2. Partial symmetry and convexity with respect to a line

Proof. This follows from a coupling argument. Let $Z_{t}$ start at $a\in S$ , and let $H_{\Delta}$ be its hitting time of the line $\Delta$ . Form the process $\tilde Z_t$ by the rule

\begin{equation*} \tilde Z_t = \begin{cases}{\sigma}_{\Delta}(Z_t) & \text{if } t < H_{\Delta}, \\Z_t & \text{if } t \geq H_{\Delta}. \end{cases} \end{equation*}

It follows from the strong Markov property and the reflection invariance of Brownian motion that $\tilde Z_t$ is a Brownian motion. Clearly $\tau_U = \tilde \tau_U$ on the set $\{\tau_U \geq H_{\Delta}\}$ , and our conditions on S imply $\tau_U \leq \tilde \tau_U$ on the set $\{\tau_U < H_{\Delta}\}$ . Furthermore, if $\sigma_{\Delta}(S)$ is strictly contained in $U \setminus (S \cup \Delta)$ , then $Z_t$ has some positive probability of leaving U before $\tilde Z_t$ does; this is implied for instance by [Reference Bass6, Theorem I.6.6]. The result follows.

This theorem allows us in essence to exclude the symmetric side of any partial symmetry axis for U in our search for pth centers. The only exception to this rule is when $\sigma_{\Delta}(S) = U \setminus (S \cup \Delta)$ , i.e. when ${\Delta}$ is a symmetry axis of U. However, in most cases a symmetry axis will contain all pth centers. To see why this is so, we need another definition.

Definition 2.2. Let $\Delta$ be a symmetry axis of U. We say that U is $\Delta$ -convex if

\begin{equation*}tz+(1-t)\sigma_{\Delta}(z)\in U\quad\text{for all $(z,t)\in U\times[0,1]$.}\end{equation*}

In other words, for every $z\in U$ , the segment joining z and $\sigma_{\Delta}(z)$ remains inside U.

It is clear that any convex U is $\Delta$ -convex for any symmetry axis $\Delta$ , and a less trivial example can be given by $\{-f(x) < y < f(x)\}$ , where f is a positive continuous function on the real line, which is ${\Delta}$ -convex with ${\Delta} = {\mathbb R}$ . A domain which is not ${\Delta}$ -convex with respect to a symmetry axis can be given by

\begin{equation*}W_{\varepsilon} = \{-{\varepsilon}<y<{\varepsilon}, |x|<1\} \cup \biggl\{|z-1|< \frac{1}{2}\biggr\} \cup \biggl\{|z+1| < \frac{1}{2}\biggr\}\end{equation*}

with ${\varepsilon} < {{1}/{2}}$ ; this has the real and imaginary axes as symmetry axes but is not ${\Delta}$ -convex with respect to the imaginary axis (though it is with respect to the real line). As will be seen below, this domain also shows why ${\Delta}$ -convexity is required in the following proposition.

Proof. Let $a \in U \setminus {\Delta}$ , and let

\begin{equation*} \hat a = \frac{1}{2} a + \frac{1}{2} {\sigma}_{\Delta}(a) \end{equation*}

be the orthogonal projection of a onto ${\Delta}$ . Let L be the line parallel to ${\Delta}$ which passes through the point

\begin{equation*} \frac{1}{2} a + \frac{1}{2} \hat a. \end{equation*}

Speaking informally, this is the line halfway between a and ${\Delta}$ . ${\Delta}$ -convexity implies that L is a partial symmetry axis of U, and if S is the component of $U \setminus L$ containing a then ${\sigma}_L(S)$ is strictly contained in $U \setminus (S \cup L)$ . It therefore follows from Theorem 2.1 that $\mathbb{E}_a[\tau_U^p] < \mathbb{E}_{\hat a}[\tau_U^p]$ . The result follows.

