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The oscillation of $\omega_f(A)$ of $f$ on a set $A$ to be the number $$\omega_f(A)=\sup\limits_{x,y\in A}|f(x)-f(y)|=M_A(f)-m_A(f).$$

The following equality is where I'm scratching my head a bit: $$\sup\limits_{x,y\in A}|f(x)-f(y)|=M_A(f)-m_A(f).$$
Where $M_A(f)=\sup\limits_{x\in A} f(x)$, $m_Af(x)=\inf\limits_{x\in A} f(x)$, and $f$ is a bounded function on $A$.

So here is my attempt to prove the equality. Putting how we defined $M_A(f)$ and $m_A(f)$ together: $$M_A(f)-m_A(f)=\sup\limits_{x\in A} f(x)-\inf\limits_{x\in A} f(x).$$

Since $f$ is bounded: $\inf\limits_{x \in A} f(x)=-\sup\limits_{x \in A} -f(x)$.

So, $$M_A(f)-m_A(f)=\sup\limits_{x\in A} f(x)+\sup\limits_{x \in A} -f(x).$$ $$M_A(f)-m_A(f)=\sup\limits_{x\in A} f(x)-\sup\limits_{x \in A} f(x).$$

I'm wondering if this is right so far or if I've drifted off to far left field.

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$\sup(-f(x))$ is not $-\sup f(x)$. The correct equality is $\sup(-f(x))=-\inf f(x)$. (You have used this one in the first part of your proof.) –  Martin Sleziak Nov 3 '12 at 8:25
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You're doing fine, right up until your last line.

Next, I'd rewrite as $$M_A(f)-m_A(f)=\sup\limits_{x\in A}f(x)+\sup\limits_{y\in A}-f(y).$$ Note also that we can drop the absolute value bars in the definition of $\omega_f(A)$. (Why?) Think you can get the rest of the way from there?

Alternately, you can prove that $$M_A(f)=m_A(f)+\omega_f(A),$$ which you might find to be a simpler task.

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I'm actually confused about why I can remove the absolute value bars. –  emka Nov 4 '12 at 9:05
    
Note that $|f(x)-f(y)|$ is equal to either $f(x)-f(y)$ or $f(y)-f(x)$, regardless of our choice of $x,y\in A$. so any value that $|f(x)-f(y)|$ takes can be obtained by $f(x)-f(y)$ instead (swapping $x$ and $y$, if necessary). Put another way, $f(x)-f(y)$ takes on "twice as many values" (in a sense) as $|f(x)-f(y)|$ does, but they have the same supremum. –  Cameron Buie Nov 4 '12 at 9:57
    
At the point where $\sup-f(y)$ how do I pull out the negative sign...or to put it another way, is the sup an operator which I can factor out. –  emka Nov 5 '12 at 1:26
    
Indeed, yes! That's precisely what we need. Having rewritten it with a $y$ instead of an $x$, we have $\begin{eqnarray*} \sup_{x\in A}\bigl(f(x)\bigr)+\sup_{y\in A}\bigl(-f(y)\bigr) & = & \sup_{x\in A}\left(f(x)+\sup_{y\in A}\bigl(-f(y)\bigr)\right)\\ & = & \sup_{x\in A}\left(\sup_{y\in A}\bigl(f(x)+-f(y)\bigr)\right)\\ & = & \sup_{x\in A}\sup_{y\in A}\bigl(f(x)-f(y)\bigr)\\ & = & \sup_{x,y\in A}\bigl(f(x)-f(y)\bigr).\end{eqnarray*}$ –  Cameron Buie Nov 5 '12 at 1:34
    
I was hesitant about treating it like a distributable operator. –  emka Nov 5 '12 at 2:02
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That can't be right, because you've just proved that $M_A(f)-m_A(f)=0$ (you can't change $\sup -f(x)$ into $-\sup f(x)$).

Let $S=\{|f(x)-f(y)| : x,y \in A\}$. It's not hard to show that $M_A(f)-m_A(f)$ is an upper bound of $S$: $$\mbox{*** $M_A$ and $m_A$ go here ***} \le f(x)-f(y) \le \mbox{*** $M_A$ and $m_A$ go here ***}$$

Now suppose that $c$ is an upper bound of $S$. Here's a guide without any quantifiers:

$$f(x)-f(y) \le c \\ f(x)-c \le f(y) \\ f(x)-c \le m_A(f) \\ f(x) \le m_A(f)+c \\ M_A(f) \le m_A(f)+c $$

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