Volume growth, spectrum and stochastic completeness of infinite graphs

Abstract

We study the connections between volume growth, spectral properties and stochastic completeness of locally finite weighted graphs. For a class of graphs with a very weak spherical symmetry we give a condition which implies both stochastic incompleteness and discreteness of the spectrum. We then use these graphs to give some comparison results for both stochastic completeness and estimates on the bottom of the spectrum for general locally finite weighted graphs.

Appendix A: Reducing subspaces and commuting operators

We study symmetries of selfadjoint operators. These symmetries are given in terms of bounded operators commuting with the selfadjoint operator in question. We present a general characterization in Theorem A.1. With Lemma A.5, we then turn to the question of how symmetries of a symmetric non-negative operator carry over to its Friedrichs extension. Finally, we specialize to the situation in which the bounded operator is a projection onto a closed subspace. The main result of this appendix, Corollary A.8, characterizes when a selfadjoint operator commutes with such a projection.

While these results are certainly known in one form or another, we have not found all of them in the literature in the form discussed below. In the main body of the paper they will be applied to Laplacians on graphs. However, they are general enough to be applied to Laplace-Beltrami operators on manifolds as well.

A subspace $U$ of a Hilbert space is said to be invariant under the bounded operator $A$ if $A$ maps $U$ into $U$.

Theorem A.1

Let $L$ be a selfadjoint non-negative operator on the Hilbert space $\mathcal H$ and $A$ a bounded operator on $\mathcal H$. Then, the following assertions are equivalent:

• (i) $D(L)$ is invariant under $A$ and $L A x= A Lx$ for all $x\in D(L)$.

• (ii) $D(L^{1/2})$ is invariant under $A$ and $L^{1/2} A x= A L^{1/2}x$ for all $x\in D(L^{1/2})$.

• (iii) $1_{[0,t]} (L) A = A 1_{[0,t]} (L)$ for all $t\ge 0$.

• (iv) $e^{- t L} A = A e^{- t L}$ for all $t\ge 0$.

• (v) $(L + \alpha )^{-1} A = A (L+ \alpha )^{-1}$ for all $\alpha >0$.

• (vi) $g(L) A = A g(L)$ for all bounded measurable $g : [0,\infty ) \longrightarrow \mathbb{C }$.

 

Proof

This is essentially standard. We sketch a proof for the convenience of the reader. We first show that (iii), (iv), (v) and (vi) are all equivalent:

(iii) $\Longrightarrow$ (iv): This follows by a simple approximation argument.

(iv) $\Longrightarrow$ (v): This follows immediately from $(L + \alpha )^{-1} = \int _0^\infty e^{- t \alpha } e^{- t L } dt$ (which, in turn, is a direct consequence of the spectral calculus).

(v) $\Longrightarrow$ (vi): The assumption (v) together with a Stone/Weierstrass-type argument shows that $g(L) A = A g(L)$ for all continuous $g : [0,\infty )\longrightarrow \mathbb{C }$ with $g(x) \rightarrow 0$ for $x\rightarrow \infty$. Now, it is not hard to see that the set

$$\begin{aligned} \{ f: [0,\infty ) \longrightarrow \mathbb{C }\ | \ f \text{ measurable} \text{ and} \text{ bounded} \text{ with} f(L) A = A f(L)\} \end{aligned}$$

is closed under pointwise convergence of uniformly bounded sequences. This gives the desired statement (vi).

(vi) $\Longrightarrow$ (iii): This is obvious.

We now show (ii) $\Longrightarrow$ (i) $\Longrightarrow$ (v) and (vi) $\Longrightarrow$ (ii).

(ii) $\Longrightarrow$ (i): This is clear as $L = L^{1/2} L^{1/2}$.

