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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – JZS
    Commented Dec 29, 2022 at 15:36
  • 1
    $\begingroup$ 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 $\nu=1$. $\endgroup$ Commented Dec 29, 2022 at 16:48

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