How to verify the relation between the multiplication of supremum?

Question

Let $f$ and $g$ be functions mapping from $\mathbb{R}$ to $\mathbb{R}_+ /\{0\}$.

I'm wondering that is there a relation between $\sup_{x \in \mathbb{R}} \frac{f(x)}{g(x)}$ and $\frac{\sup_{x \in \mathbb{R}} f(x) }{\sup_{x \in \mathbb{R}} g(x)}$?

Is it $\sup_{x \in \mathbb{R}} \frac{f(x)}{g(x)} \leq \frac{\sup_{x \in \mathbb{R}} f(x) }{\sup_{x \in \mathbb{R}} g(x)}$? or $\sup_{x \in \mathbb{R}} \frac{f(x)}{g(x)} \geq \frac{\sup_{x \in \mathbb{R}} f(x) }{\sup_{x \in \mathbb{R}} g(x)}$?

Thanks in advance!

Answer 1

Simple counter-example

Let us look at $f$ and $g$ which are 1 on $\mathbb R$ except on 0 where $f(0)=0.2$ and $g(0)=0.1$.

Then $\sup_{x \in \mathbb{R}} \frac{f(x)}{g(x)}=2$ and $\frac{\sup_{x \in \mathbb{R}} f(x) }{\sup_{x \in \mathbb{R}} g(x)}=1$

Now inverse the role of $f$ and $g$ and you get $\sup_{x \in \mathbb{R}} \frac{f(x)}{g(x)}=0.5$ and $\frac{\sup_{x \in \mathbb{R}} f(x) }{\sup_{x \in \mathbb{R}} g(x)}=1$

Finally we cannot conclude anything!

Edit: If we consider smooth functions,

let us take $f(x)=1-0.8\exp(-x^2/2)$ and $g(x)=1-0.9\exp(-x^2/2)$

Then $\sup_{x \in \mathbb{R}} \frac{f(x)}{g(x)}=2$ and $\frac{\sup_{x \in \mathbb{R}} f(x) }{\sup_{x \in \mathbb{R}} g(x)}=1$

And again, you can inverse the role of $f$ and $g$ to obtain the inequality in the reverse direction.