[proofplan]
We encode the exponential moments in the logarithmic moment generating function $\psi(\lambda)=\log\mathbb E[e^{\lambda X}]$. The integrability assumptions justify differentiating $\psi$ and rewriting the entropy bound as a differential inequality for $\psi(\lambda)/\lambda$. Integrating this inequality from $0$ to $\lambda$ gives the centered Laplace bound. The one-sided tail estimate then follows from the elementary indicator bound applied to $e^{\lambda(X-\mathbb E[X])}$ and optimization over $\lambda\geq 0$.
[/proofplan]
[step:Define the logarithmic moment generating function and justify differentiability]
Since the assumptions at $\lambda=0$ give $\mathbb E[|X|]<\infty$, the mean $\mathbb E[X]$ is finite. Define the moment generating function $M:[0,\infty)\to(0,\infty)$ by
\begin{align*}
M(\lambda)=\mathbb E[e^{\lambda X}].
\end{align*}
Define the logarithmic moment generating function $\psi:[0,\infty)\to\mathbb R$ by
\begin{align*}
\psi(\lambda)=\log M(\lambda).
\end{align*}
Fix $\lambda_0>0$. Choose $\delta\in(0,\lambda_0)$. For $\lambda\in(\lambda_0-\delta,\lambda_0+\delta)$, the pointwise derivative of $\lambda\mapsto e^{\lambda X}$ is $X e^{\lambda X}$, and
\begin{align*}
|X|e^{\lambda X}\leq |X|e^{(\lambda_0+\delta)X}\mathbb 1_{\{X\geq 0\}}+|X|e^{(\lambda_0-\delta)X}\mathbb 1_{\{X<0\}}.
\end{align*}
The right-hand side is integrable because $\mathbb E[|X|e^{(\lambda_0+\delta)X}]<\infty$ and $\mathbb E[|X|e^{(\lambda_0-\delta)X}]<\infty$. The difference quotients for $r\mapsto e^{rX}$, with $r\in(\lambda_0-\delta,\lambda_0+\delta)$, converge pointwise to $Xe^{\lambda X}$ and are bounded by the same integrable [random variable](/page/Random%20Variable) after possibly shrinking the interval around the fixed point $\lambda$. Hence the [Dominated Convergence Theorem](/theorems/4) applied to these difference quotients gives, for every $\lambda>0$,
\begin{align*}
M'(\lambda)=\mathbb E[Xe^{\lambda X}].
\end{align*}
Since $M(\lambda)>0$, the chain rule gives
\begin{align*}
\psi'(\lambda)=\frac{\mathbb E[Xe^{\lambda X}]}{\mathbb E[e^{\lambda X}]}.
\end{align*}
At the origin, for $0<\lambda\leq 1$,
\begin{align*}
\left|\frac{e^{\lambda X}-1}{\lambda}\right|\leq |X|e^X\mathbb 1_{\{X\geq 0\}}+|X|\mathbb 1_{\{X<0\}},
\end{align*}
and the right-hand side is integrable. The [Dominated Convergence Theorem](/theorems/4) gives
\begin{align*}
\lim_{\lambda\downarrow 0}\frac{M(\lambda)-1}{\lambda}=\mathbb E[X].
\end{align*}
Since $M(0)=1$ and $\log(1+h)/h\to 1$ as $h\to 0$, it follows that
\begin{align*}
\lim_{\lambda\downarrow 0}\frac{\psi(\lambda)}{\lambda}=\mathbb E[X].
\end{align*}
[/step]
[step:Rewrite the entropy hypothesis as a differential inequality]
For $\lambda>0$, the random variable $e^{\lambda X}$ is non-negative and satisfies
\begin{align*}
\mathbb E[e^{\lambda X}]<\infty
\end{align*}
and
\begin{align*}
\mathbb E[e^{\lambda X}\log(e^{\lambda X})]=\lambda\mathbb E[Xe^{\lambda X}]<\infty.
\end{align*}
Using the definition of entropy and $M(\lambda)=e^{\psi(\lambda)}$, we compute
\begin{align*}
\operatorname{Ent}(e^{\lambda X})=\lambda\mathbb E[Xe^{\lambda X}]-M(\lambda)\log M(\lambda).
\end{align*}
Substituting $\mathbb E[Xe^{\lambda X}]=M(\lambda)\psi'(\lambda)$ and $\log M(\lambda)=\psi(\lambda)$ gives
\begin{align*}
\operatorname{Ent}(e^{\lambda X})=e^{\psi(\lambda)}\bigl(\lambda\psi'(\lambda)-\psi(\lambda)\bigr).
