Let $f$ be a non-negative continuous function on a compact metric space $Z$ that satisfies $$\max_{x\in Z } f(x) \cdot \min_{x\in Z } f(x) = 1.$$ Then $$\max_{x\in Z } \log(f(x)) = \left\lVert \log(f) \right\rVert_\infty.$$
I tried to fix my proof after 5xum pointed out a flaw and gave me a hint on how to fix it. The only step I'm not absolutely sure about is where I claim that
$$\max_{x\in Z} \left\lvert \log(f(x)) \right\rvert = \max_{x\in Z} (- \log(f(x))) = - \min_{x\in Z} \log(f(x)).$$
Here's the edited proof:
By definition, we have $$\left\lVert \log(f) \right\rVert_\infty = \sup_{x\in Z} \left\lvert \log(f(x)) \right\rvert.$$
Since $f$, $\log$ and the absolute value function are continuous functions on compact sets, we have
$$\sup_{x\in Z} \left\lvert \log(f(x)) \right\rvert = \max_{x\in Z} \left\lvert \log(f(x)) \right\rvert.$$
We therefore have to show that $$\max_{x\in Z} \log(f) = \max_{x\in Z} \left\lvert \log(f(x)) \right\rvert.$$
We first show that $\max_{x\in Z} \log(f) = \left\lvert \max_{x\in Z} \log(f(x)) \right\rvert.$ Assume that for all $x\in Z$, we have $0\leq f(x) < 1$. This would imply $$0\leq \max_{x\in Z } f(x) \cdot \min_{x\in Z } f(x) < 1,$$ which contradicts our assumption about $f$. Therefore, there must be at least one $x_0\in Z$ such that $f(x_0) \geq 1$, which implies $\log f(x_0) \geq 0$. It follows that $\max_{x\in Z} \log(f) \geq 0$ and thus $\max_{x\in Z} \log(f) = \left\lvert \max_{x\in Z} \log(f(x)) \right\rvert.$
For the next step, observe that $\left\lvert \max_{x\in Z} \log(f(x)) \right\rvert \leq \max_{x\in Z} \left\lvert \log(f(x)) \right\rvert$ and thus $\max_{x\in Z} \log(f)\leq \max_{x\in Z} \left\lvert \log(f(x)) \right\rvert$. To show equality, assume that $\max_{x\in Z} \log(f) < \max_{x\in Z} \left\lvert \log(f(x)) \right\rvert$. Then, since
$$\max_{x\in Z} \left\lvert \log(f(x)) \right\rvert = \max_{x\in Z} (- \log(f(x))) = - \min_{x\in Z} \log(f(x)),$$
we have $\max_{x\in Z} \log(f) < - \min_{x\in Z} \log(f(x))$. But the assumption that $\max_{x\in Z } f(x) \cdot \min_{x\in Z } f(x) = 1$ implies $\log \max_{x\in Z } f(x) = - \log \min_{x\in Z } f(x)$, and thus, since $\log$ is monotonic, that $\max_{x\in Z } \log f(x) = - \min_{x\in Z } \log f(x)$. Therefore, we must have
$$\max_{x\in Z} \log(f) = \max_{x\in Z} \left\lvert \log(f(x)) \right\rvert.$$
I don't think the proof is correct.
Why would $$\sup_{x\in Z} \log(f(x)) = \log \sup_{x\in Z} (f(x)) \geq \log (1) = 0$$
imply
$$\sup_{x\in Z} \log(f) = \sup_{x\in Z} \left\lvert \log(f(x)) \right\rvert?$$
I don't see the implication here...