Extending a uniformly continuous function

[radou]

Homework Statement

So, let (X, dx) and (Y, dy) be metric spaces, and let Y be complete. Let A be a subset of X, and f : A --> Y a uniformly continuous function. Prove that f can be uniquely extended to a uniformly continuous function g : Cl(A) --> Y.

The Attempt at a Solution

My first idea was to use Theorem 30.1. in Munkres.

If x is in Cl(A), since X is a metric space, there is a sequence of points xn of A converging to x. Further on, since f is continuous, the sequence f(xn) converges to f(x). But is x is not in A, i.e. if it is in Cl(A)\A, f(x) may not be defined. This is a problem I don't know how to remedy.

Uniform continuity of f implies Cauchy-continuity, i.e. for any Cauchy sequence xn in A, the sequence f(xn) is Cauchy in Y. Since Y is complete, f(xn) must converge.

I feel I'm close, but I can't add the pieces together. :uhh:

 

[micromass]

Well, you are VERY close

We wish to define g(x), if x is in cl(A). Assume first that we know our g, what properties can we derive?

Well, if x is in cl(A), then there exists a sequence (x_n) in A which converges to x. Since g is continuous, we have that

$$g(x)=\lim_{n\rightarrow +\infty}{g(x_n)}=\lim_{n\rightarrow +\infty}{f(x_n)}$$

So IF our g would exist, then the above limit would equal g(x). This gives us an idea of definition of g! Namely, what if we would DEFINE g(x) as

$$g(x)=\lim_{n\rightarrow +\infty}{g(x_n)}$$

There are however, some things we have to check:
1) the above definition is good: assume that we would have another sequence that converges to x, does the value of g(x) remain the same?
2) g=f on A
3) g is uniformly continuous

Hope that helped. And I hope I didn't give away to much

 

[radou]

Well, you are VERY close

We wish to define g(x), if x is in cl(A). Assume first that we know our g, what properties can we derive?

Well, if x is in cl(A), then there exists a sequence (x_n) in A which converges to x. Since g is continuous, we have that

$$g(x)=\lim_{n\rightarrow +\infty}{g(x_n)}=\lim_{n\rightarrow +\infty}{f(x_n)}$$

So IF our g would exist, then the above limit would equal g(x). This gives us an idea of definition of g! Namely, what if we would DEFINE g(x) as

$$g(x)=\lim_{n\rightarrow +\infty}{g(x_n)}$$

There are however, some things we have to check:
1) the above definition is good: assume that we would have another sequence that converges to x, does the value of g(x) remain the same?
2) g=f on A
3) g is uniformly continuous

Hope that helped. And I hope I didn't give away to much

This idea (defining g as the limit) occurred to me already, but I wasn't sure if it's correct. Along with the idea that g(x) = f(x) for x in A. Of course, since X is Hausdorff (since it is a metric space), sequences converge to at most one limit, so it's well-defined. The only thing which remains is to check uniform continuity - I'll think about it a bit and repost later.

Thanks!

 

[micromass]

Hmm, that the function g is well-defined is not that obvious, I think. You say that

$$g(x)=\lim_{n\rightarrow +\infty}{f(x_n)}$$

But who says that the sequence (f(x_n)) even converges?

And if you have another sequence (x'_n) that converges to x, then we have

$$g(x)=\lim_{n\rightarrow +\infty}{f(x_n)}~\text{and}~g(x)=\lim_{n\rightarrow +\infty}{f(x^\prime_n)}$$

So it still needs to be shown that

$$\lim_{n\rightarrow +\infty}{f(x_n)}=\lim_{n\rightarrow +\infty}{f(x^\prime_n)}$$

All these things are not difficult, but it shows you why we demanded that f is uniformly continuous and Y is complete...

 

[radou]

Well, since X is metrizable, if xn --> x, then f(xn) --> f(x). So it should converge to some point of Y, right? As I pointed out, I used Theorem 30.1.(b) of Munkres here, but it's a bit different - in the theorem f(x) is indeed well-defined, since f is defined on all of X!

Let xn and xn' be sequences in A converging to x. Assume f(xn) --> y1 and f(xn') --> y2. Using Theorem 30.1.(b) again, since xn --> x and xn' --> x, we conclude that f(xn) --> f(x) = y1 and f(xn') --> f(x) = y2, so it follows that y1 = y2. Something's wrong here I think, it's too easy to hold...

 

[micromass]

No, you cannot use theorem 30.1.(b). For the simple reason that f(x) is not defined.

