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!

  • $\begingroup$ Whenever we consider continuous functions with domain a compact Hausdorff space $X$, we can obtain the inequality $\sup_{x \in X} \frac{f(x)}{g(x)} \geq \frac{\sup_{x \in X} f(x) }{\sup_{x \in X} g(x)}$ $\endgroup$ – Aweygan Jun 14 '17 at 2:33
  • $\begingroup$ Thanks @Aweygan. Could you briefly show us how to prove the assertion you mentioned before? Many thanks again. $\endgroup$ – Paradiesvogel Jun 14 '17 at 2:54
  • $\begingroup$ It's not too hard to see that $\sup_{x\in X}f(x)g(x)\leq\sup_{x\in X}f(x)\sup_{x\in X}g(x)$, so $\sup_{x\in X}\frac{f(x)}{g(x)}g(x)\leq\sup_{x\in X}\frac{f(x)}{g(x)}\sup_{x\in X}g(x)$. Additionally, if $f$ and $g$ are continuous on $\mathbb R$ and vanish at infinity, then we have the same result. $\endgroup$ – Aweygan Jun 14 '17 at 2:59

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.


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