Asymptotics of relative heat traces and determinants on open surfaces of finite area

Abstract

The goal of this article is to prove that on surfaces with asymptotically cusp ends the relative determinant of pairs of Laplace operators is well defined. We consider a surface with cusps $(M,g)$ and a metric $h$ on the surface that is a conformal transformation of the initial metric $g$. We prove the existence of the relative determinant of the pair $(\Delta _{h},\Delta _{g})$ under suitable conditions on the conformal factor. The core of the paper is the proof of the existence of an asymptotic expansion of the relative heat trace for small times. We find the decay of the conformal factor at infinity for which this asymptotic expansion exists and the relative determinant is defined. Following the paper by B. Osgood, R. Phillips, and P. Sarnak about extremal of determinants on compact surfaces, we prove Polyakov’s formula for the relative determinant and discuss the extremal problem inside a conformal class. We discuss necessary conditions for the existence of a maximizer.

Appendices

Appendix A

In this appendix, we give the proof of Lemma 2.4. We prove the estimate of $K_{1,h}$. The estimate of $K_h$ then follows by a standard gluing parametrix construction. We use the notation introduced in Sect. 4.1 and Proposition 4.4. Let us recall Eq. (4.7):

$$\begin{aligned} K_{1,h}(z,w,t)=\sum _{m\in \mathbb{Z }} k_{h}({\widetilde{z}}, {\widetilde{w}}+m,t), \end{aligned}$$

where $\pi ({\widetilde{z}})= z, \pi ({\widetilde{w}})= w$, and ${\widetilde{z}}=(x_1,y_1)$ and ${\widetilde{w}}=(x_2,y_2)$ can be chosen so that $0\le x_i \le 1$.

We know that $d_{h}(z,w) = \inf _{m\in \mathbb{Z }} d_{{\hat{h}}}({\widetilde{z}},{\widetilde{w}}+m)\le d_{{\hat{h}}}({\widetilde{z}},{\widetilde{w}}+m)$ for all $m\in \mathbb{Z }$. Then using the estimate in Eq. (4.6) with constant $c_1>0$ corresponding to the metric $h$, we obtain

$$\begin{aligned} K_{1,h}(z,w,t)&\ll t^{-1} \sum _{m\in \mathbb{Z }} \exp {\left( -{\frac{c_{1}d_{{\hat{h}}}^{2}({\widetilde{z}}, {\widetilde{w}} +m)}{t}}\right)}\\&\le t^{-1} e^{-{\frac{c_{1}d_{h}^{2}(z,w)}{2t}}} \sum _{m\in \mathbb{Z }} e^{-{\frac{c_{1}d_{{\hat{h}}}^{2}({\widetilde{z}}, {\widetilde{w}} +m)}{2t}}}\\&\le t^{-1} e^{-{\frac{c_{1}d_{h}^{2}(z,w)}{2t}}} \left( e^{-{\frac{c_{2}d_{{\hat{g}}}^{2}({\widetilde{z}}, {\widetilde{w}})}{2t}}} + \sum _{m\ne 0} e^{-{\frac{c_{2}d_{{\hat{g}}}^{2}({\widetilde{z}}, {\widetilde{w}} +m)}{2t}}} \right) \end{aligned}$$

Now we use the formula for the hyperbolic distance to estimate it; for $m\ne 0$, we have

$$\begin{aligned} d_{{\hat{g}}}((x_1, y_1), (x_2 + m, y_2))&= \cosh ^{-1} \left(1 + {\frac{(x_1 - x_2 - m)^{2}+ (y_1-y_2)^2}{2 y_1 y_2}}\right)\\&\ge \log \left(1 + {\frac{(x_1 - x_2 - m)^{2}}{2y_1 y_2}}\right) \ge \log \left(1 + {\frac{(|m|-1)^2}{2y_1 y_2}}\right) \end{aligned}$$

