# Invertibility of an operator of the form $I-B$

Suppose I have an operator on $L^2(\mathbb{R}^3)$ of the form $A=I-B$, where:

1) $B$ is bounded on $L^2(\mathbb{R}^3)$, moreover $\Vert B\Vert_{L^2\rightarrow L^2}=1$,

2) $B$ has a positive integral kernel,

3) $A$ is injective.

Can i conclude that $A$ is invertible on $L^2(\mathbb{R}^3)$?

• @Anton: I'm confused, if the integral kernel of $B$ is in $L^2$ then $B$ is a compact operator, and since $I-B$ is injective by Freedholm alternative we get that it's invertible. I'm missing something? Neverthless in my specific example the integral kernel of $B$ only belongs to $L^{2,\infty}$ – Capublanca Feb 10 '17 at 9:10
• Anton, if the kernel is $L^2$, $B$ is Hilbert-Schmidt hence compact. Then $A$ is Fredholm of index zero. Since it is injective, it is invertible. – Denis Serre Feb 10 '17 at 9:10
• Capublanca, you beat me for 20 seconds. – Denis Serre Feb 10 '17 at 9:11
• @DenisSerre: Yeah, by photofinish :D – Capublanca Feb 10 '17 at 9:14

This is never true if the kernel of $B$ is symmetric and (say) positive-definite, as follows from the spectral theorem. In this case, $A$ has dense range for $\overline{\operatorname{ran}A} = (\operatorname{ker}A)^\perp=L^2(\mathbb R^3)$, but it is not surjective and $A^{-1}$ is unbounded. That $A$ is injective excludes $0$ from being an eigenvalue of $A$, but $0$ is still in the spectrum of $A$.

Indeed, $\sigma(B)\subseteq [0,1]$ with $1$ belonging to the spectrum of $B$ (since $\|B\|_{L^2\to L^2} =1$) implies that $\sigma(A)\subseteq [0,1]$ with $0$ belonging to the spectrum of $A=I-B$.

Example. Take the Gauss-Weierstrass kernel $(4\pi)^{-3/2}e^{-|x-y|^2/4}$ as kernel of $B$. There are two possibilities to see that $1$ is not an eigenvalue:

1) It holds $Bf= k\ast f$ with $k(x) = (4\pi)^{-3/2}e^{-|x|^2/4}$. If $f$ was a (normalized) eigenfunction, then $\bigl(\hat k(\xi)-1\bigr)\,\hat{f}(\xi)=0$ a.e., with $\hat{f}$ being the Fourier transform of $f$. But $\hat k(\xi) = e^{-|\xi|^2}$, so $\hat f(\xi) = 0$ a.e., a contradiction.

2) It holds $B= e^\Delta$, i.e., $B$ is the solution operator of the initial-value problem for the heat equation at time $1$. If $f$ was an eigenfunction of $B$, then $f$ would be an eigenfunction of the generator $\Delta$ (with eigenvalue $0$ as $e^0=1$), which again is a contradiction.

• i think there is some problem with your argument. Take $B$ a compact operator with norm 1. In that case $A$ is invertible by Freedholm theory. – Capublanca Feb 10 '17 at 9:30
• If $B$ is compact, then $1$ is an eigenvalue of $B$ and $A$ is not injective. – ifw Feb 10 '17 at 9:34
• You still need to give an example where 1 is not an eigenvalue of B. – user1688 Feb 10 '17 at 11:07
• @Anton $1$ cannot be an eigenvalue of $B$, as $A$ is supposed to be injective. Do you mean an operator $B$ meeting all the assumptions? – ifw Feb 10 '17 at 11:53
• Yes. That's why I said "example". – user1688 Feb 10 '17 at 12:59

The answer is no and an example follows by amenability of $\mathbb{R}^n$ as a locally compact group.

Let $\mu$ be a compactly supported probability measure defined by a density function $f$ with respect to the Lebesgue measure and let $T$ be the operator given by the kernel $k(x,y)=f(y-x)$ (i.e. $T$ acts by convolution with $f$). $T$ has norm 1.

For $B_n=$ the ball of radius $n$ in $\mathbb{R}^3$ the operator satisfies $T\chi_{B_n}-\chi_{B_n}\to 0$ in norm. Thus the range of $I-T$ is not closed.

If the support of $f$ contains a neighborhood of the the origin then $I-T$ is injective since constant functions are not in $L^2$. However $T$ is not invertible.

I'm not sure if my unswer would be useful, but I have the following thing in mind.

1. If B - is self-adjoint (symmetric) then it has a countable number of eigenvectors $e_j$ with eigenvalues which form a basis in $L^2(\mathbb{R}^3)$.
2. Since $A$ is claimed to be injective then $B$ cannot have $\lambda=1$ as an eigenvalue.
3. Then we can rewrite equation $(I-B)x=y$ in the following way: \begin{equation} \sum\limits_j (I-B)e_jx_j = \sum\limits_{j}e_jy_j, \end{equation} and consequently define the quasi-inverse: \begin{equation} x_j = \dfrac{y_j}{1-\lambda_j}, \, j = \overline{1,\infty}. \end{equation} Now the question is if $x = (x_j)\in L^2(\mathbb{R}^3)$? The answer is not necessarily if $\lambda_j$ have a convergent subsequence $\lambda_{j(k)}$: \begin{equation} \lambda_{j(k)}\rightarrow 1, \, k\rightarrow \infty. \end{equation} To prove this I will do the following: since $B$ has an eigenvalue basis, I will define the following symmetric operator $Q:L^2(\mathbb{R}^3)\rightarrow L^2(\mathbb{R}^3)$: \begin{equation} Qe_j = \left(1-\dfrac{1}{j+1}\right)e_j, \end{equation} where $\{e_j\}$ is the spectral basis for $B$. It is easy to see that $Q$ is symmetric, self-adjoint, positive and \begin{equation} \|Q\| = 1. \end{equation} Also $I-Q$ is injective. Indeed, if $I-Q$ is not injective then for some $x\neq 0$ it holds: $$Qx = x \rightarrow \|x-Qx\|^2_{L^2} = \sum\limits_{j=1}^{\infty} \dfrac{x^2_j}{(j+1)^2} > 0,$$ which makes a contradiction.

Applying the same process for $I-Q$ we obtain that series $\{x_j(y_j)\}_j$ will explode in infinity for some chosen $y$. Hence we have an example when operator I-Q is not invertible, but all conditions are satisfied.