Equivalent maximization of $\log (f(x)) $ when $ f(x) $ is non-convex

Question

I would like to know whether the following is equivalent. $$ \max_{x,y,z} \log (f(x,y,z)) = \max_{x,y,z} f(x,y,z) $$

In principle, I need to solve $ \max_{x,y,z} \log (f(x,y,z)) $ where $ f(x,y,z) $ is a non-convex function that maps $ x,y,z \in \mathbb{C} $ to a real value such that $ f(x,y,z) \geq 0 $. Since $ \log(.) $ increases monotonically, my presumption is that I can equivalently solve $ \max_{x,y,z} f(x,y,z) $. Would this be correct? How can I prove that this assumption holds?

Answer 1

It’s true that the $x,y,z\in A$ for which $\ln(f(x,y,z))$ is maximal (in $A$), also make $f(x,y,z)$ maximal (in $A$).

We can prove this as following: As every $f(x,y,z)\in\mathbb{R}^{+}$, we see that $f(x,y,z)\geq f(x’,y’,z’) \Leftrightarrow \ln(f(x,y,z))\leq\ln(f(x’,y’,z’))$ (as $\ln x$ increases monotonically in $\mathbb{R}^{+}_0)$.

Hence, the $x,y,z\in A$ that maximize $\ln(f(x,y,z))$ (so for all $x’,y’,z’\in A$, $\ln(f(x,y,z))\geq\ln(f(x’,y’,z’))$), also maximize $f(x,y,z)$ (so for all $x’,y’,z’\in A$, $f(x,y,z)\geq f(x’,y’,z’)$), as those inequalities are equivalent.

However, this doesn’t mean that $$ \max_{x,y,z} \ln (f(x,y,z)) = \max_{x,y,z} f(x,y,z) $$ As the LHS is the $\ln$ of the RHS.

Answer 2

First of all, I think you meant to write $$ \mathop{\textrm{argmax}}_{x,y,z} \log(x,y,z) = \mathop{\textrm{argmax}}_{x,y,z} f(x,y,z) $$

Otherwise, the monotonicity of $\log$ ensures the result because you have the equivalence $$ f(x^*,y^*,z^*) \ge f(x,y,z) \Leftrightarrow \log(f(x^*,y^*,z^*)) \ge \log(f(x,y,z)). $$