I don't understand uniform continuity

[Arian.D]

I don't understand uniform continuity :(

I don't understand what uniform continuity means precisely. I mean by definition it seems that in uniform continuity once they give me an epsilon, I could always find a good delta that it works for any point in the interval, but I don't understand the significance of this definition and I don't know how I could study if a function is uniformly continuous on an interval or not.
For some functions it's easy, for example I've already proved that the function f(x)=ax+b is uniformly continuous over R for any a,b, because I could always choose delta to be epsilon/a and it doesn't depend on the point that I'm writing down the definition of continuity for it.
I think I need to see more examples.

If someone writes down a lot of examples here and proves that some of them are uniformly continuous while the others aren't I'll be happy :). I need to see how the definition works in real-situation examples. I hope I'm not asking too much.

Thanks in advance.

 

[SteveL27]

I don't understand what uniform continuity means precisely. I mean by definition it seems that in uniform continuity once they give me an epsilon, I could always find a good delta that it works for any point in the interval, but I don't understand the significance of this definition and I don't know how I could study if a function is uniformly continuous on an interval or not.
For some functions it's easy, for example I've already proved that the function f(x)=ax+b is uniformly continuous over R for any a,b, because I could always choose delta to be epsilon/a and it doesn't depend on the point that I'm writing down the definition of continuity for it.
I think I need to see more examples.

If someone writes down a lot of examples here and proves that some of them are uniformly continuous while the others aren't I'll be happy :). I need to see how the definition works in real-situation examples. I hope I'm not asking too much.

Thanks in advance.

In plain old continuity, you give me an x and and epsilon; I'll give you back a delta. If you gave me a different x and the same epsilon, I'd give you a different delta. Given a fixed epsilon, delta is a function of x.

In uniform continuity, you give me an epsilon and I'll give you back a delta that does not depend on x.

In other words in continuity, the function can "stretch" in such a way that the delta that works for some epsilon at x, fails for a different x. But with uniform continuity, the same delta works for epsilon regardless of x.

Perhaps that helps.

(ps) Also it will make a lot more sense if you think about a particular function that's continuous but not uniformly continuous. f(x) = 1/x is a good example. Do you see why it fits the definition of continuity but not uniform continuity?

 

[mathwonk]

try this

you asked about uniform continuity. it just means that if you want the values f(x) - f(y) to be smaller than e, you only need the distance between x and y to be smaller than another certain distance d.

 

[Arian.D]

Perhaps that helps.

(ps) Also it will make a lot more sense if you think about a particular function that's continuous but not uniformly continuous. f(x) = 1/x is a good example. Do you see why it fits the definition of continuity but not uniform continuity?

That does help. Thank you.

No, I don't see why it fits the definition of continuity but not uniform continuity. I know that 1/x is continuous everywhere except at x=0, but in Calculus they taught us to study continuity only at one point, not on a whole interval. They defined we say f is continuous on [a,b] if it's continuous at every point within the interval but they never said anything about how to show that by definition.

try this

you asked about uniform continuity. it just means that if you want the values f(x) - f(y) to be smaller than e, you only need the distance between x and y to be smaller than another certain distance d.

The pdf you gave has errors in it. For instance, It's not true that |a+h| <= |a| - |h|. a=3 and h=1 is a counterexample to this, but it helped me a lot. Thanks.well, I tried to prove that some elementary functions are uniformly continuous over R. Please check my proofs:1) y=ax+b is uniformly continuous over R.
\forall \epsilon &gt; 0, \exists \delta &gt;0 : |x-y|&lt;\delta \implies |f(x)-f(y)|&lt;\epsilon
\forall \epsilon &gt; 0, \exists \delta &gt;0 : |x-y|&lt;\delta \implies |(ax+b) - (ay+b)| &lt; \epsilon
|a(x-y)|&lt;\epsilon \implies |a||x-y| &lt; \epsilon \implies |x-y| &lt; \frac{\epsilon}{|a|}
Therefore if we set \delta = \frac{\epsilon}{|a|} the definition works and it's obvious that delta is not dependent on y.

