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If ${\theta }_r$ have circular spacing at least $\delta$, thenMultiply the exponential sum by $e(-Mt)$, apply the Sobolev–Gallagher inequality on disjoint arcs of length $\delta$, and sum. The finite-interval Parseval identities and Cauchy-Schwarz inequality bound the derivative contribution by $2\pi N\sum |a_n{|}^2$.

Let $K(\Delta )={\max }_t\#\{r:\Vert {\theta }_r-t\Vert \le \Delta /2\}$. Without a separation assumption,For $0<\Delta \le 1$, each point belongs to at most $K(\Delta )$ integration arcs in the Sobolev–Gallagher inequality. For $\Delta \ge 1$, $K(\Delta )$ is the total number of points, and the Cauchy-Schwarz inequality bound $|S{|}^2\le N\sum |a_n{|}^2$ suffices.

The points are $\delta$-spaced if their circular spacing satisfies $\Vert {\theta }_r-{\theta }_s\Vert \ge \delta$ for $r\ne s$, where $\Vert t\Vert$ is distance to the nearest integer. Ordinary distance on the real line would be insufficient because the complex exponential is periodic.

Let $S(t)={\sum }_{M<n\le M+N}{a}_{n}e(nt)$ and $F(t)=e(-Mt)S(t)$. Multiplication by this unit-modulus factor leaves $|S(t)|$ unchanged and places the frequencies of $F$ in $1,\dots ,N$. Put $E=\sum |a_n{|}^2$. The permitted Sobolev–Gallagher inequality, in the form needed here, isFor $0<\delta \le 1$, the arcs about the ${\theta }_r$ have disjoint interiors on the circle group. Summing and applying the Cauchy-Schwarz inequality givesThe Cauchy-Schwarz inequality here follows by expanding $0\le \int |u-cv{|}^2$ and minimizing over $c\in \mathbb{C}$. For completeness, the finite-interval Parseval identities follow by expanding the squares: ${\int }_0^{1}e(kt)dt$ is one at $k=0$ and zero at every other integer $k$. Thus $\int |F{|}^2=E$ and $\int |F\prime {|}^2\le 4{\pi }^2{N}^{2}E$. We obtain the exponential-sum large sieve boundIf $\delta >1$, there is at most one point, and the direct Cauchy-Schwarz inequality bound $|S{|}^2\le NE$ proves the same assertion.

Use the same Sobolev–Gallagher inequality on arcs of length $\Delta$. They now overlap, but every point belongs to at most $K(\Delta )$ arcs, by the definition of that local multiplicity. For $0<\Delta \le 1$, summing the integrals therefore givesThis is the local-multiplicity large sieve. If $\Delta \ge 1$, every point is within circular distance $\Delta /2$ of every center, so $K(\Delta )=R$. The direct Cauchy-Schwarz inequality bound ${\sum }_r|S({\theta }_r){|}^2\le RNE$ gives the requested estimate in this remaining case.