Uniform continuity done right

Naturally, the title is a tribute to Sheldon Axler.

Let X and Y be metric spaces.

Definition 1. A map f\colon X\to Y is uniformly continuous if for every \epsilon>0 there exists \delta>0 such that d_X(a,b)<\delta \implies d_Y(f(a),f(b))<\epsilon.

Definition 2. A map f\colon X\to Y is uniformly continuous if there exists T>0 and a function \omega\colon [0,T)\to [0,\infty) such that \omega(0)=0, \omega is continuous at 0, and d_Y(f(a),f(b))\le \omega(d_X(a,b)) whenever d_X(a,b)<T.

The definitions are evidently equivalent, but the second one introduces the function \omega, called a modulus of continuity, and this merits further investigation. There is an obvious way to define one such function: \omega_1 (t)=\sup \lbrace d_Y(f(a),f(b)) \colon d_X(a,b) \le t\rbrace; this is the smallest, but not necessarily the most useful modulus of continuity for f. The restriction to some interval [0,T) is necessary because in general, \omega_1(t) may be infinite for large values of t. For example, define f\colon \mathbb Z\to\mathbb Z by f(n)=n^2; this is a uniformly continuous function but \omega_1(t)=\infty for all t\ge 1.

Note that \omega_1 is nondecreasing. Hence (by taking a smaller T if necessary), it is bounded. We would like to have more than boundedness: for one thing, it would be nice if the modulus of continuity was continuous itself. We can achieve this and more, by replacing \omega_1 with its concave majorant:

\displaystyle \omega_2(t)=\inf\lbrace \alpha\, t+\beta \colon \omega_1(s)\le \alpha\, s+\beta \text{ for all } s\in [0,T) \rbrace

This looks complicated, so a sketch is in order: the black curve is the original, ugly \omega_1, and the red curve (drawn only when different from black) is \omega_2.

Concave envelope

Since our family of linear functions \alpha\, t +\beta is bounded below by 0 on [0,\infty), its infimum is finite. It’s easy to check that \omega_2 is concave, and therefore continuous on [0,\infty). Since \omega_1\le \omega_2 by definition, we still have the inequality d_Y(f(a),f(b))\le \omega_2(d_X(a,b)). But for this inequality to be useful, we need to know that \omega_2(0)=0. To this end, we must show that for any \beta>0 there exists \alpha such that \omega_1(s)\le \alpha\, s+\beta for all s. The only reasonable thing to try is \alpha =\sup_{s>0} (\omega_1(s)-\beta)/s : if this is finite, it works. And \alpha is finite because \omega_1-\beta is negative for small s.

Thus, we can give the third definition of uniform continuity, equivalent to the previous two.

Definition 3. A map f\colon X\to Y is uniformly continuous if there exists T>0 and a concave function \omega\colon [0,T)\to [0,\infty) such that \omega(0)=0 and d_Y(f(a),f(b))\le \omega(d_X(a,b)) whenever d_X(a,b) <T.

Such \omega is its own modulus of continuity; this follows from subadditivity \omega(t+s)\le \omega(t)+\omega(s), which in turn follows from concavity.

Armed with the above, we should be able to easily prove the following

Theorem. Let E be a subset of a metric space X. Any uniformly continuous bounded function f\colon E\to\mathbb R can be extended to a function F\colon X\to\mathbb R with the same modulus of continuity, supremum, and infimum.

The part about same supremum and infimum is trivial: once you have some extension F, truncate it by setting \tilde F(x)=\min(\sup f, \max(\inf f, F(x))).

The boundedness of f cannot be dispensed with: for instance, f(n)=n^2 is a uniformly continuous function from \mathbb Z to \mathbb R, but it cannot be extended to a uniformly continuous function F\colon \mathbb R\to\mathbb R. (Why?)

Proof. The boundedness of f allows us to have T=\infty in Definition 3. For x\in X we define

\displaystyle F(x)=\sup_{y\in E}\lbrace f(y)-\omega(d(x,y)) \rbrace

which is finite because f is bounded. Also, F is uniformly continuous because \omega is:

\displaystyle F(x)-F(x')\le \sup_{y\in E} \lbrace \omega(d(x',y))-\omega(d(x,y)) \rbrace \le \omega(d(x,x'))

Finally, F(x)=f(x) for x\in E because in this case y=x achieves the supremum in the definition of F(x). QED

P.S. Probably the best known recent paper in which moduli of continuity play the primary role is Global well-posedness for the critical 2D dissipative quasi-geostrophic equation by Kiselev, Nazarov and Volberg. It led to an interview, not to mention a bunch of citations.

One thought on “Uniform continuity done right

  1. Just wanted to point out that the theorem (with minor additions) includes the Mcshane extension of Lipschitz functions as a special case.

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