# Smooth circle/divisor problem?

Consider the characteristic function f of a disc of radius r. The classical circle problem asks for a bound on the error term in the expression

$\sum_{(x,y) \in \mathbb{Z}^2} f(x,y) = closed expression + error term$,

where the closed expression happens to be the area of the circle. (The divisor problem is the analogous question for the characteristic function of the domain under a hyperbola $y = r/x$.) Hardy and Landau showed that the error term is no smaller than r^{1/2} or so.

Question: what happens if we let f be a smoothed version of the characteristic function of a disc? Does the error term become conjecturally much smaller (how much smaller?)? What has been proven in that case?

(Obviously the answer will depend on the extent and type of smoothing - how strongly? Would the classical problem be equivalent to a smoothing where the decay happens within an annulus of constant width?)

The error term becomes very much smaller with smoothing. Even with a $C^1$ smoothing function there's a drastic effect:

$\sum_{n\leq X}r(n)(1-n/X)=\frac{\pi}{2}X+O(X^{-1/4})$.

See e.g. p. 74 of Iwaniec-Kowalski. If $f \in C^{\infty}(\mathbf{R}_{>0})$ has compact support and total mass one, and is identically one on an interval $(0,\epsilon)$, then

$\sum_{n}r(n)f(n/X)=\pi X+O(X^{-A})$

for any $A$. This is an easy exercise with Mellin transforms.

• ... or with the Poisson summation formula. – Terry Tao Jun 27 '11 at 22:16
• Of course! But for some reason I find "drag a contour as far to the left as you wish" more 'physically' satisfying than "integrate by parts $N$ times". – David Hansen Jun 27 '11 at 23:01
• This is nice - but it corresponds to smoothing at the same scale as the radius. I have been convinced that the non-smooth problem corresponds to smoothing at unit scale. What about smoothing at scales larger than 1 but smaller than r? – H A Helfgott Jun 28 '11 at 9:09

Seems to indicate that $r^{1/2}$ (no log terms) is easy for the smoothed problem.