## Smooth Sums to Sharp Sums 1

In this note, I describe a combination of two smoothed integral transforms that has been very useful in my collaborations with Alex Walker, Chan Ieong Kuan, and Tom Hulse. I suspect that this particular technique was once very well-known. But we were not familiar with it, and so I describe it here.

In application, this is somewhat more complicated. But to show the technique, I apply it to reprove some classic bounds on $\text{GL}(2)$ $L$-functions.

This note is also available as a pdf. This was first written as a LaTeX document, and then modified to fit into wordpress through latex2jax.

## Introduction

Consider a Dirichlet series

$$\begin{equation}

D(s) = \sum_{n \geq 1} \frac{a(n)}{n^s}. \notag

\end{equation}$$

Suppose that this Dirichlet series converges absolutely for $\Re s > 1$, has meromorphic continuation to the complex plane, and satisfies a functional equation of shape

$$\begin{equation}

\Lambda(s) := G(s) D(s) = \epsilon \Lambda(1-s), \notag

\end{equation}$$

where $\lvert \epsilon \rvert = 1$ and $G(s)$ is a product of Gamma factors.

Dirichlet series are often used as a tool to study number theoretic functions with multiplicative properties. By studying the analytic properties of the Dirichlet series, one hopes to extract information about the coefficients $a(n)$. Some of the most common interesting information within Dirichlet series comes from partial sums

$$\begin{equation}

S(n) = \sum_{m \leq n} a(m). \notag

\end{equation}$$

For example, the Gauss Circle and Dirichlet Divisor problems can both be stated as problems concerning sums of coefficients of Dirichlet series.

One can try to understand the partial sum directly by understanding the integral transform

$$\begin{equation}

S(n) = \frac{1}{2\pi i} \int_{(2)} D(s) \frac{X^s}{s} ds, \notag

\end{equation}$$

a Perron integral. However, it is often challenging to understand this integral, as delicate properties concerning the convergence of the integral often come into play.

Instead, one often tries to understand a smoothed sum of the form

$$\begin{equation}

\sum_{m \geq 1} a(m) v(m) \notag

\end{equation}$$

where $v(m)$ is a smooth function that vanishes or decays extremely quickly for values of $m$ larger than $n$. A large class of smoothed sums can be obtained by starting with a very nicely behaved weight function $v(m)$ and take its Mellin transform

$$\begin{equation}

V(s) = \int_0^\infty v(x) x^s \frac{dx}{x}. \notag

\end{equation}$$

Then Mellin inversion gives that

$$\begin{equation}

\sum_{m \geq 1} a(m) v(m/X) = \frac{1}{2\pi i} \int_{(2)} D(s) X^s V(s) ds, \notag

\end{equation}$$

as long as $v$ and $V$ are nice enough functions.

In this note, we will use two smoothing integral transforms and corresponding smoothed sums. We will use one smooth function $v_1$ (which depends on another parameter $Y$) with the property that

$$\begin{equation}

\sum_{m \geq 1} a(m) v_1(m/X) \approx \sum_{\lvert m – X \rvert < X/Y} a(m). \notag

\end{equation}$$

And we will use another smooth function $v_2$ (which also depends on $Y$) with the property that

$$\begin{equation}

\sum_{m \geq 1} a(m) v_2(m/X) = \sum_{m \leq X} a(m) + \sum_{X < m < X + X/Y} a(m) v_2(m/X). \notag

\end{equation}$$

Further, as long as the coefficients $a(m)$ are nonnegative, it will be true that

$$\begin{equation}

\sum_{X < m < X + X/Y} a(m) v_2(m/X) \ll \sum_{\lvert m – X \rvert < X/Y} a(m), \notag

\end{equation}$$

which is exactly what $\sum a(m) v_1(m/X)$ estimates. Therefore

$$\begin{equation}\label{eq:overall_plan}

\sum_{m \leq X} a(m) = \sum_{m \geq 1} a(m) v_2(m/X) + O\Big(\sum_{m \geq 1} a(m) v_1(m/X) \Big).

\end{equation}$$

Hence sufficient understanding of $\sum a(m) v_1(m/X)$ and $\sum a(m) v_2(m/X)$ allows one to understand the sharp sum

$$\begin{equation}

\sum_{m \leq X} a(m). \notag

\end{equation}$$