since $-1 \le x_1-x_2 \le 1$ and $(|m|-1)^2 \le (x_1-x_2 - m)^2 \le (|m|+1)^2$, if $|m|\ne 0$. We proceed now to estimate the series in the same way as in (4.11), but we do not need to restrict the values of $y_1$ and $y_2$ to $[1,a]$ any more. We keep the value $y_1 y_2$ in the estimates instead of using the bound $a^2$

$$\begin{aligned} \sum _{|m|\ge 2} e^{-{\frac{c_{2}d_{{\hat{g}}}^{2}({\widetilde{z}}, {\widetilde{w}} + m)}{t}}}&\le \sum _{|m|\ge 1} e^{-{\frac{c_{2} \log (1 + {\frac{m^{2}}{2y_1 y_2}})^{2}}{2t}}}\\&\ll \int \limits _{1}^{\infty } e^{-{\frac{c_{2} \log (1 + {\frac{u^{2}}{2y_1 y_2}})^{2}}{2t}}} du \\&\ll y_1^{1/2} y_2^{1/2} (1+\sqrt{t}e^{ct}) \le C(\tau )y_1^{1/2} y_2^{1/2} \end{aligned}$$

for some constant $C(\tau )$ that depends on $\tau , 0<t\le \tau$. Putting all the terms together, we obtain

$$\begin{aligned} K_{1,h}(z,w,t)&\ll t^{-1} e^{-{\frac{c_{1}d_{h}^{2}(z,w)}{2t}}} \left( 2 + e^{-{\frac{c_{2}d_{{\hat{g}}}^{2}({\widetilde{z}}, {\widetilde{w}})}{2t}}} + \sum _{m\ne 0} e^{-{\frac{c_{1} \log (1 + {\frac{m^{2}}{2y_1 y_2}})^{2}}{2t}}} \right)\\&\ll t^{-1} y_1^{1/2} y_2^{1/2} e^{-{\frac{c_{1}d_{h}^{2}(z,w)}{2t}}}. \end{aligned}$$

For the derivatives of the heat kernel, we apply the results by Cheng et al. in [8, Theorems 6, 7], to $(\mathbb{H },{\hat{h}})$ that has bounded geometry. The fist two derivatives of the heat kernel $K_{1,h}$ can be estimated in the same way as we did for the heat kernel. As the authors point out in [8], the constant in each estimate will depend on the curvature of $M$ and its covariant derivatives.

Appendix B

1.1 B1 Observation

In the proof of Theorem 3.1, we repeatedly make use of the following elementary facts:

• (1) For any $a>0$, and $b,n ,m \in \mathbb{R }$, we have that

  $$\begin{aligned} \int \limits _{n}^{m}e^{-ax^{2}-bx}dx = {\frac{e^{b^{2}/4a}}{\sqrt{a}}} \int \limits _{\sqrt{a}(n+{\frac{b}{2a}})}^{\sqrt{a}(m+{\frac{b}{2a}})} e^{-v^{2}}dv \le {\frac{\sqrt{\pi } e^{b^{2}/4a}}{\sqrt{a}}}. \end{aligned}$$

• (2) For any $c>0, 0<t\le T, k,\ell \ge 0$ with $k+\ell >2$ we have

  $$\begin{aligned} \int \limits _{1}^{\infty } \int \limits _{1}^{\infty } y^{-k} y^{\prime -\ell } e^{-\frac{c}{t}\log (y/y^{\prime })^{2}} dy dy^{\prime }\le \sqrt{t}e^{(1-k)^{2}t/c}. \end{aligned}$$

  (7.1)

• (3) Let $\varphi \in C^{\infty }(M), \psi =e^{-2\varphi }-1$ and ${\widetilde{\psi }}=e^{2\varphi }-1$. If $\varphi \vert _{Z}(y,x), \Delta _{g}\varphi \vert _{Z}(y,x)$ and $\vert \nabla _{g}\varphi \vert _{g}\vert _{Z} (y,x)$ are $O(y^{-k})$ as $y\rightarrow \infty$, then so are $\psi \vert _{Z}(y,x), \Delta _{g}\psi \vert _{Z}(y,x), \vert \nabla _{g}\psi \vert _{g}\vert _{Z} (y,x)$ and the analogs functions corresponding to ${\widetilde{\psi }}$.

