2
$\begingroup$

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.

  • 2
    $\sup(-f(x))$ 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.)2012-11-03

2 Answers 2

1

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.

  • 0
    Understandable. If we were looking at something of the form $\sup_{x\in A}\bigl(f(x)+g(x)\bigr),$ for instance, we **can't** (in general) break that up into $\sup_{x\in A}f(x)+\sup_{x\in A}g(x).$ The latter will be *at least as big* as the former, but need not be equal. For an example where they aren't equal, consider $A=\Bbb R$, $f(x)=e^{-x^2}$, and $g(x)=e^{-(x-1)^2}$.2012-11-05
2

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 $