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

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 ?

• Thanks for your comment. The fact that log Sobolev inequalities has at least "18" different proofs (based on optimal transport, isoperimetric inequality, and etc) really tells me that you will have different route towards the advertised conclusion, so I specifically want to "avoid" the usual proof using eigen-expansions. Mar 4, 2021 at 17:37
• No worries! I am glad that you can possibly help me on this problem Mar 4, 2021 at 17:44
• Are Fourier expansions OK? Mar 4, 2021 at 18:17
• Look at the proof of en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions Isn't it beautiful? Now try proving it without Fourier series. This is of the same nature to the inequality you discuss, though simpler Mar 4, 2021 at 23:30
• @TeresaLisbon Thanks! No worries! I am glad to discuss with you as well! Mar 10, 2021 at 17:09

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!

• A hammer and a half, but I like hammers! +1 Mar 5, 2021 at 12:13
• Thanks Teresa! I will welcome any other different proofs Mar 5, 2021 at 17:40
• Sure, still taking a look through your reference and some of mine. Not looking promising right now, but I will still have a good go at this one. Mar 5, 2021 at 17:43

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 $$C,\Lambda:=-\lambda>0$$.

• Thanks! But at the end we want to show that $\min F(x,u,u') \geq 0$ right? Does the last identity shows this point? Mar 9, 2021 at 17:40
• @FeiCao Unfortunately not, as I don't see a way to evaluate the log-sine integral. As this shows, standard use of calculus of variations may not always be able to derive the optimum of a functional. It might be an interesting discussion in itself to show the inequality $$\int_0^1\log\left(\Lambda+\frac{\sqrt C\sin(\pi x+D)}{\cos D\tanh(\pi\sqrt C/2)}\right)\,dx\le\frac{\Lambda-1}2+\frac{\Lambda\sinh\pi\sqrt C}{2\pi\sqrt C}$$ directly. Mar 10, 2021 at 10:41
• Thanks! I agree the last inequality deserves an separate and detailed discussions! Since no one else answered it, I am going to accept your answer Mar 10, 2021 at 17:11
• @FeiCao I have posted the inequality question here (math.stackexchange.com/questions/4058979/…) with your question linked. Mar 12, 2021 at 12:05
• Thanks! Appreciated! Mar 18, 2021 at 1:29