# Convexity preservation and global optimality

This is a question I've had a tough time getting a good answer to.

Consider the problem to minimize $f(x)$. Assume $f$ is differentiable and nice in every way, but we do not know if $f$ is convex.

A common approach is to instead minimize $g(y)$ where $y=f(x)$. Oftentimes, $g(y)$ will be convex and we are assured that any local minimum is a global minimum, call it $x^*$. But it is well known that convexity of $g(f(x))$ does NOT imply convexity of $f(x)$. So how do we know that $x^*$ globally minimizes $f(x)$?

If $g$ is monotone and $g(f(x))$ is convex, then $f(x)$ is always convex, right? But does this imply that all non-monotonic manipulations of $f$ are 'disallowed'?

$\exp(\log(x))$ is convex, $\exp$ is monotone, but $\log(x)$ is not convex.
• Yes but say we want to minimize $f(x)$ and we'd like to solve an equivalent problem. Suppose $exp(f(x))$ happens to be convex. If we find a unique minimizer of $exp(f(x))$, how would we know whether this is a unique minimizer of $f(x)$ as well? Dec 19, 2014 at 11:51
• You know it because $f(x)<f(y)$ implies $\exp(f(x))< \exp(f(y))$ and vice versa. The conceptual difficulty you seem to be having revolves around the notion of equivalence between optimization models. Boyd & Vandenberghe is a good reference on that concept, too. Dec 19, 2014 at 21:41