# If $f(x) \leq g(x)$ is it true that $\sup_I(f(x))\leq \sup_I(g(x))$?

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?

• Yes. Why not prove it yourself? Use the definition of $\sup$. – GEdgar Jul 5 '20 at 22:27

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. $$s_f\leq s_g$$.

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 $$\sup_I f \le \sup_I g.$$