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!
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 $P(Z < \epsilon - \delta)>0$ $$\gamma > 0 \Rightarrow\Bbb{E}[\exp( \gamma Z)]\geq e^{\gamma (\epsilon + \delta)} \Bbb{P}(Z> \epsilon+ \delta) $$ $$\gamma < 0 \Rightarrow\Bbb{E}[\exp( \gamma Z)]\geq e^{\gamma (\epsilon - \delta)} \Bbb{P}(Z< \epsilon - \delta) $$
$$J(\gamma) = \gamma \epsilon - \log \Bbb{E}[\exp( \gamma Z)] \leq \begin{cases} \gamma \epsilon - \gamma (\epsilon + \delta )-\log\big(\Bbb{P}(Z> \epsilon+ \delta) \big)\xrightarrow[\gamma \to +\infty]{} - \infty \\ \gamma \epsilon - \gamma (\epsilon - \delta ) -\log \big(\Bbb{P}(Z> \epsilon+ \delta)\big) \xrightarrow[\gamma \to -\infty]{} - \infty \end{cases} $$