The convex conjugate also known as Legendre–Fenchel transformation of a convex function $f:\mathbb{R}\to\mathbb{R}\cup\{+\infty\}$ is function $f^\ast:\mathbb{R}\to\mathbb{R}\cup\{+\infty\}$ definite by $$ f^{\ast}(x^\ast)=\sup_{x\in\mathbb{R}}\{ x\cdot x^\ast -f(x)\} $$ Let $\{a_k\}_{n\in\mathbb{N}}$ and $\{b_k\}_{n\in\mathbb{N}}$ numerical sequences such that $\sum_{k=1}^\infty e^{a_k\cdot x+b_k}<\infty$. Let the function $ f(x)=\lim_{n\to \infty}\left(-\frac{1}{n}\log \sum_{k=1}^n e^{a_k\cdot x+b_k}\right). $ By a simple calculations and by Holder's inequality we prove that $f(x)$ is convex. Then Legendre-Fenchel transformation is well defined.
Question. What is the convex conjugate of the function $f(x)=\lim_{n\to \infty}\left(-\frac{1}{n}\log \sum_{k=1}^n e^{a_k\cdot x+b_k}\right)$?
A more explicit way, I would calculate the supremum below \begin{align} f^{\ast}(x^\ast) = & \sup_{x\in\mathbb{R}} \left\{ x\cdot x^\ast + \lim_{n\to \infty}\left(\frac{1}{n}\log \sum_{k=1}^n e^{a_k\cdot x+b_k}\right) \right\}. \\ \end{align}