# Show $A_0 + S$ is invertible, where $S$ is the Riesz projection of bounded $A_0$

Let $$A_0$$ be a bounded linear operator on a Hilbert space $$H$$. Suppose $$0$$ is an isolated point of the spectrum of $$A_0$$. Let $$S$$ be the corresponding Riesz projection, namely, $$S = -\frac{1}{2\pi i} \int_{C_\varepsilon} (A_0 - \lambda)^{-1}d\lambda,$$ where $$C_\varepsilon$$ is a circle of small radius $$\varepsilon$$, centered at $$0$$, contained entirely in the resolvent set of $$A_0$$. By Cauchy's Theorem, the definition of $$S$$ is independent of such $$\varepsilon$$. Assume also that $$A_0 S = 0.$$

It's straightforward to see that $$\ker A_0 \subseteq \text{ran}\, S$$. So additionally assuming $$A_0 S = 0$$ gives $$\ker A_0 = \text{ran}\, S$$.

I would like to show that $$A_0 + S : H \to H$$ has a bounded inverse.

Under the given assumptions, this conclusion is asserted to be true in the note by Jensen and Nenciu, Rev. Math. Phys. 16(5) (2004).

If I additionally assume that $$A_0$$ is self-adjoint, it follows that $$A_0 S = 0$$, $$S$$ is an orthogonal projection, and that $$A_0 : (\ker A_0)^\perp \to (\ker A_0)^\perp$$ has bounded inverse. Proofs of these facts may be found in Chapter 6 of Hislop and Sigal's book on spectral theory. Having these properties readily gives that $$A_0 + S : H \to H$$ is bijective and thus invertible by the open mapping theorem.

But I have not managed to reach the conclusion assuming only that $$A_0 S = 0$$. Hints, solutions, or references are greatly appreciated.

This is true in general Banach spaces $$X$$ (see Dunford-Schwartz vol I for the general theory I use). The assumption $$A_0S=0$$ says that $$0$$ is a simple pole of the resolvent so that $$X=Ker (A_0) \oplus Im (A_0)$$ and the splitting is given by the projection $$S$$, that is $$Ker (A_0)=S(X)$$, $$Im(A_0)=Ker (S)$$. Then $$A_0$$ is invertible from $$Im(A_0)$$ into itself and $$S$$ is the identity on $$Ker (A_0)$$ and identically zero on $$Im (A_0)$$ and then $$A_0+S$$ is invertible.
• Thank you for the answer. Would it be possible for you provide a more specific reference to the theory you use? Is there a particular chapter or section of Dunford-Schwartz that you refer to? The last piece I cannot show is that the kernel of $S$ is contained in the image of $A_0$, and for this I suppose I will need to see how $A_0S =0$ implies the resolvent has a simple pole.
• I am using the results in DS I, Chapter 7, section 3. See in particular Theorem 18 (pag. 573 in my edition). In your case $\nu=1$. Commented Dec 29, 2022 at 16:48