# 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. 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. 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.

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