Short-time behavior of the heat kernel and Weyl’s law on ${{\mathrm{RCD}}}^*(K,N)$ spaces

Abstract

In this paper, we prove pointwise convergence of heat kernels for mGH-convergent sequences of ${{\mathrm{RCD}}}^{*}(K,N)$-spaces. We obtain as a corollary results on the short-time behavior of the heat kernel in ${{\mathrm{RCD}}}^*(K,N)$-spaces. We use then these results to initiate the study of Weyl’s law in the ${{\mathrm{RCD}}}$ setting.

Appendix: refinements of Karamata’s theorem

In this section we prove Theorem 2.5 and its one-sided versions mentioned in Remark 2.6. We follow the proofs in Theorems 10.2 and 10.3 of [46], borrowing also the terminology “Abelian”, “Tauberian” from there.

Throughout this section $\nu$ is a nonnegative and $\sigma$-finite Borel measure on $[0,+\infty )$. The results will then be applied to the case when $\nu :=\sum _i\delta _{\lambda _i}$.

Lemma 5.1

For all $t>0$ one has

$$\begin{aligned} \int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x)=\int _0^\infty t\nu ([0,y])e^{-ty}\mathsf {d}y. \end{aligned}$$

(5.1)

Proof

By Cavalieri’s formula and the change of variables $r=e^{-ty}$ we get

$$\begin{aligned} \int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x)=\int _0^1\nu \left( \left\{ x:\ e^{-tx}\ge r\right\} \right) \mathsf {d}r= \int _0^\infty te^{-ty}\nu \left( \left\{ x:\ e^{-tx}\ge e^{-ty}\right\} \right) \mathsf {d}y \end{aligned}$$

and we conclude, since $\{x:\ e^{-tx}\ge e^{-ty}\}=[0,y]$. $\square$

We start with the Abelian case, easier when compared to the Tauberian one.

Theorem 5.2

(Abelian theorem) Assume that there exist $\gamma \in [0,+\infty )$ and $C \in [0,+\infty )$ such that

$$\begin{aligned} \lim _{a \rightarrow +\infty }\frac{\nu ([0, a])}{a^{\gamma }}=C. \end{aligned}$$

(5.2)

Then

$$\begin{aligned} \lim _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x)=C \Gamma (\gamma +1). \end{aligned}$$

(5.3)

More generally,

$$\begin{aligned} \limsup _{a \rightarrow +\infty }\frac{\nu ([0, a])}{a^{\gamma }}\le C<+\infty \quad \Longrightarrow \quad \limsup _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x)\le C \Gamma (\gamma +1) \end{aligned}$$

(5.4)

and

$$\begin{aligned} \liminf _{a \rightarrow +\infty }\frac{\nu ([0, a])}{a^{\gamma }}\ge c\quad \Longrightarrow \quad \liminf _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x)\ge c \Gamma (\gamma +1). \end{aligned}$$

(5.5)

Proof

Let $F(a):=\nu ([0, a])$ and $G(a):=(a+1)^{-\gamma }F(a)$. Then (5.2) yields

$$\begin{aligned} \lim _{a \rightarrow +\infty }G(a)=C. \end{aligned}$$

(5.6)

In particular $\sup _a G(a)<\infty$. Then Lemma 5.1 gives

$$\begin{aligned} t^{\gamma }\int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x) = t^{\gamma +1}\int _0^\infty e^{-tx}(x+1)^{\gamma }G(x)\mathsf {d}x= \int _0^\infty e^{-y}(y+t)^{\gamma }G\left( y/t \right) \mathsf {d}y. \end{aligned}$$

(5.7)

Since for any $t \in (0, 1]$

$$\begin{aligned} e^{-y}(y+t)^{\gamma }G(y/t) \le e^{-y}(y+1)^{\gamma } \sup _a G(a) \in L^1([0,+\infty )), \end{aligned}$$

(5.8)

applying the dominated convergence theorem to (5.7) as $t \downarrow 0$ shows (5.3) because $G(y/t) \rightarrow C$ as $t \downarrow 0$ by (5.6).

The one-sided versions (5.4), (5.5) follow by an analogous argument, using Fatou’s lemma and noticing that in the $\limsup$ case the functions in (5.8) are dominated as $t\rightarrow 0^+$ by an integrable function. $\square$

Now we deal with the Tauberian case.