Note that this proposition completely solves our problem in the case that our domain is ${\Delta}$ -convex with respect to two or more non-parallel symmetry axes, since all pth centers must lie at their point of intersection, and we have also incidentally proved the purely geometrical fact that all such symmetry axes must coincide at a unique point; more on this in the final section. Thus, for instance, all p-centers of any regular polygon, a circle, an ellipse, a rhombus, and any number of other easily constructed examples must lie at their natural centers. To see an example of a domain with intersecting symmetry axes but where the point of intersection is not a pth center, let us return to the domains $W_{\varepsilon}$ described immediately before this proposition. Proposition 2.1 implies that all pth centers lie on the real line, but it is easy to see that if we make ${\varepsilon}$ sufficiently small then 0, the intersection point of the two symmetry axes, will not be a pth center (clearly $\tau_{W_{\varepsilon}}(1) \geq \tau_{\{|z-1|<1/2\}}(1)$ , so that $\mathbb{E}_1[\tau_{W_{\varepsilon}}^p]$ always remains greater than a positive constant, but $\tau_{W_{\varepsilon}}(0)$ decreases monotonically to 0 a.s. as ${\varepsilon} \searrow 0$ , so that $\mathbb{E}_1[\tau_{W_{\varepsilon}}^p] \searrow 0$ ).

Let us now look at an example that shows the use of the results proved up to this point, but also their limitations. Suppose U is the isosceles right-angled triangle with vertices at $-1,1,$ and i. The imaginary axis is an axis of symmetry, and U is ${\Delta}$ -convex with respect to this axis, so all pth centers must lie on the imaginary axis. The line $\{y={{1}/{2}}\}$ is a partial symmetry axis for U, with $U \cap \{y > {{1}/{2}}\}$ the symmetric side, so all p-centers must lie on $\{x=0, y \leq {{1}/{2}}\}$ . Now common sense tells us that the pth centers cannot be too close to the real axis as well, because this is a boundary component, but there is no good partial symmetry axis to apply to conclude that rigorously. The way out of this difficulty is to extend our method of reflection to curves more general than straight lines. For this, we will need to utilize the conformal invariance of Brownian motion, via the following famous theorem of Lévy (see [Reference Bass6] or [Reference Mörters and Peres15]).

Theorem 2.2. If f is a holomorphic function, then $f(Z_{t})$ is a time-changed Brownian motion. More precisely, $f(Z_{\kappa^{-1}(t)})$ is a Brownian motion where

\begin{equation*}\kappa(t)\,:\!=\int_{0}^{t}|f'(Z_{s})|^{2} \,{\textrm{d}} s \quad\text{for $t\geq0$.}\end{equation*}

This allows us to extend Theorem 2.1 as follows.

Proposition 2.2. Suppose U is a domain with an axis of symmetry $\Delta$ , and suppose f is a conformal map defined on U with the property that $|f'(z)| \geq |f'({\sigma}_\Delta(z))|$ for all $z \in A$ , where A is one component of $U \backslash \Delta$ and ${\sigma}_\Delta$ is the symmetry over $\Delta$ . Then, for any $a \in A$ , we can find Brownian motions $Z_t$ starting at f(a) and $\tilde Z_t$ starting at $f({\sigma}_\Delta(a))$ defined on the same probability space such that $\tau_{f(U)}(f(a)) \geq \tilde \tau_{f(U)}(f({\sigma}_\Delta(a)))$ a.s. (where $\tilde \tau_U$ is the exit time from U of $\tilde Z$ ). In particular,

\begin{equation*}\mathbb{E}_{f(a)}\big[\tau_{f(U)}^p\big] \geq \mathbb{E}_{f({\sigma}_\Delta( a))}\big[\tau_{f(U)}^p\big].\end{equation*}

If there is any point in A at which $|f'(z)| > |f'({\sigma}_\Delta(z))|$ then

\begin{equation*}\mathbb{E}_{f(a)}\big[\tau_{f(U)}^p\big] > \mathbb{E}_{f({\sigma}_\Delta(a))}\big[\tau_{f(U)}^p\big]\end{equation*}

(for this statement we recall the assumption that $\mathbb{E}_{w}[\tau_{f(U)}^p]< \infty$ for any $w \in f(U)$ ).

