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Math Help - Proof relating to boundedness

  1. #1
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    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 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.
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  2. #2
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    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.
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  3. #3
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    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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