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For coefficients supported on $N$ consecutive integers,The star restricts to primitive Dirichlet characters. Their finite Fourier transform of a primitive Dirichlet character has normalization of absolute value $\sqrt{q}$. Orthogonality of Dirichlet characters therefore bounds the weighted contribution for one modulus by ${\sum }_{\begin{array}{c}a\mathrm{mod}q\\ (a,q)=1\end{array}}|{\sum }_n{a}_{n}e(an/q){|}^2$. Sum over $q$ and use the exponential-sum large sieve on the distinct reduced fractions of denominators at most $Q$.

A suitable large sieve statement is the following. If ${\alpha }_j$ are separated modulo one by at least $0<\delta \le 1$ and $a_n$ is supported on $N$ consecutive integers, thenThe sharper classical constant $N-1+{\delta }^{-1}$ also suffices, but is unnecessary for the requested bound. This is the exponential-sum large sieve.

Put $M=|A|$, $S(\alpha )={\sum }_{n\in A}e(n\alpha )$, and $A_p(r)=\#\{n\in A:n\equiv r(\mathrm{mod}p)\}$. By the orthogonality of roots of unity,At most $p-k(p)$ of these residue counts are nonzero. The Cauchy-Schwarz inequality gives $M^2\le (p-k(p)){\sum }_r{A}_p(r{)}^2$, and the frequency $h=0$ contributes exactly $M^2$. Thus, if $k(p)<p$,If $k(p)=p$ for some prime, $A$ is empty and the result is immediate.

All fractions $h/p$, $1\le h<p$, $p\le X$, are distinct reduced fractions. Their circular spacing is at least $X^{-2}$, because two distinct fractions of these denominators differ modulo one by at least $1/(p{p}^{\prime })$. Apply the large sieve with $a_n=1_A(n)$ and $\delta =X^{-2}$, and sum the preceding lower bounds:For $M>0$ and a positive sieve denominator, division provesWhen the denominator is zero this asserts no restriction; that degenerate case is interpreted as a vacuous bound.

For the least-nonresidue application, first establish the positive density of power-smooth integers. Fix $\eta =1/k$, with an integer $k\ge 2$, and let $H$ tend to infinity. Choose ordered primes, with repetitions allowed, in the intervalTheir product $d$ satisfies $H^{1-\eta }\le d\le H$. Every number $md\le H$ has $m\le H^{\eta }$, so is $H^{\eta }$-smooth. By partial summation of the Prime number theorem, or the Mertens theorem for reciprocal primes,The number of ordered representations $(m,{\ell }_1,\dots ,{\ell }_k)$ is at leastbecause $\lfloor H/d\rfloor \ge H/(2d)$ for $1\le d\le H$.

This representation count must not be confused with a count of distinct integers. A number at most $H$ has at most $\lfloor 1/(\eta -{\eta }^2)\rfloor$ prime factors in that interval, counted with multiplicity. There are therefore only a bounded number, depending on $\eta$, of ordered $k$-tuples which can occur as divisors of it; the cofactor is then fixed. Dividing by this bounded multiplicity provesfor a fixed ${\kappa }_{\eta }>0$ and all sufficiently large $H$.

Now work on a dyadic prime interval $Q<p\le 2Q$ and let ${\mathcal{E}}_Q$ be the primes there with $n(p)>p^{0.01}$. Take $H=Q^2$ and $\eta =1/200$, so $H^{\eta }=Q^{0.01}$. Let $A$ be the $Q^{0.01}$-smooth integers at most $Q^2$. The density lemma gives $|A|\ge \kappa Q^2$ for an absolute $\kappa >0$.

For each $p\in {\mathcal{E}}_Q$, every prime factor of a member of $A$ is at most $Q^{0.01}\le p^{0.01}<n(p)$ and is smaller than $p$. Hence it is a nonzero quadratic residue modulo $p$. Multiplicativity of the Legendre symbol makes every member of $A$ a nonzero quadratic residue modulo $p$. It therefore avoids all $(p-1)/2$ quadratic-nonresidue classes. In the sieve inequality set $k(p)=(p-1)/2$ at exceptional primes and zero at every other prime up to $X=2Q$. For odd $p$, $k(p)/p\ge 1/3$, soThus each dyadic interval has at most $36/\kappa$ exceptional primes. The finitely many small intervals can be absorbed into this absolute bound. Summing the $O(\log N)$ intervals up to $N$ proves the large-sieve logarithmic exceptional-prime bound for least quadratic nonresidues:The least nonresidue is normally defined for odd primes; any convention at the prime two changes the count by at most one.

For coefficients $b_n$ supported on an interval of length $H$, write $B=\sum b_n$ and $B_p(a)={\sum }_{n\equiv a(\mathrm{mod}p)}{b}_n$. The variance form of the large sieve statesThe constant is absolute. One may take the explicit right side $(Q^2+2\pi H)\sum |b_n{|}^2$, by orthogonality of roots of unity and the exponential-sum large sieve proved in Question 2.

Take $H=\lfloor N^2\rfloor$, $Q=N$, and $b_n$ the indicator function of the $N^{\epsilon }$-smooth numbers up to $N^2$. Applying the given smooth-number density with parameter $\epsilon /2$ giveswith a harmless adjustment of the constant for integer endpoints. If an odd prime $p\le N$ has least quadratic nonresidue $n(p)>N^{\epsilon }$, then $p>N^{\epsilon }$: a quadratic nonresidue always occurs among $1,\dots ,p-1$. Every prime factor of every selected smooth number is thus a nonzero quadratic residue modulo $p$. By the multiplicativity of the Legendre symbol, every selected number is a nonzero quadratic residue modulo $p$.

There are $(p-1)/2$ nonzero quadratic nonresidue classes, and $B_p(a)=0$ on all of them. Their contribution to the variance is at leastIf $R$ denotes the number of exceptional primes, the variance form of the large sieve, with $\sum |b_n{|}^2=B$, yields $R{B}^2\ll N^{2}B$. ConsequentlyThis is the bounded exceptional primes for least quadratic nonresidues argument. Using an interval of length $N^2$ is what matches the $Q^2$ term; an interval of length $N$ would not give a bounded exceptional set.

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.

For coefficients $b_n$ supported on an interval of $H$ consecutive integers, set $B=\sum b_n$ and $B_p(a)={\sum }_{n\equiv a(\mathrm{mod}p)}{b}_n$. ThenThe orthogonality of roots of unity gives $p{\sum }_a|B_p(a)-B/p{|}^2={\sum }_{r=1}^{p-1}|{\sum }_n{b}_{n}e(rn/p){|}^2$. The distinct fractions $r/p$ have circular spacing at least $Q^{-2}$; now apply the exponential-sum large sieve.