Consider the spherical shell (annulus)
$$A(R,r) = \{ x \in \mathbb{R}^3 : R \leq |
x|\leq R+r \}.$$ Think of the limit $R \to \infty$.
Assume that $r$ depends on $R$ as $r(R) = R^{-\delta}$. We are interested in the distribution of lattice points in $A(R,r)$.


From results on the Gauss circle problem in three dimensions (see e.g. [Ivic, Krätzel, Kühleitner, and Nowak - Lattice points in large regions and related arithmetic functions: Recent developments in a very classic topic](https://arxiv.org/abs/math/0410522)) I know that the number of lattice points in the ball $B(R)$ is given by the volume of the ball, up to an error (the lattice point discrepancy) which is bounded by $\mathcal{O}(R^{\frac{42}{32}+\epsilon})$, for all $\epsilon > 0$. So by taking the difference we can obtain the number of lattice points in the annulus for $\delta >0$ not too large. We find that the number of lattice points is of order $R^{2}r$ (surface of the sphere times width of the shell).


In [Bourgain, Sarnak, and Rudnick - Local statistics of lattice points on the sphere](https://arxiv.org/abs/1204.0134), I found the reference to [Duke - Hyperbolic distribution problems and half-integral weight Maass forms](https://doi.org/10.1007/BF01393993) & [Golubeva–Fomenko - Asymptotic distribution of lattice points on the three-dimensional sphere](https://doi.org/10.1007/BF02342921), showing that the the lattice points exactly on the sphere are uniformly distributed, if $R^2 \neq 0,4,7 \bmod 8$. (However there are only order $R^{1-\epsilon}$ lattice points exactly on the sphere.)


My question is: Is anything known about the distribution of lattice points in the thin spherical shell? For example, consider a spherical cap on $B(R)$ and 'fatten' it up to a radius $R+r$. We now have a segment of the annulus (if you want, the intersection of a cone with the annulus) and ask whether the number of lattice points in this segment is to leading order given by the area of the spherical cap times $r$. If yes, does this stay true if the solid angle defining the cap goes to zero (not too fast) as $R \to \infty$? If it is not true for all such segments, could I at least say that it is true for most segments, or for a specific sequence of radii R?


Another possible way of asking could be: consider summing a continuous function over $x/ |x|$, $x \in A(R,r)$ (and normalize by the number of points). Does the sum converge to the integral w.r.t. the uniform measure on the unit sphere?


Is anyone aware of results in this direction?



Are lattice points in thin spherical shells uniformly distributed?
Yes, they are equidistributed as long as $\delta<11/16$ and $r=R^{-\delta}$ and $R\to\infty$. Without loss of generality, we shall assume that $\delta>-1$ (i.e. $r<R$).


To see this, let $\mathcal{F}\subset S^2$ be a fixed convex region with piecewise smooth boundary on the unit sphere. Let $\mu(\mathcal{F})$ be the normalized area of $\mathcal{F}$, where normalization is such that $\mu(S^2)=1$. Let $r(n)$ be the number of representations $n=|\mathbf{x}|^2$ with $\mathbf{x}\in\mathbb{Z}^3$, and let $r(n,\mathcal{F})$ be the number of representations with the additional constraint that $\mathbf{x}/|\mathbf{x}|\in\mathcal{F}$.


Let $k\in\mathbb{N}$ be fixed. By Theorem 1 in [the 1990 Inventiones paper of Duke and Schulze-Pillot](https://eudml.org/doc/143750),
\begin{align\*}\sum\_{\substack{R^2\leq n\leq (R+r)^2\\4^k\nmid n}}r(n,\mathcal{F})&=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum\_{\substack{R^2\leq n\leq (R+r)^2\\4^k\nmid n}}r(n)\\&=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum\_{R^2\leq n\leq (R+r)^2}r(n)+O\Bigl(\sum\_{\substack{R^2\leq n\leq (R+r)^2\\4^k\mid n}}r(n)\Bigr).\end{align\*}
On the right hand side, by a result of [Heath-Brown (1999)](https://ora.ox.ac.uk/objects/uuid:4b17126d-c3a0-4827-8fa2-0ad82872d17e),
$$\sum\_{R^2\leq n\leq (R+r)^2}r(n)=\bigl(4\pi+o(1)\bigr)R^2r.$$
For $4^k\mid n$, we have that $r(n)=r(n/4^k)$, hence Heath-Brown's result also yields
$$\sum\_{\substack{R^2\leq n\leq (R+r)^2\\4^k\mid n}}r(n)=\sum\_{4^{-k}R^2\leq m\leq 4^{-k}(R+r)^2}r(m)=\bigl(4\pi\cdot 8^{-k}+o(1)\bigr)R^2r.$$
To summarize so far,
$$\sum\_{\substack{R^2\leq n\leq (R+r)^2\\4^k\nmid n}}r(n,\mathcal{F})=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum\_{R^2\leq n\leq (R+r)^2}r(n)+\bigl(O(8^{-k})+o(1)\bigr)R^2r.$$
Combining this with the earlier two displays (consequences of Heath-Brown's result), we infer that
\begin{align\*}\sum\_{R^2\leq n\leq (R+r)^2}r(n,\mathcal{F})&=\bigl(\mu(\mathcal{F})+o(1)\bigr)\sum\_{R^2\leq n\leq (R+r)^2}r(n)+\bigl(O(8^{-k})+o(1)\bigr)R^2r\\[6pt]&=\bigl(\mu(\mathcal{F})+O(8^{-k})+o(1)\bigr)\sum\_{R^2\leq n\leq (R+r)^2}r(n).
\end{align\*}
In other words,
$$\limsup\_{R\to\infty}\left|\frac{\sum\_{R^2\leq n\leq (R+r)^2}r(n,\mathcal{F})}{\sum\_{R^2\leq n\leq (R+r)^2}r(n)}-\mu(\mathcal{F})\right|=O(8^{-k}).$$
As $k\in\mathbb{N}$ is arbitrary, and the left hand side is independent of $k$, the left hand side is zero. That is,
$$\lim\_{R\to\infty}\frac{\sum\_{R^2\leq n\leq (R+r)^2}r(n,\mathcal{F})}{\sum\_{R^2\leq n\leq (R+r)^2}r(n)}=\mu(\mathcal{F}).$$


