Dual of concave function is convex

Question

If $U(x)$ is strictly increasing and strictly concave and $lim_{x \rightarrow \infty}$ U'(x) = 0, prove that its dual: $$U^{*}(y) = max_x \{U(x) - xy\}$$ is convex.

Does anyone know how to prove this?

Thanks!

Answer 1

In convex analysis, the convex conjugate of a function $f$ is usually defined to be \begin{equation*} f^*(y) = \sup_x \, \langle y, x \rangle - f(x). \end{equation*} With this definition, $f^*$ is convex because a supremeum of convex functions is convex. (And for each $x$, the function $y \mapsto \langle y, x \rangle - f(x)$ is convex.)

The function $U^*$ as you have defined it is convex by a similar argument. We don't need any assumptions on $U$ beyond $U:\mathbb R^n \to [-\infty,\infty]$ to reach this conclusion.