Recently, I came across the following question while studying Fredholm operator. Recall an operator $S$ on a Hilbert space $\mathcal H$ is said to be Fredholm if $Range(S)$ is closed along with both $ker S$ and $ker(S^*)$ is finite dimensional. My question is as follows:

Let $T$ be a bounded operator on separable complex Hilbert space $\mathcal H$ such that it's spectrum $\sigma(T) \subseteq \overline{\mathbb D}.$ Assume $(T-wI)$ is fredholm operator for every $w\in \mathbb D.$ Does it imply $\varphi(T)-w$ is also fredholm for every $w\in \mathbb D$ and for every automorphism $\varphi$ of the unit disc $\mathbb D?$


1 Answer 1


If by automorphism of the disk you mean a biholomorphic map, yes. Then $\varphi$ is of the form $$ \varphi(z)=\frac{z-a}{\overline az-1} $$ for some $a\in\mathbb D$. A computation shows, that then $$ \varphi(T)-w=\left( T-\frac{w-a}{1-w\overline a}\right)(1-w\overline a)(\overline aT-1)^{-1} $$ is a product of a Fredholm times a scalar $\ne 0$ times an invertible operator, hence Fredholm.


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