Theorem 5.3

(Tauberian theorem) Assume that there exist $\gamma \in [0,+\infty )$ and $D \in [0,+\infty )$ such that

$$\begin{aligned} \lim _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )} e^{-tx}\mathsf {d}\nu (x)=D. \end{aligned}$$

(5.9)

Then

$$\begin{aligned} \lim _{a \rightarrow +\infty }\frac{\nu ([0, a))}{a^{\gamma }}=\frac{D}{\Gamma (\gamma +1)}. \end{aligned}$$

(5.10)

Proof

If $\gamma =0$, then applying the monotone convergence theorem to (5.9) shows (5.10), hence we can assume $\gamma >0$. For any $t \in (0, 1]$ let $\nu _t,\, \mu$ be Borel measures on $[0,+\infty )$ be, respectively, defined by

$$\begin{aligned} \nu _t(A):=t^{\gamma }\nu (t^{-1}A),\qquad \mu (A):=\int _Ax^{\gamma -1}\mathsf {d}x \end{aligned}$$

(5.11)

for any Borel subset A. Then (5.10) is equivalent to

$$\begin{aligned} \lim _{t \rightarrow 0^+}\nu _t([0, 1))=\frac{D}{\Gamma (\gamma )}\mu ([0, 1)) \end{aligned}$$

(5.12)

because

$$\begin{aligned} \nu _t([0, 1))=t^{\gamma }\nu \left( [0, t^{-1})\right) \quad \text {and}\quad \mu ([0, 1))=\int _0^1x^{\gamma -1}\mathsf {d}x=\frac{1}{\gamma }=\frac{\Gamma (\gamma )}{\Gamma (\gamma +1)}. \end{aligned}$$

(5.13)

In order to prove (5.12), we will show

$$\begin{aligned} \lim _{t \rightarrow 0^+}\int f(x)\mathsf {d}\nu _t(x)=\frac{D}{\Gamma (\gamma )}\int f(x)\mathsf {d}\mu (x) \end{aligned}$$

(5.14)

for any $f \in C_c([0,+\infty ))$ as follows.

Let $\hat{\nu }_t:=e^{-x}\mathsf {d}\nu _t(x)$ and $\hat{\mu }:=e^{-x}\mathsf {d}\mu (x)$ be the corresponding weighted measures on $[0,+\infty )$. Then (5.9) with Lemma 5.1 yields

$$\begin{aligned} \lim _{t \rightarrow 0^+}\hat{\nu }_t([0,+\infty ))=\lim _{t\rightarrow 0^+}\int e^{-x}\mathsf {d}\nu _t(x) =\lim _{t \rightarrow 0^+}\int e^{-tx}t^{\gamma }\mathsf {d}\nu (x)=\frac{D}{\Gamma (\gamma )}\hat{\mu }([0,+\infty )). \end{aligned}$$

(5.15)

In particular

$$\begin{aligned} \sup _{t<1}\hat{\nu }_t([0, +\infty ))<+\infty . \end{aligned}$$

(5.16)

More strongly, (5.9) yields

$$\begin{aligned} \lim _{t \rightarrow 0^+}\int g(x)\mathsf {d}\hat{\nu }_t(x)=\frac{D}{\Gamma (\gamma )}\int g(x)\mathsf {d}\hat{\mu }(x) \end{aligned}$$

(5.17)

for any polynomial g(x) in $e^{-x}$ (i.e., $g(x)=\sum _{i=1}^Na_ie^{-ix}$). Because

$$\begin{aligned} \lim _{t \rightarrow 0^+}\int e^{-kx}\mathsf {d}\hat{\nu }_t(x)&=\lim _{t \downarrow 0}\int e^{-(k+1)x}\mathsf {d}\nu _t(x) \\&=\lim _{t \rightarrow 0^+}\int e^{-(k+1)tx}t^{\gamma }\mathsf {d}\nu (x) \\&=\frac{D}{(k+1)^{\gamma }}\nonumber =\frac{D}{\Gamma (\gamma )}\int e^{-kx}\mathsf {d}\hat{\mu }(x). \end{aligned}$$

Let $C_0([0,+\infty ))$ be the set of continuous functions f on $[0,+\infty )$ such that $f(x) \rightarrow 0$ as $x \rightarrow +\infty$. Then since the set of polynomials in $e^{-x}$ is dense in $C_0([0,+ \infty ))$ with respect to the norm $\sup |f|$, applying the Stone–Weierstrass theorem to $(C_0([0,+\infty )), \sup |\cdot |)$ with (5.16) shows that (5.17) is satisfied for any $g \in C_0([0,+\infty ))$, which implies (5.14).

We are now in a position to prove (5.12) by using (5.14). Indeed, it is well known that the weak convergence implies $\nu _t(E)\rightarrow D\mu (E)/\Gamma (\gamma )$ for any compact set $E\subset [0,+\infty )$ with $\mu (\partial E)=0$. Choosing $E=[0,1]$ we obtain (5.14). $\square$

Remark 5.4

The difficulty to obtain a one-sided version out of the previous proof, as we did for the Abelian case, can also be explained as follows: if we consider the push forward $\sigma _t$ of the measures $\hat{\nu _t}$ under the map $x\mapsto e^{-x}$, the argument above shows that all moments of all weak limit points of $\sigma _t$ are uniquely determined. Hence, since a finite Borel measure in [0, 1] is uniquely determined by its moments, uniqueness follows. If we replace the assumption (5.9) by a bound on the $\liminf$ or the $\limsup$, we find only an inequality between the moments of the measures, which does not seem to imply, in general, the corresponding inequality for the measures.

