$ \sup_{x \in A}\log(f(x))=\log(\sup_{x \in A}f(x)) $?

Question

Consider a function $f:A\subseteq \mathbb{R} \rightarrow (0,\infty) $. Is $$ \sup_{x \in A}\log(f(x))=\log(\sup_{x \in A}f(x)) $$ true?

I believe the answer is yes but I would like to have a formal confirmation.

Answer 1

You have to show that $\log(\sup_{x\in A}f(x))$ is the least upper bound of $B=\{\log(f(x)):x\in A\}$. I'll assume that “log” refers to the natural logarithm (it works the same for logarithms in any base $>1$).

For simplicity, let $M=\sup_{x\in A}f(x)$ and assume first that $M\in\mathbb{R}$.

1. $\log M$ is an upper bound for $B$. Indeed, if $t\in A$, $f(x)\le M$, so $\log(f(x))\le \log M$

2. $\log M$ is the least upper bound for $B$. Indeed, if $N$ is an upper bound for $B$, then $\log(f(x))\le N$, for any $x\in A$. Therefore $f(x)\le e^N$ and so $e^N\ge M$, which implies $N\ge \log M$.

Fill in the details and do the case when the supremum is infinity.