The second order differential equation $\Delta v+2v\ln v+(\lambda+1)v=0$ is not difficult to solve in one dimensional case, and this is what the authors of the paper need to proceed by induction. If you have a second order ODE of the type $F(v'',v)=0$ then assuming that $v'(t)=w(v(t))$ for some function $w$, we obtain $v''(t)=w'(v(t))v'(t)=w'(v(t))w(v(t))$. Thus the initial ODE takes the form $F(w'(s)w(s),s)=0$ which is the first order ODE and then you try to solve it.
The way I understand how to deal with functional inequalities (and in particular with Log-Sobolev inequality) goes like this (besides of variational calculus approach when it happens to be difficult). First lets make the following observation:
- The inequality holds for all test functions (i.e., the smooth, compactly supported nonnegative functions).
This is a strict requirement. It should already give you a condition on an integrand with whom you compose test functions. For example, if the inequality $\int_{0}^{1}B(\varphi(t))dt\leq B\left(\int_{0}^{1}\varphi(t)dt\right)$ holds for all test functions $\varphi$ then this is equivalent to the fact that $B(x)$ is a convex function. You may argue in the similar way for the Log-Sobolev inequality:
- Let $g=f^{2}$ then the Log-Sobolev inequality can be rewritten as follows
$$ \int_{\mathbb{R}^{n}} M(g,\|\nabla g\|)d\mu \leq M\left( \int_{\mathbb{R}^{n}}gd\mu,0 \right) \quad \text{for all} \quad g> 0, \quad g\in C^{\infty}_{c}(\mathbb{R}^{n}) \quad(1) $$ where $M(x,y)=x\ln x - \frac{y^{2}}{2x}$ (since $d\mu$ is the standard Gaussian measure we have that $c=1$ in the inequality).
Now we expect that (1) should mimic to some concavity type condition on $M(x,y)$. A subtle observation is that (1) is always implied by the following concavity type condition:
\[M_{y} \leq 0 \quad \text{and} \quad \begin{pmatrix} M_{xx}+\frac{M_{y}}{y} & M_{xy} \\
M_{xy} & M_{yy} \end{pmatrix} \leq 0. \quad (2)\]
Clearly $M(x,y)=x\ln x - \frac{y^{2}}{2x}$ satisfies this condition therefore the Log-Sobolev inequality holds. Also Bobkov's inequality can be recovered in the same way (and many others).
Unfortunately (1) does not imply (2), so it is not if and only if condition. If somebody wants if and only if condition then: Let $n\geq 2$. (2) holds if and only if $P_{t} M(f,\|\nabla f\|)\leq M(P_{t}f, \| \nabla P_{t} f\|)$ for all test functions $f$ and for all $t\geq 0$, where $P_{t}$ is the Ornstein--Uhlenbeck semigroup $$ P_{t} f (x) = \int_{\mathbb{R}^{n}}f(xe^{-t}+y\sqrt{1-e^{-2t}})d\mu(y). $$ And then (1) follows by taking $t\to \infty$.
After reading Ryan O'Donnell's answer I am wondering (but I have never thought about this) whether (2) implies two-point inequality i.e., for Boolean functions and the fact that you can proceed by induction to use CLT:
$$ \frac{1}{2}\left[ M\left( a,\left|\frac{a-b}{2} \right|^{2}\right)+ M\left( b,\left|\frac{a-b}{2} \right|^{2}\right)\right] \leq M\left( \frac{a+b}{2},0\right)? $$ One has to write some Taylor's expansion and check it.