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 Response to Uniform continuity done right

atishmitra says:

2012-05-18 at 13:16

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

Uniform continuity done right

One Response to Uniform continuity done right

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