4

We have defined continuity as $\forall \epsilon, \exists \delta > 0 s.t. |x-c| < \delta, x \in D \implies |f(x) - f(c)| < \epsilon$

and uniform continuity as $\forall \epsilon, \exists \delta > 0 s.t. |x-y| < \delta, x,y \in D \implies |f(x) - f(y)| < \epsilon$

The only difference I see is that we have $x,y$ in the domain rather than just $x$, but what does that really change? Furthermore, we have a theorem about the continuity on a closed domain implies uniform continuity, why does this work? I understand the proof but I'm just not sure why it works.

user144464
  • 387

3 Answers3

6

Actually, the definitions of continuity and uniform continuity are:

Continuity: $$\forall c\in D: \forall\epsilon > 0\exists \delta > 0: \forall x\in D: |x-c|<\delta\implies|f(x)-f(y)|<\epsilon$$

Uniform continuity: $$\forall\epsilon > 0\exists \delta > 0: \forall x,y\in D: |x-y|<\delta\implies|f(x)-f(y)|<\epsilon$$

Which is a huge difference.

A function is continuous if, for every point $c$, if I pick a value $x$ close to $c$, then $f(x)$ will have to be close to $c$.

A function is uniformly continuous if for any pair of points $x,y$ that is close enough, the values $f(x)$ and $f(y)$ will be close as well.

For example, let $\epsilon$ be $0.001$. Take the function $f(x)=\frac1x$ on $(0,1)$. If I take any point $c$ on $(0,1)$, then if I take the value $x$ close to $c$, then $\frac{1}{x}$ will be close to $\frac{1}{x}$. Take $c=\frac{1}{1000}$. If $x=\frac{1,000,001}{1,000,000,000} = c+\frac{1}{1,000,000,000}$, then $f(x) = \frac{1,000,000,000}{1,000,001}\approx999.999$ which is close to $f(c)=1000$. This means that if $x$ is $\frac{1}{1,000,000,000}$ close to $c$, then $f(x)$ will be $\epsilon$-close to $f(c)$.

On the other hand, I can still find a pair of values $x,y$ which are arbitrarily close and the values $f(x)$ and $f(y)$ will not be. For example, taking $x=\delta$ and $y=2\delta$, you have that $f(x)-f(y)=\frac{1}{\delta}-\frac{1}{2\delta} = \frac{1}{2\delta}$ which can be very large (if $\delta = \frac{1}{1000}$, then $f(x)-f(y)$ for this pair of values is $500$).

5xum
  • 123,496
  • 6
  • 128
  • 204
2
  1. You are wrong, there is another big difference. In the definition of continuity, the point $c$ is fixed before picking $\varepsilon>0$. In the definition of uniform continuity, there is no special point, and $x$ and $y$ are considered after picking $\varepsilon >0$ and its companion $\delta>0$. In other words, continuity refers to a particular point (a pointwise definition), while uniform continuity refers to a whole set.
  2. Continuity on a compact set implies uniform continuity. The line $(-\infty,+\infty)$ is a closed set, but it is easy to define a continuous function on $\mathbb{R}$ that is not uniformly continuous.
Siminore
  • 35,136
  • Thanks, you say that there are a lot of examples where a function is continuous but not uniformally continuous. The thing is, I can't see why that would hold. If someone asked me to graph a function that was not continuous I'd have no problems doing it, but for not uniform continuous I wouldn't know how to do it. – user144464 May 12 '14 at 07:53
  • Consider for instance $x \mapsto \sin (x^2)$: as $x \to +\infty$, you can find points that lie arbitrarily close to each other, but whose images are not close. – Siminore May 12 '14 at 10:02
1

Firstly, the first definition you gave was for continuity at $c$, for continuity on the whole interval $D$ we need the following to hold:

$\forall\epsilon\gt 0,\exists \delta\gt 0,|x-c|\lt \delta , x\in D\Rightarrow |f(x)-f(c)|\lt\epsilon$

For each $c\in D$.

But you can notice that the definition of uniform continuity is nicely contained in the below definition:

$\forall\epsilon\gt 0,\exists \delta\gt 0,|x-y|\lt \delta , x,y\in D\Rightarrow |f(x)-f(y)|\lt\epsilon$

Now we discuss the differences, they are subtle in terms of terminology, but very different in nature.

If we consider standard continuity, if $c$ changes, then $\forall\epsilon\gt 0$ we need a new "$\delta$" for the definition to hold, so the continuity depends on which part of the interval we are in. This already points to non uniformity, as things are changing throughout our interval.

Whereas for uniform continuity $\forall\epsilon\gt 0,\exists \delta\gt 0$, such that if any two points in the whole interval have less than $\delta$ distance from each other, then the function evaluated at those points will be less than $\epsilon$ distance apart.

These is stronger than standard continuity as our $\delta$ only depends on $\epsilon$, and it does not depend on which points in the interval we consider.

Ellya
  • 11,783