Abstract Nonsense

Crushing one theorem at a time

Extending Uniformly Continuous Functions


Point of post: In this post I will prove a theorem which, informally, says that every uniformly continuous function f:X\to Y where \left(X,d_X\right),\left(Y, d_Y\right) are metric spaces, Y complete, may be extended to a uniformly continuous function f:\overline{X}\to Y. I do this not because this theorem in particularly useful (although, it has its uses) or difficult but because I have not been thrilled with any of the proofs I have seen so far, as far as the explanation of the motivation and technique go. I will forgo the usual structures for proofs for a more didactic approach.

Theorem: Let X be a subspace of some metric space \left(Z,\rho_1\right) and \left(Y,\rho_2\right) a complete metric space. Then, if f:X\to Y is a uniformly continuous function there exists a unique extension \tilde{f}:\overline{X}\to Y which is uniformly continuous.

Motivation

So what’s the idea? What kind of function can we, in general, construct from this f which will be uniformly continuous on \overline{X}. Well, believe it or not the idea is simple enough. Look at the points we want to define the function on, namely the points in \overline{X}-X. Each of these points, x say, has a sequence \{x_n\}_{n\in\mathbb{N}} which converges to it. But, note that in particular each such sequence \{x_n\}  is Cauchy. But, by the uniform continuity of f the sequences \left\{f(x_n)\right\}_{n\in\mathbb{N}} is Cauchy. But! Since Y is complete we know that this sequence converges to some point y. So, what if we said that the extension of f at x is this limit y? This is the general approach. We will define this idea rigorously, show that it’s a well-defined function, prove it’s uniformly continuous, and show that in essence this is the only way to do it.

So, you might ask “Why does Y have to complete? Why can’t f just be continuous?” These are legitimate questions. While I can’t claim that there aren’t weaker conditions on f and Y for which this is true (although, the following is related, though more parallel than a generalization viz. Tietze’s Extension Theorem) I can show that we can’t forgo either completely. In particular (consider all domain spaces as subspaces of \mathbb{R} with the usual metric/topology) we can’t get rid of uniform continuity since the function

\displaystyle f:(0,1)\to\mathbb{R}:x\mapsto \frac{1}{x}

is continuous, and has a complete codomain space yet can’t be continuous extended to \overline{(0,1)}=[0,1] since we’d have to “define” f(0) to be such that

\displaystyle f(0)=\lim f\left(\frac{1}{n}\right)=\lim n

We also can’t forget the completeness of the codomain space. For this consider the map

\displaystyle f:\left\{\frac{1}{n}:n\in\mathbb{N}\right\}\to\left\{\frac{1}{n}:n\in\mathbb{N}\right\}:x\mapsto x

this is evidently uniformly continuous. But, we can’t extend f to a uniformly continuous mapping from \overline{\left\{\frac{1}{n}:n\in\mathbb{N}\right\}}=\left\{\frac{1}{n}:n\in\mathbb{N}\right\}\cup\{0\} otherwise we’d have that the codomain is the continuous image of a compact set and thus compact, but this is impossible. Thus, neither of the two conditions may be entirely removed.

Proof:

Defining the function

We take the intuitive idea we discussed above and put this into rigorous form exactly what this function \tilde{f} should be.

We may clearly assume that X is not closed otherwise we’re done. So, let x\in\overline{X}-X. Then, there exists a sequence \{x_n\}_{n\in\mathbb{N}} in X such that x_n\to x. So, clearly since \{x_n\}_{n\in\mathbb{N}} is convergent in \overline{X} it must be Cauchy. Thus, since f is continuous we know that \{f(x_n)\}_{n\in\mathbb{N}} is Cauchy. So, since Y is complete we know that f(x_n)\to y_x for some y\in Y. So, let

\tilde{f}(x)=\begin{cases}f(x) & \mbox{if}\quad x\in X\\ y_x & \mbox{if}\quad x\in \overline{X}-X\end{cases}

Showing that the function is well-defined

As I state in the first sentence of the following paragraph it’s not at all clear (initially at least) that \tilde{f} is well-defined. In other words, what’s to say we couldn’t take take sequences in X converging to some point x and get two different values with our “formula”? Saying this one more way, what’s to say that the value at x\in \overline{X}-X of \tilde{f} is independent of the choice of sequence as our formulation of it suggests? This is precisely what we will show below. Namely that even if one starts with two different sequences converging to x the “formula” will give the same value for \tilde{f}(x).

