Erdős Problem 988: spherical cap discrepancy
Statement and attribution
For $P=\{p_1,\ldots,p_n\}\subset S^2$, let $D(P)=\sup_C|\#(P\cap C)-n\sigma(C)|$, where $\sigma$ is normalized surface area and $C$ ranges over spherical caps. We prove the stronger bound \[
D(P)\ge c n^{1/4}.
\] The qualitative question was answered by Schmidt,
Irregularities of Distribution IV, Invent. Math. 7 $1969$, 55--82. The proof below reconstructs the distance-energy method underlying Stolarsky's invariance principle and its discrepancy applications. All identities and the lower bound needed here are proved below; no novelty is claimed.
The cap integral equals a distance deficit
Parameterize caps by $C(z,t)=\{x\in S^2:x\cdot z\ge t\}$, with $z\in S^2$ and $-1\le t\le1$, and set $\nu=\sum_i\delta_{p_i}-n\sigma$. Since $\nu(S^2)=0$, \[
E:=\int_{S^2}\int_{-1}^{1}\nu(C(z,t))^2\,dt\,d\sigma(z)
=-\frac12\iint\!\left[\int_{S^2}\int_{-1}^{1}
(1_{C(z,t)}(x)-1_{C(z,t)}(y))^2\,dt\,d\sigma(z)\right]d\nu(x)d\nu(y).
\] This follows by expanding the square: the two terms involving only $x$ or only $y$ vanish after integration against $\nu\otimes\nu$. For fixed $z$, the inner integral is $|(x-y)\cdot z|$. Rotational symmetry and the uniform distribution of one coordinate of $z$ on $[-1,1]$ show that its average is $|x-y|/2$. Also \[
\int_{S^2}|x-y|\,d\sigma(y)
=\frac12\int_{-1}^{1}\sqrt{2-2u}\,du=\frac43.
\] Consequently \[
E=\frac14\left(\frac43n^2-\sum_{i,j}|p_i-p_j|\right).
\tag{1}
\] All integrands are bounded, so Fubini is legitimate even for this signed finite measure. The parameter space has measure $2$, whence $E\le2D(P)^2$.
Positive polynomial kernels
Put $T(x,y)=(1+x\cdot y)/2$. For every integer $k\ge0$, \[
Q_k:=\iint T(x,y)^k\,d\nu(x)d\nu(y)\ge0.
\tag{2}
\] Indeed, expanding $(1+x\cdot y)^k$ and then each power of $x\cdot y$ expresses $Q_k$ as a sum of nonnegative coefficients times squares $\left(\int x_1^{a_1}x_2^{a_2}x_3^{a_3}\,d\nu(x)\right)^2$. For fixed $x$, $T(x,y)$ is uniform on $[0,1]$ under $\sigma(y)$; hence \[
Q_k=\sum_{i,j}T(p_i,p_j)^k-\frac{n^2}{k+1}.
\tag{3}
\] Every summand in the double sum is nonnegative, and its diagonal contributes $n$. For $k\ge2n$, this gives $Q_k\ge n/2$.
The high-degree tail forces a deficit
The binomial expansion, including its uniformly convergent endpoint, is \[
\sqrt{1-t}=1-\sum_{k\ge1}a_k t^k,\qquad
a_k=\frac{\binom{2k}{k}}{4^k(2k-1)}>0,\qquad
\sum_{k\ge1}a_k=1.
\] Uniform convergence follows from positivity and summability of the coefficients. Since $|x-y|=2\sqrt{1-T(x,y)}$, $1$--$3$ imply \[
\frac43n^2-\sum_{i,j}|p_i-p_j|
=2\sum_{k\ge1}a_kQ_k
\ge n\sum_{k\ge2n}a_k.
\] There is no cancellation from the discarded terms because of $2$. Let $b_k=\binom{2k}{k}/4^k$, with $b_0=1$. Then $a_k=b_{k-1}-b_k$, so the last tail equals $b_{2n-1}$. Moreover \[
\log b_m=\sum_{j=1}^{m}\log(1-1/(2j))
\ge-\frac12\sum_{j=1}^{m}\frac1j-\frac14\sum_{j=1}^{m}\frac1{j^2}
\ge-\frac12\log(m+1)-C,
\] using $\log(1-u)\ge-u-u^2$ for $0\le u\le1/2$. Thus $b_{2n-1}\ge c/\sqrt n$. Combining everything gives $2D(P)^2\ge E\ge c\sqrt n$, as required.
Conclusion and sources
The minimum discrepancy tends to infinity, with the proved quantitative rate $cn^{1/4}$. The argument also allows repeated points. It does not claim a matching upper bound for the supremum norm. References:
https://www.erdosproblems.com/988; Schmidt,
https://doi.org/10.1007/BF01418774; K. B. Stolarsky,
Sums of distances between points on a sphere II, Proc. Amer. Math. Soc. 41 $1973$, 575--582,
https://doi.org/10.1090/S0002-9939-1973-0333995-9.
COMPLETION ESTIMATE
COMPLETION: 100%