# Oscillation of a closed subinterval

1. Nov 5, 2012

### chaotixmonjuish

Given $\epsilon > 0$, suppose $\omega_f(x) < \epsilon$ for each $x \in [a,b]$. Then show there is $\delta > 0$ such that for every closed interval $I \in [a,b]$ with $l(I)< \delta$ we have $\omega_f(I) < \epsilon$.

My first approach to this was trying to think of it as an anaglous to the definition of continuity. However, it would appear, at least to me, that we are talking about a "uniformly" oscillating interval. I'm just not sure how to prove this. My first idea was simply, since I know that things are $\epsilon$ spaced, I could always pick intervals small enough. I'm not sure how to do this rigorously.

Last edited by a moderator: Nov 5, 2012
2. Nov 5, 2012

### HallsofIvy

The first thing you will need to do is define "$\omega_f(x)$" and "$\omega_f(I)$". They are, of course, the "oscillation" of f at x and on interval I, but what are the precise definitions?

3. Nov 5, 2012

### chaotixmonjuish

The definitions of oscillation of a set and point respectively:

$\omega_f(A)=\sup\limits_{x,y\in A}f(x)-f(y)|$

$\omega_f(x)=\inf\{\omega_f(x)=\inf\{\omega_f(x-\epsilon, x+\epsilon)\cap A) : \epsilon > 0\}$

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