Let $P:\ell^1(\mathbb{Z}^d) \rightarrow \ell^1(\mathbb{Z}^d)$
be given by


$$(Pz)(x)=\sum\_{y \tilde \ x} \frac{1}{2d} z(y)$$


where the tilde indicates that $y$ is a neighboured vertex of $x.$


I would like to know: Let $\lambda \notin [-1,1]$ is it true that 
$\operatorname{ran}(P-\lambda)$ is dense in $\ell^1.$ In other words, does this operator have no residual spetrum outside $[-1,1]?$ 


Background: If we look at $P$ on $\ell^2(\mathbb{Z}^d)$ this operator is even self-adjoint and thus there is no residual spectrum at all. 



Stochastic operator on $\ell^1$ has dense range
Yes, it is true (if we replace $2^d$ to $2d$). If the range of $P-\lambda$ is not dense, there exists a bounded linear functional $\eta\in \ell^\infty=(\ell^1)^\*$ which vanishes on this range. In other words, there exists a not identically zero bounded function $\eta$ on $\mathbb{Z}^d$ which satisfies $P^\*\eta=P\eta=\lambda\eta$. Consider two cases.


1) $|\lambda|>1$. Note that since $\eta(x)=\lambda^{-1}(2d)^{-1}\sum\_{y\sim x} \eta(y)$, the sum of coeffcients on the right is $\lambda^{-1}$. Iterating this, we see that each $\eta(x)$ is a linear combination of other values of $\eta$ with sum of coefficients $\lambda^{-n}$ for any $n$. Since $\lambda^{-n}$ tends to 0 and $\eta$ is bounded, we conclude that $\eta\equiv 0$, a contradiction.


2) $\lambda$ is not real. Denote $\mu=2d \lambda$. Take large $N$ and denote $\Omega\_N=[-N,N]^d\cap \mathbb{Z}^d$. Consider the sum $S\_n=\sum\_{x\in \Omega\_N,y\sim x} \eta(y)\overline{\eta(x)}$. It equals, if we replace $\sum\_{y\sim x} \eta(y)$ to $\mu \eta(x)$, to $S\_n=\mu\sum\_{x\in \Omega\_n}|\eta(x)|^2$. On the other hand, all internal edges have real contribution to this sum. For boundary edges use an estimate $2|\eta(y)\overline{\eta(x)}|\leqslant |\eta(x)|^2+|\eta(y)|^2$. Then denoting $\mu=u+iv$ and taking the imaginary part, we get $|v|\sum\_{x\in \Omega\_n} |\eta(x)|^2\leqslant 2\sum\_{x\in \Omega\_{n+1}\setminus \Omega\_{n-1}} |\eta(x)|^2$. Now denote $a\_n=\sum\_{x\in \Omega\_n} |\eta(x)|^2$. This sequence is positive for large $n$, increases and satisfies an inequality $a\_{n+1}-a\_{n-1}\geqslant ca\_n$ for certain positive $c=|v|/2$. It follows by induction that $a\_n\geqslant c\_0 \rho^n$, where $\rho >1$ is a root of $\rho ^2-1=c\rho$. On the other hand, since $\eta$ is bounded, $a\_n=O(n^d)$. A contradiction.


