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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?

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  • $\begingroup$ Yes. Why not prove it yourself? Use the definition of $\sup$. $\endgroup$ – GEdgar Jul 5 '20 at 22:27
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Obviously this is true. Justification: suppose $f(x)\leq g(x)$ for all $x$ in a set $E$. Set $s_f=\sup_{x\in E}f(x)$ and $s_g:=\sup_{x\in E}g(x)$. Fix $x_0\in E$. Then $f(x_0)\leq g(x_0)\leq \sup_{x\in E}g(x)=s_g$. So $f(x_0)\leq s_g$ for any $x_0$ ($x_0$ was fixed, but arbitrary). So $s_g$ is an upper bound for $\{f(x): x\in E\}$. Therefore the least upper bound of this set, which is precisely $s_f$ is less than $s_g$, i.e. $s_f\leq s_g$.

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Note that $\sup_I g$ is an upper bound for $\{f(x) \mid x \in I\}.$ (Why?)
Since $\sup_I f$ is the least such upper bound (by definition), we have that $$\sup_I f \le \sup_I g.$$

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