The Wilton-type bounds involving half-integral weight cusp forms

Question

There is a basic question which puzzles me for a while, and maybe look naive for some experts here. The question is the following:

Let $f(z)=\sum_{n\ge 1} a_f(n) n^{k/2-1/4}e(nz)\in S_{k+1/2}(4N)$ be a newfrom of half-weight $k + 1/2$ for $\Gamma_0(4N)$. Does any one know if we have

$$ \sum_{n\le X} a_f(n)e(n\alpha)\ll_{f,\varepsilon}X^{1/2+\varepsilon} \;?? \label{1}\tag{1} $$ As is well-known, if $f$ is a $GL_2$-newform, this is called the Wilton-type bound. It is not clear for me that this estimate holds for half-integral weight cusp forms.

On the other hand, if the estimate \eqref{1} is not true, whether or not one instead has $$ \sum_{n\le X} a_f(n)e(n\alpha)\ll_{f,\varepsilon}X^{1-\delta} $$ for some constant $\delta>0$?

So, if any expert here know some relevant results on this question, please show some guides or corresponding references, many thanks.

Thanks in advance!

Answer 1

Let $f\in S_{k+1/2}(4N)$ be half-integral weight cusp form of weight $k+1/2$ which has a Fourier series expansion $f(z) = \sum_{n\geq 1} \hat{f}(n) e(nz)$ where the Fourier coefficients are bounded by $$\hat{f}(n)\ll n^{\left(\frac{k+1/2}{2}-\frac{1}{4}\right)}d(n) = n^{k/2}d(n).$$ The exponent can't be improved if $f$ is coming from the subspace of theta functions (see the 1987 Inventiones paper of Iwaniec). So, normalizing the Fourier coefficients we can rewrite the Fourier expansion as $$f(z)= \sum_{n \geq 1}\lambda_{f}(n)n^{k/2}e(nz).$$ Now for any $\alpha\in \mathbb{R}$, our job is to obtain a non-trivial bound for the following exponential sum weighted by normalized Fourier coefficients $$\sum_{n\leq X}\lambda_{f}(n)e(n\alpha).$$ Observe that we can write the following exponential sum as $$\sum_{n\leq X} \lambda_{f}(n)n^{k/2}e(n\alpha) = (f*D_X)(\alpha) = \int_{0}^{1}f(z+\alpha)D_{X}(z) \ dx$$ where $D_X$ is the Dirichlet kernel given by $$D_X(z) = \sum_{n\leq X} e(-nz)\ll \frac{e^{2\pi Xy}}{\left|1-e(-z)\right|}.$$ Since $f$ is a cups form so, $f(z)\ll y^{-\frac{2k+1}{4}}$ uniformly in $x$. Now plugging these two bounds in above expression we get $$\sum_{n\leq X}\lambda_{f}(n)n^{k/2}e(n\alpha)\ll y^{-\frac{2k+1}{4}}e^{2\pi Xy}\int_{0}^{1}\frac{dx}{\left|1-e(-z)\right|}\ll y^{-\frac{2k+1}{4}}e^{2\pi Xy}\log\left(2+y^{-1}\right).$$ Now choosing $y=1/X$ we have $$\sum_{n\leq X}\lambda_{f}(n)n^{k/2}e(n\alpha)\ll X^{\frac{k}{2}+\frac{1}{4}}\log(2X)$$ and from this bound one can easily establish the following non-trivial bound for the required sum $$\sum_{n\leq X} \lambda_{f}(n)e(n\alpha)\ll X^{1/4}\log(2X).$$