Maximization of log-sum-exp function

Question

I am reading about robust optimization and there is a claim:

$$ \max_{\mu^-\leq\mu\leq \mu^+} \ln \left( \exp\left\{w+\ln\left(\frac{1+\mu}{2}\right)\right\} +\exp\left\{-w+\ln\left(\frac{1-\mu}{2}\right)\right\} \right) $$ is equivalent to $$ \max_{-1\leq u\leq1} \{wu-\phi(u)\}, $$ with $$ \phi(u)= \begin{cases} \frac{1}{2}[(1+u)\ln\left(\frac{1+u}{1+\mu^-}\right) +(1-u)\ln\left(\frac{1-u}{1-\mu^-}\right)], & -1\leq u\leq\mu^-, \\ 0, & \mu^-\leq u \leq \mu^+, \\ \frac{1}{2}[(1+u)\ln\left(\frac{1+u}{1+\mu^+}\right) +(1-u)\ln\left(\frac{1-u}{1-\mu^+}\right)], & \mu^+\leq u\leq1. \end{cases} $$ when $-1\leq \mu^-\leq\mu^+\leq1$.

Could anyone help me prove this? There is a hint using the equality $$ \ln(\exp(x_1)+...+\exp(x_n))=\max_y\left\{x^Ty-\sum_{i=1}^{n}y_i\ln y_i:y\geq0,\sum_i y_i=1\right\}. $$ But I don't know how to use this equality.

Answer 1

This question is answered mostly by simple algebra and calculus, so here is a sketch of how the second expression is derived.

Set $y_{1}=u$ and $y_2=1-u$ in the conjugate function expression for the log-exp-sum. Simplify what is obtained after using the conjugate function expression to replace the log-exp-sum in the original problem, freely removing any constant terms (relative to the optimization variables $u$ and $\mu$) since one is only interested in equivalence and exchange the order of maximizing over $\mu$ and $u$ after dividing all terms by two. At this point the product $wu$ arises as a term which is optimized over and it does not depend on $\mu$ so the inner max over $\mu$ can be distributed inside (to become a minimization after the minus is distributed). Finally a case by case analysis of optimization over $\mu$ for the inner expression and for different regions $ -1\leq u\leq\mu^-,\mu^-\leq u \leq \mu^+, \mu^+\leq u\leq1 $ gives the expression for $\phi(u)$.