The following problem optimization problem arose in a project I am working on with a student. I would like to minimize the quantity:


$$ M=\frac{1}{12} + \int\_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right)^2 Q(x)^2 \ dx - 4 \left[ \int\_0^\frac{1}{2} \left( \tfrac{1}{2}-x \right) Q(x) \ dx\right]^2 $$
over all continuously differentiable functions $Q$ subject to the conditions that $Q(\tfrac{1}{2})=0$ and
$$ \int\_0^\frac{1}{2} Q(x)^2 \ dx = 1.$$


Is it possible to explicitly solve such a problem? (The presence of the constant $1/12$ should ensure that $M$ is positive for all $Q$.)



An optimization problem
I'll change the variable $y=\frac 12-x$ to make typing easier. Since, as Peter already observed, the condition $Q(0)=0$ is worthless and since $\frac 1{12}$ is just an additive constant, we are just to minimize $\int\_0^{1/2}y^2Q^2-4\left(\int\_0^{1/2}yQ\right)^2$ under the condition $\int\_0^{1/2}Q^2=1$. Since everything is about pure quadratic forms, we just need to find the least $a>0$ such that $\int\_0^{1/2}y^2Q^2-4\left(\int\_0^{1/2}yQ\right)^2+a^2\int\_0^{1/2}Q^2\ge 0$. Then $-a^2$ ($+\frac 1{12}$, if you want to keep that term) is the answer. That is just a Cauchy-Schwarz type thingy and the answer is given by $\int\_0^{1/2}\frac{4y^2}{y^2+a^2}dy=1$ with $Q$ proportional (with the coefficient to determine from the $\int\_0^{1/2}Q^2=1$ condition) to $\frac {2y}{y^2+a^2}$ (so the condition that is worthless is automatic as well). 


The integral determining $a$ is $2-4a\arctan\frac{1}{2a}$, so we need $a\arctan\frac 1{2a}=\frac 14$, which is something that one has to solve numerically. I've got $a\approx 0.2144889545$.