It’s not at all  apparent that \tilde{f}(x) is independent of the choice of sequence \{x_n\}_{n\in\mathbb{N}}. So, suppose that \{x_n\}_{n\in\mathbb{N}},\{x'_n\}_{n\in\mathbb{N}} are two sequences in \overline{X} with x_n,x'_n\to x\in \overline{X}-X. Then, using the process described above we see that that each sequence \{x_n\}_{n\in\mathbb{N}},\{x'_n\}_{n\in\mathbb{N}} admits limits y_x,y_{x'} respectively. We aim to show that y_x=y_{x'}. To do this let \varepsilon>0 be arbitrary. Since f(x_n)\to y_x there exists some N_1\in\mathbb{N} such that

\displaystyle N_1\leqslant n\implies \rho_2\left(f(x_n),y_x\right)<\frac{\varepsilon}{3}.

Similarly, since f(x'_n)\to y_{x'} there exists some N_2\in\mathbb{N} such that

\displaystyle N_2\leqslant n\implies \rho_2\left(f(x'_n),y_{x'}\right)<\frac{\varepsilon}{3}

Lastly, we note that since f is uniformly continuous there exists some \delta>0 such that

\displaystyle \rho_1(z_1,z_2)<\delta\implies \rho_2\left(f(z_1),f(z_2)\right)<\frac{\varepsilon}{3}

and since \rho_1(x_n,x'_n)\to 0 there exists some N_3\in\mathbb{N} such that

\displaystyle N_3\leqslant n\implies \rho_1(x_n,x'_n)<\delta\implies \rho_2\left(f(x_n),f(x'_n)\right)<\frac{\varepsilon}{3}

So, let N=\max\left\{N_1,N_2,N_3\right\} then we see that for N\leqslant n

\displaystyle \rho_2\left(y_x,y_{x'}\right)\leqslant \rho_2\left(f(x_n),y_x\right)+\rho_2\left(f(x'_n),y_{x'}\right)+\rho_2\left(f(x_n),f(x'_n)\right)<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon

and since \varepsilon was arbitrary it follows that \rho_2\left(y_x,y_{x'}\right)=0 and so y_x=y_{x'} as required. Thus, the function \tilde{f}:\overline{X}\to Y is well-defined.

The function \tilde{f} is actually uniformly convergent

Ok, so we have a mapping \tilde{f}:\overline{X}\to Y. So, now what? We need to show that this mapping is actually uniformly continuous as we promised. Clearly this is non-obviously (initially) either. Every extension need not be continuous. What if we just said that g(x)=y_0 for all x\in \overline{X}-X and for some fixed y_0\in Y. This would admit an extension of f, but one which is almost undoubtedly discontinuous.

We now claim that \tilde{f}:\overline{X}\to Y is uniformly continuous. To see this let \varepsilon>0 be arbitrary. Then, there exists some \delta>0 such that

\displaystyle \rho_1(x,y)<\delta\implies \rho_2\left(f(x),f(y)\right)<\frac{\varepsilon}{3}

for all x,y\in X. We claim that \frac{\delta}{3} is sufficient. To prove this we break it into three cases

Case 1: x,y\in X

If this this is the case we’re clearly done by construction.

Case 2: y\in X,x\in \overline{X}-X. Since x\in \overline{X}-X we may choose some sequence \{x_n\}_{n\in\mathbb{N}} such that x_n\to x. So, choose x_{n_0}\in X such that \rho_1(x_{n_0},x)<\frac{2\delta}{3} and \rho_2\left(f(x_n),\tilde{f}(x)\right)<\frac{2\varepsilon}{3} (this is possible since we may choose N_1 such that \rho_1(x_{N_1},x)<\frac{2\delta}{3} since x_n\to x and we may choose N_2 such that \rho_2\left(f(x_{N_2}),\tilde{f}(x)\right)<\frac{2\varepsilon}{3} since f(x_n)\to \tilde{f}(x). So, we may take n_0=\max\left\{N_1,N_2\right\}). Then,

\displaystyle \rho_1(x_n,y)\leqslant \rho_1(x_n,x)+\rho_1(x,y)<\frac{2\delta}{3}+\frac{\delta}{3}=\delta

and so \rho_2\left(f(x_n),f(y)\right)<\frac{\varepsilon}{3} since x_n,y\in X. Thus,