(i) $\Longrightarrow$ (v): Obviously, (i) implies $A (L + \alpha ) x = (L + \alpha ) A x$ for all $\alpha \in \mathbb{R }$ and $x\in D(L)$. As $(L+\alpha )$ is injective for $\alpha >0$ we infer for all such $\alpha$ that

$$\begin{aligned} (L + \alpha )^{-1} A = A (L+ \alpha )^{-1}. \end{aligned}$$

(vi) $\Longrightarrow$ (ii): For every natural number $n$ the operator $L^{1/2} 1_{[0,n]} (L) = (id^{1/2} 1_{[0,n]}) (L)$ is a bounded operator commuting with $A$ by (vi). Let $x\in D(L^{1/2})$ be given and set $x_n :=1_{[0,n]} (L) x$. Then, $x_n$ belongs to $D(L^{1/2})$. Moreover, as $1_{[0,n]} (L)$ is a projection, we obtain by (vi) that

$$\begin{aligned} A x_n = A 1_{[0,n]} (L) x = 1_{[0,n]} (L) A 1_{[0,n]} (L) x =1_{[0,n]} (L) A x_n. \end{aligned}$$

In particular, $A x_n$ belongs to $D(L^{1/2})$ as well. This gives, by (vi) again, that

$$\begin{aligned} L^{1/2} A x_n = L^{1/2} 1_{[0,n]} (L) A x_n = A L^{1/2} 1_{[0,n]}(L) x. \end{aligned}$$

As $x$ belongs to $D(L^{1/2})$, we infer that $L^{1/2} 1_{[0,n]} (L) x$ converges to $L^{1/2} x$. Moreover, $A x_n$ obviously converges to $A x$. As $L^{1/2}$ is closed, we obtain that $A x$ belongs to $D(L^{1/2})$ as well and $L^{1/2} A x = L^{1/2} A x$ holds. $\square$

Remark

 

• (a) The method to prove (v) $\Longrightarrow$ (vi) can be strengthened as follows: Let $L$ be a selfadjoint operator with spectrum $\Sigma$. Let $\mathcal B (\Sigma )$ be the algebra of all bounded measurable functions on $\Sigma .$ A sequence $(f_n)$ in $\mathcal B (\Sigma )$ is said to converge to $f\in \mathcal B (\Sigma )$ in the sense of (♣) if the $(f_n)$ are uniformly bounded and converge pointwise to $f$. Let $F$ be a subset of $\mathcal B$ such that $f(L) A = A f(L)$ holds for all $f\in F$. If the smallest subalgebra of $\mathcal B$ which contains $F$ and is closed under convergence with respect to (♣) is $\mathcal B$, then $g(L) A = A g(L)$ for all $g\in \mathcal B$.

• (b) If $L$ is an arbitrary selfadjoint operator then the equivalence of (i), (iii) and (vi) is still true and the semigroup in (iv) can be replaced by the unitary group and the resolvents in (v) can be replaced by resolvents for $\alpha \in \mathbb{C }\setminus \mathbb{R }$ (as can easily be seen using (a) of this remark).

 

Definition A.2

Let $L$ be a selfadjoint non-negative operator on a Hilbert space $\mathcal H$ and $A$ a bounded operator on $\mathcal H$. Then, $A$ is said to commute with $L$ if one of the equivalent statements of the theorem holds.

Corollary A.3

Let $L$ be a selfadjoint non-negative operator on a Hilbert space $\mathcal H$ and $A$ a bounded operator on $\mathcal H$. Then, $A$ commutes with $L$ if and only if its adjoint $A^*$ commutes with $L$.

Proof

Take adjoints in (iii) of the previous theorem. $\square$

A simple situation in which the previous theorem can be applied is given next.

Proposition A.4

Let $L$ be a selfadjoint non-negative operator on the Hilbert space $\mathcal H$ and let $A$ be a bounded operator on $\mathcal H$. Let, for each natural number $n$, a closed subspace $\mathcal H _n$ of $\mathcal H$ be given with $A \mathcal H _n \subset \mathcal H _n$ and $\overline{\cup _{n} \mathcal H _n} = \mathcal H$. If, for each $n$, there exists a selfadjoint non-negative operator $L_n$ from $\mathcal H _n$ to $\mathcal H _n$ with $A L_n = L_n A$ and

$$\begin{aligned} (L_{n+ k} + \alpha )^{-1} x \rightarrow (L + \alpha )^{-1} x, \;k\rightarrow \infty , \end{aligned}$$

for all natural numbers $n$, $x\in \mathcal H _n$ and $\alpha >0$, then $A L = LA$ holds. A corresponding statement holds with resolvents replaced by the semigroup.

