This is a part of an exercise that I'm doing, in Durrett's Probability book.
Let $X$ be a r.v which is not constant. Let $\phi(\theta)=E\exp(\theta X)<\infty$ for $\theta\in(-\delta,\delta),$ and let $\psi(\theta)=\log \phi(\theta).$ Prove that $\psi$ is strictly convex.
I wanted to write $\psi$'' but it's not always well defined, because $\phi'$ is not always well defined. To calculate $\phi'$, we derive inside the expectation, so $\phi'(\theta)=E(X\exp(\theta X))$, but nothing garantees that $E(X\exp(\theta X))$ is finite.
I also tried to write the classic definition of convex functions $\psi(\lambda \theta_1+(1-\lambda)\theta_2)<\lambda\psi(\theta_1)+(1-\lambda)\psi(\theta_2),$ but it doesn't work either.
I hope that someone can help me solve it. Thanks!