## Lightness, hyperspace, and lower oscillation bounds

When does a map ${f\colon X\to Y}$ admit a lower “anti-continuity” bound like ${d_Y(f(a),f(b))\ge \lambda(d_X(a,b))}$ for some function ${\lambda\colon (0,\infty)\to (0, \infty)}$ and for all ${a\ne b}$? The answer is easy: ${f}$ must be injective and its inverse must be uniformly continuous. End of story.

But recalling what happened with diameters of connected sets last time, let’s focus on the inequality ${\textrm{diam}\, f(E)\ge \lambda (\textrm{diam}\, E)}$ for connected subsets ${E\subset X}$. If such ${\lambda}$ exists, the map f has the LOB property, for “lower oscillation bound” (oscillation being the diameter of image). The LOB property does not require ${f}$ to be injective. On the real line, ${f(x)=|x|}$ satisfies it with ${\lambda(\delta)=\delta/2}$: since it simply folds the line, the worst that can happen to the diameter of an interval is to be halved. Similarly, ${f(x)=x^2}$ admits a lower oscillation bound ${\lambda(\delta) = (\delta/2)^2}$. This one decays faster than linear at 0, indicating some amount of squeezing going on. One may check that every polynomial has the LOB property as well.

On the other hand, the exponential function ${f(x)=e^x}$ does not have the LOB property, since ${\textrm{diam}\, f([x,x+1])}$ tends to ${0}$ as ${x\to-\infty}$. No surprise there; we know from the relation of continuity and uniform continuity that things like that happen on a non-compact domain.

Also, a function that is constant on some nontrivial connected set will obviously fail LOB. In topology, a mapping is called light if the preimage of every point is totally disconnected, which is exactly the same as not being constant on any nontrivial connected set. So, lightness is necessary for LOB, but not sufficient as ${e^x}$ shows.

Theorem 1: Every continuous light map ${f\colon X\to Y}$ with compact domain ${X}$ admits a lower oscillation bound.

Proof. Suppose not. Then there exists ${\epsilon>0}$ and a sequence of connected subsets ${E_n\subset X}$ such that ${\textrm{diam}\, E_n\ge \epsilon}$ and ${\textrm{diam}\, f(E_n)\to 0}$. We can assume ${E_n}$ compact, otherwise replace it with its closure ${\overline{E_n}}$ which we can because ${f(\overline{E_n})\subset \overline{f(E_n)}}$.

The space of nonempty compact subsets of ${X}$ is called the hyperspace of ${X}$; when equipped with the Hausdorff metric, it becomes a compact metric space itself. Pass to a convergent subsequence, still denoted ${\{E_n\}}$. Its limit ${E}$ has diameter at least ${\epsilon}$, because diameter is a continuous function on the hyperspace. Finally, using the uniform continuity of ${f}$ we get ${\textrm{diam}\, f(E) = \lim \textrm{diam}\, f(E_n) = 0}$, contradicting the lightness of ${f}$. ${\quad \Box}$

Here is another example to demonstrate the importance of compactness (not just boundedness) and continuity: on the domain ${X = \{(x,y)\colon 0 < x < 1, 0 < y < 1\}}$ define ${f(x,y)=(x,xy)}$. This is a homeomorphism, the inverse being ${(u,v)\mapsto (u, v/u)}$. Yet it fails LOB because the image of line segment ${\{x\}\times (0,1)}$ has diameter ${x}$, which can be arbitrarily close to 0. So, the lack of compactness hurts. Extending ${f}$ to the closed square in a discontinuous way, say by letting it be the identity map on the boundary, we see that continuity is also needed, although it’s slightly non-intuitive that one needs continuity (essentially an upper oscillation bound) to estimate oscillation from below.

All that said, on a bounded interval of real line we need neither compactness nor continuity.

Theorem 2: If ${I\subset \mathbb R}$ is a bounded interval, then every light map ${f\colon I\to Y}$ admits a lower oscillation bound.

Proof. Following the proof of Theorem 1, consider a sequence of intervals ${(a_n, b_n)}$ such that ${b_n-a_n\ge \epsilon}$ and ${\textrm{diam}\, f((a_n,b_n))\to 0}$. There is no loss of generality in considering open intervals, since it can only make the diameter of the image smaller. Also WLOG, suppose ${a_n\to a}$ and ${b_n\to b}$; this uses the boundedness of ${I}$. Consider a nontrivial closed interval ${[c,d]\subset (a,b)}$. For all sufficiently large ${n}$ we have ${[c,d]\subset (a_n,b_n)}$, which implies ${\textrm{diam}\, f([c,d])\le \textrm{diam}\, f((a_n,b_n))\to 0}$. Thus ${f}$ is constant on ${[c,d]}$, a contradiction. ${\quad \Box}$

The property that distinguishes real line here is that nontrivial connected sets have nonempty interior. The same works on the circle and various tree-like spaces, but fails for spaces that don’t look one-dimensional.

## 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.

## Rectangular boxes: handle with care

A rectangular box (aka parallelepiped) looks like a sturdy object:

But this particular box, with dimensions 7.147 by 6.021 by 4.095, took me the better part of an hour to figure out.

It was a part of a numerical methods assignment: find the dimensions of a box with given volume ${V}$, surface area ${S}$, and diameter ${D}$ (i.e., space diagonal). Algebraic approach leads to pretty ugly expressions, which is the motivation for a numerical method. Specifically, the task was to apply the Newton-Raphson method to the map

$\displaystyle F(x,y,z) = \begin{pmatrix} xyz-V \\ 2(xy+yz+xz)-S \\ x^2+y^2+z^2-D^2 \end{pmatrix}$

Of course, I understood that not every triple ${(V,S,D)}$ is attainable. Also realized that the Jacobian of ${F}$ is degenerate when two of the coordinates coincide, which is a problem for the method. So I thought: let’s generate some random ${x,y,z}$ values that are not too close to one another, and give students the resulting parameters ${V,S,D}$.

