Prove that $0\le\int_0^1\log(u){\rm d}x+\frac1{2\pi^2}\int_0^1\frac1{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x$

Question

I want to prove a special case of a functional inequality stated in a book. To be specific, let $u(x)$ being a positive, differentiable function on $[0,1]$ with unit mass (i.e., $\int_0^1 u(x){\rm d}x = 1$), then we want to show that $$0 \leq \int_0^1 \log(u){\rm d}x + \frac{1}{2\pi^2}\int_0^1 \frac{1}{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x.$$ In a book, the author resorts to the construction of certain solutions of one-dimensional heat equation and used the eigen-expansion (hence the arguments are very lengthy and involved). Since I am only interested in this special inequality stated above, may I know whether there exists a simpler and more elegant proof ?

Answer 1

Here are some observations. Define the functional $$F(x,u,u')=\int_0^1\log u+\frac{u'^2}{2\pi^2u^2}\,dx$$ where $u$ satisfies the integral constraint $1=\int_0^1u(x)\,dx$ (so $u$ is a p.d.f.). We will assume that $F$ is finite over its stationary path. Consequently, define the functional to be minimised $$G(x,u,u')=\int_0^1\log u+\frac{u'^2}{2\pi^2u^2}+\lambda u\,dx$$ where $\lambda$ is a Lagrange multiplier. The Euler-Lagrange equation $G_{u'}'=G_u$ yields $$\frac d{dx}\frac{u'}{\pi^2u^2}=\lambda+\frac1u\implies uu''-2u'^2-\pi^2u^2(\lambda u+1)=0.$$ Let $v=u'^2$. This gives $v^*=2u'(u')^*=2u''$ where $v^*=dv/du$ so the differential equation becomes a first-order ODE $$v^*-\frac4uv=2\pi^2u(\lambda u+1).$$ It is easy to derive the solution $v=\pi^2(Cu^4-2\lambda u^3-u^2)$ for some constant $C$ which leads to $$u'=\pm\pi u\sqrt{Cu^2-2\lambda u-1}.$$ Separation of variables yields the solution $$\arctan\frac{\lambda u+1}{\sqrt{Cu^2-2\lambda u-1}}=\pi x+D$$ for some constant $D$ and after simplification, $$((\lambda^2+C)\cos^2(\pi x+D)-C)u^2+2\lambda u+1=0.$$ Solving the quadratic gives the stationary path $$u_s(x)=\frac{-\lambda-\sqrt{\lambda^2+C}\sin(\pi x+D)}{(\lambda^2+C)\cos^2(\pi x+D)-C}=-\frac1{\lambda+\sqrt{\lambda^2+C}\sin(\pi x+D)}$$ where the negative root is chosen to satisfy $u(x)>0$ on $[0,1]$ and $|F|<\infty$. Substituting this into the integral constraint gives $$1=\frac2{\pi\sqrt C}\left[\operatorname{arctanh}\frac{\sqrt{\lambda^2+C}+\lambda\tan(\pi x+D)/2}{\sqrt C}\right]_0^1$$ so that $$\sqrt{\lambda^2+C}=-\frac{\sqrt C}{\cos D\tanh(\pi\sqrt C/2)}.$$ Hence \begin{align}\min F(x,u,u')&=\int_0^1\log u_s+\frac12(Cu_s^2-2\lambda u_s-1)\,dx\\&=-\lambda-\frac12+\int_0^1\log u_s+\frac C2u_s^2\,dx\\&=-\frac{\lambda+1}{2}+\frac{\lambda\sqrt{\lambda^{2}+C}\cos D}{\pi\left(\left(\lambda^{2}+C\right)\cos^2D-C\right)}+\int_0^1\log u_s\,dx\\&=\frac{\Lambda-1}2+\frac{\Lambda\sinh\pi\sqrt C}{2\pi\sqrt C}-\int_0^1\log\left(\Lambda+\frac{\sqrt C\sin(\pi x+D)}{\cos D\tanh(\pi\sqrt C/2)}\right)\,dx\end{align} where $C,\Lambda:=-\lambda>0$.

Answer 2

OK, after looking back at the various different proofs of the classical log Sobolev inequality, there is at least one way we can vanquish this. It turns out it is just a special case of the inequality stated (and proved) in section 2.1. of the lecture notes "Entropy Methods and Related Functional Inequalities" written by Daniel Matthes. So here is the hammer:
Assume that $\phi:\mathbb{R}_+ \to \mathbb{R}$ is convex such that $(\phi'')^{-\frac{1}{2}}$ is concave, let $\psi$ be such that $(\psi'(s))^2 = \phi''(s)$. Then we have that $$ \int_0^1 \phi(u){\rm d}x - \phi\left(\int_0^1 u{\rm d}x\right) \leq \frac{1}{2\pi^2}\int_0^1 (\psi(u)_x)^2{\rm d}x \quad (*)$$ holding for all smooth, positive functions $u$ on $[0,1]$. Notice that if $\phi(s) = \frac 12 s^2$ and $\psi(s) = s$, $(*)$ becomes the usual Poincare inequality $$\int_0^1 u^2{\rm d}x - \left(\int_0^1 u{\rm d}x\right)^2 \leq \frac{1}{\pi^2} \int_0^1 (\partial_x u)^2{\rm d}x$$ in Daniel's notes the choice is $\phi(s) = s\log(s)$ with $\psi(s) = 2s^{\frac 12}$, leading to the well-known log Sobolev inequality, here we just need to take $\phi(s) = -\log(s)$ with $\psi(s) = \log(s)$, and the game is over. Remark: I definitely welcome any other different approaches!