I'm reading 2.6 of Durrett's book. If we define $\phi(t)=\mathbb{E}[e^{tX}]$ and for $a>\mu$, where $\mu$ is the expectation of corresponding random variable, and we define $$I(a)=\sup_{t>0} \ (at-\log \phi(t))\text{.}$$ I showed that $I(a)$ is convex. Suppose that $\mu=0$. Is it true that $I(0)=0$ and $I$ is strictly positive for $a>0$ and strictly increasing on $[0,\infty)$?
It's obvious that if $t=a=0$ then $at-\log \phi(t)=0$, but I couldn't show $\sup\ (-\log \phi(t))=0$ .
By Jensen's inequality $Ee^{tX} \geq e^{E(tX)}=1$. So $\phi (t) \geq 1$ and $-\log \, \phi(t) \leq 0$. Since $at-log \, \phi (t) $ is $\leq 0$ (when $a=0$) and vanishes when $t=0$ its supremum equals $0$.