[proofplan]
We first record the differentiability of the logarithmic [Laplace transform](/page/Laplace%20Transform) on the open exponential-moment interval and compute its derivative by differentiating under the expectation. The entropy identity is then an algebraic rewriting of $\operatorname{Ent}(e^{\beta F})$ in terms of $\Lambda_F$ and $\Lambda_F'$. The one-sided Laplace estimate follows by integrating the differential inequality for $\Lambda_F(\beta)/\beta$ from $0$ to $\beta$. Finally, the negative-parameter estimate is reduced to the positive-parameter case by applying the same argument to the [random variable](/page/Random%20Variable) $-F$.
[/proofplan]
[step:Differentiate the logarithmic Laplace transform on the exponential moment interval]
Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space on which the real-valued random variable $F:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is defined. Define the moment generating function $M_F:I\to(0,\infty)$ by
\begin{align*}
M_F(\gamma)=\mathbb E[e^{\gamma F}].
\end{align*}
Fix $\gamma_0\in I$. Choose $\varepsilon>0$ such that $[\gamma_0-\varepsilon,\gamma_0+\varepsilon]\subset I$. For $\gamma\in[\gamma_0-\varepsilon/2,\gamma_0+\varepsilon/2]$, define the constant $C_{\gamma_0,\varepsilon}:=2/(e\varepsilon)$. If $x\ge 0$, then $x e^{\gamma x}\le C_{\gamma_0,\varepsilon}e^{(\gamma_0+\varepsilon)x}$ because $\sup_{x\ge0}x e^{-\varepsilon x/2}=2/(e\varepsilon)$. If $x<0$, then $|x|e^{\gamma x}\le C_{\gamma_0,\varepsilon}e^{(\gamma_0-\varepsilon)x}$ by the same supremum after setting $y=-x$. Therefore the pointwise bound
\begin{align*}
|F|e^{\gamma F}\le C_{\gamma_0,\varepsilon}\left(e^{(\gamma_0+\varepsilon)F}+e^{(\gamma_0-\varepsilon)F}\right)
\end{align*}
holds on $\Omega$. The right-hand side is integrable by the exponential-moment hypothesis. Hence the [Dominated Convergence Theorem](/theorems/4) justifies differentiating under the expectation locally on $I$, and
\begin{align*}
M_F'(\gamma)=\mathbb E[F e^{\gamma F}]
\end{align*}
for every $\gamma\in I$. Since $M_F(\gamma)>0$, the logarithmic Laplace transform $\Lambda_F=\log M_F$ is differentiable and
\begin{align*}
\Lambda_F'(\gamma)=\frac{\mathbb E[F e^{\gamma F}]}{\mathbb E[e^{\gamma F}]}.
\end{align*}
[guided]
We need a genuine derivative formula, not just a formal calculation. Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space on which $F:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is defined. Define the map $M_F:I\to(0,\infty)$ by assigning to each $\gamma\in I$ the value
\begin{align*}
M_F(\gamma)=\mathbb E[e^{\gamma F}].
\end{align*}
The codomain is $(0,\infty)$ because $e^{\gamma F}>0$ almost surely and the expectation is finite by hypothesis.
Fix $\gamma_0\in I$. Since $I$ is open, there is $\varepsilon>0$ such that the closed interval $[\gamma_0-\varepsilon,\gamma_0+\varepsilon]$ is contained in $I$. The purpose of choosing this smaller interval is to create integrable exponential functions on both sides of $\gamma_0$, which will dominate the derivative of $e^{\gamma F}$.
For $\gamma\in[\gamma_0-\varepsilon/2,\gamma_0+\varepsilon/2]$, the function $x\mapsto |x|e^{\gamma x}$ is controlled by the two endpoint exponentials with an explicit constant. Define $C_{\gamma_0,\varepsilon}:=2/(e\varepsilon)$. If $x\ge0$, then $\gamma\le\gamma_0+\varepsilon/2$, so
\begin{align*}
x e^{\gamma x}\le x e^{(\gamma_0+\varepsilon/2)x}=x e^{-\varepsilon x/2}e^{(\gamma_0+\varepsilon)x}\le C_{\gamma_0,\varepsilon}e^{(\gamma_0+\varepsilon)x}.