To show that (f(x_n)) is continuous, you can't say that $$f(x_n)\rightarrow f(x)$$, because f(x) does not exist!

But how do we prove that f(x_n) converges to something? Simple: we show that f(x_n) is a Cauchy sequence... This will use that f is uniformly continuous...



The second part will use uniform continuity again. For this, you will have to use the following facts:
1) If (x_n) and (y_n) are two sequences such that d(x_n,y_n)-->0 and if f is uniform continuous, then d(f(x_n),f(y_n))-->0.

2) Let (x_n) and (y_n) be two sequences such that d(x_n,y_n)-->0. If x_n--> x and y_n-->y, then x=y.

3) If x_n--> x and y_n--> x, then d(x_n,y_n)-->0.


Hope this helps!

 

[radou]

OK, if xn --> x in A, let's first show that f(xn) is a Cauchy sequence. Since Y is complete, this sequence must converge.

We wish to show that for ε > 0, there exists some N such that for all m, n >= N, |f(xm) - f(xn)| < ε. Since f is uniformly continuous, for any ε > 0 there exists a δ > 0 such that for all x, y such that |x - y| < δ, we have |f(x) - f(y)| < ε. Since xn -- > x, it is a Cauchy sequence in A (this is really easy to show). Using uniform continuity of f, choose, for ε > 0, a δ > 0. Now, for δ, choose N such that m, n >= N implies |xm - xn| < δ. It follows that |f(xm) - f(xn)| < ε, for all n, m >= N.

I'll work out the second part shortly to show our function is well-defined.

 

[radou]

Btw, my notation isn't consistent, I should have used dy for the metric in Y instead of absolute values, but the point is the same.. :)

 

[micromass]

Yes, this is already correct!

 

[radou]

OK, now, step by step. Btw, for easier notation, I'll simply use absolute values instead of the metric dx on X.

1) Let xn --> x and yn --> x. Let ε > 0 be given. For ε/2, choose N1 and N2 such that for n >= N1 we have |xn - x| < ε/2 and for n >= N2 we have |yn - x| < ε/2. By the triangle inequality, for N = max {N1, N2}, we have |xn - yn| < ε. Hence |xn - yn| --> 0.

2) Now, the converse. Let xn and yn be sequences such that zn = xn - yn --> 0. We wish to prove that the limits of xn and yn must be the same. Assume xn --> x and yn --> y. It is an elementary rule that lim zn = lim (xn - yn) = lim xn - lim yn = 0, and hence x = y.

3) (I'll use the correct notation in this concluding argument) Let xn and yn be sequences in A such that dx(xn, yn) --> 0. Since f is uniformly continuous, for ε > 0 there exists δ > 0 such that for all x, y such that dx(x, y) < δ we have dy(f(x), f(y)) < ε. For this δ there exists a positive integer N such that whenever n > = N, we have dx(xn, yn) < δ (since dx(xn, yn) --> 0). Now, for any n >= N, we have dy(f(xn), f(yn)) < ε, and hence dy(f(xn), f(yn)) --> 0. By 2), it follows that lim f(xn) = lim f(yn).

We have now shown that our function g(x) is well-defined.

Edit: g(x) := lim f(xn), for any x in Cl(A)

 

[micromass]

OK, this is entirely correct.

So, you still have two things two show:

1) g(x)=f(x) for x in A. This shows that g is indeed an extension of f. This should be very easy to show!

2) g is uniformly continuous. This is probably a bit harder to show...


The following may be useful to you: the (3) from your previous post is actually equivalent to uniform continuity!
Thus the following are equivalent:
- f is inform continuous
- for every two sequences (x_n) and (y_n) holds: if d(x_n,y_n)-->0, then d(f(x_n),f(y_n))-->0

This equivalence can be interesting, because it allows you to prove uniform continuity by using sequences. And I always find sequences easier than functions...

 

[radou]

Yes, you're right. Actually, I'm constantly ignoring these little details for some reason, instead of being entirely consistent.

So, let x be in A. Since this implies x is in Cl(A), there is a sequence in A such that xn --> x. We defined g(x) as g(x) = lim f(xn). We wish to show, for x in A, g(x) = f(x), i.e. lim f(xn) = f(x). But this is almost trivial, since f is continuous, xn --> x implies f(xn) --> f(x), and this is exactly what we need. If x is in Cl(A), we have shown in the upper posts that g(x) is well defined, so I think this is settled.

This is okay, right?

Now, let's see about uniform continuity of g. We wish to show that, for ε > 0, there exists δ > 0 such that for all x, y in Cl(A) such that dx(x, y) < δ, we have dy(g(x), g(y)) = dy(lim f(xn), lim f(yn)) < ε. This is the set-up of the problem, right? I'm just making sure so that I don't miss simething later on because of this.

