A generalized conservation property for the heat semigroup on weighted manifolds

Abstract

In this text we study a generalized conservation property for the heat semigroup generated by a Schrödinger operator with nonnegative potential on a weighted manifold. We establish Khasminskii’s criterion for the generalized conservation property and discuss several applications.

Appendices

Appendix A. Local regularity theory

In this appendix we collect local regularity results for the operator $\mathcal {L}_{\rho ,V}$. They are consequences of the well-known local elliptic and local parabolic regularity theory in Euclidean spaces. In order to obtain versions on weighted manifolds, one just needs to localize the operators accordingly. For the case when $V = 0$ and $\rho = 1$, this can be found e.g. in [10, Chapters 6 and 7]. Since we assume $V \ge 0$, the proofs given there can be carried trough verbatim in our situation (otherwise some slight modifications would be needed). In other words, for the reader who is well-acquainted with local regularity theory the following lemmas are simple exercises, while other readers may find the proofs in [10].

The following Sobolev embedding theorem is a version of [10, Theorem 7.1].

Lemma A.1

Let $f \in L^2_{\mathrm{loc}}(M)$. If for each $m \ge 0$ we have $(\mathcal {L}_{\rho ,V})^m f \in L^2_{\mathrm{loc}}(M)$, then $f \in C^\infty (M)$. Moreover, if $(f_n)$ is a sequence in $L^2_{\mathrm{loc}}(M)$ such that for each $m \ge 0$ also $(\mathcal {L}_{\rho ,V})^m f_n \in L^2_{\mathrm{loc}}(M)$ and $(\mathcal {L}_{\rho ,V})^m f_n \rightarrow (\mathcal {L}_{\rho ,V})^m f$ in $L^2_{\mathrm{loc}}(M)$, then $f_n \rightarrow f$ locally uniformly.

The following hypoellipticity statement is a version of [10, Theorem 7.4].

Lemma A.2

Let $f \in C^\infty (M)$ and let $u \in \mathcal {D}'((0,T) \times M)$ satisfy

$$\begin{aligned} \partial _t u - \mathcal {L}_{\rho ,V}u = f. \end{aligned}$$

Then $u \in C^\infty ((0,T) \times M)$.

Appendix B. Convergence of semigroups

Let $\Omega \subseteq M$ an open subset. We let $(T_t^\Omega )$ denote the $L^2(\Omega ,\rho \mu )$ semigroup of the Dirichlet form

$$\begin{aligned} Q^\Omega _{\rho ,V}(f,g) = \int _\Omega \left\langle \nabla f,\nabla g\right\rangle d \mu + \int _\Omega Vfg d\mu \end{aligned}$$

with domain $D(Q^\Omega _{\rho ,V}) = W_0^1(\Omega ,\rho + V)$. We extend it to $f \in L^2(M,\rho \mu )$ by letting

$$\begin{aligned} T_t^\Omega f := {\left\{ \begin{array}{ll} T^\Omega _t f|_\Omega &{}\text {on } \Omega \\ 0 &{}\text {on } M{\setminus } \Omega . \end{array}\right. } \end{aligned}$$

Similarly, we denote by $(G_\alpha ^\Omega )$ the resolvent of $Q^\Omega _{\rho ,V}$ extended by 0 outside of $\Omega$.

Lemma B.1

Let $(\Omega _n)$ an ascending sequence of open subsets of M with $\bigcup _n \Omega _n = M$. Then, for every $t > 0$ and $\alpha >0$ we have $T_t^{\Omega _n} \rightarrow T^{\rho ,V}_t$ and $G_\alpha ^{\Omega _n} \rightarrow G^{\rho ,V}_\alpha$ strongly in $L^2(M)$, as $n \rightarrow \infty$.

Proof

We prove that $Q_{\rho ,V}^{\Omega _n}$ converges to $Q_{\rho ,V}$ in the generalized Mosco sense, see. [2, 8 Appendix] for a definition. The desired statement then follows from [2, Theorem 8.3].

We denote by $\pi _n:L^2(M,\rho \mu ) \rightarrow L^2(\Omega _n,\rho \mu )$ the restriction $f \mapsto f|_{\Omega _n}$. Its adjoint $E_n := \pi _n^*$ extends a function in $L^2(\Omega _n,\rho \mu )$ to M by letting it equal to 0 outside of $\Omega _n$. For verifying generalized Moso convergence, we need to prove the following statements.

• (a) For $f_n \in L^2(\Omega _n,\rho \mu )$, $f \in L^2(M,\rho \mu )$ with $E_n f_n \rightarrow f$ weakly in $L^2(M,\rho \mu )$ the inequality

  $$\begin{aligned} Q_{\rho ,V}(f) \le \liminf _{n \rightarrow \infty }Q^{\Omega _n}_{\rho ,V}(f_n) \end{aligned}$$

  holds [with the convention $Q_{\rho ,V}(g) = \infty$ if $f \not \in D(Q_{\rho ,V})$].

