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Let $X$ be a Banach space and $K:X\to X$ be a compact operator. If $I+K$ is injective then it is onto and hence the inverse $(I+K)^{-1}$ is bounded. What kind of qualitative or quantitative assumptions on $K$ are necessary in order to find an explicit bound on the operator norm of $(I+K)^{-1}$?

In such generality, there is probably no answer (or there are too many), so let me be more specific and explain the situation I've encountered this. The Banach space is either $L^\infty$ or a weighted $L^2$ with weight $(1+|x|)^s$ on $\mathbb{R}^n$, these cases are all of interest. Call $H_0$ the operator $-\Delta$, $V\ $ the multiplication by a function $V(x)$, and $H=H_0+V$. This should all be properly defined but I leave the definitions aside so to accommodate as many additional assumptions as necessary. A standard procedure in spectral theory is to express the resolvent operator $R(z)=(H-z)^{-1}$ as a perturbation of $R_0(z)=(H_0-z)^{-1}$ through the formula $$ R_0(z)=R(z)(I+VR_0(z)). $$ Now the main step: assume we know that $VR_0$ is a compact operator and $I+VR_0$ is injective (this typically amounts to requiring that $H$ has no eigenvalues, or resonances). Then we can invert $I+VR_0$ and write $$ R(z)=R_0(z)(I+VR_0(z))^{-1} $$ which is a very powerful formula, but, alas!, not quantitative. Is there a way to bound the norm of the inverse based on any reasonable assumption on $H_0$ and $V$? Notice that the operator $K=VR_0$ is very explicit, one can write an explicit integral kernel and extract any kind of detailed information on it. Additionally, one can make any sort of assumption on the perturbed operator $H=H_0+V$ (e.g., about the distance of the eigenvalues from the continuous spectrum etc.)

EDIT: I might add that in the example the main point is to obtain a uniform estimate for the norm of $R(z)$ as $z$ approaches the real axis, which contains the spectrum of $H$. Otherwise it might seem odd to look for an estimate on $R(z)$, which is already a well defined and bounded operator on $L^2$, with known bounds. But this procedure gives a reasonable definition of $R(z)$ also for real $z$, certainly not as a bounded operator on $L^2$ but on different spaces.

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  • $\begingroup$ Such a qualitative theory exists, in the following sense: You are describing a natural phenomenon that many people have to study. In order to do so, they develop methods. If you want to call them a theory, is a matter of taste. I wouldn't because these methods aren't simple and one needs to fine tune them on a case by case basis. One thing on relies on is what Florian mentions that one needs to control the distance of the spectrum of K to -1. The other thing, I can think of is that there are often better bounds available on P (1 + K)^{-1} Q then on (1 + K)^{-1}, where P,Q are projections. $\endgroup$
    – Helge
    Commented Sep 30, 2010 at 16:06
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    $\begingroup$ I have some old notes on this problem at math.ucla.edu/~tao/preprints/Expository/compact_operator.dvi . If you assume some upper bound on the metric entropy of the image of $K$, and also bound the operator norm of $K$, one can control $(1+K)^{-1}$ quite well after projecting away a bounded dimensional bad subspace as Helge indicates, and this implies some (worse) bound on $(1+K)^{-1}$ itself involving the distance to the spectrum. $\endgroup$
    – Terry Tao
    Commented Oct 27, 2022 at 20:00

4 Answers 4

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Let $S$ denote the spectrum of $K$ (which consists of eigenvalues and 0). Then $\|(I+K)^{-1}\| = d(S,-1)^{-1}$. The statement $-1\notin S$ is equivalent to requiring that $K$ is injective, so your request for a "quantitative version" of the inversion formula is equivalent to a quantitative version of the injectivity of $I+K$, namely a lower bound on $d(S,-1)$.

A simple criterion would be: if $\|K\|\le C < 1$ then $d(S,-1)\ge 1-C$.

I think this is the best that can be said in the general case without knowing the operators.

Edit: As Jeff points out, this is valid only for normal $K$ acting on a Hilbert space. For general operators however it's still true that $\|K\|\le C < 1$ implies that $(I+K)$ is invertible and $\|(I+K)^{-1}\|\le (1-C)^{-1}$.

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    $\begingroup$ The norm identity is $\| (I+K)^{-1} \| = d(S,-1)$ is true for normal operators, but is not true in general! A simple example is already given by $K$ a Jordan block of size $n$ and eigenvalue $\lambda$. The norm in that case is of order $1/|\lambda +1|^n$ as $\lambda \rightarrow -1$. $\endgroup$
    – Jeff Schenker
    Commented Sep 30, 2010 at 15:24
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A bit similar but different to the answer of Florian. This is not a solution to the problem, but it is a start.

Suppose $\|K\|\le 1$ and let $r(K)$ denote the spectral radius of $K$. If $r(K)<1$ then certainly $H=(1-K)^{-1}$ exists. The most natural way to find (norm-)bounds on $H$ would be to bound $K^n$ and then use block summation of the geometric series. To bound $K^n$ we would need something more than the spectral radius formula, we would need some uniform spectral radius formula.

For further development I would start here (where the problem is considered in a general commutative Banach algebra):

Jan-Erik Björk: On the spectral radius formula in Banach algebras. Pacific J. Math. Volume 40, Number 2 (1972), 279-284.

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  • $\begingroup$ Yes it really is! Also, compactness is a good to have in these problems. $\endgroup$
    – AD - Stop Putin -
    Commented Oct 7, 2010 at 13:09
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Suppose $N=\|(I+K)^{-1}\|$. In order to calculate $N$ one can do the following. Write $$N^2= \|(I+K)^{-1}(I+K^*)^{-1}\| = \|(1+K+K^*+K^* K)^{-1}\| = 1/(1+\lambda),$$ where $\lambda$ is the smallest eigenvalue of $K+K^*+K K^*$. Try to calculate it using your explicit formulas.

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    $\begingroup$ It would be easier to compute the eigenvalues of K and check which one is closer to −1, but is is exactly like computing the norm of $(I+K)^{−1}$ directly. $\endgroup$
    – Piero D'Ancona
    Commented Sep 30, 2010 at 13:35
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The second resolvent equation you are studying is an important tool in quantum scattering theory.

See M. Reed and B. Simon, "Methods of Modern Mathematical Physics", Vol IV, Academic Press

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