• (4) For $a,b,c\!>\!0$, the function $f(t) \!=\! t^{-a} e^{-c t^{-b}}$ is bounded on $(0,\infty )$ and $\lim _{t\rightarrow 0} f(t)\!=\!0$.

1.2 B2 Proof of the bounds of the integrals $J_1, J_2$ and $J_3$ in Proposition 4.5

Let us start with $J_1$ that is given by Eq. (4.18):

$$\begin{aligned} J_{1}&= \int \limits _{0}^{t}\int \limits _{Z_{a}}\int \limits _{[1,\frac{4a}{5}]\times S^{1}} \psi _{2}(z) (K_{1,h}(z,z^{\prime },s) + p_{h,D}(z,z^{\prime },s))e^{2\varphi (z^{\prime })}\\&\psi (z^{\prime }) \Delta _{Z,g} (K_{1,g}(z^{\prime },z,t-s) + p_{1,D}(z^{\prime },z,t-s)) \ dA_{g}(z^{\prime })\ dA_{g}(z)\ ds. \end{aligned}$$

Note that on this region $a\le y<\infty$ and $1\le y^{\prime } \le {\frac{4a}{5}}, \log (y/y^{\prime })$ is bounded away from zero. Using the estimates of the heat kernels and their derivatives, we obtain

$$\begin{aligned} \vert J_{1}\vert&\ll \int \limits _{0}^{t}\int \limits _{a}^{\infty } \int \limits _{1}^{\frac{4a}{5}} s^{-1} (t-s)^{-2} y \left(e^{-{\frac{c\log (y/y^{\prime })^{2}}{s}}}+ e^{-{\frac{c\log (y)^{2}}{s}}} e^{-{\frac{c\log (y^{\prime })^{2}}{s}}}\right)\\&y^{\prime -k+1} \left(e^{-{\frac{c\log (y/y^{\prime })^{2}}{t-s}}} + e^{-{\frac{c\log (y)^{2}}{t-s}}} e^{-{\frac{c\log (y^{\prime })^{2}}{t-s}}}\right) {\frac{dy^{\prime }}{y^{\prime 2}}}{\frac{dy}{y^{2}}} ds\\&\ll a t^{-2} \int \limits _{0}^{t/2}\int \limits _{a}^{\infty } s^{-1} y^{-1} \left(e^{-{\frac{c\log (5y/4a)^{2}}{s}}}+ e^{-{\frac{c\log (y)^{2}}{s}}}\right) dy ds\\&+\,\, a t^{-1} \int \limits _{t/2}^{t} \int \limits _{a}^{\infty } (t-s)^{-2} y^{-1} \left(e^{-{\frac{c\log (5y/4a)^{2}}{t-s}}} + e^{-{\frac{c\log (y)^{2}}{t-s}}}\right) dy ds. \end{aligned}$$

Since $y\ge a > {\frac{5}{4}}$, we have an estimate in $s$:

$$\begin{aligned} e^{-{\frac{c\log (5y/4a)^{2}}{s}}} + e^{-{\frac{c\log (y)^{2}}{s}}} \le e^{-{\frac{c\log (5/4)^{2}}{2s}}} \left(e^{-{\frac{c\log (5y/4a)^{2}}{2s}}} + e^{-{\frac{c\log (y)^{2}}{2s}}}\right) \end{aligned}$$

and $\int _{a}^{\infty }y^{-1}e^{-{\frac{c\log (5y/4a)^{2}}{2s}}}dy = \int _{\frac{5}{4}}^{\infty }v^{-1}e^{-{\frac{c\log (v)^{2}}{2s}}}dv\ll \sqrt{s}$. We get a similar estimate for $t-s$, and together these give

