Continuity and diameters of connected sets

The definition of uniform continuity (if it’s done right) can be phrased as: {f\colon X\to Y} is uniformly continuous if there exists a function {\omega\colon (0,\infty)\to (0,\infty)}, with {\omega(0+)=0}, such that {\textrm{diam}\, f(E)\le \omega (\textrm{diam}\, E)} for every set {E\subset X}. Indeed, when {E} is a two-point set {\{a,b\}} this is the same as {|f(a)-f(b)|\le \omega(|a-b|)}, the modulus of continuity. Allowing general sets {E} does not change anything, since the diameter is determined by two-point subsets.

Does it make a difference if we ask for {\textrm{diam}\, f(E)\le \omega (\textrm{diam}\, E)} only for connected sets {E}? For functions defined on the real line, or on an interval of the line, there is no difference: we can just consider the intervals {[a,b]} and obtain

{|f(a)-f(b)|\le \textrm{diam}\, f([a,b]) \le \omega(|a-b|)}

as before.

However, the situation does change for maps defined on a non-convex domain. Consider the principal branch of square root, {f(z)=\sqrt{z}}, defined on the slit plane {G=\mathbb C\setminus (-\infty, 0]}.

sqrt
Conformal map of a slit domain is not uniformly continuous

This function is continuous on {G} but not uniformly continuous, since {f(-1 \pm i y) \to \pm i } as {y\to 0+}. Yet, it satisfies {\textrm{diam}\, f(E)\le \omega(\textrm{diam}\, E)} for connected subsets {E\subset G}, where one can take {\omega(\delta)=2\sqrt{\delta}}. I won’t do the estimates; let’s just note that although the points {-1 \pm i y} are close to each other, any connected subset of {G} containing both of them has diameter greater than 1.

Capture3
These points are far apart with respect to the inner diameter metric

In a way, this is still uniform continuity, just with respect to a different metric. Given a metric space {(X,d)}, one can define inner diameter metric {\rho} on {X} by letting {\rho(a,b)} be the infimum of diameters of connected sets that contain both {a} and {b}. This is indeed a metric if the space {X} is reasonable enough (i.e., any two points are contained in some bounded connected set). On a convex subset of {\mathbb R^n}, the inner diameter metric coincides with the Euclidean metric {d_2}.

One might think that the equality {\rho=d_e} should imply that the domain is convex, but this is not so. Indeed, consider the union of three quadrants on a plane, say {A = \{(x,y) \colon x > 0\text{ or }y > 0\}}. Any two points of {A} can be connected by going up from whichever is lower, and then moving horizontally. The diameter of a right triangle is equal to its hypotenuse, which is the Euclidean distance between the points we started with.

Capture2
A non-convex domain where inner diameter metric is the same as Euclidean

Inner diameter metric comes up (often implicitly) in complex analysis. By the Riemann mapping theorem, every simply connected domain {G\subset \mathbb C}, other than {\mathbb C} itself, admits a conformal map {f\colon G\to \mathbb D} onto the unit disk {D}. This map need not be uniformly continuous in the Euclidean metric (the slit plane is one example), but it is uniformly continuous with respect to the inner diameter metric on {G}.

Furthermore, by normalizing the situation in a natural way (say, {G \supset \mathbb D} and {f(0)=0}), one can obtain a uniform modulus of continuity for all conformal maps {f} onto the unit disk, whatever the domain is. This uniform modulus of continuity can be taken of the form {\omega(\delta) = C\sqrt{\delta}} for some universal constant {C}. Informally speaking, this means that a slit domain is the worst that can happen to the continuity of a conformal map. This fact isn’t often mentioned in complex analysis books. A proof can be found in the book Conformally Invariant Processes in the Plane by Gregory Lawler, Proposition 3.85. A more elementary proof, with a rougher estimate for the modulus of continuity, is on page 15 of lecture notes by Mario Bonk.

The definition of uniform continuity is wrong

Consider a map f\colon X\to Y where X and Y are metric spaces.

The standard (wrong) definition says: f is uniformly continuous if for every \epsilon>0 there exists \delta>0 such that d_X(x_1,x_2)<\delta \implies d_Y(f(x_1),f(x_2))<\epsilon.

According to this definition, every function f\colon \mathbb Z\to \mathbb R is uniformly continuous. What exactly is uniform about the function f(n)=2^{2^n}\sin n? I don't see it.

The right definition is: f is uniformly continuous if there exists a function \omega \colon [0,\infty)\to [0,\infty) such that \omega(0+)=0 and d_Y(f(x_1),f(x_2))\le \omega(d_X(x_1,x_2)) for all x_1,x_2\in X.

What is the difference, you ask? For nice spaces, such as intervals, there is none. But for uglier spaces: highly nonconvex, disconnected, etc the second definition is more restrictive, or, the way I put it, the first definition is overly permissive (e.g., toward the function f(n)=2^{2^n}).

