I am trying to solve the following problem:
Prove the following estimate:
$$\sum_{\vert t_i\vert<T}\vert u_j(z)\vert^2\ll T^2$$ by choosing appropriate $h$ in the Selberg's (pre)-trace formula.
I was told to pick $h$ to be a bump function where $h\equiv 1$ on $[-T,T]$ and with support in $[-2T,2T]$. Thus, we decided to use $$ h(x)=\begin{cases} 1 & x\in[-T,T]\\ \frac{2T+x}{T} & x\in[-2T,-T]\\ \frac{2T-x}{T} & x\in [T,2T] \\ 0 & \text{else} \end{cases} $$
Now we have that Selberg's pre-trace formula says that $$ k(x,y)=\sum h(t_j)u_j(x)\overline{u_j(y)} $$ Thus, by setting $x=y$, we will get that the right-hand side will be $\sum_{\vert t_i\vert<T}\vert u_j(z)\vert^2+$ some other stuff associated to $[-2T,-T]$ and $[T,2T]$. We currently don't know how to deal with the other stuff, so we decided to focus on the other side of the trace formula, and use the Harish-Chandra Selberg transform. That is we have that $$k(t)=-\frac{1}{\pi}\int_t^\infty \frac{dQ(w)}{\sqrt{w-t}}$$ and $$g(u)=\frac{1}{2\pi}\int_{-\infty}^\infty h(r)e^{-iru}dr$$ where $Q(w)=g(u)$ with $w=e^u+e^{-u}-2=2\cosh(u)-2$. I believe we need to take $t=0$ here as $k(t)$ will be a function of the distance, and since we want $k(x,y)$, we should have that the distance is $0$. Thus, we first started off by computing $g(u)$ in this formula with our given $h$, just using Wolfram alpha we ended up getting $$g(u)=\frac{\cos(Tu)-\cos(2Tu)}{\pi Tu^2}$$ Now to get $k(0)$ rather than integrating with respect to $dQ(w)$, we wanted to change into $u$, coordinates, so we calculated that $$ \frac{\partial g}{\partial u}(u)=\frac{-Tu(\sin(Tu)-2\sin(2Tu))-2\cos(Tu)+2\cos(2Tu)}{\pi Tu^3}$$ Thus, we will get that $$k(0)=-\frac{1}{\pi}\int_0^\infty (\frac{-Tu(\sin(Tu)-2\sin(2Tu))-2\cos(Tu)+2\cos(2Tu)}{\pi Tu^3})(\frac{du}{\sqrt{2\cosh(u)-2}}) $$
Honestly, we have no clue how to continue from here, nor whether we are on the right track. Any help/hints will be greatly appreciated. Primarily what do we do with the other terms on the one side of the equation and how do we bound this? Or if we just done goofed the whole thing what function should we take instead. Thanks.
Additional Remark/Idea: Note that thanks to the comment below, I now see that my given $h$ does not satisfy the conditions for the Selberg trace formula. I also had a look at these notes where Theorem 3 gives a different statement of Selberg's pre-trace formula as
$$\sum_{j=0}^\infty h(\rho_j)\vert u_j(z)\vert^2=\frac{1}{4\pi}\int_{-\infty}^\infty h(\rho)\tanh(\pi\rho)\rho d\rho+\sum_{\gamma\in\Gamma-\{1\}}k(\gamma z,z)$$
Now my thought is if I choose $h$ to be a smooth compactly supported bump function such that $h\equiv 1$ on $[-T,T]$ such that $h$ has exponential decay to $0$ before the next $\rho_j$, then the left hand side of the equation is precisely $$ \sum_{\vert t_i\vert<T}\vert u_j(z)\vert^2 $$ while the right-hand side will become $$ \frac{1}{4\pi}\int_{-\infty}^\infty h(\rho)\tanh(\pi\rho)\rho d\rho+\sum_{\gamma\in\Gamma-\{1\}}k(\gamma z,z) $$ Now using the fact that $\tanh$ is a bounded function and that $h(\rho)$ will be a bump function from $[-T,T]$, the integral will be $O(T^2)$, but now I don't see how to deal with the sum of the point-pair invariant things. I don't know if this will just be constant or how it is affected as we change $T$.