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

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.

• 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 $c$. – Alex Ravsky Jun 8 at 19:38