\end{align*}
If $x<0$ and $y=-x>0$, then $\gamma\ge\gamma_0-\varepsilon/2$, so
\begin{align*}
|x|e^{\gamma x}=y e^{-\gamma y}\le y e^{-(\gamma_0-\varepsilon/2)y}=y e^{-\varepsilon y/2}e^{-(\gamma_0-\varepsilon)y}\le C_{\gamma_0,\varepsilon}e^{(\gamma_0-\varepsilon)x}.
\end{align*}
Combining the two cases gives, for every $x\in\mathbb R$,
\begin{align*}
|x|e^{\gamma x}\le C_{\gamma_0,\varepsilon}\left(e^{(\gamma_0+\varepsilon)x}+e^{(\gamma_0-\varepsilon)x}\right).
\end{align*}
Substituting $x=F(\omega)$ gives the pointwise bound
\begin{align*}
|F(\omega)|e^{\gamma F(\omega)}\le C_{\gamma_0,\varepsilon}\left(e^{(\gamma_0+\varepsilon)F(\omega)}+e^{(\gamma_0-\varepsilon)F(\omega)}\right)
\end{align*}
for every $\omega\in\Omega$ outside a null set on which $F$ may be represented arbitrarily. The right-hand side is integrable with respect to $\mathbb P$ because $\gamma_0+\varepsilon$ and $\gamma_0-\varepsilon$ belong to $I$.
Now differentiate the map $\gamma\mapsto e^{\gamma F(\omega)}$ pointwise. Its derivative is $F(\omega)e^{\gamma F(\omega)}$, and the domination just obtained is integrable locally in $\gamma$. Therefore the [Dominated Convergence Theorem](/theorems/4) permits differentiation to pass through the expectation, giving
\begin{align*}
M_F'(\gamma)=\mathbb E[F e^{\gamma F}]
\end{align*}
for every $\gamma\in I$. Since $M_F(\gamma)>0$, the logarithm is differentiable at $M_F(\gamma)$, and the chain rule gives
\begin{align*}
\Lambda_F'(\gamma)=\frac{M_F'(\gamma)}{M_F(\gamma)}=\frac{\mathbb E[F e^{\gamma F}]}{\mathbb E[e^{\gamma F}]}.
\end{align*}
[/guided]
[/step]
[step:Rewrite the entropy as the derivative of $\Lambda_F(\beta)/\beta$]
Fix $\beta\in I\setminus\{0\}$. Since $\beta\in I$, the previous step gives $\mathbb E[|F|e^{\beta F}]<\infty$, so $\mathbb E[e^{\beta F}\log(e^{\beta F})]$ is finite. By the definition of entropy,
\begin{align*}
\operatorname{Ent}(e^{\beta F})=\mathbb E[e^{\beta F}\beta F]-\mathbb E[e^{\beta F}]\log\mathbb E[e^{\beta F}].
\end{align*}
Using $\Lambda_F(\beta)=\log\mathbb E[e^{\beta F}]$ and $\mathbb E[F e^{\beta F}]=\Lambda_F'(\beta)\mathbb E[e^{\beta F}]$, we obtain
\begin{align*}
\operatorname{Ent}(e^{\beta F})=\mathbb E[e^{\beta F}]\left(\beta\Lambda_F'(\beta)-\Lambda_F(\beta)\right).
\end{align*}
Dividing by $\mathbb E[e^{\beta F}]>0$ gives
\begin{align*}
\frac{\operatorname{Ent}(e^{\beta F})}{\mathbb E[e^{\beta F}]}=\beta\Lambda_F'(\beta)-\Lambda_F(\beta).
\end{align*}
Finally, differentiating the quotient $\Lambda_F(\beta)/\beta$ for $\beta\ne0$ gives
\begin{align*}
\beta^2\frac{d}{d\beta}\left(\frac{\Lambda_F(\beta)}{\beta}\right)=\beta\Lambda_F'(\beta)-\Lambda_F(\beta).
\end{align*}
This proves the identity.