There is a parallel with the definition of a quasisymmetric map, which is the version of “uniformly continuous” for relative distances. This definition was worked out after the concept of a general metric space was well understood, which is why mathematicians were able to get it right. Uniform continuity predates metric spaces: it was defined for functions on an interval, and then the same definition was applied to more general spaces whether it fit or not.

Here is a statement which is true with the right definition of uniform continuity and false with the wrong one.

Theorem. For a metric space X, the following are equivalent:

  1. X is compact
  2. every continuous function f\colon X\to\mathbb R is uniformly continuous
  3. every map f\colon X\to Y, for every metric space Y, is uniformly continuous

Proof. 1\implies 3. Given t\ge 0, consider the set E_t=\{(x_1,x_2)\in X\times X\colon d_X(x_1,x_2)\le t\}. This is a closed subset of the compact set X\times X, hence compact. Since the map G(x_1,x_2):=d(f(x_1),f(x_2)) is continuous, it follows that G(E_t) is a bounded subset of [0,\infty). Define \omega(t)=\sup G(E_t). One can show that \omega(0+)=0, for example, by covering the diagonal of X\times X with appropriate open neighborhoods.

3\implies 2 is trivial.

2\implies 1. We must show that if X is either not complete or not totally bounded, then there exists a continuous function f\colon X\to \mathbb R which is not uniformly continuous. First, suppose X is not complete. Pick p\in \overline X\setminus X where \overline X is the completion of X. The function f(x)=1/d_{\overline X}(x,p) is continuous but not bounded on \{x\in X\colon d_{\overline X}(x,p)\le 1\}. Since the right definition of uniform continuity requires f to be bounded on bounded sets, f is not uniformly continuous.

Next, suppose that X is not totally bounded. Then there exists \epsilon>0 and an infinite sequence x_n such that d_X(x_n,x_m)\ge \epsilon whenever n\ne m. Let f(x)=\sum_n n\max(\epsilon/2 - d(x,x_n), 0). This function is continuous on X but is not uniformly continuous.
QED

At a first glance one might think than the difference between the two notions boils down to preserving bounded sets. This is not so, since any function f\colon \mathbb Z\to \mathbb R is bounded on bounded subsets of \mathbb Z, yet not every such function is uniformly continuous according to the right definition.

When \omega needs to be emphasized, one can say that f is \omega-continuous. In particular, f is Lipschitz continuous if it is \omega-continuous with \omega(t)=Ct for some C. Also, f is \alpha-Hölder continuous if it is \omega-continuous with \omega(t)=O(t^{\alpha}) as t\to 0. (I don’t think many people really want to impose the Hölder condition at large distances.) Finally, f is Dini-continuous if it is \omega-continuous with \int_0 \frac{\omega(t)}{t}\,dt<\infty.

Almost norming functionals, Part 2

Let E be a real Banach space with the dual E^*. Fix \delta\in (0,1) and call a linear functional e^*\in E^* almost norming for e if |e|=|e^*|=1 and e^*(e)\ge \delta. In Part 1 I showed that in any Banach space there exists a continuous selection of almost norming functionals. Here I will prove that there is no uniformly continuous selection in \ell_1.

Claim. Let S be the unit sphere in \ell_1^n, the n-dimensional \ell_1-space.  Suppose that f\colon S\to \ell_{\infty}^n is a map such that f(e) is almost norming e in the above sense. Then the modulus of continuity \omega_f satisfies \omega_f(2/n)\ge 2\delta.

(If an uniformly continuous selection was available in \ell_1, it would yield selections in \ell_1^n with a modulus of continuity independent of n.)

Proof. Write f=(f_1,\dots,f_n). For any \epsilon\in \{-1,1\}^n we have n^{-1}\epsilon \in S, hence

\sum\limits_{i=1}^n \epsilon_i f_i(n^{-1}\epsilon)\ge n\delta for all \epsilon\in \{-1,1\}^n. Sum over all \epsilon and change the order of summation:

\sum\limits_{i=1}^n \sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon)\ge n2^n\delta

There exists i\in\{1,2,\dots,n\} such that

\sum\limits_{\epsilon}\epsilon_i f_i(n^{-1}\epsilon) \ge 2^n \delta

Fix this i from now on. Define \tilde \epsilon to be the same \pm vector as \epsilon, but with the ith component flipped. Rewrite the previous sum as

\sum\limits_{\epsilon} -\epsilon_i f_i(n^{-1}\tilde \epsilon)\ge 2^n\delta

and add them together:

\sum\limits_{\epsilon}\epsilon_i [f_i(n^{-1}\epsilon)-f_i(n^{-1}\tilde \epsilon)]\ge 2^{n+1}\delta

Since \|n^{-1}\epsilon-n^{-1}\tilde \epsilon\|=2/n, it follows that 2^n \omega_f(2/n) \ge 2^{n+1}\delta, as claimed.