 Solution

Put $C_a=\sqrt{15/(16a^5)}$, with $a>0$. Direct integration confirms the normalization:The kinetic energy and potential energy terms in the dimensionless Rayleigh quotient areThus ${\lambda }_0\le 5/(2a^2)+a^2/7$. Differentiating with respect to $a$ gives $-5/a^3+2a/7=0$, so the unique minimizing width is $a=(35/2{)}^{1/4}$. The two energy contributions are equal there, and the compact parabolic trial bound for the quantum harmonic oscillator isThe trial is continuous and zero at $\pm a$, so its weak first derivative is $-2C_{a}x$ inside and zero outside. The jumps of that derivative do not add delta terms to the first derivative, and its squared norm is finite. The energy-form calculation is therefore valid even though the trial does not belong to the operator domain of the harmonic oscillator. Applying the second-derivative operator naively and omitting its endpoint distributions would be incorrect.