1
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.