Global properties of Dirichlet forms in terms of Green’s formula

Abstract

We study global properties of Dirichlet forms such as uniqueness of the Dirichlet extension, stochastic completeness and recurrence. We characterize these properties by means of vanishing of a boundary term in Green’s formula for functions from suitable function spaces and suitable operators arising from extensions of the underlying form. We first present results in the framework of general Dirichlet forms on $\sigma$-finite measure spaces. For regular Dirichlet forms our results can be strengthened as all operators from the previous considerations turn out to be restrictions of a single operator. Finally, the results are applied to graphs, weighted manifolds, and metric graphs, where the operators under investigation can be determined rather explicitly, and certain volume growth criteria can be (re)derived.

Appendix A. Construction of the sequence \((e_n)\) and a lemma on sequences

The subsequent two results are certainly known in one way or other. We include a proof in this appendix in order to keep the paper self-contained and as they may be useful for further references as well.

Proposition A.1

Let Q be a Dirichlet form on a $\sigma$-finite space (X, m). Then, the following holds:

• (a) There exists a sequence $(e_n)$ in D(Q) with $0\le e_n \le 1$ and $e_n \rightarrow 1$ m-a.e.

• (b) If Q is regular, the sequence $(e_n)$ from (a) can be chosen in $D(Q)\cap C_c (X)$.

• (c) If Q is regular and recurrent, then $(e_n)$ from (b) can be chosen to satisfy $e_n \rightarrow 1$ in the sense of $Q_e$, i.e., $Q(e_n)\rightarrow 0$.

Proof

(a) By our assumptions there exists an increasing sequence of sets of finite measure $(B_k)$ such that $X = \bigcup _k B_k$. Because D(Q) is dense in $L^2(X,m)$, we can choose $f_n \in D(Q)$ satisfying

$$\begin{aligned} \Vert f_n - \chi _{B_n}\Vert _{2} \rightarrow 0, \text { as }n \rightarrow \infty . \end{aligned}$$

Let $e_n = (0 \vee f_n)\wedge 1$. Since Q is a Dirichlet form, we infer $e_n \in D(Q).$ Furthermore by construction we see $0 \le e_n \le 1$ and

$$\begin{aligned} \Vert e_n - \chi _{B_n}\Vert _{2} \le \Vert f_n - \chi _{B_n}\Vert _{2}. \end{aligned}$$

We want to show that $(e_n)$ possesses a subsequence converging to 1 $m-$almost surely. For $k,n \in \mathbb {N}$ and $\delta > 0$ let

$$\begin{aligned} A_{k,n,\delta } = \{x \in B_k: |e_n(x)-1| \ge \delta \}. \end{aligned}$$

By the Markov-inequality we observe for $n \ge k$

$$\begin{aligned} m(A_{k,n,\delta }) \le \delta ^{-2} \Vert (1-e_n)\chi _{B_k} \Vert _2^2 \le \delta ^{-2}\Vert \chi _{B_n} - e_n \Vert _2^2. \end{aligned}$$

This allows us to choose a subsequence $e_{n_l}$, such that for any k

$$\begin{aligned} \sum _{l=1}^{\infty } m(A_{k,n_l,l^{-1}}) < \infty . \end{aligned}$$

Let $N = \bigcup _k \bigcap _{j \ge 1} \bigcup _{l\ge j}A_{k,n_l,l^{-1}}$ and $x \in X{\setminus } N$. It is easily verified that $e_{n_l}(x) \rightarrow 1$ as $l \rightarrow \infty$. To prove (a), it remains to show $m(N) = 0$, which can be checked directly by computing

$$\begin{aligned} m(N) \le \sum _k \lim _{j \rightarrow \infty } m \Big (\bigcup _{l \ge j} A_{k,n_l,l^{-1}}\Big ) \le \sum _k \limsup _{j \rightarrow \infty } \sum _{n\ge j} m(A_{k,n_l,l^{-1}}) =0 \end{aligned}$$

(b) Because of the regularity of Q, we know that $C_c(X) \cap D(Q)$ is dense in D(Q) (hence, in $L^2(X,m)$) with respect to $L^2(X,m)$ convergence. Thus, in the proof of (a) we can replace $f_n \in D(Q)$ by $g_n \in C_c(X) \cap D(Q)$ to obtain (b).