2) y=sin(x) is uniformly continuous over R.
We start with a tautology: \forall x \in \mathbb{R} : |sinx| \leq |x|. Since we have |\sin(x)-\sin(y)|=|2 \sin(\frac{x-y}{2}) \cos(\frac{x+y}{2})| and we have \forall x \in \mathbb{R} : |\cos(x)| \leq 1 we conclude that:
\forall x,y \in \mathbb{R} : |\sin(x)-\sin(y)| \leq 2|\frac{x-y}{2}| = |x-y| \implies |\sin(x)-\sin(x)|=|x-y|

Now for any given \epsilon if we set \delta = \frac{\epsilon}{2} we'll have:
\forall \epsilon&gt;0, \exists\delta&gt;0, \forall x,y \in \mathbb{R}: |x-y|&lt;\delta \implies |sin(x)-sin(y)| \leq |x-y| &lt; \delta=\frac{\epsilon}{2} &lt; \epsilon

3) y=cos(x) is uniformly continuous over R.
The proof is exactly the same, this time we use the fact that |\cos(x)-\cos(y)| \leq |x-y|.

4) y=√x is uniformly continuous over R^≥0.
Suppose that x,y are two non-negative real numbers, without loss of generality we could assume that x≥y≥0.
Now if we set x-y&lt;\epsilon^2 we'll have x&lt;\epsilon^2 + y. Now we'll have:
\sqrt{x} &lt; \sqrt{\epsilon^2 + y} \leq \sqrt{\epsilon^2} + \sqrt{y} \implies \sqrt{x} - \sqrt{y} &lt; \epsilon
But x - y \leq 0 \implies \sqrt{x} - \sqrt {y} \leq 0 since √x is monotonically increasing. therefore we could write x-y=|x-y| (by hypothesis) and √x - √y = |√x - √y|. What we've shown so far is if we set \delta = \epsilon^2 we have:
\forall\epsilon&gt;0,\exists \delta&gt;0, \forall x,y\geq0: |x-y|&lt; \delta \implies |\sqrt{x} - \sqrt{y}|&lt; \epsilon

I've also concluded these results:

1) If f and g are uniformly continuous over [a,b], so is (f+g).
Proof:
\forall \epsilon&gt;0, \exists \delta&gt;0: |x-y|&lt;\delta \implies |f(x)-f(y)|&lt;\epsilon
\forall \epsilon&gt;0, \exists \delta&#039;&gt;0: |x-y|&lt;\delta&#039; \implies |g(x)-g(y)|&lt;\epsilon
Now if \delta and \delta&#039; could be chosen independently from y, their minimum is also independently chosen. (obvious fact!). Therefore:
\forall \epsilon&gt;0 \exists d&gt;0: |x-y|&lt;d \implies |(f+g)(x)-(f+g)(y)| \leq |f(x)-f(y)|+|g(x)-g(y)| &lt; 2\epsilon.
That ends the proof.

2) If f and g are uniformly continuous over [a,b], (f.g) could be uniformly continuous or not.

Proof:
y=x is uniformly continuous over R, but x² is not uniformly continuous over R.
y=√x is uniformly continuous over [0,∞), y=x is also uniformly continuous over [0,∞).
So both cases could happen.

3) If f and g are uniformly continuous over [a,b], so is (cf). (c is a constant).
The proof is very easy.Now I got the following questions, I would be very glad if someone could answer them here or give me a hint to do it on my own:

1) Is e^x uniformly continuous over R?
2) If f is uniformly continuous over [a,b], what could we say about its inverse?
3) Are sinh(x) and cosh(x) uniformly continuous over R?
4) Could we say if a function is uniformly continuous from its Taylor series?

Thanks.

 

[mathwonk]

thanks for the correction. fortunately the incorrect statement is not used, and when it is used a few lines later it is used correctly.

those notes are not polished, they are just the actual classroom lecture notes that i wrote up at night and handed out everyday. as such i am glad they helped.

 

[micromass]

You might want to prove the following two nice theorems:

1) If f:[a,b]\rightarrow \mathbb{R} is continuous, then it is uniformly continuous. This theorem can be extended by taking instead of [a,b] any compact domain.

2) f is uniformly continuous if and only if for every two sequence (x_n)_n and (y_n)_n hold that |x_n-y_n|\rightarrow 0 implies |f(x_n)-f(y_n)|\rightarrow 0

This last one can be used to show that f(x)=e^x is not uniformly continuous on \mathbb{R}. For example, take x_n=n and y_n=n+\frac{1}{n}

Important is also:

3) If f is uniformly continuous and if (x_n)_n is a Cauchy sequence, then (f(x_n))_n is a Cauchy sequence.

Another nice obeservation:

4) If f is differentiable and if f^\prime is bounded, then f is uniformly continuous.

 

[mathwonk]

as to e^x being uniformly continuous over all of R, that seems unlikely doesn't it? look at

e^(x+h). Is there some small value of h that will insure this is always within, say 1, of e^x, for all x?

i.e. fix h, and try to find x so that e^(x+h) is a good bit bigger than e^x.try to notice the links between micromass's proposed theorems and some of your questions.

