Gross's log Sobolev inequality proof with variational calculus? For $f\in C^{1}(\mathbb{R}^{n})$, Gross's logarithmic Sobolev inequality says that
$$\int f^{2} \log f^{2}\,d\mu -\int f^{2}\,d\mu \log\left(\int f^{2}\,d\mu\right)\leq \frac{2}{c}\int |\nabla f|^{2}d\mu,$$
where $d\mu=\frac{1}{\pi^{n/2}}e^{-|x|^{2}}dx$ for some $c>0$. 
Any suggestions on a variational type proof? Can someone explain the variational argument proposed in the linked paper at pg 1268?
Attempt 
Proof
We first prove it for $n=1$ and then finish by induction. We may assume $\int u^{2}d\mu=1$ and let $f(t)=v e^{t^{2}/2}$ then it suffices to prove
$$J(v):=\int_{0}^{\infty} \left(\frac{|\nabla v|^{2}}{2}-v^{2}\ln(|v|)\right)\,dt\geq \frac{\sqrt{\pi}}{4}$$
constrained to $\int_{0}^{\infty} v^{2}\,dt=\frac{\sqrt{\pi}}{2}$. But $\Delta v+2v\ln(v)+(\lambda+1)v=0$, where $\lambda$ is the Lagrange multiplier, doesn't look easy to solve.
In the paper below, a different variational calculus argument is proposed, which I still don't understand. 

Adams, R. A.; Clarke, Frank H.
  "Gross's logarithmic Sobolev inequality: a simple proof."
  Amer. J. Math. 101 (1979), no. 6, 1265–1269. MR 548880 DOI 10.2307/2374139

Thank you
 A: Probably you mean Gross's Log-Sobolev Inequality?  I cannot help with understanding any variational calculus proofs, but there are other proofs I find easier to understand.
Actually, I prefer obtaining the Hypercontractive Inequality first, $\|P_t f\|_q \leq \|g\|_p$ provided $e^{-2t} \leq \frac{p-1}{q-1}$, and then recovering Log-Sobolev by taking $q = 2$, squaring both sides, and differentiating at $t = 0$.  (See, e.g., exercise 10.23 at http://analysisofbooleanfunctions.org )  As for the Hypercontractive Inequality itself, the most traditional short way to prove it (due to Gross, but also independently to Bonami previously) is to prove it first for Boolean functions (by induction on $n$) and then to pass to the Gaussian setting using the Central Limit Theorem.  Janson's book Gaussian Hilbert Spaces (Chapter 5) does this nicely (or you can see Chapters 9, 10, 11 at the aforementioned web site).  One can also prove Log-Sobolev directly by induction in the Boolean case (see, e.g., Exercise 10.26).  
If one only cares about the Gaussian setting, there are other short routes to Hypercontractivity (and hence Log-Sobolev).  There is the "Neveu method"; although it's short, it uses stochastic calculus and is thus not very elementary.  On the other hand, a very direct and simple proof is given in Ledoux's recent paper "Remarks on Gaussian Noise Sensitivity...", based on early ideas by Hu and by Mossel--Neeman.
A: The following should answer to your question: if you denote $f^{2}=g$ then the log-Sobolev inequality can be rewritten as follows:
$$
\int_{\mathbb{R}^{n}} \left( g \ln g - \frac{1}{2c}\frac{|\nabla g|^{2}}{g}  \right)d\mu \leq \left( \int_{\mathbb{R}^{n}} g d\mu \right) \ln \left(\int_{\mathbb{R}^{n}} g d\mu  \right).
$$
Consider the case $n=1$. Lets start from a slightly general  optimization problem:
\begin{align*}
B(t,x,z) \stackrel{\mathrm{def}}{=} \sup_{f \in C^{1}(\mathbb{R})}\left\{\int_{-\infty}^{t} F(f,f')d\mu, \;  f(t)=x, \; \int_{-\infty}^{t} f d\mu=z \right\}.  \quad (1)
\end{align*}
where $d\mu=\varphi(x) dx$ is a nice measure with density $\varphi(x)$ (in your case it will be the Gaussian measure), $F(x,y)$ is a fixed  smooth enough function (in your case  $F(x,y)=x\ln x - \frac{1}{2c} \frac{y^{2}}{x}$).
Clearly if you find explicitly $B(t,x,z)$ then log-Sobolev inequality simply is a statement that 
$$
B(\infty, x, z) \leq z \ln z \quad \forall x \geq 0
$$
for the function $F(x,y)$ mentioned above.  But how to find $B$? 
It is the fundamental principle in optimization theory that $B$ should satisfy a Hamilton--Jacobi--Bellman equation. I will briefly summarize it in the following two lemmas. 
The next lemma shows that sometimes you do not have to find $B$ precisely. 