(c) By recurrence there exists a sequence $h_n \in D(Q)$ such that $0 \le h_n \le 1$, $h_n \rightarrow 1$ pointwise m-almost surely and

$$\begin{aligned} \lim _{n \rightarrow \infty } Q(h_n)= 0. \end{aligned}$$

By regularity of Q we can choose $\tilde{e}_n \in C_c(X) \cap D(Q)$ satisfying

$$\begin{aligned} \Vert \tilde{e}_n-h_n \Vert _Q \rightarrow 0 \text { as } n\rightarrow \infty . \end{aligned}$$

Let $e_n = (0 \vee h_n)\wedge 1$ such that $0\le e_n \le 1$. We will show, that $e_n$ has a subsequence converging to 1 m-almost everywhere and

$$\begin{aligned} \lim _{n\rightarrow \infty }Q(e_n) = 0. \end{aligned}$$

Let $A_{k,n,\delta }$ be sets defined as in the proof of (a). The first assertion follows as above, using

$$\begin{aligned} m(A_{k,n,\delta })&\le \delta ^{-2} \Vert (e_n - 1)\chi _{B_k}\Vert _2^2 \\&\le \delta ^{-2} \Vert (\tilde{e}_n - 1)\chi _{B_k}\Vert _2^2\\&\le \delta ^{-2} \left[ \Vert (\tilde{e}_n - h_n)\chi _{B_k}\Vert _2 + \Vert (1- h_n)\chi _{B_k}\Vert _2 \right] ^2 \\&\le \delta ^{-2} \left[ \Vert (\tilde{e}_n - h_n)\Vert _2 + \Vert (1- h_n)\chi _{B_k}\Vert _2 \right] ^2. \end{aligned}$$

The second statement can be deduced by

$$\begin{aligned} Q(e_n)^ {1/2} \le Q(\tilde{e}_n)^ {1/2} \le Q(\tilde{e}_n-h_n)^ {1/2} + Q(h_n)^ {1/2} \le \Vert \tilde{e}_n-h_n\Vert _Q + Q(h_n)^ {1/2}. \end{aligned}$$

This finishes the proof.

Lemma A.2

Let $(a_{n,m})_{n,m \in \mathbb {N}}$ be a sequence of real numbers satisfying $a_{n+1,m} \ge a_{n,m}$ for each $n,m \in \mathbb {N}$. Then,

$$\begin{aligned} \liminf _{n\rightarrow \infty }\liminf _{m\rightarrow \infty } a_{n,m} \le \liminf _{m\rightarrow \infty }\liminf _{n\rightarrow \infty } a_{n,m}. \end{aligned}$$

Proof

Suppose $\liminf _{m\rightarrow \infty }\liminf _{n\rightarrow \infty } a_{n,m} < \infty$. Let $\varepsilon > 0$ be arbitrary. Choose an increasing sequence of indices $(m_l)$ such that

$$\begin{aligned} \liminf _{n\rightarrow \infty } a_{n,m_l} \le \liminf _{m \rightarrow \infty } \liminf _{n\rightarrow \infty } a_{n,m} + \varepsilon \end{aligned}$$

for each $l \ge 1$. This and the monotonicity in n imply

$$\begin{aligned} a_{n,m_l} \le \liminf _{m \rightarrow \infty } \liminf _{n\rightarrow \infty } a_{n,m} +\varepsilon . \end{aligned}$$

As $(a_{n,m_l})_{l \ge 1}$ is a particular subsequence of $(a_{n,m})_{m \ge 1}$, we infer

$$\begin{aligned} \liminf _{m\rightarrow \infty } a_{n,m} \le \lim _{l \rightarrow \infty } \liminf _{n\rightarrow \infty } a_{n,m_l} + \varepsilon \end{aligned}$$

which proves the claim.