• (b) For every $f \in D(Q_{\rho ,V})$ there exist $f_n \in D(Q_{\rho ,V}^{\Omega _n})$ with $E_n f_n \rightarrow f$ strongly in $L^2(M,\rho \mu )$ and

  $$\begin{aligned} \limsup _{n\rightarrow \infty } Q_{\rho ,V}^{\Omega _n}(f_n) \le Q_{\rho ,V}(f). \end{aligned}$$

The closedness of $Q_{\rho ,V}$ implies that it is lower semicontinuous with respect to weak convergence. Hence, $E_n f_n \rightarrow f$ weakly in $L^2(M,\rho \mu )$ yields

$$\begin{aligned} Q_{\rho ,V}(f) \le \liminf _{n\rightarrow \infty }Q_{\rho ,V}(E_n f_n). \end{aligned}$$

It follows from the definition of $Q_{\rho ,V}^{\Omega _n}$ that $Q_{\rho ,V}(E_n f_n) = Q^{\Omega _n}_{\rho ,V}(f_n)$. This proves (a).

For proving (b) we use that $C_c^\infty (M)$ is dense in $D(Q_{\rho ,V})$ with respect to the form norm. For given $f \in D(Q_{\rho ,V})$ let $(f_n)$ a sequence in $C_c^\infty (M)$ that converges to f with respect to $\Vert \cdot \Vert _{W^1}$. From this sequence we can build up a sequence $(g_n)$ with ${\mathrm{supp}}\, g_n \subseteq \Omega _n$ and $g_n \rightarrow f$ with respect to $\Vert \cdot \Vert _{W^1}$ as follows. For $n \in \mathbb {N}$ we define

$$\begin{aligned} k_n := \max \left\{ k\le n \mid {\mathrm{supp}}\, f_k \subseteq \Omega _n\right\} \end{aligned}$$

and set $g_n := f_{k_n}$. By construction we have ${\mathrm{supp}}\, g_n \subseteq \Omega _n$. The sequence $(k_n)$ is increasing and since the $(f_n)$ have compact support, it diverges. We obtain $g_n \rightarrow f$ with respect to $\Vert \cdot \Vert _{W^1}$. These considerations imply $\pi _n g_n \in D(Q^{\Omega _n}_{\rho ,V})$ and

$$\begin{aligned} \limsup _{n\rightarrow \infty } Q^{\Omega _n}_{\rho ,V}(\pi _n g_n) = \limsup _{n\rightarrow \infty } Q_{\rho ,V}(g_n) = Q_{\rho ,V}(f). \end{aligned}$$

This finishes the proof. $\square$

Appendix C. Two lemmas on measurable choices

The following lemmas are certainly well known to experts. Since we could not find proper references, we include their proofs for the convenience of the reader. Let $0 < T \le \infty$. By $\lambda$ we denote the Lebesgue measure on (0, T). Let $u:(0,T) \rightarrow L^2(M,\mu )$. A measurable function $\tilde{u} \in L_\mathrm {loc}^1((0,T) \times M,\lambda \otimes \mu )$ such that $u(t) = \tilde{u}_t$ in $L^2(M,\mu )$ for $\lambda$-a.e. $t \in (0,T)$ is called a locally integrable version of u. Note that by Fubini’s theorem this is well-defined, i.e., it is independent of the choice of the representative of $\tilde{u}$.

Recall that in this paper $\left\langle \cdot ,\cdot \right\rangle _\mu$ denotes the $L^2(M,\mu )$-inner product and that $\left\langle \cdot ,\cdot \right\rangle$ denotes the dual pairing between test functions and distributions.

Lemma C.1

Let $0 < T \le \infty$ and let $u:(0,T) \rightarrow L^2(M,\mu )$ continuous. Then there exists a locally integrable version of u.

Proof

Since the Borel-$\sigma$-algebra of M is countably generated, $L^2(M,\mu )$ is separable. Let $(f_k)_{k \ge 1}$ be a countable orthonormal basis for $L^2(M,\mu )$. Moreover, let $I_n \subseteq (0,T)$ increasing compact intervals with $\bigcup _n I_n = (0,T)$. For $k,n \in \mathbb {N}$ the maps

$$\begin{aligned} g_{n,k} : (0,T) \times M \rightarrow \mathbb {R},\, (x,t) \mapsto \left\langle u(t),f_k\right\rangle _\mu 1_{I_n}(t) f_k(x) \end{aligned}$$

are clearly measurable. We consider

$$\begin{aligned} u_{n,l}:= \sum _{k = 1}^l g_{n,k}. \end{aligned}$$

Parseval’s inequality in $L^2(M,\mu )$ and the strong continuity of u imply

$$\begin{aligned} \int _{0}^T \int _M |u_{n,l}|^2 d \mu d\lambda \le \int _{0}^T 1_{I_n}(t) \Vert \tilde{u}(t)\Vert _2^2 d \lambda (t) < \infty , \end{aligned}$$

uniformly. Hence, for each $n \in \mathbb {N}$ the limit $u_n := \lim _{l\rightarrow \infty } u_{n,l}$ exists in $L^2((0,T) \times M,\lambda \otimes \mu )$. For $n \ge m$ the functions $u_n$ and $u_m$ only differ on $I_n {\setminus } I_m$; indeed we have $u_m = u_n 1_{I_m \times M}$. Hence, the limit $\tilde{u} = \lim _{n\rightarrow \infty } u_{n}$ exists in $L^1_{\mathrm{loc}}((0,T) \times M,\lambda \otimes \mu )$ and satisfies $\tilde{u}1_{I_n \times M} = u_n$. Parseval’s identity and the properties of $\tilde{u}$ yield

