is known as the "closed range theorem".

The image of dual operator $T^*$ is the annihilator of the kernel of $T$

1 Answers1