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For a prime $p$, the divisor sum defining the real weight is $\varphi (0)-\varphi (\log p/\log X)$. If $p>X^{1/3}$, the second term vanishes and the first is one. Thus $F(p)=1$, while $F(n)\ge 0$ for every integer because $\varphi$ is real and the sum is squared.

Put $h(x)=e^x\varphi (x)$. With the Fourier transform convention in the question, Fourier inversion givesThe function $h$ is smooth and has compact support. Using integration by parts $k$ times, with zero endpoint terms, provesChoose an integer $k\ge A$ to obtain $|\psi (t)|{\ll }_A|t{|}^{-A}$ for $|t|\ge 1$. In particular every polynomially weighted absolute integral of $\psi$ is finite.

For the double integral, useDifferentiating the second formula gives $\int \psi (t)(1+it)e^{-(1+it)u}dt=-{\varphi }^{\prime }(u)$. absolute convergence from the rapid decay permits Fubini's theorem. The derivative energy constant for a smooth sieve cutoff is thereforeThere is no complex conjugate in this integral: the two factors both become $-{\varphi }^{\prime }(u)$, which is real. The constant is positive; in fact the Cauchy-Schwarz inequality applied to $\varphi (1/3)-\varphi (0)=-1$ gives $c_{\varphi }\ge 3$.

Here is the basic structure of the smooth divisor-square sieve asymptotic, with enough uniform estimates to specify the error. Write $L=\log X$ and $R=X^{1/3}$. Expand the square and count multiples of the least common multiple $[d,e]$ in an arbitrary interval $I$ of length $X$:The sum is effectively restricted to $d,e<R$. Each count is $X/[d,e]+O(1)$, independent of the interval's position, so the total error is $O_{\varphi }(X^{2/3})$.

The Fourier representation of a smooth Selberg weight readsHence, with $z^{\prime }=(1+i{t}^{\prime })/L$,The four local terms correspond to the prime dividing neither divisor, only the first, only the second, or both. This is the Euler product for a smoothed divisor-square correlation. Factor it aswhereThe local factors of $H$ are $1+O(p^{-2+4\eta })$ when $|\mathrm{Re}z|,|\mathrm{Re}{z}^{\prime }|\le \eta$ for small fixed $\eta <1/4$. Thus $H$ is a holomorphic function near $(0,0)$, and evaluating each local factor at zero gives exactly one. Consequently $H(0,0)=1$ and $H(z,z^{\prime })=1+O(|z|+|z^{\prime }|)$ there.

The pole expansion $\zeta (1+w)=w^{-1}+O(1)$ now yieldswhen $|t|,|t^{\prime }|\le L^{1/4}$. Integrating the error against the rapidly decreasing transforms gives $O_{\varphi }(L^{-2})$. The complementary tails are harmless uniformly: absolute values in the original divisor series giveAny prescribed power decay of ${\int }_{|t|>L^{1/4}}|\psi (t)|dt$ is available, so these tails and the tails of the limiting kernel are $O_{\varphi }(L^{-2})$ after choosing sufficiently many applications of integration by parts. The same absolute bound justifies the original exchanges of infinite sums and integrals. Using the evaluated double integral,We conclude, uniformly over all interval locations,This proof does not assume that the interval starts near the origin.

Finally choose this one fixed smooth cutoff. primes above $R$ in $(X_0,X_0+X]$ contribute one each to the nonnegative weight, and at most $R$ primes lie below $R$. Thereforefor all sufficiently large $X$, uniformly in $X_0$. The bounded range $2\le X\le X_1$ follows by increasing the fixed constant and using the trivial interval bound $X+1$. Thus the short-interval prime upper bound from a smooth divisor weight is

PutIf $Y<n\le Y+X$ is prime, then $n>D=X^{1/10}$. Since $f$ is supported on $[-1,1]$, the only divisor of $n$ contributing to $W(n)$ is $d=1$, and $W(n)=f(0)=1$. Every summand $W(n{)}^2$ is nonnegative, so

LetSince $g$ is the Fourier transform of $x\mapsto e^{x}f(x)$, the Fourier inversion theorem gives the Fourier representation of a smooth Selberg weightExpanding the square, interchanging the absolutely convergent sums and integrals, and usingwe obtainThe error is $O(D^2)$ because the support of $f$ restricts both divisors to $d_j\le D$. The main factor isand its Euler product is

It remains to estimate the integral using the assumed zeta-factor bound. The transform $g$ is a Schwartz function, since $e^{x}f(x)$ is smooth and compactly supported. We may therefore truncate to $|t_1|,|t_2|\le T$, losing an arbitrarily large negative power of $T$. Uniformly in the needed truncated range, the standard estimates near the pole of the Riemann zeta function giveandAfter multiplication, one net factor $(\log D{)}^{-1}$ remains. The polynomial factors in $t_1,t_2$ are integrable against the rapidly decreasing $g(t_1)g(t_2)$, and choosing $T$ as a sufficiently large power of $\log D$ makes the tails negligible. HenceSince $D=X^{1/10}$ and $D^2=X^{1/5}\ll X/\log X$,This proves the short-interval prime upper bound from a smooth divisor weight.

For a real smooth cutoff $\varphi$ supported on $[-1/3,1/3]$ with $\varphi (0)=1$, define $F_X(n)=({\sum }_{d\mid n}\mu (d)\varphi (\log d/\log X){)}^2$. Its sum over any interval of length $X$ is $c_{\varphi }X/\log X+O_{\varphi }(X/{\log }^{2}X+X^{2/3})$, uniformly in the interval location. Here $c_{\varphi }$ is the derivative energy constant for a smooth sieve cutoff. Expanding the square yields an interval-counting error $O(X^{2/3})$ and an Euler product for a smoothed divisor-square correlation; its zeta function pole provides the main term. Every prime exceeding $X^{1/3}$ has weight one, giving the short-interval prime upper bound from a smooth divisor weight.