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The original theorem is a as follows:

THEOREM Let $f$ be continuous on $[a,b]$. Then for any $\epsilon>0$ there exists a finite partition of $[a,b]$ such that the oscillation of $f$ on each subinterval is smaller than $\epsilon$.

The proof is as follows.

PROOF We argue by contradiction. Suppose there is an $\epsilon_0$ such that no finite partition of $[a,b]$ exists such that the oscillation is less than $\epsilon_0$. We use a bisection argument. Since the theorem fails on $[a,b]$, it must fail on $[a,c]$ or $[c,b]$, where $c$ is the midpoint of the interval. Call $[a_1,b_1]$ the interval where it fails. If it fails on both, we agree to choose the left interval. Continue the process, call $[a_{n+1},b_{n+1}]$ the half of the interval $[a_{n},b_{n}]$ where the theorem fails. Note that on each $[a_n,b_n]$ the oscillation must be $\geq \epsilon_0$. Let $A$ be the set consisting of $a,a_1,a_2,\dots$, and let $\alpha=\sup A$. Then $\alpha\in[a,b]$, so that $f$ is continuous at $\alpha$. Then, there is an interval $(\alpha-\delta,\alpha+\delta)$ where the oscillation of $f$ is less than $\epsilon_0$. But we must have $[a_n,b_n]\subset (\alpha-\delta,\alpha+\delta)$ for some $n$ with $(b-a)/2^n<\delta$, so that the oscillation of $f$ in $[a_n,b_n]$ is also less than $\epsilon_0$. This contradiction proves the thereom.

It has already been prove that if $f$ is continuous on $[a,b]$ then exist $c$ and $d$ on that interval such that $$f(c)=\inf f $$ and $$f(d)=\sup f$$

where $\inf f =\inf \{f(x):a\leq x\leq b\}$ and $\sup f =\sup\{f(x):a\leq x\leq b\}$. According to Apostol, we define the oscilation of $f$ on $[a,b]$ by $$f(d)-f(c)$$

Now, I need want to prove the part that is in boldface. Given any $\epsilon >0$ if $\alpha\in [a,b]$, then there exists a $\delta >0$ such that the oscillation of $f$ on $(\alpha-\delta,\alpha+\delta)$ can be made smaller than $\epsilon$.

He hasn't defined what the oscillation is for open intervals, and he talks about the maximum and minimum of $f$, calling them $M(f)$ and $m(f)$ to set the oscillation to be $M(f)-m(f)$. I'm guessing on closed intervals, this is $f(d)-f(c)$ as before. $(1)$ doesn't hold for open intervals, like $(\alpha-\delta,\alpha+\delta)$. How can I prove this? Any hints? What is a more concrete definition of the oscillation of $f$ on an interval $I$?

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If by continuity you fix $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ for all $y$,$|x-y|<\delta$, haven't you shown the oscillation on that interval $(x-\delta,x+\delta)$ is less than $2\epsilon$? Or am I oversimplifying and misunderstanding the task? –  rschwieb Sep 9 '12 at 23:42
    
Yes, I thought of that. –  Pedro Tamaroff Sep 9 '12 at 23:54
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1 Answer

up vote 3 down vote accepted

The oscillation of a function $f$ on a set $X$ is defined as:

$$ \omega(f, X) = \sup_{x, y \in X} \left|f(x) - f(y)\right| $$

If $f$ is continuous on $[a, b]$, then it's uniformly continuous and for each $\epsilon$ we can find a $\delta$ so that, for all $x,y$,

$$ |x - y| < \delta \Rightarrow \left|f(x) - f(y)\right| < \epsilon $$

In particular:

$$ \forall x, y \in \left(\alpha - \frac{\delta}{2}, \alpha + \frac{\delta}{2}\right) : \left|f(x) - f(y)\right| < \epsilon $$

Hence, the oscillation of $f$ on this neighborhood is also smaller than $\epsilon$.

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That's a problem since what I wrote above is supposed to be used in the proof of uniform continuity of $f$. More precisely, we're proving that if $f$ is continuous on $[a,b]$ then for every $\epsilon>0$ there exists a partition $P$ of $[a,b]$ such that the oscilation of $f$ on each subinterval is less than $\epsilon$. This is proven by contradiction with a "bisection" argument. (+1) for the defintion though. –  Pedro Tamaroff Sep 10 '12 at 0:14
    
@PeterTamaroff - Uniform continuity on a closed interval can be proved by showing that: 1) closed intervals are compact, 2) continuous functions on compact sets are uniformly continuous. I guess you're looking for a more direct proof then? –  Ayman Hourieh Sep 10 '12 at 0:19
    
No compactness considerations are allowed here, since it is "just" calculus =). If you can take a look at Apostol's Calc 1, it is page 187 on my edition. –  Pedro Tamaroff Sep 10 '12 at 0:20
    
Added quantifiers. Hope you don't mind. –  Pedro Tamaroff Oct 12 '12 at 0:32
    
@PeterTamaroff Not at all! –  Ayman Hourieh Oct 12 '12 at 1:32
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