 

[Arian.D]

You might want to prove the following two nice theorems:

1) If f:[a,b]\rightarrow \mathbb{R} is continuous, then it is uniformly continuous. This theorem can be extended by taking instead of [a,b] any compact domain.

2) f is uniformly continuous if and only if for every two sequence (x_n)_n and (y_n)_n hold that |x_n-y_n|\rightarrow 0 implies |f(x_n)-f(y_n)|\rightarrow 0

This last one can be used to show that f(x)=e^x is not uniformly continuous on \mathbb{R}. For example, take x_n=n and y_n=n+\frac{1}{n}

Important is also:

3) If f is uniformly continuous and if (x_n)_n is a Cauchy sequence, then (f(x_n))_n is a Cauchy sequence.

Another nice obeservation:

4) If f is differentiable and if f^\prime is bounded, then f is uniformly continuous.

Wow. These theorems are very useful. I'm not sure if I could prove them on my own because I'm not so much into Analysis, but I guess some of them are proved in a good (but devilish) analysis book like baby Rudin perhaps.
The fourth one is very useful I think, it somehow answers my second question.

Theorem: If f is differentiable on [a,b] and bounded and is not zero in the interval, f^-1 is uniformly continuous. (I guess).
My proof: well, suppose we have y=f(x), if the inverse exists, we have x=f^-1(y), by implicit differentiation we get: 1 = y&#039; . (f^{-1})^\prime(y).
Therefore:
(f^{-1})^\prime(y) = \frac{1}{f^\prime(y)}
So if f is differentiable and f' is bounded and is not zero in the interval, the inverse function is uniformly continuous.
Right?

as to e^x being uniformly continuous over all of R, that seems unlikely doesn't it? look at

e^(x+h). Is there some small value of h that will insure this is always within, say 1, of e^x, for all x?

i.e. fix h, and try to find x so that e^(x+h) is a good bit bigger than e^x.try to notice the links between micromass's proposed theorems and some of your questions.

I don't understand what you're trying to get at. Would you explain more please?

 

[micromass]

Wow. These theorems are very useful. I'm not sure if I could prove them on my own because I'm not so much into Analysis, but I guess some of them are proved in a good (but devilish) analysis book like baby Rudin perhaps.
The fourth one is very useful I think, it somehow answers my second question.

Theorem: If f is uniformly continuous on R, so is its inverse. (I guess).
My proof: well, suppose we have y=f(x), if the inverse exists, we have x=f^-1(y), by implicit differentiation we get: 1 = y&#039; . (f^{-1})^\prime(y).
Therefore (f^{-1})^\prime(y) = \frac{1}{f^\prime(y)}
So if f is differentiable and f' is bounded and is not zero in the interval, the inverse function is uniformly continuous.
Right?

No. It's not because something is uniformly continuous and has an inverse that it is differentiable! You can't use derivatives here.

I don't think it's true by the way, look at \sqrt{x} which is uniformly continuous and x^2 which is not.

If you ask about uniform continuous on [a,b], then it's true.

 

[Arian.D]

No. It's not because something is uniformly continuous and has an inverse that it is differentiable! You can't use derivatives here.

I don't think it's true by the way, look at \sqrt{x} which is uniformly continuous and x^2 which is not.

You're very quick to check the replies. I corrected the statement, check it again please.

 

[micromass]

You're very quick to check the replies. I corrected the statement, check it again please.

That seems alright, but it can be generalized.

It is true that:

Theorem: if f is a continuous bijection on [a,b], then the inverse function is uniformly continuous

The proof depends on some other results:

Theorem: A continuous bijection with domain [a,b] has a continuous inverse.

Theorem: A continuous function maps a closed interval to a closed interval.

Theorem: A continuous map on [a,b] is uniformly continuous.

 

[Arian.D]

That seems alright, but it can be generalized.

It is true that:

Theorem: if f is a continuous bijection on [a,b], then the inverse function is uniformly continuous

The proof depends on some other results:

Theorem: A continuous bijection with domain [a,b] has a continuous inverse.

Theorem: A continuous function maps a closed interval to a closed interval.

Theorem: A continuous map on [a,b] is uniformly continuous.

so, by this theorem, e^x is uniformly continuous because lnx is a continuous bijection on (0,∞). (I don't know what you mean by a continuous bijection though, I guessed you mean a map which is continuous and bijective).

 

[micromass]

so, by this theorem, e^x is uniformly continuous because lnx is a continuous bijection on (0,∞). (I don't know what you mean by a continuous bijection though, I guessed you mean a map which is continuous and bijective).