Proof. Let $Z_{t}$ be a Brownian motion starting at a, and let $\tilde Z_{t}$ be defined as in Theorem 2.1. According to Theorem 2.2, the processes $f(Z_t)$ and $f(\tilde Z_{t})$ are time-changed Brownian motions, and the time changes are given by $\kappa^{-1}(s)$ and $\tilde \kappa^{-1}(s)$ , respectively, where

\begin{equation*}\left.\begin{aligned}\kappa(t) & =\int_{0}^{t}\mid f'(Z_{t})\mid^{2}\,{\textrm{d}} t\\\tilde \kappa(t) & =\int_{0}^{t}\mid f'(\tilde Z_{t})\mid^{2}\,{\textrm{d}} t\end{aligned}\right\} \; \;t<\tau_{U}.\end{equation*}

Now our assumptions imply $|f'(Z_t)| \geq |f'(\tilde Z_{t})|$ a.s. for all $t < \tau_U$ , and thus $\kappa(t) \geq \tilde \kappa(t)$ a.s. for all $t < \tau_U$ . It follows from this that $\tau \geq \tilde \tau$ a.s., where $\tau$ and $\tilde \tau$ are the exit times from f(U) of the Brownian motions $f(Z_{\kappa^{-1}(s)})$ and $f(\tilde Z_{\tilde \kappa^{-1}(s)})$ , which begin at f(a) and $f({\sigma}_\Delta(a))$ , respectively. The result follows.

We can obtain a corollary that will be useful for the isosceles triangle and in other cases by taking

\begin{equation*}f(z) = z_0 + R\biggl(\frac{z+i}{z-i}\biggr)\quad \text{for $z_0 \in {\mathbb C}$ and $R>0$,} \end{equation*}

which takes the real axis to the circle $C = \{|z-z_0|=R\}$ , the upper half-plane to the outside of C, and the lower half-plane to the inside. We have

\begin{equation*} |f'(z)| = \frac{2R}{|z-i|^2}, \end{equation*}

and it is easy to check that $|f'(z)| > |f'(\bar z)|$ for all z in the upper half-plane. Applying Proposition 2.2 and working through the implications yields the following.

Note that here ${\sigma}_C$ denotes reflection over the circular arc C, that is,

\begin{equation*} {\sigma}_C(z) = z_0 + \frac{R^2}{\overline{z-z_0}}. \end{equation*}

We remark further that the singularity that f has at i does not cause a problem in this result, because $\mathbb{E}_0(Z_t = i \mbox{ for some } t \geq 0) = 0$ , so a.s. a Brownian motion starting at 0 will not hit the singularity in any case. The compact set $\{B_t\colon 0\leq t \leq \tau_U\}$ is therefore bounded away from i a.s., and the result goes through.

Let us now apply this corollary to the isosceles right-angled triangle U shown in Figure 1. If C is the circle passing through $-1$ and 1 which intersects the real axis at angles of ${{\pi}/{8}}$ , then the reflection of the set $\mathcal{A}=\mathcal{I} \cap U$ will be the region $\mathcal{B}$ in the upper half-plane bounded by C and the circle which passes through $-1$ and 1 and intersects the real axis at angles of ${{\pi}/{4}}$ ; this can be seen by noting that the transformation ${\sigma}_C$ preserves angles and also preserves the class of circles on the Riemann sphere (which includes lines, interpreted as circles through $\infty$ ).

Figure 1: Reflection over a circle.

As this region lies within U, we conclude that no pth centers lie within $\mathcal{A}$ . A bit of Euclidean geometry shows that C intersects the imaginary axis at $\csc({{\pi}/{8}}) - \cot({{\pi}/{8}}) \approx 0.20$ , and coupled with our observations above we see that all pth centers must lie on the imaginary axis between the points $0.2i$ and $0.5i$ . In fact, the upper bound of $0.5i$ can be improved by using the angle bisector of the angles at 1 or $-1$ ; see Example 3.2. The reader may also have observed that the reflected circular domain does not do a good job of filling the triangle, and therefore it stands to reason that the lower bound may also be improved; more on this in the final section.

Finally, the following result can be useful when U is mapped to itself by an antiholomorphic function $\bar f$ (this means that the conjugate of $\bar f$ , f(z), is holomorphic). We will denote the derivative of this function (with respect to $\bar z$ ) by $\overline{f'(z)}$ . An example of this is when U is an annulus, as will be explored in Section 3.

