Let $X$ be a random variable with $\mathbb{E}X=0$, $-1 \le X \le 1$, $\text{Var} (X)=\sigma ^2$.
I've been trying to prove the Bernstein's inequality, and I have to show the following, but don't know how to deal with those $\sigma ^2$ on the RHS.
$$\mathbb{E}(e^X)\le \frac{1}{1+\sigma ^2} e^{-\sigma ^2}+\frac{\sigma ^2}{1+\sigma ^2} e$$
I wish I can get some help.
Let us define $$ F(x) = e^x-Ax^2-Bx. $$ where $$ A=\frac{e-(\sigma^2+2)e^{-\sigma^2}}{(1+\sigma^2)^2}, $$ and $$ B=\frac{2\sigma^2e-(\sigma^4+2\sigma^2-1)e^{-\sigma^2}}{(1+\sigma^2)^2}. $$The constants $A,B$ are chosen so that $$ F(-\sigma^2) = F(1), \quad F'(-\sigma^2)=0, \quad F''(-\sigma^2)<0. $$ Since $x\mapsto e^x$ is convex and $F''(-\sigma^2)<0$, the equation $F'(x) = e^x -2Ax-B=0$ has only $2$ roots $x=-\sigma^2$ and $x=x_0>-\sigma^2$. Since $$F'(x)\begin{cases}>0,\quad x\in(-\infty,-\sigma^2)\\<0,\quad x\in (-\sigma^2,x_0)\\>0,\quad x\in (x_0,\infty)\end{cases},$$ it follows that $$ F(x) \le F(-\sigma^2)=F(1)\quad\forall x\in [-1,1]. $$ Hence by taking expectation on $F(X)$, we have $$ E[e^X-AX^2-BX]=E[e^X]-A\sigma^2 \le F(1), $$ or equivalently $$ E[e^X]\le A\sigma^2 +F(1)= A(\sigma^2-1)-B+e. $$ After some calculation, we get $$ A(\sigma^2-1)-B+e=\frac{e^{-\sigma^2}}{1+\sigma^2}+\frac{\sigma^2 e}{1+\sigma^2} $$ and the inequality follows.