$$\begin{aligned} \vert J_{1}\vert&\ll a t^{-2} \int \limits _{0}^{t/2} s^{-1} e^{-{\frac{c\log (5/4)^{2}}{2s}}} \int \limits _{\frac{5}{4}}^{\infty } y^{-1} e^{-{\frac{c\log (y)^{2}}{2 s}}} dy ds \\&\quad + \, a t^{-1} \int \limits _{t/2}^{t} (t-s)^{-2} e^{-{\frac{c\log (5/4)^{2}}{2(t-s)}}} \int \limits _{\frac{5}{4}}^{\infty } y^{-1} e^{-{\frac{c\log (y)^{2}}{2(t-s)}}} dy ds\\&\ll a t^{-2} \int \limits _{0}^{t/2} s^{-1/2} e^{-{\frac{c\log (5/4)^{2}}{2s}}} ds + a t^{-1} \int \limits _{t/2}^{t} (t-s)^{-3/2} e^{-{\frac{c\log (5/4)^{2}}{2(t-s)}}} ds\\&\ll a t^{-2} e^{-{\frac{c\log (5/4)^{2}}{4t}}} \int \limits _{0}^{t/2} ds + a t^{-1} e^{-{\frac{c\log (5/4)^{2}}{2t}}} \int \limits _{t/2}^{t} ds \ll a (t^{-1}+1)e^{c_{1}/t} \ll a e^{-{\frac{c^{\prime }}{t}}}, \end{aligned}$$

for some constants $c_{1}, c^{\prime } >0$, where we also used part $(4)$ of Observation .

For $J_{2}$, we had reduced the problem to the following estimate:

$$\begin{aligned} \vert J_{2}\vert \ll \int \limits _{t/2}^{t} \Vert M_{\chi _{Z_{\frac{4a}{5}}}} M_{\psi } \Delta _{Z,g}e^{-(s/2)\Delta _{Z,g}}M_{\phi }^{-1}\Vert _{2} \Vert M_{\phi }e^{-(s/2)\Delta _{Z,g}}\Vert _{2} ds. \end{aligned}$$

Now we proceed to estimate each of the HS norms appearing as integrand on the right-hand side as follows:

$$\begin{aligned}&\Vert M_{\chi _{Z_{\frac{4a}{5}}}} M_{\psi } \Delta _{Z,g}e^{-(s/2)\Delta _{Z,g}}M_{\phi }^{-1}\Vert _{2}^{2}\\&\quad = \int \limits _{Z_{\frac{4a}{5}}}\int \limits _{Z} \vert \psi (z) \Delta _{Z,g} K_{Z,g}(z,z^{\prime },s/2)\phi (z^{\prime })^{-1}\vert ^{2} dA_{g}(z^{\prime })dA_{g}(z)\\&\quad \ll \int \limits _{\frac{4a}{5}}^{\infty } \int \limits _{1}^{\infty } y^{-2k} y y^{\prime } s^{-4} \left(e^{-\frac{4c}{s} (\log (y/y^{\prime }))^{2}} + e^{-\frac{4c}{s} (\log (yy^{\prime }))^{2}}\right)y^{\prime } {\frac{dy^{\prime }}{y^{\prime 2}}} {\frac{dy}{y^{2}}}\\&\quad = s^{-4} \int \limits _{\frac{4a}{5}}^{\infty } \int \limits _{1}^{\infty } y^{-2k-1} e^{-\frac{4c}{s} (\log (y^{\prime }/y))^{2}}\ dy^{\prime } dy\\&\quad \quad \quad +\,\, s^{-4} \int \limits _{\frac{4a}{5}}^{\infty } \int \limits _{1}^{\infty } y^{-2k-1} e^{-\frac{4c}{s} (\log (y^{\prime }))^{2}}\ dy^{\prime } dy. \end{aligned}$$