Proposition 2.3. Let $\overline{f}\colon U\longrightarrow U$ be antiholomorphic, and consider the two sets

\begin{equation*}\begin{alignedat}{1}\Omega & \,:\!=\{z\in U\mid |\bar f'(z)|<1\}{,}\\\Lambda & \,:\!=\{z\in U\mid |\bar f'(z)|=1\}.\end{alignedat}\end{equation*}

If $\bar f(U\setminus\Omega)\subset \Omega$ and $\bar f_{|\Lambda}=id_{\Lambda}$ , then all pth centers are contained in $\Omega$ .

Proof. Let $Z_{t}$ be a Brownian motion starting at $z\in U\setminus\Omega$ and let $H_{\Lambda}$ be its hitting time of $\Lambda$ , and consider $W_{t}$ the Brownian motion derived from $f(Z_{t})$ , that is,

\begin{equation*}W_{t}\,:\!=f(Z_{\kappa^{-1}(t)}) {,}\end{equation*}

where

\begin{equation*}\kappa(t)=\int_{0}^{t}|\bar f'(Z_{s})|^{2}\,{\textrm{d}} s.\end{equation*}

Now we are going to construct two Brownian motions $\widetilde{Z}_{t}$ and $\widetilde{W}_{t}$ starting respectively at z and $w\,:\!=f(z)$ , such that $\widetilde{W}_{t}$ leaves U before $\widetilde{Z}_{t}$ , as follows.

1. (1) If $H_{\Lambda}<\tau_{U}^{{\scriptscriptstyle Z}}$ , then run an independent Brownian motion, say $B_{t}$ , starting at $Z_{H_{\Lambda}}$ , and set

   \begin{equation*}\begin{alignedat}{1}\widetilde{Z}_{t} & \,:\!=Z_{t}\textbf{1}_{\{t\leq H_{\Lambda}\}}+B_{t-H_{\Lambda}}\textbf{1}_{\{H_{\Lambda}<t\}}{,}\\\widetilde{W}_{t} & \,:\!=W_{t}\textbf{1}_{\{t\leq\kappa(H_{\Lambda})\}}+B_{t-\kappa(H_{\Lambda})}\textbf{1}_{\{\kappa(H_{\Lambda})<t\}}{,}\end{alignedat}\end{equation*}
   where $\widetilde{W}_{t}$ and $\widetilde{Z}_{t}$ are well-defined Brownian motions as
   \begin{equation*}\widetilde{W}_{\kappa(H_{\Lambda})}=W_{\kappa(H_{\Lambda})}=f(Z_{H_{\Lambda}})=Z_{H_{\Lambda}}=\widetilde{Z}_{H_{\Lambda}}.\end{equation*}
   Now note that
   \begin{equation*}\tau_{U}^{{\scriptscriptstyle \widetilde{Z}}}=H_{\Lambda}+\inf\{t,\,B_{t}\notin U\mid B_{0}=Z_{H_{\Lambda}}\} \leq\kappa(H_{\Lambda})+\inf\{t,\,B_{t}\notin U\mid B_{0}=Z_{H_{\Lambda}}\} =\tau_{U}^{{\scriptscriptstyle \widetilde{W}}}.\end{equation*}

2. (2) If $H_{\Lambda}\geq\tau_{U}^{{\scriptscriptstyle Z}}$ , then just set $\widetilde{Z}_{t}=Z_{t}$ and $\widetilde{W}_{t}=W_{t}$ . Therefore

   \begin{equation*}\tau_{U}^{{\scriptscriptstyle \widetilde{Z}}}\leq\kappa(\tau_{U}^{{\scriptscriptstyle \widetilde{Z}}})=\tau_{U}^{{\scriptscriptstyle \widetilde{W}}}.\end{equation*}

In both cases we have

\begin{equation*}\tau_{U}^{{\scriptscriptstyle \widetilde{Z}}}\overset{{\scriptstyle a.s}}{\leq}\tau_{U}^{{\scriptscriptstyle \widetilde{W}}}.\end{equation*}

Hence

\begin{equation*}\mathbb{E}_{z}((\tau_{U}^{{\scriptscriptstyle \widetilde{Z}}})^p)\leq \mathbb{E}_{w}((\tau_{U}^{{\scriptscriptstyle \widetilde{W}}})^p),\end{equation*}

which ends the proof.

3. Applications

In this section we work through a series of examples that show how our results may be applied.

