The ranges of $T+iI$ and of $T-iI$.

Question

In Rudin's book: Functional analysis. P386.exercises 9.

Let $H^2$ be the space of all holomorphic function $f(z)=\sum_{n=0}^{\infty} c_{n}z^n$ in the open unit disc that satisfy $$\Vert f\Vert ^2=\sum_{n=0}^{\infty} \vert c_{n}\vert^2<\infty.$$ Define $V\in \mathscr{B}(H^2)$ by $(Vf)(z)=zf(z)$. And operator $T$ in $H^2$ given by $$ (Tf)(z)=i\frac{1+z}{1-z}f(z).$$

I have proved $V$ is the Cayley transform of the symmetric operator $T$. Actually, $$(Tf)(z)=i\sum_{n=0}^{\infty}\Big(2\sum_{k=0}^{n-1}c_{k}+c_{n}\Big)z^n.$$ And I know that the $H^2$ is isomorphic to $l^2$ via $f \to \{c_{n}\}$.

But I do not know the reasons for the ranges of $T+iI$ and of $T-iI$; Rudin tells one is $H^2$ and one has codimension 1. Thanks for any advance details on this question.

Answer 1

You can calculate explicitly: $$ \frac{1+z}{1-z}+1=\frac2{1-z},\ \ \ \ \frac{1+z}{1-z}-1=\frac{2z}{1-z}. $$ So $T+iI$ is multiplication by $2i/(1-z)$, which is invertible (multiply by $(1-z)/2i$), while $T-iI$ is multiplication by $2iz/(1-z)$, which is not invertible; its range is $zH^2$, which has codimension $1$.