 

[micromass]

Yes, you're right. Actually, I'm constantly ignoring these little details for some reason, instead of being entirely consistent.

So, let x be in A. Since this implies x is in Cl(A), there is a sequence in A such that xn --> x. We defined g(x) as g(x) = lim f(xn). We wish to show, for x in A, g(x) = f(x), i.e. lim f(xn) = f(x). But this is almost trivial, since f is continuous, xn --> x implies f(xn) --> f(x), and this is exactly what we need. If x is in Cl(A), we have shown in the upper posts that g(x) is well defined, so I think this is settled.

This is okay, right?

Yes, this is good. I did it this way: take x in A, then the sequence defined by x_n=x (thus: the constant sequence) is a sequence that converges to x. By definition of g, we then have

$$g(x)=\lim_{n\rightarrow +\infty}{f(x_n)}=\lim_{n\rightarrow +\infty}{f(x)}=f(x)$$

But your argument is as good.

Now, let's see about uniform continuity of g. We wish to show that, for ε > 0, there exists δ > 0 such that for all x, y in Cl(A) such that dx(x, y) < δ, we have dy(g(x), g(y)) = dy(lim f(xn), lim f(yn)) < ε. This is the set-up of the problem, right? I'm just making sure so that I don't miss simething later on because of this.

Yes, this is the set-up. But if this becomes to complicated, maybe you can apply my restatement of uniform continuity with sequences (see last post), I have a feeling it is easier with this... But you can do it this way to...

 

[radou]

Yes, this is good. I did it this way: take x in A, then the sequence defined by x_n=x (thus: the constant sequence) is a sequence that converges to x. By definition of g, we then have

$$g(x)=\lim_{n\rightarrow +\infty}{f(x_n)}=\lim_{n\rightarrow +\infty}{f(x)}=f(x)$$

But your argument is as good.

Hm, but is it necessary to explicitly refer to the constant sequence?

Can't we just use Theorem 30.1., without referring to a specific sequence? I ask because I feel there still might be something I don't understand here.

Yes, this is the set-up. But if this becomes to complicated, maybe you can apply my restatement of uniform continuity with sequences (see last post), I have a feeling it is easier with this... But you can do it this way to...

OK, I'll think about it tomorrow.

 

[micromass]

Hm, but is it necessary to explicitly refer to the constant sequence?

Can't we just use Theorem 30.1., without referring to a specific sequence? I ask because I feel there still might be something I don't understand here.

No, referring to the constant sequence is not necessary at all! I just gave you another way of proving this. But your argument is good to!

 

[radou]

No, referring to the constant sequence is not necessary at all! I just gave you another way of proving this. But your argument is good to!

OK.

Actually, this is a pretty instructive and interesting problem, which it definitely wasn't at first glance!

 

[micromass]

Well, maybe I'll even make the question more interesting if I tell you why this exercise is useful.
Let me indicate just one of it's applications:

Let $$X=\mathbb{R}^\mathbb{[0,1]}$$ be the set of all functions with domain [0,1]. Equip this set with uniform convergence. Let $$A=\mathcal{C}([0,1]\mathbb{R})$$ be the set of all continuous functions with domain [0,1]. Let $$Y=\mathbb{R}$$ the usual real line.

We define the following uniform continuous function:
$$T:\mathcal{C}([0,1],\mathbb{R})\rightarrow \mathbb{R}:f\rightarrow \int f$$

By exercise 2, the function T can be extended to a function $$U:\overline{\mathcal{C}([0,1],\mathbb{R})}\rightarrow \mathbb{R}$$.

So, if we can integrate continuous functions, then exercise 2 can be used to be able to integrate more functions. Specifically, every function which can be approximated (uniformly) by continuous functions, can be integrated.

Now, all this is not very exciting. But there are ways to generalize exercise 2. And this gives rise to the concept of Lebesgue-integration, which is a very important concept in mathematics. But you'll see more of this if you study functional analysis...

 

[radou]

Well, maybe I'll even make the question more interesting if I tell you why this exercise is useful.
Let me indicate just one of it's applications:

Let $$X=\mathbb{R}^\mathbb{[0,1]}$$ be the set of all functions with domain [0,1]. Equip this set with uniform convergence. Let $$A=\mathcal{C}([0,1]\mathbb{R})$$ be the set of all continuous functions with domain [0,1]. Let $$Y=\mathbb{R}$$ the usual real line.

