Logarithm of the maximum equals maximum of logarithm Does anybody know how to prove that the maximum of the logarithm of a given function equals the logarithm of the maximum of the function? That is, if $f$ is a function of $x$, how to prove that
$$\log(\max(f(x)))=\max(\log(f(x)))$$
It seems to be a obvious equality, and I have checked it numerically, but is there a simple way to prove it?
Thank you

Hint: Use the fact that the logarithm is an increasing function. Let $y$ be a point such that $f(y) = \max(f(x))$. Then for any $x\neq y$ we have $f(x) \le f(y)$ and thus
$$
\log(f(x)) \le \log(f(y)) = \log(\max(f(x))).
$$
Can you finish the proof?