\displaystyle \rho_2\left(\tilde{f}(x),\tilde{f}(y)\right)\leqslant \rho_2\left(\tilde{f}(x),f(x_n)\right)+\rho_2\left(f(x_n),f(y)\right)<\frac{2\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon

Case 3: Suppose that x,y\in \overline{X}-X. Then, there exists sequences \{x_n\}_{n\in\mathbb{N}},\{y_n \}_{n\in\mathbb{N}} in \overline{X} with x_n\to x,y_n\to y.  So, note that we may choose x_{n_0},y_{n_1} such that \rho_1\left(x_{n_0},x\right),\rho_1\left(y_{n_1},y\right)<\frac{\delta}{3} and \rho_2\left(f(x_{n_0}),\tilde{f}(x)\right),\rho_2\left(f(y_{n_0}),f(y)\right)<\frac{\varepsilon}{3}. Next note that by construction

\displaystyle \rho_1\left(x_{n_0},y_{n_1}\right)\leqslant \rho_1\left(x_{n_0},x\right)+\rho_1\left(y_{n_1},y\right)+\rho_1\left(x,y\right)<\frac{\delta}{3}+\frac{\delta}{3}+\frac{\delta}{3}=\delta

and thus

\displaystyle \rho_2\left(f(x_{n_0}),f(y_{n_1})\right)<\frac{\varepsilon}{3}

Thus,

\displaystyle \begin{aligned}\rho_2\left(\tilde{f}(x),\tilde{f}(y)\right) &\leqslant \rho_2\left(\tilde{f}(x),f\left(x_{n_0}\right)\right)+\rho_2\left(\tilde{f}(y),f(y_{n_1})\right)+\rho_2\left(f(x_{n_0}),f(y_{n_1})\right)\\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}\\ &=\varepsilon\end{aligned}

Thus, since all the cases have been covered we may conclude definitively that \tilde{f} is uniformly continuous.

Uniqueness

So we have proved that f:X\to Y admits a uniformly continuous extension \tilde{f}:\overline{X}\to Y which is uniformly continuous. But, we claimed that such an extension is unique. As always, there is no (immediately apparent) reason why we couldn’t have extended this function in two ways, say by a constant difference, and arrive at two distinct uniformly continuous extensions. In fact, we may extend f to \tilde{f}_1,\tilde{f}_2 by declaring that for x\in \overline{X}-X we have that \tilde{f}_1(x)=y_1,\tilde{f}_2(x)=y_2  for some (assuming that \text{card }Y\geqslant 2, but this is a pretty safe assumption) distinct y_1,y_2\in Y. Clearly then \tilde{f}_1,\tilde{f}_2 are two distinct extensions of f. But, there is no guarantee (and as we shall show no possible way) that both of these (let alone either of them) are uniformly continuous.

We shall now prove that such an extension is unique. In fact, we will show that if X,\overline{X},Y are as above and g:X\to Y is continuous there exists at most one continuous extension \tilde{g}:\overline{X}\to Y. To prove this let, \tilde{g}_1,\tilde{g}_2 be two continuous extensions of g. To prove that \tilde{g}_1=\tilde{g}_2 it suffices to show that \tilde{g}_1(x)=\tilde{g}_2(x) for a fixed but arbitrary x\in\overline{X}. Since X is dense in \overline{X} there exists a sequence \{x_n\}_{n\in\mathbb{N}} in X with x_n\to x. Thus, by continuity.

\tilde{g}_1(x)=\tilde{g}_1\left(\lim x_n\right)=\lim\tilde{g}_1(x_n)=\lim\tilde{g}_2(x_n)=\tilde{g_2}\left(\lim x_n\right)=\tilde{g}_2(x)

from where it follows that \tilde{g}_1=\tilde{g}_2

\text{ }

Remark: This follows the more general principle that there can exist at most one extension from f:X\to Y to \widetilde{f}:\overline{X}\to Y if X is an arbitrary topological space and Y is Hausdorff. This follows since the ‘agreement’ set A(f,\widetilde{f})=\left\{x\in \overline{X}:f(x)=\widetilde{f}(x)\right\} is dense (since it contains X) and closed (since A(f,\widetilde{f}))=(f\oplus\widetilde{f})^{-1}(\Delta_Y) where f\oplus \widetilde{f}:\overline{X}\to Y\times Y:x\mapsto (f(x),\widetilde{f}(x)) and \Delta_Y is the diagonal.