No. I specifically said: a continuous bijection on [a,b]. Things like (0,∞) or \mathbb{R} are not closed intervals. The theorems I listed are all false for functions whose domains are not closed intervals.

And yes, a continuous bijection is something that is continuous and bijective.

 

[Arian.D]

No. I specifically said: a continuous bijection on [a,b]. Things like (0,∞) or \mathbb{R} are not closed intervals. The theorems I listed are all false for functions whose domains are not closed intervals.

And yes, a continuous bijection is something that is continuous and bijective.

well, right, I think you're interested in compact sets which are closed intervals in \mathbb{R}. One another reason that analysis is not my field of interest.

Okay. Why uniform continuity is interesting? Does it have any interesting results that convince us to distinguish it from simple continuity?

 

[micromass]

One of the interesting results is:

If f:A\rightarrow \mathbb{R} is uniformly continuous, then it has a continuous extension f:\overline{A}\rightarrow \mathbb{R}.

This is a very useful theorem as it allows you to do nice things. For example, it can be used to define integrals.

 

[Arian.D]

One of the interesting results is:

If f:A\rightarrow \mathbb{R} is uniformly continuous, then it has a continuous extension f:\overline{A}\rightarrow \mathbb{R}.

This is a very useful theorem as it allows you to do nice things. For example, it can be used to define integrals.

Sounds nice.

Were my proofs correct? Does the set of all uniformly continuous functions form a vector space? (I tried to show that its closed under function addiction and scalar multiplication).

After all these nice theorems, I'm still wondering whether e^x is uniformly continuous or not. Mathwonk said something, but I didn't quite understand him/her :(.

 

[micromass]

Were my proofs correct? Does the set of all uniformly continuous functions form a vector space? (I tried to show that its closed under function addiction and scalar multiplication).

That is correct.

After all these nice theorems, I'm still wondering whether e^x is uniformly continuous or not. Mathwonk said something, but I didn't quite understand him/her :(.

It is not. Mathwonk and I gave you proofs.

 

[Arian.D]

It is not. Mathwonk and I gave you proofs.

Where did you propose a proof?
I didn't understand what Mathwonk said unfortunately :( It was a bit over the top for my brain I think. I kinda guessed what he was trying to get at, but I didn't understand it precisely.

 

[micromass]

Here:

2) f is uniformly continuous if and only if for every two sequence (x_n)_n and (y_n)_n hold that |x_n-y_n|\rightarrow 0 implies |f(x_n)-f(y_n)|\rightarrow 0

This last one can be used to show that f(x)=e^x is not uniformly continuous on \mathbb{R}. For example, take x_n=n and y_n=n+\frac{1}{n}

 

[mathwonk]

I wAS JUST THINKING you would expand e^(x+h) as e^x.e^h. Then subtracting we get

e^(x+h) - e^x = e^x.e^h - e^x = e^x(e^h - 1).

now uniform contimnuity would say you can always make this as small as you want,

just by making h small. I.e., some small h would have to make this small for all x.but any h > 0 at all will have e^h - 1 > 0. Then with h chosen, for big enough x, the product

e^x(e^h - 1) will be really big.

If you still don't get it, you still need to think hard about what uniform continuity means.

 

[Arian.D]

I wAS JUST THINKING you would expand e^(x+h) as e^x.e^h. Then subtracting we get

e^(x+h) - e^x = e^x.e^h - e^x = e^x(e^h - 1).

now uniform contimnuity would say you can always make this as small as you want,

just by making h small. I.e., some small h would have to make this small for all x.but any h > 0 at all will have e^h - 1 > 0. Then with h chosen, for big enough x, the product

e^x(e^h - 1) will be really big.

If you still don't get it, you still need to think hard about what uniform continuity means.

Now it's clear. It gets a bit hard to understand you when you don't use latex codes but your post was helpful. Thanks.

2) f is uniformly continuous if and only if for every two sequence (x_n)_n and (y_n)_n hold that |x_n-y_n|\rightarrow 0 implies |f(x_n)-f(y_n)|\rightarrow 0

This last one can be used to show that f(x)=e^x is not uniformly continuous on \mathbb{R}. For example, take x_n=n and y_n=n+\frac{1}{n}

Well, so by your theorem, we expect lim_{n\to\infty}e^n - e^{n+1/n} to go to zero, but it could be shown by calculus that this thing goes to infinity as n gets larger.

lim_{n\to\infty} \frac{1-e^{1/n}}{e^{-n}} Using a generalized version of L'hopital's rule we get:
lim_{n\to\infty} \frac{\tfrac{1}{n^2}e^{1/n}}{-e^{-n}}
but this thing goes to infinity. That's why the theorem you suggested implies that e^x is not uniformly continuous.

