Naturally, the title is a tribute to Sheldon Axler.

Let and be metric spaces.

**Definition 1.** A map is *uniformly continuous* if for every there exists such that .

**Definition 2.** A map is *uniformly continuous* if there exists and a function such that , is continuous at , and whenever .

The definitions are evidently equivalent, but the second one introduces the function , called a modulus of continuity, and this merits further investigation. There is an obvious way to define one such function: ; this is the smallest, but not necessarily the most useful modulus of continuity for . The restriction to some interval is necessary because in general, may be infinite for large values of . For example, define by ; this is a uniformly continuous function but for all .

Note that is nondecreasing. Hence (by taking a smaller 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 with its *concave majorant*:

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

Since our family of linear functions is bounded below by on , its infimum is finite. It’s easy to check that is concave, and therefore continuous on . Since by definition, we still have the inequality . But for this inequality to be useful, we need to know that . To this end, we must show that for any there exists such that for all . The only reasonable thing to try is : if this is finite, it works. And is finite because is negative for small .

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

**Definition 3.** A map is *uniformly continuous* if there exists and a concave function such that and whenever .

Such is its own modulus of continuity; this follows from *subadditivity* , which in turn follows from concavity.

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

**Theorem**. *Let be a subset of a metric space . Any uniformly continuous bounded function can be extended to a function with the same modulus of continuity, supremum, and infimum. *

The part about same supremum and infimum is trivial: once you have some extension , truncate it by setting .

The boundedness of cannot be dispensed with: for instance, is a uniformly continuous function from to , but it cannot be extended to a uniformly continuous function . (Why?)

**Proof**. The boundedness of allows us to have in Definition 3. For we define

which is finite because is bounded. Also, is uniformly continuous because is:

Finally, for because in this case achieves the supremum in the definition of . 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.

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