3
$\begingroup$

The spectral radius of Markov operator for the lattice graph $\mathbb{Z}^n$ is $1$

The lattice graph $\mathbb{Z}^n$ is defined as $V=\mathbb{Z}^n , E=\{ \{\vec{x} , \vec{y}\} : \vec{x},\vec{y}\in \mathbb{Z}^n , ||\vec{x}-\vec{y}||_1:=\sum_1^n|x_i-y_i|=1 \}$.

Markov's Operator for a function $\varphi:V\rightarrow\mathbb{C}$ , $\varphi\in L^2(\mathbb{Z}^n,\nu)$ ($\nu(v):=deg(v)$ for $v\in V=\mathbb{Z}^n$) is defined as $M\varphi(x) = \frac{1}{deg(x)}\sum_{y\in B(x,1)\setminus{\{x\}}} \varphi(y)$.

I wish to show that the spectral radius of $M$ with respect to $L^2(\mathbb{Z}^n,\nu)$ is $1$.

It is known that Markov operator is a self-adjoint operator, thus the specatral radius $\rho = ||M|| = \sup_{\varphi \neq 0} \frac{|\langle M\varphi ,\varphi \rangle|}{||\varphi||^2}$ (when $ \langle \varphi,\varphi\rangle = \sum_{\vec{x} \in\mathbb{Z}^n}deg(\vec{x})\varphi(x)\overline{\varphi(x)} $ ), it is also known that $\rho \le 1$.

So in order to prove that $\rho = 1$ , it is enough to define a sequence $\{\varphi_n\}_{n=1}^\infty \subset L^2(\mathbb{Z}^n,\nu)$ such that $\lim_{n \rightarrow \infty} \frac{|\langle M\varphi_n ,\varphi_n \rangle|}{||\varphi_n||^2} = 1$ (when $\varphi_n \neq 0$ for all $n\in \mathbb{N}$).


Edit : the next method can't work, because if it works, one may use the same $\{\varphi_n\}_1 ^\infty$ definition for $2n$-regular tree $T_{2n}$ and by famous theorem by Kesten (Harry) the spectral radius of Markov's averaging operator for $d$ regular tree is $\frac{2\sqrt{d-1}}{d}$ , in this case $\frac{2\sqrt{2n-1}}{2n}$ which can't be $1$ in case $n\neq 1$ (i.e for $\mathbb{Z}$).


I thought of using the next method to define $\varphi_n$, we choose a local "star" $x^{*}$ in $\mathbb{Z}^n$, denoting $B(x^*,1)\setminus\{x^*\}$ with $\{y_1,y_2,...,y_{2n}\}$ (as for the edges definition, $deg(x^*) = 2n$ ). we define $\varphi(z) = 0 $ for all $z\notin B(x^*,1)$, so:

$\frac{|\langle M\varphi,\varphi \rangle|}{||\varphi||^2} = \frac{|\sum_{y\in B(x^*,1)}\deg(y)M\varphi(y) \cdot \overline{\varphi(y)}|}{\sum_{y\in B(x^*,1)}\deg(y)|\varphi(y)|^2} = $

$ = \frac{|\sum_{y\in B(x^*,1)}\deg(y) \frac{1}{\deg(y)}\sum_{z\in B(y,1)}\varphi(z) \cdot \overline{\varphi(y)}|}{2n\sum_{y\in B(x^*,1)}|\varphi(y)|^2} = $

$ = \frac{|\sum_{y\in B(x^*,1)}\sum_{z\in B(y,1)}\varphi(z) \cdot \overline{\varphi(y)}|}{2n(|\varphi(x^*)|^2 + |\varphi(y_1)|^2 +... + |\varphi(y_{2n})|^2 )} = $

$ = \frac{|[\varphi(x^*)\varphi(y_1) + \varphi(x^*)\varphi(y_2) + ...\varphi(x^*)\varphi(y_{2n})] + [ \varphi(y_1)\varphi(x^*) + \varphi(y_2)\varphi(x^*) + ... + \varphi(y_{2n})\varphi(x^*)]|}{2n(|\varphi(x^*)|^2 + |\varphi(y_1)|^2 +... + |\varphi(y_{2n})|^2 )} = $

$ = \frac{2|\varphi(x^*)\varphi(y_1) + \varphi(x^*)\varphi(y_2) + ...\varphi(x^*)\varphi(y_{2n})|}{2n(|\varphi(x^*)|^2 + |\varphi(y_1)|^2 +... + |\varphi(y_{2n})|^2 )} = $

$ = \frac{1}{n} \frac{|\varphi(x^*)\varphi(y_1) + \varphi(x^*)\varphi(y_2) + ...\varphi(x^*)\varphi(y_{2n})|}{(|\varphi(x^*)|^2 + |\varphi(y_1)|^2 +... + |\varphi(y_{2n})|^2 )} = $

Now I tried to work out how to choose the values of $\varphi$ such that the expression $\frac{|\varphi(x^*)\varphi(y_1) + \varphi(x^*)\varphi(y_2) + ...\varphi(x^*)\varphi(y_{2n})|}{(|\varphi(x^*)|^2 + |\varphi(y_1)|^2 +... + |\varphi(y_{2n})|^2 )}$ will be equal to $n$ , by that I may define infinitely many $\varphi_n$ which satisfies $\{\varphi_n\} \rightarrow _{n\rightarrow \infty} 1$ as wished. But I don't succeed to do so.

Is there any claim that such a combination of complex numbers exists and by this, to avoid a specific definition of the function $\varphi$?

off course any other approach or reference would be appreciated.

$\endgroup$
  • $\begingroup$ I think that the idea from your deleted answer is OK. In order to show that $\frac{|\langle M\varphi_n,\varphi_n \rangle|}{||\varphi_n||^2}$ tends to $0$ is suffices to remark that $M\varphi_n(x)=x$ for all $x\in\Bbb Z^d$ but a set of size $O(n^{d-1})=o(\|\varphi_n\|)$, on which both $|M\varphi_n(x)|$ and $|\varphi_n(x)|$ are bounded by $c$. $\endgroup$ – Alex Ravsky Jun 8 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.