# Calculating the convex conjugate of the function $f(x)=\lim_{n\to \infty}\left(-\frac{1}{n}\log \sum_{k=1}^n e^{a_k\cdot x+b_k}\right)$.

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}

• I'm confused about the scaling here. Since $L(x)=\sum_{k>0}\exp(a_kx+b_k)<\infty$, doesn't the argument of the $\log$ have a finite limit, and $\log(\ldots)/n\to 0 ?$ – kimchi lover Jul 17 '17 at 12:21
• @kimchilover This depends on the values ​​of $a_k$ and $b_k$. We can have $\lim_{n\to\infty}\sum_{0<k\leq n}\exp(a_kx+b_k)=\infty$ and $\lim_{n\to\infty}(1/n)\sum_{0<k\leq n}\exp(a_kx+b_k)<\infty$. – MathOverview Jul 17 '17 at 14:15
• But your division by $n$ happens outside the logarithm. So I am still as perplexed as ever. – kimchi lover Jul 17 '17 at 14:20
• @kimchilover Oops, I wanted to say $\lim_{n\to\infty}\sum_{0<k\leq n}\exp(a_kx+b_k)=\infty$ and $\lim_{n\to\infty}\left[\sum_{0<k\leq n}\exp(a_kx+b_k)\right]^{1/n}<\infty$. – MathOverview Jul 17 '17 at 14:24
• I get it now. The $a_k$ and $b_k$ also depend on $n$, sort of suppressed in the notation. Can't your $\sum_k \exp(a_k x + b_k)$ be the moment generating functions of any discrete random variable taking on $n$ values? Such as binomially distributed with parameters $p$ and $n-1$, for which the $\log(\cdot)/n$ limit recovers the log m.g.f. of the Bernouilli summand, yielding a different answer for each possible $p?$ So are you really asking, how does $f^*(x)$ depend on the sequences of $a$ and $b$ sequences? – kimchi lover Jul 17 '17 at 14:45