# Modulus of Continuity

By: Rhiannon Begley and Shuyi Chen

### Introduction

In class we have begun to and will continue to discuss functions, their continuity, and their limits at a particular point. Although we can determine the continuity and limits of functions with the tools we have learned in class, we cannot determine how quickly the function is converging to its limit at that point. In this paper, we will discuss a concept called a modulus of continuity that is used to quantify the rate of convergence of a function at a particular point of continuity.

### Definitions

We will begin by simply stating the definitions we will work with in the remainder of this report.

Definition: A function $\sigma :[0,\infty)\rightarrow [0,\infty)$ is called a modulus of continuity if $\sigma(0)=0$ and $\lim_{d\rightarrow 0^{+}}\sigma(d)=0$.

Definition: $f: A\rightarrow \mathbb{R}$ is continuous at $c\in A$ with (local) modulus of continuity $\sigma$ if there exists a $\delta >0$ so that $|f(x)-f(c)|\leq \sigma (|x-c|)$ for $|x-c|<\delta$.

### The Modulus of Continuity as a Rate of Convergence

Let’s break this down by looking at a sequence $(x_n)=(c+\frac{1}{n})$. This sequence provides a sampling of points in a small $\delta$ neighborhood around $c$. This is converging to $c$ at a rate of $\frac{1}{n}$. Now consider our function $f$. The rate at which $f(x_n)$ converges to $f(c)$ is $f(\frac{1}{n})$. Compare that to the rate at which $\sigma(|x_n-c|)$ converges to 0. This rate is $\sigma(|\frac{1}{n}|)$. So, for this $\delta$ neighborhood around $c$, if $f(\frac{1}{n}\leq \sigma(|\frac{1}{n}|)$, then $\sigma$ is a local modulus of continuity for $f$.

We can also visualize this for different possible $\sigma$ we will discuss later. If the function remains inside the cones (the yellow highlighted areas), then $\sigma$ is a local modulus of continuity for that function at that point.

### Examples

To help illustrate this concept of moduli of continuity, we will consider the common family of moduli (the power functions)

$\sigma_p(d):=d^p,\qquad\text{for }p>0.$

In order to show $\sigma_p$ is a moduli of continuity for a function, we need to produce a $\delta$ so that $|f(x)-f(c)|\leq \sigma_p (|x-c|)$ when $|x-c|<\delta$.

Example 1: $f(x)=a$
p=2: Consider $\sigma_2=d^2$
Then $|f(x)-f(c)|=|a-a|=0\leq \sigma_2 (|x-c|)=(x-c)^2$. Since $0\leq (x-c)^2$ always holds, any $\delta >0$ will suffice so that $|f(x)-f(c)|\leq \sigma_2 (|x-c|)$ for $|x-c|<\delta$.
Thus $\sigma_2$ is a local modulus of continuity for $f$ at $c$.

p=1: Consider $\sigma_1=d$
Then $|f(x)-f(c)|=|a-a|=0\leq \sigma_1 (|x-c|)=|x-c|$. Since $0\leq |x-c|$ always holds, any $\delta >0$ will suffice so that $|f(x)-f(c)|\leq \sigma_1 |x-c|$ for $|x-c|<\delta$. Thus $\sigma_1$ is a local modulus of continuity for $f$ at $c$.

Similarly, the function $f(x)=a$ is continuous at $c\in A$ with (local) modulus of continuity $\sigma_p$ no matter what $p$ we choose. This makes sense intuitively, since $f$ is a constant function. It is already at its limit point no matter what $x$ in the domain we choose, so it will always converge to its limit quicker than, or at least at the same rate, as any other function we could choose.

Example 2$f(x)=x^2$
p=1: Consider $\sigma_1=d$

We will think about this example using the picture above. The double-edged cone area is where $|f(x)-f(c)|\leq \sigma_1 |x-c|$ for $|x-c|<\delta$. If the $f(x)$ (where $x$ is in $|x-c|<\delta$) falls into the double-edges cones, it pretends the local modulus of continuity.
When $c\in(-1,1)$, all of the points in $f(x)=x^2$ fall into the double-edged cones. Thus $\sigma_1$ is a local moduli of continuity for $f$ when looking at $c\in(-1,1)$.

When $c\not\in(-1,1)$,say $x=2$, the graph of the function $f(x)=x^2$ (the green highlighted part) does not go inside the double-edged cones. Thus $\sigma_1$ is not a local moduli of continuity for $f$ when looking at $c\not\in(-1,1)$.

### Connection to Derivatives

A simple way to see whether the graph of function falls into the cones that describe the modulus of continuity is to compare the slope of the function at the point $c$ with the slope of the modulus of continuity.
For example: when $x=2$ ,The function is differentiable and has a slope of 4 at $x=2$. The function is growing locally with a rate of 4. But we are comparing it with $\sigma_1=d$ with a growth rate of 1. The graph of the function stays outside of the cone.
So for any slope that is larger than 1, $\sigma_1$ is not a local modulus of continuity for that function at that point.

Similarly, considering $\sigma_1(d)=md$. The m is the slope of the cones that describe the modulus of continuity. Any derivative of the function at a point that is greater than m means $\sigma_1$ is not a local modulus of continuity for that function at that point. So in general, $\sigma_1=md$ is a local modulus of continuity for $f$ at $c$ if $m\geq f'(c)$.

### References

https://arxiv.org/pdf/math/0607672.pdf

https://www.revolvy.com/main/index.php?s=Modulus%20of%20continuity&item_type=topic

http://www.stat.cmu.edu/~cshalizi/754/notes/lecture-08.pdf

http://fourier.eng.hmc.edu/e176/lectures/NM/node3.html

#### One Response

1. Jeremy LeCrone says:

Very Good Exploration of these concepts. Thank you for the blog post.