With ${x=7.147}$, ${y=6.021}$, and ${z=4.095}$ the parameters are ${V=176.216}$, ${S=193.91}$, and ${D=10.203}$. Sure, a little rounding can’t hurt when numbers are of this size and we are quite far from the critical points of ${F}$. So I put ${V=176}$, ${S=194}$ and ${D=10}$ in one of the versions of the assignment.

But the Newton-Raphson method would not converge… because no such box exists! The rounding did hurt after all.

This motivated me to describe all attainable triples ${(V,S,D)}$ explicitly, which ended up being less of a chore than I expected. It helps to realize that ${(x+y+z)^2 = D^2+S}$, which reduces the search to the intersection of the sphere ${x^2+y^2+z^2=D^2}$ with the plane ${x+y+z=\sqrt{D^2+S}}$. This is a circle (called ${C}$ below), and the allowed range for ${V}$ is between the minimum and maximum of ${xyz}$ on ${C}$.

This goes into Calculus 3 territory. Using Lagrange multipliers with two constraints looks like a tedious job. Instead, I decided to parametrize ${C}$. Its center is ${(c,c,c)}$ where ${c=\dfrac13\sqrt{D^2+S}}$. The radius is ${\displaystyle r = \sqrt{ D^2 - \frac13(D^2+S)}=\frac13 \sqrt{6D^2-3S}}$. We also need an orthonormal basis of the subspace ${x+y+z=0}$: the vectors

$\displaystyle \frac{1}{\sqrt{6}} \langle 2, -1, -1\rangle \quad \text{and}\quad \frac{1}{\sqrt{2}} \langle 0, 1, -1\rangle$

do the job.

So, the circle ${C}$ is parametrized by

$\displaystyle x = c+\frac{2r}{\sqrt{6}} \cos t \\ y = c-\frac{r}{\sqrt{6}} \cos t +\frac{r}{\sqrt{2}} \sin t \\ z = c-\frac{r}{\sqrt{6}} \cos t -\frac{r}{\sqrt{2}} \sin t$

This is not as bad as it looks: the product ${xyz}$ simplifies to

$\displaystyle xyz = c^3 - \frac{cr^2}{2} + \frac{r^3\sqrt{6}}{18} \cos 3t$

which tells us right away that the volume ${V}$ lies within

$\displaystyle c^3 - \frac{cr^2}{2} \pm \frac{r^3\sqrt{6}}{18}$

In terms of the original data ${S,D}$ the bounds for ${V}$ take the form

$\displaystyle \frac{5S-4D^2}{54}\sqrt{S+D^2} \pm \frac{\sqrt{2}}{54} (2D^2-S)^{3/2}$

(And of course, ${V}$ cannot be negative even if the lower bound is.) It is easy to see that ${2D^2-S\ge 0}$ with equality only for a cube; however ${2D^2-S}$ can be relatively small even when the box does not look very cube-ish. For the box pictured above, the tolerance ${\frac{\sqrt{2}}{54} (2D^2-S)^{3/2}}$ is approximately ${1.4}$; after rounding ${S\approx 194 }$ and ${D\approx 10}$ this drops to ${0.38}$, and the desired volume of ${176}$ is way out of the allowable range ${181\pm 0.38}$.

Yes, the set of attainable triples ${(V,S,D)}$ is quite thin. Parallelepipeds are fragile objects: handle them with care.

## The geometry of situation: diameter vs radius

Mathematicians and engineers are disinclined to agree about anything in public: should the area of a circle be described using the neat formula $\pi r^2$ or in terms of the more easily measured diameter as $\frac{1}{4} \pi d^2$, for example?
J. Bryant and C. Sangwin, How Round is Your Circle?

I will argue on behalf of the diameter but from a mathematician’s perspective. The diameter of a nonempty set $A\subset \mathbb R^2$ is

$\displaystyle \mathrm{diam}\, A = \sup_{a,b\in A} |a-b|$

Whether $\mathrm{diam}\, \varnothing$ should be $0$ or $-\infty$ I’ll leave for you to decide. The radius of $A$ can be defined as

$\displaystyle \mathrm{rad}\, A = \inf_{x\in \mathbb R^2}\sup_{a\in A}|x-a|$

For a circle — whether this word means $\mathbb S^1$ or $\mathbb D^2$ — these definitions indeed agree with the diameter and radius. The example of $\mathbb S^1$ shows that in the definition of the radius we should not require $x\in A$.

The problem of determining the radius of a given set was posed in 1857 by J.J.Sylvester in Quarterly Journal of Pure and Applied Mathematics. Thanks to Google Books, I can reproduce his article in its entirety:

Suppose that $f\colon A\to \mathbb R^2$ is a map of $A$ that is nonexpanding/short/metric/1-Lipschitz or whatyoucallit: $| f(a)- f(b) | \le |a-b|$ for all $a,b\in A$. Clearly, the diameter does not increase: $\mathrm{diam}\, f(A)\le \mathrm{diam}\,A$. What happens to the radius is not nearly as obvious…

It turns out that the radius does not increase either. Indeed, by Kirszbraun’s theorem $f$ can be extended to a 1-Lipschitz map of the entire plane, and the extended map tells us where the center of a bounding circle should go. Kirszbraun’s theorem is valid in $\mathbb R^n$ for every $n$, as well as in a Hilbert space. Hence, nonexpanding maps do not increase the radius of any subset of a Hilbert space.

However, general normed vector spaces are different…

The example given below is wrong; the map is not 1-Lipschitz. I keep it for historical record.