I find the following steps in a paper:
Let $Y_1,Y_2,...,Y_n$ be a martingale difference sequence, (which means there is a martingale $X_0,X_1,...$ and $Y_i=X_i-X_{i-1}$) with $-a_k\le Y_k\le 1-a_k$ for each $k$, for suitable constants $a_k$. Let $S_n=S_{n-1}+Y_n$, then $E(e^{hS_n})=E(e^{hS_{n-1}}E(e^{hY_n}|\mathcal{F_{n-1}}))\le E(e^{hS_{n-1}})((1-a_n)e^{-ha_n}+a_ne^{h(1-a_n)})$.
I am confused with the last step. By observing $(1-a_n)(-ha_n)+a_nh(1-a_n)=0$. Here are some facts may be helpful: $E(Y_n|\mathcal{F_{n-1}})=0$. $\phi(x)=e^{hx}$ is a convex function and Jensen's inequality: if $\psi$ is a convex function, then $\psi(E(X|\mathcal{F}))\le E(\psi(X)|\mathcal{F})$.
The difficulty I meet is the direction of Jensen's inquality is not we want and it seems we need to assume $0\le a_k\le 1$ to use property of convex functions.
Let us use $-a_n\leq Y_n\leq 1-a_n$ and convexity of $\psi(x)=e^{hx}$. For all $x$ inside $[a,b]$ holds: $$ x=\dfrac{b-x}{b-a}a+\dfrac{x-a}{b-a}b, $$ that with $a=-a_n$, $b=1-a_n$, $b-a=1$ leads to $$ x=(1-a_n-x)(-a_n)+(x+a_n)(1-a_n). $$ Convexity give $$ \psi(x)=\psi\bigl((1-a_n-x)(-a_n)+(x+a_n)(1-a_n)\bigr)\leq (1-a_n-x)\psi(-a_n)+(x+a_n)\psi(1-a_n). $$ Then apply conditional expectation $E(\psi(Y_n)|\mathcal F_{n-1})$ and get desired inequality.