Proof

By assumption we have $A (L + \alpha )^{-1}x = (L + \alpha )^{-1} A x$ for all $x$ from the dense set $\cup _{n} \mathcal H _n$. By boundedness of the respective operators we infer $A (L + \alpha )^{-1} = (L + \alpha )^{-1} A$ and the statement follows from the previous theorem. $\square$

The previous theorem deals with symmetries of a selfadjoint operator $L$. Often, the selfadjoint operator arises as the Friedrichs extension of a symmetric operator. We next study how symmetries of a symmetric operator carry over to its Friedrichs extension. Specifically, we consider the following situation:

• (*) Let $\mathcal H$ be a Hilbert space with inner product $\langle \cdot , \cdot \rangle$. Let $L_0$ be a symmetric operator on $\mathcal H$ with domain $D_0$. Let $Q_0$ be the associated form, i.e., $Q_0$ is defined on $D_0 \times D_0$ via $Q_0 (u,v) :=\langle L_0 u, v\rangle .$ Assume that $Q_0$ is non-negative, i.e., $Q_0 (u,u)\ge 0$ for all $u\in D_0$. Then, $Q_0$ is closable. Let $Q$ be the closure of $Q_0$, $D(Q)$ the domain of $Q$ and $L$ the Friedrichs extension of $L_0$, i.e., $L$ is the selfadjoint operator associated to $Q$.

Lemma A.5

Assume $(*)$. Let $A$ be a bounded operator on $\mathcal H$ with $D_0$ invariant under $A$ and $A^*$, $A L_0 x = L_0 A x$ and $A^*L_0 x = L_0 A^*x$ for all $x\in D_0$. Then, the following assertions are equivalent:

• (i) $D(L)$ is invariant under $A$ and $A L x = L A x$ for all $x\in D(L)$.

• (ii) $D(Q)$ is invariant under $A$ and $A^*$ and $Q (A x, y) = Q (x, A^*y)$ for all $x,y\in D(Q)$.

• (iii) There exists a $C\ge 0$ with both $Q_0 (A x, Ax) \le C Q_0 (x,x)$ and $Q_0 (A^*x, A^*x) \le C Q_0(x,x)$ for all $x\in D_0$.

 

Proof

(iii) $\Longrightarrow$ (ii): By $A L_0 x = L_0 A x$ for all $x\in D_0$ we infer that $Q_0 (Ax, y) = Q_0 (x, A^*y)$ for all $x,y\in D_0$. As $Q$ is the closure of $Q_0$, it now suffices to show that both $(A u_n)$ and $(A^*u_n)$ are a Cauchy sequences with respect to the $Q$-norm, whenever $(u_n)$ is a Cauchy sequence with respect to the $Q$-norm in $D_0$. This follows directly from (iii).

(ii) $\Longrightarrow$ (i): Let $x \in D(L)$ be given. Then, $x$ belongs to $D(Q)$ and, by (ii), $Ax$ belongs to $D(Q)$ as well. Thus, we can calculate for all $y\in D(Q)$

$$\begin{aligned} Q (A x, y) = Q (x, A^*y) = \langle L x, A^*y\rangle = \langle A L x, y\rangle . \end{aligned}$$

This implies that $A x\in D(L)$ and $L A x = A L x$. Hence, we obtain (i).

(i) $\Longrightarrow$ (iii): From Theorem A.1 and (i) we infer that $L^{1/2} A x = A L^{1/2}x$ for all $x\in D(L^{1/2})$. Now, for $x\in D_0$ it holds that

$$\begin{aligned} Q_0 (x,x) = \langle L_0 x, x\rangle = \langle L^{1/2} x, L^{1/2} x\rangle = \Vert L^{1/2} x\Vert ^2. \end{aligned}$$

By $A x \in D_0$ for $x\in D_0$ a direct calculation gives

$$\begin{aligned} Q_0 (A x, A x) = \Vert L^{1/2} A x\Vert ^2 = \Vert A L^{1/2} x\Vert ^2 \le \Vert A\Vert ^2 \Vert L^{1/2} x\Vert ^2 = \Vert A\Vert ^2 Q_0 (x,x). \end{aligned}$$

A similar argument shows $Q_0 (A^*x, A^*x) \le \Vert A^*\Vert ^2 Q_0 (x,x)$. This finishes the proof. $\square$

We now turn to the special situation that $A$ is the projection onto a closed subspace. In this case, some further strengthening of the above result is possible. We first provide an appropriate definition.

