Existence of a global maximum of a function defined with the moment-generating function Can someone give me an idea how to prove the following exercise? 
Let $Z$ be a real-valued random variable whose moment-generating function $m_Z$, with $m_Z(\gamma)= E\left[ \exp(\gamma Z) \right]$, is defined on 
$\mathbb{R}$. Furthermore, assume $E\left[Z \right] = 0$.
Define $J(\gamma)= \gamma \epsilon - \log(m_Z(\gamma))$ for all $\gamma \in \mathbb{R}$ and $\epsilon > 0$.
To show:
If $P(\{Z > \epsilon\}) > 0$ and $P(\{ Z < \epsilon \}) > 0$, there exists
a global maximum of $J$.
Hint: maybe it's useful to note that: $J$ has a global maximum $\iff$ $m_{Z-\epsilon}$ has a global minimum
Many thanks in advance! 
 A: Consider $J(\gamma)= \gamma \epsilon - \log( \Bbb{E} \left[ \exp(\gamma Z) \right])$. Now calculate:
$$J'(\gamma)  = \epsilon - \frac{\Bbb{E} \left[ Z \exp(\gamma Z) \right]}{\Bbb{E} \left[ \exp(\gamma Z) \right]}$$
\begin{align*}J''(\gamma)  &=  - \frac{\Bbb{E} \left[ Z^2 \exp(\gamma Z) \right]}{\Bbb{E} \left[ \exp(\gamma Z) \right]}  + \frac{\Bbb{E} \left[ Z \exp(\gamma Z) \right]^2}{\Bbb{E} \left[ \exp(\gamma Z) \right]^2} \\&= -\frac{1}{\Bbb{E} \left[ \exp(\gamma Z) \right]^2}\bigg(\Bbb{E} \left[ Z^2 \exp(\gamma Z) \right]\Bbb{E} \left[ \exp(\gamma Z) \right]-\Bbb{E} \left[ Z \exp(\gamma Z) \right]^2\bigg) \leq 0
\end{align*}
Since by the Cauchy-Schwarz inequality:
$$\Bbb{E} \left[ Z \exp(\gamma Z) \right]^2 \leq \Bbb{E} \left[ \big(Z \exp(\gamma Z)^{1/2} \big)^2 \right]\Bbb{E} \left[ \big( \exp(\gamma Z)^{1/2} \big)^2 \right] =  \Bbb{E} \left[ Z^2 \exp(\gamma Z) \right]\Bbb{E} \left[ \exp(\gamma Z) \right]$$
Therefore $J''(\gamma)<0$ and if the maximum is attained at the interior, then it is a global maximum. To see that it indeed reaches a maximum in the interior note that $J'(0) = \epsilon >0$ (since $\Bbb{E}[Z] = 0$) and that $J(\gamma) \xrightarrow[|\gamma| \to \infty]{} -\infty$ since $\Bbb{P}(Z>\epsilon)>0$ Therefore the maximum is attained in the interior and it is a global maximum.
EDIT:  Since $\Bbb{P}(Z>\epsilon)>0$ there is a $\delta>0$ $P(Z> \epsilon + \delta)>0$ and $P(Z < \epsilon - \delta)>0$
$$\gamma > 0 \Rightarrow\Bbb{E}[\exp( \gamma Z)]\geq e^{\gamma (\epsilon + \delta)} \Bbb{P}(Z> \epsilon+ \delta) $$
$$\gamma < 0 \Rightarrow\Bbb{E}[\exp( \gamma Z)]\geq e^{\gamma (\epsilon - \delta)} \Bbb{P}(Z< \epsilon - \delta) $$
$$J(\gamma) = \gamma \epsilon - \log \Bbb{E}[\exp( \gamma Z)] \leq \begin{cases} \gamma \epsilon - \gamma (\epsilon + \delta )-\log\big(\Bbb{P}(Z> \epsilon+ \delta) \big)\xrightarrow[\gamma \to +\infty]{} - \infty \\
\gamma \epsilon - \gamma (\epsilon - \delta ) -\log \big(\Bbb{P}(Z> \epsilon+ \delta)\big) \xrightarrow[\gamma \to -\infty]{} - \infty  \end{cases}  $$
