sup of the sum of functions

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

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

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

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

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

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

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

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

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