[proofplan]
Fix an index $k$ and condition on the past $\mathcal F_{k-1}$. The boundedness assumption lets us dominate the exponential function on the interval $[-c_k,c_k]$ by its affine interpolation between the two endpoints. Taking conditional expectations then collapses the interpolation term because the martingale difference condition gives conditional mean zero. The remaining endpoint average is $(e^{\lambda c_k}+e^{-\lambda c_k})/2$, and an elementary power-series comparison bounds it by $\exp(\lambda^2 c_k^2/2)$.
[/proofplan]
[step:Treat the zero bound case separately]
Fix $k \in \{1,\dots,n\}$ and $\lambda \in \mathbb R$. From the martingale difference sequence hypothesis, $X_k$ is $\mathcal F_k$-measurable, integrable, and satisfies $\mathbb E[X_k \mid \mathcal F_{k-1}]=0$ almost surely. The estimate below is a one-step consequence of integrability, conditional mean zero, and the almost-sure bound $|X_k| \leq c_k$; the full adapted sequence structure supplies these hypotheses for each index $k$. If $c_k = 0$, then $|X_k| \leq 0$ almost surely, hence $X_k = 0$ almost surely. Therefore $e^{\lambda X_k} = 1$ almost surely, and the defining property of [conditional expectation](/page/Conditional%20Expectation) gives
\begin{align*}
\mathbb E[e^{\lambda X_k} \mid \mathcal F_{k-1}]
=
\mathbb E[1 \mid \mathcal F_{k-1}]
=
1
=
\exp\left(\frac{\lambda^2 c_k^2}{2}\right)
\end{align*}
almost surely. Hence it remains to prove the claim when $c_k > 0$.
[/step]
[step:Dominate the exponential by its affine endpoint envelope]
Assume $c_k > 0$. Define the deterministic function $\phi: [-c_k,c_k] \to \mathbb R$ by
\begin{align*}
\phi(x) := e^{\lambda x}.
\end{align*}
Since $\phi$ is convex on $[-c_k,c_k]$, every $x \in [-c_k,c_k]$ satisfies
\begin{align*}
e^{\lambda x}
\leq
\frac{c_k+x}{2c_k}e^{\lambda c_k}
+
\frac{c_k-x}{2c_k}e^{-\lambda c_k}.
\end{align*}
[guided]
We want an upper bound that becomes useful after conditioning on $\mathcal F_{k-1}$. Since $X_k$ is only known to be bounded and conditionally mean-zero, the right comparison is not a Taylor expansion with an uncontrolled remainder, but the affine chord of the convex function $x \mapsto e^{\lambda x}$ over the interval where $X_k$ takes its values.
Define the map $\phi: [-c_k,c_k] \to \mathbb R$ by
\begin{align*}
\phi(x) := e^{\lambda x}.
\end{align*}
The function $\phi$ is convex because its second derivative is $\phi''(x)=\lambda^2 e^{\lambda x} \geq 0$ for every $x \in [-c_k,c_k]$. Define the deterministic weight function $\alpha: [-c_k,c_k] \to [0,1]$ by
\begin{align*}
\alpha(x) := \frac{c_k+x}{2c_k}.
\end{align*}
Define the deterministic weight function $\beta: [-c_k,c_k] \to [0,1]$ by
\begin{align*}
\beta(x) := \frac{c_k-x}{2c_k}.
\end{align*}
For a fixed $x \in [-c_k,c_k]$, since $c_k>0$ and $-c_k \leq x \leq c_k$, we have $\alpha(x),\beta(x) \in [0,1]$ and
\begin{align*}
\alpha(x)+\beta(x)=1,
\qquad
\alpha(x)c_k+\beta(x)(-c_k)=x.
\end{align*}
Convexity of $\phi$ applied to the convex combination $x=\alpha(x)c_k+\beta(x)(-c_k)$ gives
\begin{align*}
e^{\lambda x}
=
\phi(x)
\leq
\alpha(x)\phi(c_k)+\beta(x)\phi(-c_k)
=
\frac{c_k+x}{2c_k}e^{\lambda c_k}
+
\frac{c_k-x}{2c_k}e^{-\lambda c_k}.