Proposition 5.5

Assume that for some $\gamma \in [0, +\infty )$ one has

$$\begin{aligned} \limsup _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s)\le C_0<+\infty . \end{aligned}$$

(5.18)

Then

$$\begin{aligned} \limsup _{\lambda \rightarrow +\infty }\frac{\nu ([0,\lambda ])}{\lambda ^{\gamma }}\le eC_0. \end{aligned}$$

(5.19)

Proof

Note that for any $\lambda >0$ and any $t>0$

$$\begin{aligned} \nu ([0,\lambda ])\le e^{\lambda t}\int _{[0,\lambda ]}e^{-st}\mathsf {d}\nu (s) \le e^{\lambda t}\int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s). \end{aligned}$$

(5.20)

By (5.18), for any $\epsilon >0$ there exists $t_0>0$ such that $\int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s)\le (C_0+\epsilon )t^{-\gamma }$ for any $t<t_0$. Thus (5.20) yields $\nu ([0,\lambda ])\le e^{\lambda t}(C_0+\epsilon )t^{-\gamma }$ for any $\lambda >0$ and any $t<t_0$. Letting $\lambda :=t^{-1}$ and then letting $t \downarrow 0$ shows (5.19). $\square$

Proposition 5.6

Assume that for some $\gamma \in [0,+\infty )$ one has

$$\begin{aligned} \liminf _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s)>0,\qquad \limsup _{t \rightarrow 0^+}t^{\gamma }\int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s)<+\infty . \end{aligned}$$

(5.21)

Then

$$\begin{aligned} \liminf _{\lambda \rightarrow +\infty }\frac{\nu ([0,\lambda ])}{\lambda ^{\gamma }} >0. \end{aligned}$$

(5.22)

Proof

Call $C_0>0$ the $\liminf$ and $C_1<+\infty$ the $\limsup$ in (5.21). Note that for any $\lambda >0$ and any $t>0$

$$\begin{aligned} \int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s)&=\int _{[0,\lambda ]}e^{-st}\mathsf {d}\nu (s) + \sum _{\ell =1}^{\infty }\int _{(\ell \lambda ,(\ell +1)\lambda ]}e^{-st}\mathsf {d}\nu (s) \\&\le \nu ([0,\lambda ])+ \sum _{\ell =1}^{\infty }e^{-\ell \lambda t}\nu ([0,(\ell +1)\lambda ]) \\&= \sum _{\ell =0}^{\infty }e^{-\ell \lambda t}\nu ([0,(\ell +1)\lambda ]). \end{aligned}$$

In particular, letting $\lambda :=t^{-1}$ yields

$$\begin{aligned} t^{\gamma }\int _{[0,+\infty )}e^{-st}\mathsf {d}\nu (s)\le t^{\gamma }\sum _{\ell =0}^{\infty }e^{-\ell }\nu \left( \left[ 0,\frac{\ell +1}{t}\right] \right) . \end{aligned}$$

(5.23)

Thus there exists $t_0>0$ such that for any $t<t_0$

$$\begin{aligned} 0<\frac{C_0}{2} \le t^{\gamma }\sum _{\ell =0}^{\infty }e^{-\ell }\nu \left( \left[ 0,\frac{\ell +1}{t}\right] \right) . \end{aligned}$$

(5.24)

Next let us discuss the right-hand side of (5.24). By (5.21)and Proposi tion 5.5 there exists $\hat{\lambda }>0$ such that $\nu ([0,\lambda ])\le (eC_1+1)\lambda ^{\gamma }$ for any $\lambda \ge \hat{\lambda }$. Thus for any $t>0$ with $t^{-1} \ge \hat{\lambda }$ we get

$$\begin{aligned} \nu \left( \left[ 0,\frac{\ell +1}{t}\right] \right) \le \left( eC_1+1\right) \frac{(\ell +1)^{\gamma }}{t^{\gamma }}. \end{aligned}$$

In particular

$$\begin{aligned} t^{\gamma }\sum _{\ell =k}^{\infty } e^{-\ell }\nu \left( \left[ 0,\frac{\ell +1}{t}\right] \right) \le \left( eC_1+1\right) \sum _{\ell =k}^{\infty }e^{-\ell }(\ell +1)^{\gamma } \end{aligned}$$

(5.25)

for any $k \in \mathbb {N}$ and any $t>0$ with $t^{-1}\ge \hat{\lambda }$.

For any $\delta >0$ there exists $k_0 \in \mathbb {N}$ such that $\sum _{\ell =k_0+1}^{\infty }e^{-\ell }(\ell +1)^{\gamma }<\delta$. Then, combining (5.24) with (5.25) yields

$$\begin{aligned} 0<\frac{C_0}{2}<t^{\gamma }\sum _{\ell =0}^{k_0}e^{-\ell } \nu \left( \left[ 0,\frac{\ell +1}{t}\right] \right) +\left( eC_1+1\right) \delta \end{aligned}$$

(5.26)

for any $t>0$ with $t<t_0$ and $t^{-1} \ge \hat{\lambda }$, which easily shows (5.22) choosing $\delta >0$ so small that $(eC_1+1)\delta <C_0/2$. $\square$