# Uniformly continuous (real analysis)

536 views

I’m looking at the definition and can’t wrap my head around

In: Mathematics

In simple terms: if x and y are very close, then f(x) and f(y) must also be very close for a uniformly continous f.

There exists a distance between x and y that brings the difference between f(x) and f(y) below some arbitrary limit epsilon (as small as we want)

In practise that means that the function has no infinite values, and no infinite slopes anywhere.

Typical counterexamples are 1/x (infinite value and slope at values close to 0) and the heavyside function (infinite slope at 0, if x is positive and y is negative then no matter how close you put them the difference between f(x) and f(y) is always 1 so we can’t make it as small as we want)

The opposite would be (guess) continuous by chuncks.

Take a pen. Can you draw a single line from left to right that follow your function?

* Yes, it’s uniformly continuous — cos(x)
* No, It’s not uniformly continuous — 1/x
* If you can’t draw a line for part of the region, then it’s not even continuous (don’t have an example for that one)

PS: I tried to be as ELI5 as possible, but I might be wrong given it’s not my native language, and especially for mathematics, word carry precise meaning.

The first time I ever learnt about the definition of uniformly continuous I couldn’t understand the difference with the definition of continuity. Let’s see both definitions and we will try to understand what do they mean and why they are different.

The definition of continuity is as follows:

>A function f is continuous at a point *a* if given ε > 0, there exists δ > 0 such that if *x* is a point satisfying |*x* – *a*| < δ then |f(*x*) – f(*a*)| < ε.

What this definition is saying is that a function is continuous at a point *a* if you take *x* close to *a* then f(*x*) will be close to f(*a*) (if the distance from *x* to *a* is less than δ then the distance from f(*x*) to f(*a*) will be less than ε).

There are two key things in this definition: First one is that the choice of δ will depend on ε (I think it’s obvious because if you want f(*x*) and f(*a*) to be very close (ε) then *x* and *a* will need to be very close (δ)). The second one is not that obvious and it is actually the key to uniform continuity: δ will also depend on the point *a*. This is very important so let’s look at this example:

Consider f(*x*) = *x*^2 which you should know is continuous at every real number *x*. Consider two points *a* = 0 and *b* = 200. For the point *a* let’s consider the point *c* = 1 (which is 1 unit away from *a*). Then, f(*a*) = 0^2 = 0 and f(*c*) = 1^2 = 1. So we get that f(*c*) – f(*a*) = 1. Now take *d* = 201, and now we get that f(*b*) = 200^2 = 40000 and f(*d*) = 201^2 = 40401 and thus f(*d*) – f(*b*) = 401.

In both cases we took points which were 1 unit away. However, the distance between their images varied a lot. So if we want to find points whose images are close to f(*a*) we need to get points which are somewhat closet to *a*. But if we want to find points whose images are close to f(*b*) we need get points which need to be really close to *b*. In fact, if you want to find a point e whose image is 1 unit away from f(*b*) you’d need something smaller than *e* = 200.0025, which is 0.0025 units away from b (compare it to a and c, which were only 1 unit away).

Now we know that δ depends on ε (how close we want the images to be) and on the point *a* (as shown by the example), let’s look at the definition of uniform continuity.

>A function f is uniformly continuous if given ε > 0, there exists δ > 0 such that if *x* and *y* are points satisfying |*x* – *y*| < δ then |f(*x*) – f(*y*)| < ε.

I’ll be honest, the first time I saw this definition I couldn’t see the difference, but they are *very different* although very subtle. In fact, if a function is uniformly continuous then is continuous. What’s the difference, then? Can you see it?

The key is in the first definition, given a point and ε you find some δ, which might only works for that point as we showed (by the example: the first two points were 1 unit away and their images were 1 unit away, but even though the second ones were also 1 unit away their images were 401 units away), whereas in the second one, given ε **there is a δ which works for every two points**. Quote from Wikipedia:

>Thus for continuity at each point, one takes an arbitrary point *x*, and then there must exist a distance δ […], while for uniform continuity a single δ must work uniformly for all points *x* (and *y*) […].

[Here’s the link](https://en.wikipedia.org/wiki/Uniform_continuity#Local_continuity_versus_global_uniform_continuity). By the way, the term *uniformly* comes from the fact than in standard continuity, each point will have it’s own δ but in uniform continuity there is a δ that will work for every point, making everything uniform (you’d only need a single δ).

Consider the function f(x) = 1/x defined on the open interval (0,1). This function is continuous but we will prove that it’s not uniformly continuous:

Let ε = 1. Our question is: Can we find some δ >0 such that if two points verify |*x* – *y*| < δ then |f(*x*) – f(*y*)| < 1? We will show that no matter what δ > 0 you find, there will be two points *a* and *b* such that |*a* – *b*| < δ but |f(*a*) – f(*b*)| >= 1. In fact, consider *a* = 1/*n* and *b* = 1/(*n*+1) (where *n* is a positive integer). You can show that *a* -* b* = 1/(*n* *(*n*+*1*)) which goes to 0 as *n* goes to infinity. So given δ > 0 there will be some integer *M* such that 0 < a – b < δ as long as we take *n* > *M* (I hope you can see where this comes from). However, f(*b*) – f(*a*) = f(1/(*n*+1)) – f(1/*n*) = *n* + 1 – *n* = 1 >= 1. So this function is continuous in (0,1) but not uniformly continuous.

There is a theorem that if your function is continuous and is defined on a closed inteval [a,b] then that function is uniformly continuous. The idea is that for ε > 0 every point will have it’s own δ > 0, but unlike 1/x, because continuous functions on closed intevals don’t go to infinity, there must be a minimun δ. And because each other δ will be bigger, such minimun δ will work for every point as well (if points being less than 1 unit appart works, then points being less than 0.0025 units appart will also work because 0.0025 < 1) so you just proved that for that ε and for that minimum δ you have precisely what uniform continuity states.

Continuity means that given epsilon, for any point a on the function’s domain, I can find a delta such that|x-a|<delta –> |f(x)-f(a)| < epsilon. In other words, delta may depend on the specific value of a that I’m looking at. So we can think of delta as dependent on a. I like to write it delta(a) as a reminder that it’s really a function.

Now, uniform continuity just means that delta(a) is a constant. Ie, it does not depend on the choice of a. In other words, given epsilon, I can find a delta such that |x-a|< delta –> |f(x)-f(a)| *for any* a.

Notice that in continuity, I choose a first, then find a delta that works for that specific a, whereas in uniform continuity I get to pick delta first, since it works for any and all values of a.