Let $\mu$ be the Mobius function. Davenport proved that for any $\alpha\in \mathbb{R}$, for any $N\in \mathbb{N}$ and for any $A>0$, there exists a constant $C_A$ such that $$ \left|\sum_{n=1}^{N} \mu(n)e^{2\pi i\alpha}\right|\le C_A\frac{N}{\log^AN}. $$ I would ask whether it holds that for any $\alpha\in \mathbb{R}$, for any $N,M\in \mathbb{N}$ with $M<N$ and for any $A>0$, there exists a constant $C_A$ (which is independent of $M$ and $N$) such that $$ \left|\sum_{n=M}^{N} \mu(n)e^{2\pi i\alpha}\right|\le C_A\frac{N-M}{\log^A(N-M)}. $$
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$\begingroup$ At least heuristically, the answer is no: for numbers $n$ near $x$, we expect to be able to find around $c\log x$ consecutive values of $\mu(n)$ that all equal $0$ or $1$ (with plenty of $1$s) just by randomness considerations; taking $M$ and $N$ to be the endpoints of that interval of length $\gg \log x$ and $\alpha=0$, the LHS is $\gg \log x$ while the RHS is $o(\log x)$. $\endgroup$– Greg MartinCommented May 13, 2018 at 17:11
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