The first integral in the last equality above can be estimated by fixing $y$ and making the change of variables $v = \log (y^{\prime }/y), y^{\prime } = ye^{v}, dy^{\prime } = ye^{v}dv$:

$$\begin{aligned} s^{-4}\int \limits _{\frac{4a}{5}}^{\infty } \int \limits _{-\log (y)}^{\infty } y^{-2k} e^{v}e^{{\frac{-4c}{s}}v^{2}}\ dv \ dy&\ll s^{-4} e^{{\frac{s}{4c}}} \sqrt{s} \int \limits _{\frac{4a}{5}}^{\infty } y^{-2k} \int \limits _{-\infty }^{\infty } e^{-v^{2}}\ dv \ dy \\&\ll s^{-7/2} a^{-2k+1} e^{{\frac{s}{4c}}}. \end{aligned}$$

As for the second integral, we obtain in a similar way:

$$\begin{aligned} s^{-4} \int \limits _{\frac{4a}{5}}^{\infty } \int \limits _{1}^{\infty } y^{-2k-1} e^{-\frac{4c}{s} (\log (y^{\prime }))^{2}}\ dy^{\prime } \ dy \ll s^{-7/2} e^{{\frac{s}{4c}}} a^{-2k}. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert M_{\chi _{Z_{\frac{4a}{5}}}} M_{\psi } \Delta _{Z,g}e^{-s/2\Delta _{Z,g}}M_{\phi }^{-1}\Vert _{2} \ll s^{-7/4}\left(a^{-k} + a^{-k+1/2}\right). \end{aligned}$$

For the operator $M_{\phi }e^{-(s/2)\Delta _{Z,g}}$, using Eq. (3.4) we have

$$\begin{aligned}&\Vert M_{\phi }e^{-(s/2)\Delta _{Z,g}} \Vert _{2}^{2} \\&\quad \ll \int \limits _{1}^{\infty } \int \limits _{1}^{\infty } s^{-2} y^{-1} y y^{\prime } \left(e^{-{\frac{2c}{s}}(\log (y/y^{\prime }))^{2}} + e^{-{\frac{2c}{s}}(\log (yy^{\prime }))^{2}}\right)^{2} {\frac{dy^{\prime }}{y^{\prime 2}}} {\frac{dy}{y^{2}}}\\&\quad \ll \int \limits _{1}^{\infty } \int \limits _{1}^{\infty } s^{-2} y^{\prime -1} y^{-2} \left(e^{-{\frac{4c}{s}}(\log (y/y^{\prime }))^{2}} + e^{-{\frac{4c}{s}}(\log (yy^{\prime }))^{2}}\right) {dy^{\prime }} {dy}\\&\quad \ll s^{-2}\sqrt{s} e^{s/4c} + s^{-2} \int \limits _{1}^{\infty } y^{\prime -1} e^{-{\frac{4c}{s}}(\log (y^{\prime }))^{2}} {dy^{\prime }} \ll s^{-3/2} \left(1+e^{s/4c}\right). \end{aligned}$$

Since $s\le t\le 1$ we have that $\Vert M_{\phi } e^{-(s/2)\Delta _{Z,g}} \Vert _{2} \ll s^{-3/4}$. It follows that

$$\begin{aligned} \vert J_{2}\vert \ll \int \limits _{t/2}^{t} s^{-7/4} (a^{-k} + a^{-k+1/2})\cdot s^{-3/4} ds \ll a^{-k+1/2} t^{-3/2}. \end{aligned}$$

Now, for $J_{3}$, we have

$$\begin{aligned} J_{3}&= \int \limits _{t/2}^{t}\int \limits _{Z_{a}} \int \limits _{Z_{\frac{4a}{5}}} \psi _{2}(z) K_{Z,h}(z,z^{\prime },s)e^{2\varphi (z^{\prime })} \chi _{Z_{\frac{4a}{5}}}(z^{\prime })\\&(\Delta _{Z,g}-\Delta _{Z,h})_{z^{\prime }} K_{Z,g}(z^{\prime },z,t-s) dA_{g}(z^{\prime })dA_{g}(z)ds. \end{aligned}$$