Definition A.6

Let $\mathcal H$ be a Hilbert space and $S$ a symmetric operator on $\mathcal H$ with domain $D(S)$. A closed subspace $U$ of $\mathcal H$ with associated orthogonal projection $P$ is called a reducing subspace for $S$ if $D(S)$ is invariant under $P$ and $S P x = P S P x$ for all $x\in D(S)$.

The previous definition is just a commutation condition in the form discussed above as shown in the next lemma.

Definition A.7

Let $S$ be a symmetric operator on the Hilbert space $\mathcal H$ and $P$ be the orthogonal projection onto a closed subspace $U$ of $\mathcal H$. Then, the following assertions are equivalent:

• (i) $U$ is a reducing subspace for $S$.

• (ii) $D(S)$ is invariant under $P$ and $S P x = P S x$ holds for all $x\in D(S)$.

 

Proof

The implication (ii) $\Longrightarrow$ (i) is obvious. It remains to show (i) $\Longrightarrow$ (ii): We first show $P S y = 0$ for all $y\in D(S)$ with $P y = 0$ (i.e., $y \perp U$): Choose $x\in D(S)$ arbitrarily. Then, as $Px \in D(S) \subset D(S^*)$ we obtain

$$\begin{aligned} \langle P S y, x\rangle = \langle Sy, Px\rangle = \langle y, S^*P x\rangle = \langle y, S Px\rangle = \langle y, P S P x\rangle = \langle Py, S P x\rangle = 0. \end{aligned}$$

As $D(S)$ is dense, we infer $P S y=0$. Let now $x\in D(S)$ be arbitrary. Then, $x = Px + (1-P) x$ and both $Px$ and $(1-P) x$ belong to $D(S)$. Thus, we can calculate

$$\begin{aligned} P S x = P S Px + P S (1-P) x = PS P x = S P x. \end{aligned}$$

This finishes the proof. $\square$

We now come to the main result of the appendix dealing with symmetries of symmetric operators in terms of reducing subspaces.

Corollary A.8

Assume $(*)$. Let $U$ be a closed subspace of $\mathcal H$ and $A$ the orthogonal projection onto $U$. Assume that $D_0$ is invariant under $A$. Then, the following assertions are equivalent:

• (i) $U$ is a reducing subspace for $L_0$, i.e., $L_0 A x = A L_0 x$ for all $x\in D_0$.

• (ii) $Q_0 (A x, A y) = Q_0 (A x, y) = Q_0 (x, Ay)$ for all $x,y\in D_0$.

• (iii) $D(Q)$ is invariant under $A$ and $Q (A x, A y) = Q (Ax, y) = Q (x, Ay)$ for all $x,y\in D(Q)$.

• (iv) $U$ is a reducing subspace for $L$.

• (v) $A$ commutes with $e^{-t L}$ for every $t\ge 0$.

• (vi) $A$ commutes with $(L+\alpha )^{-1}$ for any $\alpha >0$.

 

Proof

Obviously, (i) and (ii) are equivalent. The equivalence of (iii) and (iv) follows from the equivalence of (i) and (ii) in Lemma A.5. The equivalence between (iv), (v) and (vi) follows immediately from Theorem A.1. The implication (iii) $\Longrightarrow$ (ii) is clear (as $A D_0 \subseteq D_0$). It remains to show (ii) $\Longrightarrow$ (iii): A direct calculation using (ii) gives for all $x\in D_0$ that

$$\begin{aligned} Q_0 (x,x)&= Q_0 ((A + (1-A))x, x)\\&= Q_0 (Ax,x) + Q_0 ((1-A)x ,x)\\&= Q_0 (Ax,Ax) + Q_0 ((1-A)x, (1-A)x). \end{aligned}$$

This shows

$$\begin{aligned} Q_0 (Ax,Ax) \le Q_0 (x,x) \end{aligned}$$

for all $x\in D_0$. Now, the implication (iii) $\Longrightarrow$ (ii) from Lemma A.5 gives (iii). $\square$