Is this what you wanted to say?

 

[micromass]

Yeah, that's it.

 

[Arian.D]

And the last question on the topic, where can I find the proofs of your proposed theorems? In particular I'm interested to see the proof for #2 and #4 because they look very nice.

 

[mathwonk]

those two theorems are really just a little exercise to see if you understand the definition of uniform continuity, assuming you know the mean value theorem. In fact one of those theorems is actually proved in the pdf file I posted.

 

[Skrew]

In general to show that a function fails to be uniformly continuous, you are given a delta and you choose a a constant C, such that there exists x and y where |x-y| < delta but |f(x) - f(y)| >= C.

Also a function with a bounded derivative is something stronger then uniform continuity. (it begins with an L...)

 

[Arian.D]

those two theorems are really just a little exercise to see if you understand the definition of uniform continuity, assuming you know the mean value theorem. In fact one of those theorems is actually proved in the pdf file I posted.

Well, then I'll try my best to prove them on my own.
Just one question, do we need to assume x_n and y_n are convergent sequences? because today I was talking about e^x with some guy and he said that it was uniformly continuous, then I told them about that theorem and he replied back that maybe those two sequences should be convergent, so do the two sequences need to be convergent or the theorem also holds for divergent sequences? (because we used two divergent sequences to prove that e^x is NOT uniformly continuous.

In general to show that a function fails to be uniformly continuous, you are given a delta and you choose a a constant C, such that there exists x and y where |x-y| < delta but |f(x) - f(y)| >= C.

Also a function with a bounded derivative is something stronger then uniform continuity. (it begins with an L...)

Yup, That's kinda obvious, but thanks for pointing it out.

 

[micromass]

No, it holds for every two sequences.

And e^x is not uniformly continuous at all on R, so the person you talked to is wrong.

 

[Arian.D]

OK.
This is what I've done so far:

Theorem: f is uniformly continuous if and only if for any two sequences \{x_n\}_{n\in \mathbb{N}} and \{y_n\}_{n \in \mathbb{N}}, |x_n - y_n| \to 0 implies |f(x_n) - f(y_n)| \to 0
Well, I assume that f is a real valued function and its domain is a subset of R, I also assume that x_n and y_n are two real-valued sequences.

Proof: If f is uniformly continuous on R, then for any two x and y in the domain and for any given epsilon>0 we could find a delta which is independent from epsilon such that |x-y|&lt;\delta \implies |f(x)-f(y)|&lt;\epsilon. Since \{x_n\}_{n \in \mathbb{N}} and \{y_n\}_{n \in \mathbb{N}} are two real valued sequences and that definition works for any x and y real numbers in the domain we conclude that |x_n - y_n|&lt;\delta \implies |f(x_n) - f(y_n)|&lt;\epsilon provided that f(x_n) and f(y_n) are defined, which means x_n and y_n should be in the domain of f.

To show that the converse holds, I'm going to prove the contrapositive holds, suppose that |x_n - y_n| \to 0 \implies|f(x_n) - f(y_n)| \to 0 is false, therefore there exist two sequences that |x_n - y_n| \to 0 but |f(x_n) - f(y_n)| \to 0 is false.
|f(x_n) - f(y_n)| \to 0 is false is equivalent to the following proposition:
\forall\delta&gt;0, \exists \epsilon&gt;0: |x_n - y_n|&lt;\delta \implies |f(x_n) - f(y_n)| \geq \epsilon
But this is equivalent to f not being uniformly continuous (as skrew pointed out). Therefore the theorem is proved.

Is my proof correct?

 

[Skrew]

The last part is wrong.

The statement "|x_n - y_n| ->0 does not imply |f(x_n) -f(y_n)| -> 0"

Says that for every delta > 0 there exists a i such that |x_i - y_i| < delta but |f(x_i) - f(y_i)| >= constant.

In particular your constant epsilon does not change when disproveing uniform continuity, only the delta.

Also e^x is uniformly continuous on any bounded interval.

 

[SteveL27]

(because we used two divergent sequences to prove that e^x is NOT uniformly continuous.

There's an intuitive way to see that e^x is not uniformly continuous on the reals.

Visualize the graph of e^x. As you go to the right along the positive x-axis, e^x gets steeper and steeper. The graph "stretches out," so that if you go far enough to the right, you can make a very small change in x be a huge change in e^x.

Do you see that?

It's the exact same problem that one has with f(x) = 1/x. With that function, as you get closer to zero on the positive x-axis, you can make f(x) vary hugely. It's the same issue.