\end{align*}
The entropy hypothesis therefore implies
\begin{align*}
e^{\psi(\lambda)}\bigl(\lambda\psi'(\lambda)-\psi(\lambda)\bigr)\leq \frac{C\lambda^2}{2}e^{\psi(\lambda)}.
\end{align*}
Dividing by the positive quantity $\lambda^2e^{\psi(\lambda)}$ gives
\begin{align*}
\frac{\lambda\psi'(\lambda)-\psi(\lambda)}{\lambda^2}\leq \frac{C}{2}.
\end{align*}
Since
\begin{align*}
\left(\frac{\psi(\lambda)}{\lambda}\right)'=\frac{\lambda\psi'(\lambda)-\psi(\lambda)}{\lambda^2},
\end{align*}
we have, for every $\lambda>0$,
\begin{align*}
\left(\frac{\psi(\lambda)}{\lambda}\right)'\leq \frac{C}{2}.
\end{align*}
[guided]
The entropy assumption is useful because it contains exactly the combination $\lambda\psi'(\lambda)-\psi(\lambda)$, which is the numerator in the derivative of $\psi(\lambda)/\lambda$. We now make that identity explicit.
Recall the definitions used in this step: the moment generating function is the map $M:[0,\infty)\to(0,\infty)$ defined by $M(\lambda)=\mathbb E[e^{\lambda X}]$, and the logarithmic moment generating function is the map $\psi:[0,\infty)\to\mathbb R$ defined by $\psi(\lambda)=\log M(\lambda)$.
Fix $\lambda>0$. The random variable $e^{\lambda X}$ is non-negative. Its expectation is finite by hypothesis. Also,
\begin{align*}
e^{\lambda X}\log(e^{\lambda X})=\lambda X e^{\lambda X},
\end{align*}
and the integrability of $X e^{\lambda X}$ follows from $\mathbb E[|X|e^{\lambda X}]<\infty$. Hence the entropy expression is legitimate. By the definition of entropy,
\begin{align*}
\operatorname{Ent}(e^{\lambda X})=\mathbb E[e^{\lambda X}\log(e^{\lambda X})]-\mathbb E[e^{\lambda X}]\log\mathbb E[e^{\lambda X}].
\end{align*}
Since $\log(e^{\lambda X})=\lambda X$, this becomes
\begin{align*}
\operatorname{Ent}(e^{\lambda X})=\lambda\mathbb E[Xe^{\lambda X}]-M(\lambda)\log M(\lambda).
\end{align*}
We also verify the differentiability identity used in this step. Choose $\delta\in(0,\lambda)$ and set $I:=(\lambda-\delta,\lambda+\delta)$. For every $r\in I$, the pointwise derivative of $r\mapsto e^{rX}$ is $Xe^{rX}$, and
\begin{align*}
|X|e^{rX}\leq |X|e^{(\lambda+\delta)X}\mathbb 1_{\{X\geq 0\}}+|X|e^{(\lambda-\delta)X}\mathbb 1_{\{X<0\}}.
\end{align*}
The right-hand side is integrable by the hypotheses at the two positive parameters $\lambda+\delta$ and $\lambda-\delta$. Therefore the difference quotients for $M$ at $\lambda$ are dominated by an integrable random variable and converge pointwise to $Xe^{\lambda X}$, so the [Dominated Convergence Theorem](/theorems/4) gives $M'(\lambda)=\mathbb E[Xe^{\lambda X}]$. Since $\psi=\log\circ M$ and $M(\lambda)>0$, the chain rule gives $\psi'(\lambda)=M'(\lambda)/M(\lambda)$. Therefore $\mathbb E[Xe^{\lambda X}]=M(\lambda)\psi'(\lambda)$. Since $M(\lambda)=e^{\psi(\lambda)}$ and $\log M(\lambda)=\psi(\lambda)$, we get
\begin{align*}
\operatorname{Ent}(e^{\lambda X})=e^{\psi(\lambda)}\bigl(\lambda\psi'(\lambda)-\psi(\lambda)\bigr).
\end{align*}
Now insert the assumed entropy bound:
\begin{align*}
e^{\psi(\lambda)}\bigl(\lambda\psi'(\lambda)-\psi(\lambda)\bigr)\leq \frac{C\lambda^2}{2}e^{\psi(\lambda)}.
