If $f(x) \leq g(x)$ is it true that $\sup_I(f(x))\leq \sup_I(g(x))$ ?
I haven't been able to find this anywhere.I need to prove if this holds:
$\sup_I(e^{g(x)}|x^k|) \leq e$ with $x \in (-1,1)$ and $|g(x)|\in (0,x)$ , $k \in \mathbb{N}$
My try:
Since the exponential is strictly increasing and $g(x) \leq x \leq 1$:
$e^{g(x)}|x^k|\leq e.1$ and then using the property that I am not sure if it is true:
$\sup_I(e^{g(x)}|x^k|)\leq \sup_I(e)=e$
What do you thing? If the property in question is true, how to prove it?