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 $L^2(\mathbb{R}^3)$?
 A: 
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.
A: I'm not sure if my unswer would be useful, but I have the following thing in mind. 


*

*If B - is self-adjoint (symmetric) then it has a countable number of eigenvectors $e_j$ with eigenvalues which form a basis in $L^2(\mathbb{R}^3)$.

*Since $A$ is claimed to be injective then $B$ cannot have $\lambda=1$ as an eigenvalue. 

*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.
A: 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.
