Find the minimum of this functional

Question

Can anyone help me with this problem? I have to find the minimum of the following functional:

$F:L^2(0,1)\to\mathbb{R}$, such that $F(u) = \int_0^1 (1+x^2)u^2(x)dx$

on the set: $K=$ {$u \in L^2(0,1)|\int_0^1u(x)dx=1$}.

How can I proceed?

Answer 1

Suppose that $u_0\in K$ is a minimizer. Let $\alpha\in L_2$ be such that $\int_0^1 \alpha(x)dx=0$. Then $u_0+t \alpha\in K$ for all real $t$. Define $$\phi(t)=F(u_0+t\alpha)=\int_0^1 h(x)(u_0+t \alpha)^2(x)dx$$ and note that $\phi'(0)=0$, since $\phi$ is minimized at $t=0$, since $F$ is minimized at $u_0$.

We can compute $$\phi'=\frac{d}{dt}\int_0^1 h(x)[u_0^2(x)+2u_0(x)t\alpha(x)+t^2\alpha(x)^2]dx$$ and $$0=\phi'(0)=2\int_0^1 h(x)u_0(x)\alpha(x)dx.$$ Here, $h(x)=1+x^2$.

So the minimizer satisfies $\int_0^1 h(x)u_0(x)\alpha(x)dx=0$ whenever $\int_0^1 \alpha(x)dx=0$. This yields that $hu_0$ must be constant. To see this, note that the continuous linear functionals $u\mapsto \int_0^1 h(x)u_0(x)u(x)dx$ and $u\mapsto \int_0^1 u(x)dx$ have the same kernels. Here we use the fact that $u_0$ is not identically zero (since it's in $K$). So $u_0$ should be proportional to $1/h(x)$, and the proportionality constant can be determined by the requirement that $\int_0^1 u_0(x)dx=1$.