Proof relating to boundedness

**Runty** · November 16th 2010, 08:47 AM

I know a few things about how I'd go about this question, though I don't have all the pieces I need. This question probably isn't that hard.

Suppose that $f(x)$ is a bounded function on $[a,b]$ . If $M, M'$ denote the least upper bounds and $m, m'$ denote the greatest lower bounds of $f, |f|$ respectively, prove that $M'-m'\leq M-m$ .

The only materials I have right now is fairly simple stuff relating to this sort of thing. It's a pair of theories (for which I don't have the proofs on hand).
If $f$ is a bounded and integrable function on $[a,b]$ , and if $M$ and $m$ are the least upper and greatest lower bounds of $f$ over $[a,b]$ , then
$m(b-a)\leq \int_a^b f(x)dx\leq M(b-a)$ if $a\leq b$ , and
$m(b-a)\geq \int_a^b f(x)dx\geq M(b-a)$ if $b\leq a$ .
Also, since $f$ is a bounded and integrable function on $[a,b]$ , then $|f|$ is also bounded and integrable over $[a,b]$ .

Those are probably most of what would be needed to solve this, though I'm having some difficulties putting the pieces together. If I could get a hand with it, that'd be great.

**Runty** · November 17th 2010, 11:35 AM

Trying to get all the cases of $m'$ and $M'$ for $|f|$ is extremely tough, or I'm looking at it wrong.

The main issue is this: $|f|=f$ if $f\geq 0$ , and $|f|=-f$ if $f<0$ (at least according to another source). Trying to find all the least upper bounds and greatest lower bounds from those is insane, if not impossible. I don't even have a source to work off of in relation to the concept.

Really, I don't have a clue how we're supposed to do this.

**Runty** · November 18th 2010, 08:00 AM

Sorry for triple-post, but I managed to get a nearly-completed answer. It just has a couple of parts missing. My answer is split into three different cases (with the third having two sub-cases).

First, we note that $\forall x\in [a,b]$ that $m\leq f(x)\leq M$ .
Now, we note that if both $m$ and $M\geq 0$ , then $f$ is nonnegative on $[a,b]$ and hence $f=|f|$ , $M=M'$ , and $m=m'$ . In which case, the result follows.

If both $m$ and $M\leq 0$ , then $f$ is non-positive on $[a,b]$ and hence $|f|=-f$ . In this case (these require verification), $M'=-m$ and $m'=-M$ . So $M'-m'=-m-(-M)=M-m$ and the result follows.

Now we consider the case where $M>0$ and $m<0$ . Then $M'=\max\{M,-m\}$ and $m'=\min\{M,-m\}$ (these results need to be shown).

In the case of $M'=M$ and $m'=-m$ ,
$M'-m'=M+m<M$ , since $m<0$
while $M-m>M$ , since $m<0$
Thus, $M'-m'<M<M-m$

In the case of $M'=-m$ and $m'=M$ ,
$M'-m'=-m-M<-m$ , since $M>0$
while $M-m>-m+M>-m$
Thus, $M'-m'<0<M=m$
The result follows.

That's my answer for now. I just need those two verification parts in the second and third cases to finish up the proof. If there are any mistakes, however, please let me know.

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