= Solution
A precise useful form of the principle is the <type I–type II inverse principle for Möbius correlation>. Normalize $|f(n)|\leq1$, and suppose
$$
\left|\sum_{n\leq X}\mu(n)f(n)\right|\geq\delta X.
$$
For any positive integer cutoffs $U,V$ with $U+V\leq\delta X/2$, define
$$
c_d=\sum_{\substack{bc=d\\b\leq U,\ c\leq V}}\mu(b)\mu(c),
\qquad a_d=\sum_{\substack{c\mid d\\c>V}}\mu(c).
$$
Then $|c_d|,|a_d|\leq\tau(d)$, where $\tau$ is the <divisor function>, and at least one of the following two sums has modulus at least $\delta X/4$:
$$
T_{\mathrm I}=\sum_{d\leq UV}c_d\sum_{k\leq X/d}f(dk),
\qquad
T_{\mathrm {II}}=\sum_{\substack{d>V,\ w>U\\dw\leq X}}a_d\mu(w)f(dw).
$$
The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a <periodic function>: this is the “somewhat periodic” branch. The second is a <bilinear sum> with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.
For clarity, the <Vaughan identity for the Möbius function> proves this version immediately. Split $\mu=\mu_{\leq U}+\mu_{>U}$ and likewise at $V$. Since $\mu*\mu*\mathbf1=\mu$,
$$
\boxed{\mu=\mu_{\leq U}+\mu_{\leq V}
-\mu_{\leq U}*\mu_{\leq V}*\mathbf1
+\mu_{>U}*\mu_{>V}*\mathbf1.}
$$
Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm I}+T_{\mathrm {II}}$. The <triangle inequality> gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.
We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha=\sqrt2$ and $e(u)=e^{2\pi iu}$. The <Diophantine bound for the square root of two> is
$$
\|h\alpha\|\geq\frac1{4h}\qquad(h\geq1),
$$
where $\|u\|$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\geq1$, while $h\sqrt2+a\leq4h$; dividing proves the bound.
Use the preceding <Dirichlet convolution> identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the <type I sum>, the <exponential geometric sum bound> gives
$$
\left|\sum_{k\leq X/d}e(\alpha dk)\right|
\ll\min\{X/d,\|d\alpha\|^{-1}\}\ll d.
$$
Consequently
$$
|T_{\mathrm I}|\ll\sum_{d\leq P^2}\tau(d)d
\ll P^4\log(2P)=O(X^{2/5}\log X),
$$
using $\sum_{d\leq Y}\tau(d)\ll Y\log(2Y)$ from counting factor pairs.
For the <type II sum>, perform a <dyadic decomposition> into $d\in(D,2D]$, $w\in(W,2W]$, keeping $dw\leq X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the <Cauchy-Schwarz inequality> and the <divisor-square summatory bound>,
$$
|B|^2\ll D\log^3X\left(DW+W\sum_{1\leq h\leq W}\min\{D,\|h\alpha\|^{-1}\}\right).
$$
To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w'$ still form an interval, with upper endpoint $\min(2D,X/\max(w,w'))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w'$ there are $O(W)$ pairs.
The points $0,\alpha,2\alpha,\ldots,\lfloor W\rfloor\alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore gives
$$
\sum_{1\leq h\leq W}\|h\alpha\|^{-1}\ll W\log(2W).
$$
Substitution yields the <bilinear cancellation for badly approximable phases> estimate
$$
|B|\ll DW\log^2X\left(D^{-1/2}+W^{-1/2}\right)
\ll X\log^2X\,P^{-1/2}.
$$
There are $O(\log^2X)$ blocks, so $T_{\mathrm {II}}\ll X^{19/20}\log^4X$. The divisor-square bound itself can be obtained elementarily from $\tau(n)^2\leq\tau_4(n)$ prime-power by prime-power and $\sum_{n\leq Y}\tau_4(n)\leq Y(1+\log Y)^3$; no uncontrolled coefficient bound is being suppressed.
Combining the short, type I and type II terms,
$$
\left|\sum_{n\leq X}\mu(n)e(n\sqrt2)\right|
\ll X^{2/5}\log X+X^{19/20}\log^4X+X^{1/10}=o(X).
$$
Hence
$$
\boxed{\lim_{X\to\infty}\frac1X\left|\sum_{n\leq X}\mu(n)e(n\sqrt2)\right|=0.}
$$
Back to article page
