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.

fullcitation. In particular, you omitted the name of the paper's other author, Frank Clarke. $\endgroup$ – Nate Eldredge Mar 9 '15 at 18:45very specificaboutwhich partyou want to ask about, e.g. a specific line or sentence. $\endgroup$ – Nate Eldredge Mar 9 '15 at 19:52