Boundedness: Prove that M'-m' =< M-m

  • #1

Homework Statement


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

2. The attempt at a solution
(these are the only things I have; most are properties found online)
For this, we assume that [tex]f[/tex] is integrable.
If [tex]f[/tex] is a bounded and integrable function on [tex][a,b][/tex], and if [tex]M[/tex] and [tex]m[/tex] are the least upper and greatest lower bounds of [tex]f[/tex] over [tex][a,b][/tex], then
[tex]m(b-a)\leq \int_a^b f(x)dx\leq M(b-a)[/tex] if [tex]a\leq b[/tex], and
[tex]m(b-a)\geq \int_a^b f(x)dx\geq M(b-a)[/tex] if [tex]b\leq a[/tex].
Also, since [tex]f[/tex] is both a bounded and integrable function on [tex][a,b][/tex], then [tex]|f|[/tex] is also bounded and integrable over [tex][a,b][/tex].

I haven't been able to determine how to obtain the least upper and greatest lower bounds of [tex]|f|[/tex], due to how complicated the very idea is.
 

Answers and Replies

  • #2
124
0
Is the function continuous?

Note that the case that [tex]f[/tex] is negative ([tex]\le 0[/tex]) is trivial, the same when [tex]f[/tex] is positive ([tex]\ge 0[/tex]). This leaves the case that there are positive and negative function values. This might give you a lot of help.
 
  • #3
Is the function continuous?

Note that the case that [tex]f[/tex] is negative ([tex]\le 0[/tex]) is trivial, the same when [tex]f[/tex] is positive ([tex]\ge 0[/tex]). This leaves the case that there are positive and negative function values. This might give you a lot of help.

I listed the question word-for-word in the opening post, so it doesn't say anything about continuity. I guess I could assume it is continuous, but I don't know how relevant that would be.

I'm also not sure how that bit on "positive and negative function values" is supposed to be implemented (or I'm not interpreting it correctly). Could you please give a little more detail as to what you mean?

Our class is covering higher-level integration right now, and I always get tripped up on complicated proofs and the like. Then again, I was given very bad course advice...
 
  • #4
124
0
[tex]f : X \longrightarrow \textbf{R}[/tex]


it is about the mixed case where there are [tex]x, y \in X[/tex] such that [tex]f(x) < 0[/tex] and [tex]f(y) > 0[/tex]. Therefore [tex]M > 0[/tex] and [tex]m < 0[/tex].


We have [tex]0 \le M' \le \max(M, |m|)[/tex]. Combined with [tex]m < 0[/tex] and [tex]m' \ge 0 [/tex] this gives

[tex] M'- m' \le \max(M, |m|) \le M - m[/tex]
 
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