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.

  • you mean that each vertex has $2^d$ neighbours? It looks strange. – Fedor Petrov Jul 15 '17 at 17:09
  • What for $d=3$? – Fedor Petrov Jul 15 '17 at 17:27
up vote 2 down vote accepted

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.

  • thank you for the answer, but sorry, where does this estimate after "Then denoting $\mu=u+iv$, we get" come from. Would you mind spending a few more words on this? Otherwise, I could completely follow your thoughts. – BaoLing Jul 15 '17 at 18:38
  • It looks to me that the Wiener algebra on the torus is as good as on the circle, so the operator is actually invertible. Am I missing anything? – fedja Jul 15 '17 at 19:45
  • @fedja could I please ask you to elaborate on this in an answer, this does sound very interesting to me. I did suspect invertibility as well, but wanted to look for an easier conjecture first. – BaoLing Jul 15 '17 at 19:50
  • @fedja or would you mind just briefly outlining invertibility as a comment below my question? – BaoLing Jul 15 '17 at 22:40
  • @BaoLing Just google Wiener's theorem/Wiener algebra and mimic the proof you like most :-) – fedja Jul 16 '17 at 0:47

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