A function that's continuous is one that doesn't have a tear or rip in its graph. (Speaking intuitively, as I am in this post).

A function that's uniformly continuous is one that doesn't stretch too much.

So a function like 1/x or e^x is a function that is continuous, BUT it stretches out a lot. If we pulled any harder we'd tear it.

Think of the graph of a function as a piece of stretch Turkish taffy. Or silly putty if you like.

If a little pull makes a little stretch: it's uniformly continuous.

If a little pull can sometimes make a big stretch: its continuous but NOT uniformly continuous.

If a little pull tears it into two pieces: It's not continuous.

So the condition of continuity-but-not-uniform-continuity (for which there's no good word AFAIK, but there should be) is a middle state between being uniformly continuous and discontinuous.

Do you see that? Because if you get what I wrote, then you get uniform continuity. If you don't see the visualization, pushing the symbols around will not help with intuition.

Uniform continuity means that a small change in input will ALWAYS give you a small change in output. Continuity without uniform continuity means that a small change in input may give you a HUGE change in output; as with 1/x near zero, and e^x as you move to the right on the positive x-axis; but that you still won't tear the graph.

Discontinuity means that you can make a tiny change in the input and you'll rip the graph into two pieces.

By the way I would like to say that the discussion of compact sets is obscuring the basic intuition; and I think the OP should dwell on the examples of 1/x and e^x before trying to sling proofs about continuous functions on compact sets. My two cents on that one.

 

[Arian.D]

The last part is wrong.

The statement "|x_n - y_n| ->0 does not imply |f(x_n) -f(y_n)| -> 0"

Says that for every delta > 0 there exists a i such that |x_i - y_i| < delta but |f(x_i) - f(y_i)| >= constant.

In particular your constant epsilon does not change when disproveing uniform continuity, only the delta.

Also e^x is uniformly continuous on any bounded interval.

I don't understand why the last part is wrong :(. Where did 'i' come to play?
This is what I've done step by step written down in propositional calculus:

Suppose that \forall x_n, \forall y_n:|x_n - y_n| \to 0 \implies |f(x_n)-f(y_n)| \to 0 is false:
1) Negation: \exists x_n, \exists y_n: |x_n - y_n|\to 0 \land \neg(|f(x_n)-f(y_n)| \to 0)
2) conjunction elimination: \exists x_n, \exists y_n: \neg(|f(x_n)-f(y_n)| \to 0)
3) \neg(|f(x_n)-f(y_n)| \to 0) \equiv \forall\delta&gt;0, \exists\epsilon&gt;0 : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|\geq \epsilon
4) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon
5) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon \equiv \neg(\forall\epsilon&gt;0, \exists \delta&gt;0, \forall x_n, \forall y_n : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|&lt;\epsilon)

This is what my argument was in symbolic logic, I usually suck at propositional calculus but when I write things in that way I find it easier to argue with people.

 

[Arian.D]

There's an intuitive way to see that e^x is not uniformly continuous on the reals.

Visualize the graph of e^x. As you go to the right along the positive x-axis, e^x gets steeper and steeper. The graph "stretches out," so that if you go far enough to the right, you can make a very small change in x be a huge change in e^x.

Do you see that?

It's the exact same problem that one has with f(x) = 1/x. With that function, as you get closer to zero on the positive x-axis, you can make f(x) vary hugely. It's the same issue.

A function that's continuous is one that doesn't have a tear or rip in its graph. (Speaking intuitively, as I am in this post).

A function that's uniformly continuous is one that doesn't stretch too much.

So a function like 1/x or e^x is a function that is continuous, BUT it stretches out a lot. If we pulled any harder we'd tear it.

Think of the graph of a function as a piece of stretch Turkish taffy. Or silly putty if you like.

If a little pull makes a little stretch: it's uniformly continuous.

If a little pull can sometimes make a big stretch: its continuous but NOT uniformly continuous.

If a little pull tears it into two pieces: It's not continuous.

So the condition of continuity-but-not-uniform-continuity (for which there's no good word AFAIK, but there should be) is a middle state between being uniformly continuous and discontinuous.

Do you see that? Because if you get what I wrote, then you get uniform continuity. If you don't see the visualization, pushing the symbols around will not help with intuition.

Uniform continuity means that a small change in input will ALWAYS give you a small change in output. Continuity without uniform continuity means that a small change in input may give you a HUGE change in output; as with 1/x near zero, and e^x as you move to the right on the positive x-axis; but that you still won't tear the graph.

Discontinuity means that you can make a tiny change in the input and you'll rip the graph into two pieces.

