Derivative energy constant for a smooth sieve cutoff ($c_{\varphi }={\int }_0^{1/3}{\varphi }^{\prime }(u{)}^{2}du$)

Write $e^u\varphi (u)=\int \psi (t)e^{-iut}dt$. The double integral $\iint \psi (t)\psi (t^{\prime })(1+it)(1+i{t}^{\prime })/(2+i(t+t^{\prime }))dtd{t}^{\prime }$ equals the displayed constant. Express the denominator as a Laplace integral and each differentiated Fourier factor as $-{\varphi }^{\prime }(u)$, 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 $\varphi (0)=1$, $\varphi (1/3)=0$ give $c_{\varphi }\ge 3$.