\end{align*}
The factors we divide by are positive: $\lambda^2>0$ because $\lambda>0$, and $e^{\psi(\lambda)}>0$ because it is an exponential. Dividing gives
\begin{align*}
\frac{\lambda\psi'(\lambda)-\psi(\lambda)}{\lambda^2}\leq \frac{C}{2}.
\end{align*}
Finally, differentiating the quotient $\psi(\lambda)/\lambda$ gives
\begin{align*}
\left(\frac{\psi(\lambda)}{\lambda}\right)'=\frac{\lambda\psi'(\lambda)-\psi(\lambda)}{\lambda^2}.
\end{align*}
Thus, for every $\lambda>0$,
\begin{align*}
\left(\frac{\psi(\lambda)}{\lambda}\right)'\leq \frac{C}{2}.
\end{align*}
[/guided]
[/step]
[step:Integrate the differential inequality from the origin]
Fix $\lambda>0$ and define $g:(0,\infty)\to\mathbb R$ by
\begin{align*}
g(s)=\frac{\psi(s)}{s}.
\end{align*}
From the previous step, $g'(s)\leq C/2$ for every $s>0$. The function $\psi$ is differentiable on $(0,\infty)$ by the local dominated-convergence argument above, hence $g$ is differentiable on $(0,\infty)$. On each compact interval $[\varepsilon,\lambda]\subset(0,\infty)$, the same local [Dominated Convergence Theorem](/theorems/4) argument may be chosen uniformly on finitely many neighbourhoods covering $[\varepsilon,\lambda]$, so $M'$ and $\psi'$ are continuous there; consequently $g\in C^1([\varepsilon,\lambda])$ and is absolutely continuous on $[\varepsilon,\lambda]$. For $0<\varepsilon<\lambda$, integration over $[\varepsilon,\lambda]$ gives
\begin{align*}
g(\lambda)-g(\varepsilon)\leq \frac{C}{2}(\lambda-\varepsilon).
\end{align*}
Letting $\varepsilon\downarrow 0$ and using
\begin{align*}
\lim_{\varepsilon\downarrow 0}g(\varepsilon)=\mathbb E[X],
\end{align*}
we obtain
\begin{align*}
\frac{\psi(\lambda)}{\lambda}-\mathbb E[X]\leq \frac{C\lambda}{2}.
\end{align*}
Multiplying by $\lambda>0$ yields
\begin{align*}
\psi(\lambda)-\lambda\mathbb E[X]\leq \frac{C\lambda^2}{2}.
\end{align*}
Because
\begin{align*}
\psi(\lambda)-\lambda\mathbb E[X]=\log\mathbb E[e^{\lambda(X-\mathbb E[X])}],
\end{align*}
the desired Laplace bound holds for every $\lambda>0$. At $\lambda=0$, both sides equal $0$, so the bound holds for every $\lambda\geq 0$.
[/step]
[step:Apply the indicator bound and optimize the exponential parameter]
Fix $t\geq 0$. If $t=0$, then
\begin{align*}
\mathbb P(X-\mathbb E[X]\geq 0)\leq 1=\exp(0),
\end{align*}
so the desired estimate holds. Assume now that $t>0$, and fix $\lambda>0$. Since the exponential function is increasing,
\begin{align*}
\{X-\mathbb E[X]\geq t\}\subseteq \{e^{\lambda(X-\mathbb E[X])}\geq e^{\lambda t}\}.
\end{align*}
For the non-negative random variable $Y:=e^{\lambda(X-\mathbb E[X])}$ and the positive threshold $a:=e^{\lambda t}$, the indicator inequality $\mathbb 1_{\{Y\geq a\}}\leq Y/a$ gives
\begin{align*}
\mathbb P(X-\mathbb E[X]\geq t)\leq e^{-\lambda t}\mathbb E[e^{\lambda(X-\mathbb E[X])}].
\end{align*}
Using the Laplace bound from the previous step,
\begin{align*}
\mathbb P(X-\mathbb E[X]\geq t)\leq \exp\left(-\lambda t+\frac{C\lambda^2}{2}\right).
\end{align*}
Because $t>0$ and $C>0$, choose the admissible value $\lambda=t/C>0$. Then
\begin{align*}
-\lambda t+\frac{C\lambda^2}{2}=-\frac{t^2}{2C}.
\end{align*}
Therefore
\begin{align*}
\mathbb P(X-\mathbb E[X]\geq t)\leq \exp\left(-\frac{t^2}{2C}\right).
\end{align*}
This completes the proof.
[/step]