[/step]
[step:Integrate the positive-parameter entropy inequality]
Assume $b>0$, $(0,b)\subset I$, and
\begin{align*}
\operatorname{Ent}(e^{\beta F})\le c\beta^2\mathbb E[e^{\beta F}]
\end{align*}
for every $\beta\in(0,b)$. For $\beta\in(0,b)$, the identity already proved gives
\begin{align*}
\frac{d}{d\beta}\left(\frac{\Lambda_F(\beta)}{\beta}\right)\le c.
\end{align*}
Define $G:(0,b)\to\mathbb R$ by
\begin{align*}
G(\beta)=\frac{\Lambda_F(\beta)}{\beta}.
\end{align*}
Since $\Lambda_F(0)=0$ and $\Lambda_F'(0)=\mathbb E[F]$, we have
\begin{align*}
\lim_{\beta\downarrow0}G(\beta)=\mathbb E[F].
\end{align*}
Integrating $G'(\gamma)\le c$ over $\gamma\in(t,\beta)$ with $0<t<\beta<b$ gives
\begin{align*}
G(\beta)-G(t)\le c(\beta-t).
\end{align*}
Letting $t\downarrow0$ yields
\begin{align*}
\frac{\Lambda_F(\beta)}{\beta}-\mathbb E[F]\le c\beta.
\end{align*}
Multiplying by $\beta>0$ gives
\begin{align*}
\Lambda_F(\beta)\le \beta\mathbb E[F]+c\beta^2.
\end{align*}
[/step]
[step:Reduce negative parameters to the positive case for $-F$]
Assume now that $a>0$, $(-a,a)\subset I$, and the entropy bound holds for every $\beta\in(-a,a)$. The positive-parameter conclusion just proved gives
\begin{align*}
\Lambda_F(\beta)\le \beta\mathbb E[F]+c\beta^2
\end{align*}
for every $\beta\in(0,a)$.
To handle $\beta\in(-a,0)$, define the random variable $H:\Omega\to\mathbb R$ by
\begin{align*}
H=-F.
\end{align*}
Define the open interval $I_H:=(-a,a)$. For every $\gamma\in I_H$, we have $-\gamma\in(-a,a)\subset I$, so
\begin{align*}
\mathbb E[e^{\gamma H}]=\mathbb E[e^{-\gamma F}]<\infty.
\end{align*}
Thus $H$ satisfies the exponential-moment hypothesis on $I_H$. For $s\in(0,a)$,
\begin{align*}
\operatorname{Ent}(e^{sH})=\operatorname{Ent}(e^{-sF})\le cs^2\mathbb E[e^{-sF}]=cs^2\mathbb E[e^{sH}].
\end{align*}
Define the logarithmic Laplace transform $\Lambda_H:(0,a)\to\mathbb R$ by
\begin{align*}
\Lambda_H(s)=\log\mathbb E[e^{sH}].
\end{align*}
Applying the positive-parameter estimate to $H$ gives
\begin{align*}
\Lambda_H(s)\le s\mathbb E[H]+cs^2.
\end{align*}
For $\beta=-s$, this becomes
\begin{align*}
\Lambda_F(\beta)\le \beta\mathbb E[F]+c\beta^2.
\end{align*}
The same inequality is therefore valid for every $\beta\in(-a,a)$, with equality at $\beta=0$ because $\Lambda_F(0)=0$.
[/step]
[step:Center the logarithmic Laplace transform]
For every $\beta\in(-a,a)$, the centered logarithmic Laplace transform satisfies
\begin{align*}
\log\mathbb E[e^{\beta(F-\mathbb E[F])}]=\log\left(e^{-\beta\mathbb E[F]}\mathbb E[e^{\beta F}]\right)=\Lambda_F(\beta)-\beta\mathbb E[F].
\end{align*}
Using the estimate obtained in the previous step,
\begin{align*}
\Lambda_F(\beta)-\beta\mathbb E[F]\le c\beta^2.
\end{align*}
Thus
\begin{align*}
\log\mathbb E[e^{\beta(F-\mathbb E[F])}]\le c\beta^2
\end{align*}
for every $\beta\in(-a,a)$, as required.
[/step]