By the way I would like to say that the discussion of compact sets is obscuring the basic intuition; and I think the OP should dwell on the examples of 1/x and e^x before trying to sling proofs about continuous functions on compact sets. My two cents on that one.

WOW. This is a very great post indeed. It helped a lot. I think you're somehow using the theorem which micromass posted that states 'if f is differentiable and its derivative is bounded then f is uniformly continuous'. Having accepted that as true, if the derivative is bounded, we don't expect the function to dramatically change for a small input because the rate of change for any input is bounded and can't get higher than some value. so all you said makes sense to me, it's still not fully intuitive for me, I mean I can't say if a function is uniformly continuous just by looking at the graph while I could definitely say whether a function is continuous or not by looking at its graph (Yes, I know I can't if the domain is R, because we could only graph a part of the function, not the whole of it), but still...

Your post helped a lot, I'll try to create some intuition behind uniform continuity by your post, it'll help me a lot in the future.

 

[SteveL27]

WOW. This is a very great post indeed. It helped a lot. I think you're somehow using the theorem which micromass posted that states 'if f is differentiable and its derivative is bounded then f is uniformly continuous'

Very glad this helped.

But I am NOT repeat NOT caring at all about theorems. Especially about differentiability, which has nothing to do with this.

We are looking at the intuition. If I needed to teach uniform continuity to a six year old I could do it. You're stretching a big glop of taffy, or silly putty. You can stretch it very gently; or you can stretch it very roughly but without tearing it; or you can rip it apart. Uniform continuity is pulling on a piece of taffy very gently.

Non-uniform continuity is pulling on it roughly without tearing it. And discontinuity is pulling on it so hard that you tear it into two pieces.

This is not about theorems. Only the intuition. Once you get the intuition you can bang out the proofs automatically. That's the secret of doing well in real analysis. If you try to sling the symbols before you understand the concepts, you're doomed. That's because it's about ten times faster to knock out an epsilon proof when you already have a clear visualization; than it is to struggle to put together a proof with only the symbols, lacking the understanding.

And this thread has become very symbol-driven when it's clear that you are struggling with the concepts.

Differentiability is irrelevant here. Most continuous functions aren't even differentiable, so the notions of continuity and uniform continuity should be understood on their own terms.

Please forget about differentiability; and please forget about compactness. Those two topics have greatly complicated this thread, when clarity is what's needed.

Think about taffy.

 

[Arian.D]

Very glad this helped.

But I am NOT repeat NOT caring at all about theorems.

We are looking at the intuition. If I needed to teach uniform continuity to a six year old I could do it. You can stretch it gently; or you can stretch it very roughly; or you can rip it apart. Uniform continuity is pulling on a piece of taffy very gently.

This is not about theorems. Only the intuition. Once you get the intuition you can bang out the proofs automatically. That's the whole point of real analysis. If you try to sling the symbols before you understand the concepts, you're doomed.

And this thread has become very symbol-driven when it's clear that you are struggling with the concepts.

Differentiability is irrelevant here. Most continuous functions aren't even differentiable, so the notions of continuity and uniform continuity should be understood on their own terms.

Please forget about differentiability; and please forget about compactness. Think about taffy.

Very great sentences.
Well, the last part is important, |x| is not differentiable everywhere (it's not differentiable at x=0) but yet it is uniformly continuous over R. Right? I used your intuitive method.

 

[Skrew]

I don't understand why the last part is wrong :(. Where did 'i' come to play?
This is what I've done step by step written down in propositional calculus:

Suppose that \forall x_n, \forall y_n:|x_n - y_n| \to 0 \implies |f(x_n)-f(y_n)| \to 0 is false:
1) Negation: \exists x_n, \exists y_n: |x_n - y_n|\to 0 \land \neg(|f(x_n)-f(y_n)| \to 0)
2) conjunction elimination: \exists x_n, \exists y_n: \neg(|f(x_n)-f(y_n)| \to 0)
3) \neg(|f(x_n)-f(y_n)| \to 0) \equiv \forall\delta&gt;0, \exists\epsilon&gt;0 : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|\geq \epsilon
4) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon
5) \exists x_n, \exists y_n, \forall \delta&gt;0, \exists \epsilon&gt;0: |x_n-y_n|&lt;\delta \implies |f(x_n)-f(y_n)| \geq \epsilon \equiv \neg(\forall\epsilon&gt;0, \exists \delta&gt;0, \forall x_n, \forall y_n : |x_n - y_n|&lt;\delta \implies |f(x_n)-f(y_n)|&lt;\epsilon)

This is what my argument was in symbolic logic, I usually suck at propositional calculus but when I write things in that way I find it easier to argue with people.

