= Derivative energy constant for a smooth sieve cutoff
{title2=$c_\phi=\int_0^{1/3}\phi'(u)^2\,du$}
Write $e^u\phi(u)=\int\psi(t)e^{-iut}\,dt$. The double integral $\iint\psi(t)\psi(t')(1+it)(1+it')/(2+i(t+t'))\,dt\,dt'$ equals the displayed constant. Express the denominator as a Laplace integral and each differentiated Fourier factor as $-\phi'(u)$, then use <Fubini's theorem>. For a real cutoff the resulting square is positive even though the <Fourier transform> factors have no <complex conjugation>. The <Cauchy-Schwarz inequality> and $\phi(0)=1$, $\phi(1/3)=0$ give $c_\phi\geq3$.
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