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.

Rectangular boxes: handle with care

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

Rainbow sheep inside
Rainbow sheep inside

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.

Attainable (V,S,D)
Attainable (V,S,D)

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:

77 letters, 119 citations on Google Scholar

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.
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