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$$=\overline{R(T^*)}$
In Exercise, $T^*$ is bounded linear transformation on $H$ such that
\begin{align*}
(Tf,g)&=(f,T^*g)\\
\Vert T \Vert &=\Vert T^* \Vert\\
(T^*)^*&=T
\end{align*}
I showed that $R (T) ^{\perp}=\ker(T^*)$.
In infinite vector space, $R(T^*)$ may not be closed.
I want to know how to prove $\ker(T)^\perp=\overline{R(T^*)}$ by using set inclusion.
Any help is appreciated!!
Thank you!
Best Answer
I'm going to assume Robert Lewis's comment is correct, and you didn't mean "functional."
You already know $$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.