minimization of L^2 norm of the second derivative of a probability density I have a question:
Let $\rho$ denote a probability density function defined on $[0,1]$. It is twice-differentiable and has a continuous second derivative. Denote by $M$ the set of all such functions $\rho$ satisfying $\rho(0)=\rho(1)=0$. Define
$$L(\rho)= \sqrt{\int_0^1 \rho''(x)^2dx}$$ 
What is the minimizer?
If $\rho^{\ast}$ is the minimizer, then for any $\rho\in M$ and any $\lambda\in[0,1]$, we have
$$f(0)\leq f(\lambda), \ \ \forall\lambda\in[0,1]$$
where $f(\lambda):=L^2((1-\lambda)\rho^{\ast}+\lambda\rho)$. Then a little calculus gives
$$f'(0)=2\int_0^1(\rho''-\rho''^{\ast})\rho''^{\ast}\geq 0$$
But I don't know how to continue from this inequality, could someone help me? Thanks a lot!
 A: One wants to minimize the functional $U(\varrho)=\int_0^1(\varrho'')^2$ under the constraints that
$\varrho(0)=0$, $\varrho(1)=0$, $\int_0^1\varrho=1$. Assume that $\varrho$ is a minimizer, then for every $\sigma$ such that $\sigma(0)=0$, $\sigma(1)=0$, $\int_0^1\sigma=0$, one knows that $U(\varrho+t\sigma)\geqslant U(\varrho)$, for every $|t|$ small enough. Thus,
$$
\langle\nabla U(\varrho),\sigma\rangle=0.
$$
If $\varrho$ is regular enough,
$$
\langle\nabla U(\varrho),\sigma\rangle=2\int_0^1\varrho''\sigma''=2\left[\varrho''\sigma'-\varrho^{(3)}\sigma\right]_0^1+2\int_0^1\varrho^{(4)}\sigma,
$$
thus,
$$
\varrho''(1)\sigma'(1)-\varrho''(0)\sigma'(0)+\int_0^1\varrho^{(4)}\sigma=0.
$$
If $\varrho^{(4)}$ is not constant, one can find $x\ne y$ such that $\varrho^{(4)}(x)\ne\varrho^{(4)}(y)$, and $\sigma$ such that $\sigma(0)=0$, $\sigma(1)=0$, $\int_0^1\sigma=0$, $\sigma'(0)=0$, $\sigma'(1)=0$, nonzero in neighborhoods of $x$ and $y$ only and of opposite signs on these two neighborhoods. Then $\sigma$ is admissible and contradicts the identity above.
Hence $\varrho^{(4)}$ is constant. A similar reasoning shows that $\varrho'(0)=\varrho'(1)=0$. Thus, there exists some positive constant $c$ such that $\varrho(x)=cx^2(1-x)^2$, and the normalizing condition $\int_0^1\varrho=1$ shows that $c=30$.
All the above is a way to guess the minimizer. Conversely, starting from $\varrho:x\mapsto30x^2(1-x)^2$, one checks that, for every $\sigma$ such that $\sigma(0)=0$, $\sigma(1)=0$, $\int_0^1\sigma=0$, $\sigma'(0)=0$, $\sigma'(1)=0$, 
$$
U(\varrho+t\sigma)=U(\varrho)+t^2U(\sigma)\geqslant U(\varrho).
$$
These conditions on $\sigma$ describe a neighborhood of $\varrho$ in $M$ hence $\varrho$ is indeed a local minimum of $U$ on $M$.
