 Solution

Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.