The closure of range of $T^*$ and the complement of kernel of T

Question

In Hilbert space $H$, let T be bounded linear function from $H$ to $H$.

Also, the dimension of $H$ can be infinite.

$\textbf{Exercise}$ $R (T) ^{\perp}=\ker(T^*)$ and $\ker(T)^\perp$$=\overline{R(T^*)}$

In Exercise, $T^*$ is bounded linear transformation on $H$ such that \begin{align*} (Tf,g)&=(f,T^*g)\\ \Vert T \Vert &=\Vert T^* \Vert\\ (T^*)^*&=T \end{align*}

I showed that $R (T) ^{\perp}=\ker(T^*)$. In infinite vector space, $R(T^*)$ may not be closed. I want to know how to prove $\ker(T)^\perp=\overline{R(T^*)}$ by using set inclusion.

Any help is appreciated!!

Thank you!

Answer 1

I'm going to assume Robert Lewis's comment is correct, and you didn't mean "functional."

You already know $$R(T^*)^\perp = \ker((T^*)^*)=\ker(T)$$ Now take orthogonal complement of both sides, giving $$(R(T^*)^\perp)^\perp=\ker(T)^\perp$$ All that remains is to apply the result $(E^\perp)^\perp=\overline E$ which holds for any subspace $E$. If you're not familiar with this result, try to prove it from the definition of orthogonal complement.