Prove that $\sup[f(x)+g(x)]\le \sup(f(x)) + \sup (g(x))$

Question

How to prove that: $ \sup[f(x)+g(x)]\le \sup(f(x)) + \sup (g(x))$

For me, it sounds very logical and obvious but I did not find the way to prove this. I will be glad for any tips.

You need only prove that $\sup(f(x)) + \sup (g(x))$ is an upper bound of $f(x) + g(x)$ — Ben Grossmann, Mar 09 '14 at 16:06
Mostly it boils down to writing down the definition of $\sup$ and pointing out that it's satisfied. — G Tony Jacobs, Mar 09 '14 at 16:07

score 2 · Answer 1 · answered Mar 09 '14 at 16:11

2

Well $\sup(f(x))\geq f(x)$ and $\sup(g(x))\geq g(x)$ by definition of supremum,you can write it like $\sup(f(x))+\sup(g(x))\geq f(x)+g(x)$

answered Mar 09 '14 at 16:11

kingW3

score 1 · Answer 2 · answered Mar 09 '14 at 16:11

1

Hint: You know that $(f+g)(x)=f(x)+g(x)\le \sup f(x)+\sup g(x)$

answered Mar 09 '14 at 16:11

homegrown

2 Answers2