## 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 ${G=\mathbb C\setminus (-\infty, 0]}$.

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.

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.

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 $i$th 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.