Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
The Supremum and Bounded Functions
Suprema proof: prove $\sup(f+g) \le \sup f + \sup g$

I am trying to prove this inequality but I don't seem to have any lead. Any suggestions?

share|cite|improve this question

marked as duplicate by Stefan Hansen, robjohn Jan 12 '13 at 14:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer 1

up vote 10 down vote accepted

Hint: If you can prove that $f(x)\cdot g(x) \leq \sup(f)\cdot \sup(g)$, for all $x$, what you want will follow because you will have proven that $\sup(f)\cdot \sup(g)$ is an upper bound of $f\cdot g$ and therefore it's greater than the smallest upper bound.

share|cite|improve this answer
Did it. Thanks! –  Leif Ericson Jan 12 '13 at 11:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.