We define the following uniform continuous function:
$$T:\mathcal{C}([0,1],\mathbb{R})\rightarrow \mathbb{R}:f\rightarrow \int f$$

By exercise 2, the function T can be extended to a function $$U:\overline{\mathcal{C}([0,1],\mathbb{R})}\rightarrow \mathbb{R}$$.

So, if we can integrate continuous functions, then exercise 2 can be used to be able to integrate more functions. Specifically, every function which can be approximated (uniformly) by continuous functions, can be integrated.

Now, all this is not very exciting. But there are ways to generalize exercise 2. And this gives rise to the concept of Lebesgue-integration, which is a very important concept in mathematics. But you'll see more of this if you study functional analysis...

It is indeed interesting how simple exercises actually give rise to some important concepts. I lack a thorough course in analysis, so unfortunately, I can't always appreciate these exercises and see where they're important. But this will come with time.

OK, let's reformulate our problem of showing that g is uniformly continuous in a "sequential" way.

Assume xn and yn are sequences such that dx(xn, yn) --> 0 implies dy(g(xn), g(yn)) --> 0. We need to show this implies uniform continuity of g. Let ε > 0 be given. We need to show that there exists a δ > 0 such that for any x, y in Cl(A) such that dx(x, y) < δ, we have dy(g(x), g(y)) < ε.

Let ε > 0 be given. There exists some N1 such that for n >= N1, |dx(xn, yn) - 0| < ε and some N2 such that for n >= N2, |dy(g(xn), g(yn)) - 0| < ε. So for n = max {N1, N2}, we have dx(xn, yn) < ε and dy(g(xn), g(yn)) < ε. It seems it works for δ = ε and only for members of the sequences xn and yn (except finitely many of them), but how do I check it holds for all x, y in Cl(A) such that dx(x, y) < δ ? I don't really see how to solve this right away. Perhaps I'm trying to attack this in a too straightforward-manner.

 

[radou]

Btw, a bit off-topic regarding the proof of the remaining part. We could form this corrolary, based on the Tietze extension theorem:

Let X be a normal space. If f : A --> R is an uniformly continuous map, where A is a subspace of X, then f can be extended to a continuous map f* : X --> R.

I'm not sure if this extension is unique. And it's probably too much to hope it would be uniformly continuous.

 

[micromass]

Hmm, the way you're approaching it is indeed to straightforward. Maybe you should try a proof by contradiction, i.e. assume that f is not uniformly continuous, then there exists sequences (x_n) and (y_n) such that d(x_n,y_n)->0 and d(f(x_n),g(x_n)) doesn't converge to 0.

 

[micromass]

Btw, a bit off-topic regarding the proof of the remaining part. We could form this corrolary, based on the Tietze extension theorem:

Let X be a normal space. If f : A --> R is an uniformly continuous map, where A is a subspace of X, then f can be extended to a continuous map f* : X --> R.

I'm not sure if this extension is unique. And it's probably too much to hope it would be uniformly continuous.

Yes, this is indeed correct. But you're missing a hypothesis: you need that A is metrizable. Since, if A is not a metric space, then saying that "f is uniformly continuous" is not defined. Indeed, we can only define uniform continuity for metric space, not for topological spaces in general.

And of course, f* is not necessairly unique or uniformly continuous, since the Tietze extenstion theorem does not guarantee uniqueness or uniform continuity (and again: uniform continuity is not defined for a general topological space).

 

[radou]

Yes, this is indeed correct. But you're missing a hypothesis: you need that A is metrizable. Since, if A is not a metric space, then saying that "f is uniformly continuous" is not defined. Indeed, we can only define uniform continuity for metric space, not for topological spaces in general.

And of course, f* is not necessairly unique or uniformly continuous, since the Tietze extenstion theorem does not guarantee uniqueness or uniform continuity (and again: uniform continuity is not defined for a general topological space).

Ahh, of course, that's what I wanted to say, but I wrote down nonsence again! X needs to be metrizable, since a metrizable space is normal.

 

[radou]

Hmm, the way you're approaching it is indeed to straightforward. Maybe you should try a proof by contradiction, i.e. assume that f is not uniformly continuous, then there exists sequences (x_n) and (y_n) such that d(x_n,y_n)->0 and d(f(x_n),g(x_n)) doesn't converge to 0.

Wait, did you mean, assume g is not uniformly continuous, and then "there exists sequences (x_n) and (y_n) such that d(x_n,y_n)->0 and d(g(x_n),g(x_n)) doesn't converge to 0."?