Conclusion

We have thus proved that f admits a well-defined extension \tilde{f} as described in the motivation section, that this extension is uniformly continuous, and that such an extension is unique. Thus, the theorem follows. \blacksquare

References:

For a proof see page 78 of the first reference. For a general discussion of the subject matter above as a whole see either reference.

1. Simmons, George Finlay. Introduction to Topology and Modern Analysis. Malabar, FL: Krieger Pub., 2003. Print.

2. Rudin, Walter. Principles of Mathematical Analysis. New York: McGraw-Hill, 1976. Print.

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November 3, 2010 - Posted by | Analysis | , , ,

8 Comments »

  1. in the first step “Defining the function”, when you say

    “Thus, since f is continuous we know that {f(x_n)}_n is Cauchy”

    aren’t you actually using the _uniform_ continuity of f..??

    thank you

    tom

    Comment by july | September 15, 2011 | Reply

    • Hello Tom! No, that actually isn’t a case. You are probably thinking of the fact that if f is uniformly continuous then f carries Cauchy sequences to Cauchy sequences, or with some lingo, that every uniformly continuous function is ‘Cauchy continuous’. That said, Cauchy continuity is not EQUIVALENT to uniform continuity. In particular, try proving that if $f:M\to N$ is a continuous map between metric spaces and $M$ is complete try to prove that $f$ is Cauchy complete. Does that help?

      Best,
      Alex

      Comment by Alex Youcis | September 15, 2011 | Reply

      • Thank you for the kind reply… but I still do not understand…
        I agree when you say that “if f is uniformly continuous, then f carries Cauchy sequences to Cauchy sequences” and I still think that this is what you are using to show that {f(x_n)}_n is Cauchy…

        uniformly continuous -> cauchy continuous and the converse is false…

        but now the question is: does continuity imply cauchy continuity… I think the answer is no…

        I do not understand “… to prove that $f$ is Cauchy complete “? do you mean Cauchy continuous?

        I am sorry if I took so long… 😦

        ttys

        Comment by July | November 14, 2011

  2. Tom, you are one-hundred percent correct that \text{uniformly continuous}\implies\text{ Cauchy continuous} and the converse is not true. But, what I am saying is to prove the following theorem:

    \mathbf{Theorem:} \textit{If }M,N\textit{ are metric spaces with }M\textit{ complete, then any continuous map }f:M\to N\textit{ is Cauchy continuous.}

    If this is still not clear, try looking here http://en.wikipedia.org/wiki/Cauchy-continuous_function#Properties

    Let me know how it goes!

    Best,
    Alex

    Comment by Alex Youcis | November 15, 2011 | Reply

  3. ok… the thm is trivial, isn’t it? x_n cauchy –> x_n converges by completeness –> f(x_n) converges by continuity –> f(x_n) Cauchy…

    but this does not explain why you don’t need _uniform_ continuity when you say “Thus, since f is continuous we know that \{f(x_n)\}_{n\in\mathbb{N}} is Cauchy”… in any case this theorem relies heavily on the completeness of the domain, whereas we have completeness of the codomain in the extension thm….

    I still think you must use the full strength of your hypothesis there…. tell why I am wrong…

    best

    tom

    Comment by July | November 18, 2011 | Reply

    • Oh Tom, I am so sorry. It seems all this time I was misinterpreting your question after all. You are one-hundred percent correct. We need the uniform continuity of f on X to guarantee that a Cauchy sequence in X gets mapped to a Cauchy sequence in Y.

      I must apologize for all of this confusion. I will change the wording of my post right now!

      Thanks for catching that! I hope everything is ok now.

      Best,
      Alex

      Comment by Alex Youcis | November 19, 2011 | Reply

  4. ahahah.. now I am a happier person…. :))

    Congratulations for the blog, anyways… It’s very interesting… I will be around from time to time 😉

    best,
    tom

    Comment by July | November 20, 2011 | Reply

    • Haha! Glad to hear it my friend, and thanks!

      Best,
      Alex

      Comment by Alex Youcis | November 20, 2011 | Reply


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