$$\begin{aligned} \int _0^T \Vert \tilde{u}_t- u(t)\Vert _2^2 d \lambda (t)&= \lim _{n\rightarrow \infty } \int _{I_n} \Vert \tilde{u}_t - u(t)\Vert _2^2 d \lambda (t) \\&= \lim _{n\rightarrow \infty } \int _{I_n} \Vert u_n (t,\cdot ) - \tilde{u}(t)\Vert _2^2 d \lambda (t)\\&= \lim _{n\rightarrow \infty }\lim _{l\rightarrow \infty } \int _{I_n} \Vert (u_{n,l})_t - u(t)\Vert _2^2 d\lambda (t)\\&= \lim _{n \rightarrow \infty } \lim _{l\rightarrow \infty } \int _{I_n} \sum _{k = l+1}^\infty |\left\langle u(t),f_k\right\rangle _\mu |^2 d\lambda (t). \end{aligned}$$

The strong continuity of u yields

$$\begin{aligned} \sup _{t \in I_n} \sum _{k = l+1}^\infty |\left\langle u(t),f_k\right\rangle _\mu |^2 \le \sup _{t \in I_n} \Vert u(t)\Vert ^2_2 < \infty , \text { for all } l \in \mathbb {N}. \end{aligned}$$

Hence, Lebesgue’s theorem yields

$$\begin{aligned} \lim _{l\rightarrow \infty } \int _{I_n} \sum _{k = l+1}^\infty |\left\langle \tilde{u}(t),f_k\right\rangle _\mu |^2 d \lambda (t) = 0, \end{aligned}$$

and the assertion $\tilde{u}_t = u(t)$ in $L^2(M,\mu )$ for $\lambda$-a.e. $t \in (0,T)$ is proven. $\square$

We say that $u:(0,T) \rightarrow L^2(M,\mu )$ is continuously differentiable if for all $0< s < T$ the limit

$$\begin{aligned} u'(s) = \lim _{h \rightarrow 0} \frac{1}{h}(u(s + h) - u(s)) \end{aligned}$$

exists in $L^2(M,\mu )$ and $u':(0,T) \rightarrow L^2(M,\mu )$ is continuous. In the following lemma $\partial _t$ denotes the distributional time derivative on $\mathcal {D}'((0,\infty ) \times M)$.

Lemma C.2

Let $0 < T \le \infty$ and let $u:(0,T) \rightarrow L^2(M,\mu )$ be continuously differentiable. Then there exists a locally integrable version $\tilde{u}$ of u such that $\partial _t \tilde{u} \in L^1_\mathrm {loc}((0,T) \times M,\lambda \otimes \mu )$ is a locally integrable version of $u'$.

Proof

According to the previous lemma u and $u'$ have locally integrable versions $\tilde{u}$ and v, respectively. Hence, it suffices to prove $v = \partial _t \tilde{u}$. For $\varphi \in \mathcal {D}((0,\infty )\times M)$ we compute

$$\begin{aligned} \left\langle \tilde{u},\partial _t \varphi \right\rangle&= \int _0^\infty \int _M \tilde{u}_s(x) (\partial _t \varphi )_s(x) d\mu (x) d\lambda (s) \\&= \lim _{h\rightarrow 0} h^{-1} \int _0^\infty \int _M \tilde{u}_s(x) (\varphi _{s + h}(x) - \varphi _s(x))d\mu (x) d \lambda (s) \\&= \lim _{h\rightarrow 0} h^{-1} \int _0^\infty \int _M (\tilde{u}_{s-h}(x) - \tilde{u}_s(x)) \varphi _s(x)d \mu (x) d\lambda (s) \\&= \lim _{h\rightarrow 0} h^{-1} \int _0^\infty \left\langle u(s-h) - u(s),\varphi _s\right\rangle _\mu d \lambda (s) \\&= - \int _{0}^\infty \left\langle u'(s),\varphi _s\right\rangle _\mu d\lambda (s). \end{aligned}$$

For the second to last equality we used that $\tilde{u}$ is a locally integrable version of u. Moreover, for the last equality we used a standard result for differentiating under the integral sign using that u is continuously differentiable and $\varphi$ has compact support in $(0,T) \times M$. Since v is a locally integrable version of $u'$, we further obtain

$$\begin{aligned} \int _{0}^\infty \left\langle u'(s),\varphi _s\right\rangle _\mu d\lambda (s) = \int _{0}^\infty \left\langle v,\varphi _s\right\rangle _\mu d\lambda (s) = \left\langle v,\varphi \right\rangle . \end{aligned}$$

This proves $\partial _t \tilde{u} = v$, as by definition $\left\langle \partial _t \tilde{u}, \varphi \right\rangle = - \left\langle \tilde{u},\partial _t \varphi \right\rangle$. $\square$