# The Unapologetic Mathematician

## From Local Oscillation to Neighborhoods

When we defined oscillation, we took a limit to find the oscillation “at a point”. This is how much the function $f$ changes due to its behavior within every neighborhood of a point, no matter how small. If the function has a jump discontinuity at $x$, for instance, it shows up as an oscillation in $\omega_f(x)$. We now want to investigate to what extent such localized oscillations contribute to the oscillation of $f$ over a spread-out neighborhood of a point.

To this end, let $f:X\rightarrow\mathbb{R}$ be some bounded function on a compact region $X\in\mathbb{R}^n$. Given an $\epsilon>0$, assume that $\omega_f(x)<\epsilon$ for every point $x\in X$. Then there exists a $\delta>0$ so that for every closed neighborhood $N$ we have $\Omega_f(N)<\epsilon$ whenever the diameter of $d(N)$ is less than $\delta$. The diameter, incidentally, is defined by

$\displaystyle d(N)=\sup\limits_{x,y\in N}\left\{d(x,y)\right\}$

in a metric space with distance function $d$. That is, it’s the supremum of the distance between any two points in $N$.

Anyhow, for each $x$ we have some metric ball $N_x=N(x;\delta_x)$ so that

$\displaystyle\Omega_f(N_x\cap X)<\omega_f(x)+\left(\epsilon-\omega_f(x)\right)=\epsilon$

because by picking a small enough neighborhood of $x$ we can bring the oscillation over the neighborhood within any positive distance of $\omega_f(x)$. This is where we use the assumption that $\epsilon-\omega_f(x)>0$.

The collection of all the half-size balls $N\left(x;\frac{\delta_x}{2}\right)$ forms an open cover of $X$. Thus, since $X$ is compact, we have a finite subcover. That is, the half-size balls at some finite collection of points $x_i$ still covers $X$. We let $\delta$ be the smallest of these radii $\frac{\delta_{x_i}}{2}$.

Now if $N$ is some closed neighborhood with diameter less than $\delta$, it will be partially covered by at least one of these half-size balls, say $N\left(x_p;\frac{\delta_{x_p}}{2}\right)$. The corresponding full-size ball $N_{x_p}$ then fully covers $N$. Further, we chose this ball so that the $\Omega_f(N_x\cap X)<\epsilon$, and so we have

$\displaystyle\Omega_f(N)\leq\Omega(N_x\cap X)<\epsilon$

and we’re done.

December 8, 2009 - Posted by | Analysis, Calculus

1. I have been enjoying this, yet haven’t thanked you in a while. There are only a few days left for me to land interviews for local high school Math teaching jobs that start January 2010. This week, I filled out online forms for 1.5 straight hours in Yet Another Incompatible Online Teaching Job application. They emailed me a noncommittal suggestion that I phone to schedule a screening interview, and bring “all my documents.” All. Hmmmm….

Comment by Jonathan Vos Post | December 8, 2009 | Reply

2. It is a nice presentation. Perhaps it is very complicated but I think there are already Math experts know about this equation. Very difficult.

Comment by dvestv | December 9, 2009 | Reply

3. I’m not intending this to be original, dvestv. And which particular equation do you mean?

Comment by John Armstrong | December 9, 2009 | Reply

4. [...] . In each of the subintervals not containing points of , we have for all in the subinterval. Then we know there exists a so that we can subdivide the subinterval into smaller subintervals each with a [...]

Pingback by Jordan Content Integrability Condition « The Unapologetic Mathematician | December 9, 2009 | Reply

5. I’m afraid I got a little lost in the notation and didn’t quite follow the argument here. What do you mean by big omega of little x? Also, why is big omega of a local neighborhood of x ($\Omega(N_x\cap X)$ less than the oscillation at x? I would think that inequality goes the other way.

I may just be terribly confused, so my apologies if so and I just didn’t read carefully enough.

Comment by Gilbert Bernstein | December 10, 2009 | Reply

6. Your first question is do to what we mathematicians call “my bad”. Should have been $\Omega_f(N)<\epsilon$.

For the second, it’s not less than the oscillation at $x$, it’s less than $\epsilon$ (which is connected to my earlier mistake). The “trick” (artful, not deceptive, just like in those climate emails) is that I’m writing $\epsilon$ as the oscillation plus the difference between the oscillation and $\epsilon$.

Clearer now?

Comment by John Armstrong | December 10, 2009 | Reply

• Ahh, yes. That makes so much more sense now. That’s remarkable how much of a difference a typo can make for math.

Comment by Gilbert Bernstein | December 11, 2009 | Reply