The Unapologetic Mathematician

Mathematics for the interested outsider

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

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