Return to Answer

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+\int_{t}^{t+\varepsilon} (f(t)+(s-t)y,y)d\mu \right] \geq \int_{-\infty}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.

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.

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:

1. 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:

2. 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:

$$ \frac{1}{2}\left[ M\left( a,\left|\frac{a-b}{2} \right|\right)+ M\left( b,\left|\frac{a-b}{2} \right|\right)\right] \leq M\left( \frac{a+b}{2},0\right)? $$ One has to write some Taylor's expansion and check it.

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+\int_{t}^{t+\varepsilon} (f(t)+(s-t)y,y)d\mu \right] \geq \int_{-\infty}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.

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:

1. 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:

2. 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.

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:

1. 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:

2. 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:

$$ \frac{1}{2}\left[ M\left( a,\left|\frac{a-b}{2} \right|\right)+ M\left( b,\left|\frac{a-b}{2} \right|\right)\right] \leq M\left( \frac{a+b}{2},0\right)? $$ One has to write some Taylor's expansion and check it.