# Proof relating to boundedness

• Nov 16th 2010, 08:47 AM
Runty
Proof relating to boundedness
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 $\displaystyle f(x)$ is a bounded function on $\displaystyle [a,b]$. If $\displaystyle M, M'$ denote the least upper bounds and $\displaystyle m, m'$ denote the greatest lower bounds of $\displaystyle f, |f|$ respectively, prove that $\displaystyle M'-m'\leq M-m$.

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 $\displaystyle f$ is a bounded and integrable function on $\displaystyle [a,b]$, and if $\displaystyle M$ and $\displaystyle m$ are the least upper and greatest lower bounds of $\displaystyle f$ over $\displaystyle [a,b]$, then
$\displaystyle m(b-a)\leq \int_a^b f(x)dx\leq M(b-a)$ if $\displaystyle a\leq b$, and
$\displaystyle m(b-a)\geq \int_a^b f(x)dx\geq M(b-a)$ if $\displaystyle b\leq a$.
Also, since $\displaystyle f$ is a bounded and integrable function on $\displaystyle [a,b]$, then $\displaystyle |f|$ is also bounded and integrable over $\displaystyle [a,b]$.

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.
• Nov 17th 2010, 11:35 AM
Runty
Trying to get all the cases of $\displaystyle m'$ and $\displaystyle M'$ for $\displaystyle |f|$ is extremely tough, or I'm looking at it wrong.

The main issue is this: $\displaystyle |f|=f$ if $\displaystyle f\geq 0$, and $\displaystyle |f|=-f$ if $\displaystyle f<0$ (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.
• Nov 18th 2010, 08:00 AM
Runty
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 $\displaystyle \forall x\in [a,b]$ that $\displaystyle m\leq f(x)\leq M$.
Now, we note that if both $\displaystyle m$ and $\displaystyle M\geq 0$, then $\displaystyle f$ is nonnegative on $\displaystyle [a,b]$ and hence $\displaystyle f=|f|$, $\displaystyle M=M'$, and $\displaystyle m=m'$. In which case, the result follows.

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

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

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

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