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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.