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.



Maximization of log-sum-exp function
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 $\phi(u)$.