Remember that $\Delta _{Z,g}-\Delta _{Z,h} = (e^{2\varphi (z^{\prime })}-1) \Delta _{Z,h}= {\widetilde{\psi }}(z^{\prime })\Delta _{Z,h}$, so the previous equation becomes:

$$\begin{aligned} J_{3}&= \int \limits _{t/2}^{t}\int \limits _{Z_{a}} \int \limits _{Z_{\frac{4a}{5}}} \{\psi _{2}(z) K_{Z,h}(z,z^{\prime },s) \chi _{Z_{\frac{4a}{5}}}(z^{\prime }) {\widetilde{\psi }}(z^{\prime }) \\&(\Delta _{Z,h} K_{Z,g}(z^{\prime },z,t-s)) e^{-2\varphi (z)} \} \ dA_{h}(z^{\prime })\ dA_{h}(z)\ ds\\&= \int \limits _{t/2}^{t}\int \limits _{Z_{a}} \int \limits _{Z_{\frac{4a}{5}}} \{\psi _{2}(z) (\Delta _{Z,h} K_{Z,h}(z,z^{\prime },s){\widetilde{\psi }}(z^{\prime })) \chi _{Z_{\frac{4a}{5}}}(z^{\prime })\\&K_{Z,g}(z^{\prime },z,t-s) e^{-2\varphi (z)}\} \ dA_{h}(z^{\prime })\ dA_{h}(z)\ ds\\&= \int \limits _{t/2}^{t}\int \limits _{Z_{a}} \int \limits _{Z_{\frac{4a}{5}}} \{\psi _{2}(z) e^{-2\varphi (z)}K_{Z,g}(z,z^{\prime },t-s) \chi _{Z_{\frac{4a}{5}}}(z^{\prime })\\&(\Delta _{Z,h} {\widetilde{\psi }}(z^{\prime }) K_{Z,h}(z^{\prime },z,s))\}\ dA_{h}(z^{\prime })\ dA_{h}(z)\ ds. \end{aligned}$$

Writing this in terms of the corresponding operators, we obtain

$$\begin{aligned} J_{3}&= \int \limits _{t/2}^{t}\text{ Tr}(M_{\psi _{2}} M_{e^{-2\varphi }} e^{-(t-s)\Delta _{Z,g}} M_{\chi _{Z_{\frac{4a}{5}}}} \Delta _{Z,h} M_{{\widetilde{\psi }}} e^{-s\Delta _{Z,h}})ds,\\ \vert J_{3}\vert&\le \int \limits _{t/2}^{t} \Vert M_{\chi _{Z_{\frac{4a}{5}}}} \Delta _{Z,h} M_{{\widetilde{\psi }}} e^{-s\Delta _{Z,h}} \Vert _{1}\ ds. \end{aligned}$$

We are now working in $L^{2}(M,dA_{h})$ therefore to simplify notation, we do not write the subindex $h$ in the trace and the HS norms. In the same way as above, we have

$$\begin{aligned} \Vert M_{\chi _{Z_{\frac{4a}{5}}}} \Delta _{Z,h} M_{{\widetilde{\psi }}} e^{-s\Delta _{Z,h}} \Vert _{1} \le \Vert M_{\chi _{Z_{\frac{4a}{5}}}} \Delta _{Z,h} M_{{\widetilde{\psi }}} e^{-s\Delta _{Z,h}/2}M_{\phi ^{-1}}\Vert _{2} \Vert M_{\phi } e^{-s\Delta _{Z,h}/2}\Vert _{2} \end{aligned}$$

The kernel of the operator $M_{\chi _{Z_{\frac{4a}{5}}}} \Delta _{Z,h} M_{{\widetilde{\psi }}} e^{-s\Delta _{Z,h}/2}M_{\phi ^{-1}}$ is

$$\begin{aligned} \chi _{Z_{\frac{4a}{5}}}(z^{\prime })(\Delta _{Z,h}({\widetilde{\psi }}(z^{\prime }) K_{Z,h}(z^{\prime },z,s))\phi (z)^{-1}. \end{aligned}$$