Lemma
Let $M(t,x,z) :\mathbb{R}^{3} \to \mathbb{R}$ be a $C^{1}$ function. If 
\begin{align*}
F(x,y) \varphi(t) \leq M_{t} + y M_{x}+x\varphi(t) M_{z}  \quad \forall t,x,z,y \in \mathbb{R} \quad (2)
\end{align*}
then 
\begin{align*}
\int_{\mathbb{R}} F(f,f')d\mu \leq M(-\infty, 0, 0) - M\left(\infty, 0, \int_{\mathbb{R}} fd\mu\right)
\end{align*}
for any smooth compactly supported function $f$. 
Proof
Indeed,
\begin{align*}
&0 \geq \int_{t_{1}}^{t_{2}}\left[F(f(t),f'(t))\varphi(t) - \frac{d}{dt}M\left(t,f(t), \int_{-\infty}^{t} f d\mu \right) \right]dt=\\
&\int_{t_{1}}^{t_{2}}F(f,f')d\mu-\left[M\left( t_{2},f(t_{2}), \int_{-\infty}^{t_{2}}fd\mu\right) - M\left(t_{1}, f(t_{1}), \int_{-\infty}^{t_{1}}d\mu \right) \right].
\end{align*}
And the rest follows by taking $t_{1} \to -\infty$ and $t_{2} \to +\infty$. $\square$ 
The next lemma shows that actually $B$ defined in (1) does satisfy (2)
Lemma
We have 
\begin{align*}
F(x,y) \varphi(t) \leq B_{t} + y B_{x}+x\varphi(t) B_{z}  \quad \forall t,x,z,y \in \mathbb{R}  \quad (3)
\end{align*}
Proof
Take a time $t+\varepsilon$ and consider an optimizer $f$ on the interval $(-\infty, t)$. Next extend it as $f(s) = f(t)+(s-t)y$ on the interval $[t,t+\varepsilon]$. Let $z(t)=\int_{-\infty}^{t} f d\mu$.  Then we have 
\begin{align*}
&B\left[t+\varepsilon, f(t)+\varepsilon y, z(t)+\int_{t}^{t+\varepsilon} (f(t)+(s-t)y,y)d\mu \right] \geq \int_{-\infty}^{t+\varepsilon}F(f,f')d\mu=\\
&B(t,f(t),z)+\int_{t}^{t+\varepsilon} F(f(t)+(s-t)y,y)d\mu
\end{align*}
Moving $B(t,f(t),z(t))$ to the left hand side of the inequality, dividing everything by $\varepsilon$, sending $\varepsilon \to 0$, and comparing the first order terms we obtain (3 at the point $(t,f(t),z(t),y)$. Since this point can be chosen to be arbitrary we obtain the claim. $\square$
The last lemma looks very convincing but it still requires some justifications, for example, why the optimizer $f(t)$ exists? These are deep questions and we are not going to talk about this right now. 
Thus we have obtained almost a PDE on $B$ (see (3)). Clearly (3) implies that 
\begin{align*}
B_{t}+x\varphi(t) B_{z} + \inf_{y}\{ B_{x} y - F(x,y)\varphi(t)\}\geq 0 \quad (4)
\end{align*}
The last observation is that inequality (4) should be equality, otherwise we could slightly perturb $B$ and make it smaller (this also requires further justifications). 
Thus we have finally arrived to Hamilton--Jacobi--Bellman PDE. 
$$
B_{t}+x\varphi(t) B_{z} + \inf_{y}\{ B_{x} y - F(x,y)\varphi(t)\}=0 \quad (5)
$$
and any solution of (5)  or supersolution (i.e., the one which satisfies (4)) gives rise to the functional inequality 
$$
\int_{\mathbb{R}} F(f,f')d\mu \leq B(-\infty, 0, 0)-B\left(\infty,0, \int_{\mathbb{R}} f d\mu\right) \quad (6)
$$
It's funny isn't it? The measure $d\mu$ does not matter, $F(x,y)$ does not matter.. 
In the paper you mentioned I would guess that the authors guessed from Euler--Lagrange (or new from Gross' paper) the optimizers for the log-Sobolev inequality in (1) where $F(x,y)=x\ln x - \frac{1}{2c} \frac{y^{2}}{x}$ and they just plugged them into (1) and found $B$. Then it was pretty straightforward to check (4) and obtain (6). 
For me this looks like a "cheating" :D. The paper is nice and it is very nice application of Control Theory in this field. But  I believe one should start from solving  Hamilton--Jacobi--Bellman PDE (5) which is the first order nonlinear PDE, so the characteristic method should work.
A: Gross's log-Sobolev inequality is equivalent to Stam's inequality (which can be used to prove the Shannon's entropy power inequality). This inequality can in turn be interpreted as giving a sharp lower bound of the Shannon entropy of a random vector in terms of its Fisher information. Moreover, equality holds if and only if the random vector is Gaussian. So this is a variational formulation of the log-Sobolev inequality.
See for example, http://www.math.univtoulouse.fr/~ledoux/Logsobwpause.pdf.
I believe the book Elements of Information Theory by Cover and Thomas does at least the 1-dimensional case.