 

[micromass]

To be honest, it isn't really all nonsense. Indeed, every normal space (in fact: every completely regular space) can be equipped with a notion of uniform continuity. But that is some more advanced topology.

So, what you wrote down is actually correct

 

[micromass]

Wait, did you mean, assume g is not uniformly continuous, and then "there exists sequences (x_n) and (y_n) such that d(x_n,y_n)->0 and d(g(x_n),g(x_n)) doesn't converge to 0."?

Yes, I'm sorry. That is what I meant. Stupid typo's of mine

 

[radou]

To be honest, it isn't really all nonsense. Indeed, every normal space (in fact: every completely regular space) can be equipped with a notion of uniform continuity. But that is some more advanced topology.

So, what you wrote down is actually correct

Wow, that's interesting...

Yes, I'm sorry. That is what I meant. Stupid typo's of mine

No problem... I'm working on it..

 

[radou]

Hmm, the way you're approaching it is indeed to straightforward. Maybe you should try a proof by contradiction, i.e. assume that f is not uniformly continuous, then there exists sequences (x_n) and (y_n) such that d(x_n,y_n)->0 and d(f(x_n),g(x_n)) doesn't converge to 0.

Btw, another point - we said that uniform continuity of g is equivalent to dx(xn, yn) --> 0 ==> dy(g(xn), g(yn)) --> 0.

You mean, if we assume g not to be uniformly continuous, and if we'd find sequences such that dx(xn, yn) --> 0 ==> dy(g(xn), g(yn)) --> 0 holds, we'd arrive at a contradiction, right? (perhaps you made a type or I lack an understanding of logic...)

 

[micromass]

No, if g is not uniformly continuous, then there might still be sequences (x_n) and (y_n) such that d(x_n,y_n)-->0 and d(g(x_n),g(y_n))-->0.

What I'm claiming that (if g is not u.c.) there exists sequences (x_n) and (y_n) such that d(x_n,y_n)-->0 and such that d(g(x_n),g(y_n)) does not converge to 0.

The proof is actually not a proof by contradiction, but rather: a proof by contraposition. This is: a statement p => q is equivalent to (NOT q) => (NOT p).

So here: p = "for all sequences such that..." and q = "g is uniformly continuous"
So to prove p => q, it suffices to claim that "if q does not hold, then p does not hold".

 

[radou]

Ah, OK, now I understand it completely, thanks!

 

[radou]

No, sorry, but this doesn't seem to work for me.. :( Perhaps you can give me a hint...

 

[micromass]

So, g is uniform continuous if

$$\forall \epsilon >0: \exists \delta: \forall x,y: d(x,y)<\delta~\Rightarrow d(f(x),f(y))<\epsilon$$

So, g is not uniform continuous if there exists an $$\epsilon >0$$ such that

$$\forall \delta: \exists x,y: d(x,y)<\delta~\text{and}~d(f(x),f(y))\geq \epsilon$$

Now, take $$\delta=1/n$$, then

$$\exists x_n,y_n: d(x_n,y_n)<1/n~\text{and}~d(f(x),f(y))\geq \epsilon$$

This gives us the desired sequences (x_n) and (y_n) such that d(x_n,y_n)-->0 and d(f(x_n),f(y_n)) does not converge to 0.

 

[radou]

OK, if I got this right, for every ε we can find a δ and a pair of elements xn, yn such that dx(xn, yn) < δ implies dy(g(xn), g(yn)) >= ε. If we to this for every 1/n, where n is a positive integer, we arrive at a pair of sequences (xn) and (yn) whose distances converge to 0, but whose fodtances of images don't converge to 0, right?

Sorry, but sometimes I still have problems with formulating negations out of inequalities, I'm ashamed to admit.

 

[micromass]

OK, if I got this right, for every ε we can find a δ and a pair of elements xn, yn such that dx(xn, yn) < δ implies dy(g(xn), g(yn)) >= ε.

No, that is not correct. To negate a statement you need to change "forall" to "exists" and vice versa. Furthermore, the negation of "p=> q" is "p AND (NOT q)"

So, the negation is: "There exists en epsilon such that forall delta, there exists xn and yn such that dx(xn, yn) < δ and dy(g(xn), g(yn)) >= ε.

If we to this for every 1/n, where n is a positive integer, we arrive at a pair of sequences (xn) and (yn) whose distances converge to 0, but whose distances of images don't converge to 0, right?

This remains true.

 

[radou]

OK, I think I got it now. Thanks a lot. This was indeed an instructive and rigorous exercise!

 

[micromass]

Good! To be honest, this was not a very easy exercise...