sup of the sum of functions

• Mar 29th 2011, 03:12 PM
Connected
sup of the sum of functions
Let $f,g:A\to\mathbb R$ be bounded above, then $f+g$ is bounded above and $\sup(f+g)\le \sup f+\sup g.$

Okay, I think there's no much to do here, we have $f\le \sup f$ and $g\le \sup g$ so $f+g$ is bounded above and besides $f+g\le \sup f+\sup g\implies \sup(f+g)\le \sup f+\sup g.$ Is it correct?

Now if $g$ is bounded below, then $\sup f+\inf g\le \sup(f+g).$

Well since $\inf g\le g\implies f+\inf g\le f+g$ so $\sup f+\inf g\le\sup(f+g),$ and we're done. Is it correct?
• Mar 29th 2011, 05:44 PM
Drexel28
Quote:

Originally Posted by Connected
Let $f,g:A\to\mathbb R$ be bounded above, then $f+g$ is bounded above and $\sup(f+g)\le \sup f+\sup g.$

Okay, I think there's no much to do here, we have $f\le \sup f$ and $g\le \sup g$ so $f+g$ is bounded above and besides $f+g\le \sup f+\sup g\implies \sup(f+g)\le \sup f+\sup g.$ Is it correct?

Now if $g$ is bounded below, then $\sup f+\inf g\le \sup(f+g).$

Well since $\inf g\le g\implies f+\inf g\le f+g$ so $\sup f+\inf g\le\sup(f+g),$ and we're done. Is it correct?

This looks good to me.
• Mar 29th 2011, 07:57 PM
bkarpuz
$\sup_{t\in A}\{f(t)+g(t)\}\leq\sup_{t\in A,\ s\in A}\{f(t)+g(s)\}$ (releasing the variable of $g$ might greater values)
.................................. $=\sup_{t\in A}\{f(t)\}+\sup_{s\in A}\{g(s)\}$ (as the variables of $f$ and $g$ are un related, we can split it)
.................................. $=\sup_{t\in A}\{f(t)\}+\sup_{t\in A}\{g(t)\}$. ( $\sup$ does not depend on the letter of the variable)
You can do the same for $\inf$. (Happy)