\end{align*}
This is the endpoint envelope that will interact with the conditional mean-zero property.
[/guided]
[/step]
[step:Take conditional expectations and use the martingale difference condition]
Since $|X_k| \leq c_k$ almost surely, we have $e^{\lambda X_k} \leq e^{|\lambda|c_k}$ almost surely, so $e^{\lambda X_k}$ is integrable. Applying the preceding pointwise inequality to $x=X_k$ gives
\begin{align*}
e^{\lambda X_k}
\leq
\frac{c_k+X_k}{2c_k}e^{\lambda c_k}
+
\frac{c_k-X_k}{2c_k}e^{-\lambda c_k}
\end{align*}
almost surely. The [random variable](/page/Random%20Variable) on the right is bounded, hence integrable. Monotonicity and linearity of conditional expectation give
\begin{align*}
\mathbb E[e^{\lambda X_k} \mid \mathcal F_{k-1}] \leq \frac{e^{\lambda c_k}}{2c_k}\mathbb E[c_k+X_k \mid \mathcal F_{k-1}] + \frac{e^{-\lambda c_k}}{2c_k}\mathbb E[c_k-X_k \mid \mathcal F_{k-1}].
\end{align*}
Since $c_k$ is deterministic, conditional expectation gives $\mathbb E[c_k+X_k \mid \mathcal F_{k-1}] = c_k + \mathbb E[X_k \mid \mathcal F_{k-1}]$ and $\mathbb E[c_k-X_k \mid \mathcal F_{k-1}] = c_k - \mathbb E[X_k \mid \mathcal F_{k-1}]$ almost surely. Because $\mathbb E[X_k \mid \mathcal F_{k-1}]=0$ almost surely, this becomes
\begin{align*}
\mathbb E[e^{\lambda X_k} \mid \mathcal F_{k-1}] \leq \frac{e^{\lambda c_k}+e^{-\lambda c_k}}{2}
\end{align*}
almost surely.
[/step]
[step:Bound the endpoint average by the Gaussian exponential]
Define $t := \lambda c_k \in \mathbb R$. The preceding step gives
\begin{align*}
\mathbb E[e^{\lambda X_k} \mid \mathcal F_{k-1}]
\leq
\frac{e^t+e^{-t}}{2}
\end{align*}
almost surely. We prove the elementary estimate
\begin{align*}
\frac{e^t+e^{-t}}{2}
\leq
e^{t^2/2}.
\end{align*}
Using the [power series](/page/Power%20Series) for the exponential function,
\begin{align*}
\frac{e^t+e^{-t}}{2}
=
\sum_{m=0}^{\infty}\frac{t^{2m}}{(2m)!}.
\end{align*}
For $m=0$, the inequality $(2m)! \geq 2^m m!$ is equality. For every integer $m \geq 1$, grouping the factors in $(2m)!$ as pairs gives
\begin{align*}
(2m)! = \prod_{j=1}^{m}(2j-1)(2j) \geq \prod_{j=1}^{m}2j = 2^m m!.
\end{align*}
Therefore
\begin{align*}
\sum_{m=0}^{\infty}\frac{t^{2m}}{(2m)!}
\leq
\sum_{m=0}^{\infty}\frac{t^{2m}}{2^m m!}
=
\sum_{m=0}^{\infty}\frac{(t^2/2)^m}{m!}
=
e^{t^2/2}.
\end{align*}
Substituting $t=\lambda c_k$ yields
\begin{align*}
\mathbb E[e^{\lambda X_k} \mid \mathcal F_{k-1}]
\leq
\exp\left(\frac{\lambda^2 c_k^2}{2}\right)
\end{align*}
almost surely. Since $k$ and $\lambda$ were arbitrary, the asserted inequality holds for every $1 \leq k \leq n$ and every $\lambda \in \mathbb R$.
[/step]