For a constant n it is impossible that |x_n - y_n| < delta for all delta(unless x_n = y_n).

I'm not sure where you screwed up but it's not right.

 

[Arian.D]

For a constant n it is impossible that |x_n - y_n| < delta for all delta(unless x_n = y_n).

I'm not sure where you screwed up but it's not right.

hmmm, sounds you're right. Maybe I should've put up a quantifier like for any n in natural numbers...
well, except that part, do I have other mistakes? How would you rephrase my proof to make it right?

 

[Skrew]

hmmm, sounds you're right. Maybe I should've put up a quantifier like for any n in natural numbers...
well, except that part, do I have other mistakes? How would you rephrase my proof to make it right?

I think your first statement does not correctly state what you wanted it to state.

In particular, your for alls need to be in terms of sequences, not elements of the sequence.

 

[SteveL27]

Very great sentences.
Well, the last part is important, |x| is not differentiable everywhere (it's not differentiable at x=0) but yet it is uniformly continuous over R. Right? I used your intuitive method.

Yes, exactly.

If you can "see" without any need for explanation or symbols that 1/x and e^x are continuous but not uniformly continuous, that's what I'm trying to explain.

Because as x gets larger, e^x gets a LOT LOT LOT larger. And as x gets close to zero, 1/x gets a LOT LOT LOT larger. Both 1/x and e^x get stretched.
Once you see that you can intuitively see that they must fail to be uniformly continuous.

Then the epsilons and deltas will be easer.

The things people were saying about differentiability, boundedness, and the compactness of the domain, are conditions that ensure that a given function is uniformly continuous. But they are not the definition or the meaning of uniform continuity. The meaning is in the stretchiness.

 

[Arian.D]

I think your first statement does not correctly state what you wanted it to state.

Yea. me too. But I don't know how to reword my proof. So if we exclude the details, have I got the strategy right in my proof, or my proof is totally wrong?

Yes, exactly.

If you can "see" without any need for explanation or symbols that 1/x and e^x are continuous but not uniformly continuous, that's what I'm trying to explain.

Because as x gets larger, e^x gets a LOT LOT LOT larger. And as x gets close to zero, 1/x gets a LOT LOT LOT larger. Both 1/x and e^x get stretched.
Once you see that you can intuitively see that they must fail to be uniformly continuous.

Once you can see that, the epsilons and deltas will be easer.

Well, yea, I've got the intuition now I think. e^x could be written as e^x=sinh(x)+cosh(x), because e^x is not uniformly continuous, then by what I proved earlier, sinh(x) 'OR' cosh(x) should be not uniformly continuous as well. Using the intuitive method, one could see that both sinh(x) and cosh(x) are in fact not uniformly continuous and it sounds reasonable because both have e^x in their equations that make them get steeper and steeper as we go to infinity (and also because e^-x fails to cancel out the fast growth of e^x in sinh(x)). Am I right? If yes, then I've got the intuition right.

 

[Skrew]

Yea. me too. But I don't know how to reword my proof. So if we exclude the details, have I got the strategy right in my proof, or my proof is totally wrong?

The problem starts with the forall statements, the forall needs to be applied to the sequences, not the elements of the sequences.

Also a bounded function can fail to be uniformly continuous.

For example sin(1/x) on (0, 1).

 

[Arian.D]

The problem starts with the forall statements, the forall needs to be applied to the sequences, not the elements of the sequences.

Ahh, that's right. I guess my proof is flawed even more than that, because I didn't use the fact that |x_n-y_n| goes to zero.
if |x_n - y_n| \to 0 then \exists M\in\mathbb{N}&gt;0,\forall\epsilon&gt;0,\forall n\in\mathbb{N}: n\geq M \implies |f(x_n)-f(y_n)| &lt; \epsilon
maybe this is where that i could come to play, because this time when I negate this there must exist n=i that has the property you said. Am I right?

 

[Skrew]

Ahh, that's right. I guess my proof is flawed even more than that, because I didn't use the fact that |xⁿ-yⁿ| goes to zero.
if |x_n - y_n| \to 0 then \exists M\in\mathbb{N}&gt;0,\forall\epsilon&gt;0,\forall n\in\mathbb{N}&gt;M: n\geq M \implies |f(x_n)-f(y_n)| &lt; \epsilon
maybe this is where that i could come to play, because this time when I negate this there must exist n=i that has the property you said. Am I right?

yess

 

[Arian.D]

yess

So what do I need to do now to have my proof completed?
Because after this step I think other steps are justified, or not?