sup and inf

**Connected** · March 29th 2011, 02:59 PM

Let $f,g:A\to\mathbb R$ so that $f\le g,$ then prove that if $g$ is bounded above, so is $f$ and $\inf f\le \sup g.$

First part is easy, now since $\inf f\le f\le g,$ we have that $\inf f\le g\implies \inf f\le\sup g.$ Is this correct?

Now suppose $f$ is bounded below, we have $\inf f\le \inf g.$

If $f$ is bounded below so does $g$ then $\inf f\le f\le g,$ where $\inf f\le g\implies \inf f\le \inf g.$ Is it correct?

Thanks!

**bkarpuz** · March 29th 2011, 08:06 PM

Originally Posted by Connected

Let $f,g:A\to\mathbb R$ so that $f\le g,$ then prove that if $g$ is bounded above, so is $f$ and $\inf f\le \sup g.$

First part is easy, now since $\inf f\le f\le g,$ we have that $\inf f\le g\implies \inf f\le\sup g.$ Is this correct?

Correct

Originally Posted by Connected

Now suppose $f$ is bounded below, we have $\inf f\le \inf g.$

If $f$ is bounded below so does $g$ then $\inf f\le f\le g,$ where $\inf f\le g\implies \inf f\le \inf g.$ Is it correct?

Correct

In such questions, I can suggest you to always mention the domains of the variables, i.e.,
Assume that $f(t)\leq g(t)$ for all $t\in A$ , then
we have
$\inf_{t\in A}\{f(t)\}\leq f(t)\leq g(t)$ for all $t\in A$ ,
which implies that (dont forget that $\inf_{t\in A}\{f(t)\}$ is constant and is independent of $t$ )
$g$ is bounded below by a constant $\inf_{t\in A}\{f(t)\}$ , and thus,
$\inf_{t\in A}\{f(t)\}\leq\inf_{t\in A}\{g(t)\}\leq g(t)$ for all $t\in A$ .

**Connected** · March 30th 2011, 03:02 AM

Yes of course, it's just to save writing, that's all!

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