Using the decay assumptions on $\varphi$ and its derivatives, we have that

$$\begin{aligned} \vert \Delta _{Z,h}({\widetilde{\psi }} K_{Z,h})\vert ^{2}&\ll \vert {\widetilde{\psi }}\Delta _{Z,h}K_{Z,h}\vert ^{2} + \vert K_{Z,h} \Delta _{Z,h} {\widetilde{\psi }}\vert ^{2} + 2\vert \langle \nabla {\widetilde{\psi }}, \nabla K_{Z,h}\rangle \vert ^{2}\\&\ll y^{\prime -2k+1}y (s^{-4} + s^{-2} + s^{-3})\left(e^{-{\frac{c}{s}}(\log (y/y^{\prime }))^{2}} + e^{-{\frac{c}{s}}(\log (yy^{\prime }))^{2}}\right)^{2}. \end{aligned}$$

Since for $0<s<1$ we have that $s^{-4} + s^{-2} + s^{-3}\ll s^{-4}$, we can estimate the HS norm by

$$\begin{aligned}&\Vert M_{\chi _{Z_{\frac{4a}{5}}}} \Delta _{Z,h} M_{{\widetilde{\psi }}} e^{-s\Delta _{Z,h}/2}M_{\phi ^{-1}} \Vert _{2}^{2}\\&\qquad = \int \limits _{Z} \int \limits _{Z} \vert \chi _{Z_{\frac{4a}{5}}}(z^{\prime }) {\widetilde{\psi }}(z^{\prime }) \Delta _{h,z^{\prime }} K_{h}(z^{\prime },z,s/2) \phi (z)^{-1} \vert ^{2} dA_{h}(z^{\prime }) dA_{h}(z)\\&\qquad \ll s^{-4} \int \limits _{1}^{\infty } \int \limits _{\frac{4a}{5}}^{\infty } \ y^{2}\ y^{\prime -2k+1} \left(e^{-{\frac{2c}{s}}(\log (y/y^{\prime }))^{2}} + e^{-{\frac{2c}{s}}(\log (yy^{\prime }))^{2}}\right)^{2}{\frac{dy^{\prime }}{y^{\prime 2}}} {\frac{dy}{y^{2}}}\\&\qquad \ll s^{-4} \int \limits _{\frac{4a}{5}}^{\infty } \int \limits _{1}^{\infty } \left(y^{\prime -2k-1} e^{-{\frac{4c}{s}}(\log (y/y^{\prime }))^{2}} + y^{\prime -2k-1} e^{-{\frac{4c}{s}}(\log (y))^{2}}\right) \ dy \ dy^{\prime }\\&\qquad \ll (a^{-2k+1}+a^{-2k}) s^{-7/2} e^{s/4c} \ll a^{-2k+1} s^{-7/2}. \end{aligned}$$

We finally obtain

$$\begin{aligned} \Vert M_{\phi }^{-1} e^{-(s/2) \Delta _{Z,h}} {\widetilde{\psi }} \Delta _{h} \Vert _{2} \le a^{-k+1/2} s^{-7/4}. \end{aligned}$$

For the operator $e^{-(s/2) \Delta _{Z,h}}M_{\phi }$, the proof goes in the same way as for the operator $M_{\phi }e^{-(s/2) \Delta _{Z,g}}$. At the end, we obtain

$$\begin{aligned} \Vert e^{-s \Delta _{Z,h}}M_{\phi }\Vert _{2} = \left( \int \limits _{Z} \int \limits _{Z} \vert K_{Z,h}(z,z^{\prime },s/2) \phi (z^{\prime })\vert ^{2} dA_{h}(z^{\prime }) dA_{h}(z)\right)^{1/2} \ll s^{-3/4}. \end{aligned}$$

In this way,

$$\begin{aligned} \vert J_{3}\vert \ll \int \limits _{t/2}^{t} a^{-k+1/2} s^{-7/4} s^{-3/4} ds \ll a^{-k+1/2} t^{-3/2}. \end{aligned}$$