Example 3.1. Let U be the upper half-disk $\{|z|<1, \textrm{Im}(z)>0\}$ . The imaginary axis is an axis of symmetry, and U is $\Delta$ -convex with respect to this axis, so all pth centers lie on the imaginary axis. The line

\begin{equation*} \Delta=\!:\,\biggl\{\textrm{Im}(z)=\frac{1}{2}\biggr\} \end{equation*}

is clearly a partial symmetry axis with symmetric part

\begin{equation*} U \cap \biggl\{\textrm{Im}(z)>\frac{1}{2}\biggr\}, \end{equation*}

and U is $\Delta$ -convex as well. Thus all p-centers belong to the set

\begin{equation*}\biggl\{\textrm{Re}(z)=0,\ 0\leq \textrm{Im}(z)\leq\frac{1}{2}\biggr\}.\end{equation*}

Now, if we let C be the circle $\{|z+i|= \sqrt{2}\}$ , then C passes through 1 and $-1$ , making an angle of ${{\pi}/{4}}$ at each point with the real axis. If we let $\mathcal{A} = U \cap \mathcal{I}$ as before, with $\mathcal{I}$ the inside of C, then $\sigma_C(\mathcal{A}) = \mathcal{B}$ , where $\mathcal{B} = U \cap \mathcal{O}$ and $\mathcal{O}$ is the outside of C. By Corollary 2.1, no pth center lies in $\mathcal{A}$ . Thus all pth centers lie on the line segment

\begin{equation*}\biggl\{\textrm{Re}(z)=0,\ \sqrt{2}-1 \leq \textrm{Im}(z) \leq \frac{1}{2}\biggr\},\end{equation*}

which is in bold in Figure 2.

Figure 2: Reflection over a circle.

Example 3.2. Now let U be an isosceles triangle with vertices at $-1, 1$ , and Ni with $N>0$ . It will be convenient for us to index U by the angles at 1 and $-1$ , so if we let ${\theta}$ be this angle then $N = \tan {\theta}$ . Proposition 2.1 tells us that all pth centers lie on the imaginary axis. We have seen already from the example discussed in connection with Proposition 2.2 that all pth centers must lie below $({{M}/{2}})i$ , but we will now show how this can be improved. Let B be the angle bisector of one of the base angles of U. B is a partial symmetry axis of U, with symmetric side given by the component of $U \backslash B$ corresponding to the shorter side of the triangle. Thus, if ${\theta} > {{\pi}/{3}}$ , then all pth centers must lie above B, while if ${\theta} < {{\pi}/{3}}$ , then all pth centers must lie below B. Now let M be the perpendicular bisector of the edge connecting 1 to Mi (this is often referred to as the mediator). This is also a partial symmetry axis of U, and the symmetric side is the component of $U \backslash M$ which does not contain $-1$ . Thus, if ${\theta} > {{\pi}/{3}}$ , then all pth centers must lie below M, while if ${\theta} < {{\pi}/{3}}$ , then all pth centers must lie above M. Thus the intersections of M and B with the imaginary axis provide upper and lower bounds for all pth centers, although which is the upper bound and which is the lower bound depends on ${\theta}$ . Figure 3 demonstrates this phenomenon ( $\Delta$ denotes the imaginary axis).

Figure 3: The angle bisector and mediator.

Naturally they coincide at ${\theta} = {{\pi}/{3}}$ . It can be checked that, regardless of ${\theta}$ , this gives a better upper bound than ${{N}/{2}}$ , which was given by reflection over $\{\textrm{Im}(z) = {{N}/{2}}\}$ . A bit of Euclidean geometry shows that the intersection of B with the imaginary axis is at the point $\tan({{\theta}/{2}})i$ , and the intersection of M with the imaginary axis is at the point $({{1}/{\tan {\theta}}}- {{1}/{\sin 2{\theta}}})i$ . Furthermore, we always have as a lower bound the intersection of the imaginary axis and the circle passing through 1 and $-1$ , making an angle of ${{\theta}/{2}}$ with the real axis; this follows from Corollary 2.1 as above. This point is

\begin{equation*}\frac{1-\cos({{\theta}/{2}})}{\sin({{\theta}/{2}})}.\end{equation*}

Figure 4 shows these upper and lower bounds; all pth centers must lie in the regions labeled $\Omega$ .

Figure 4: Upper and lower bounds for pth centers.

Proof. Consider $f(z)={{rR}/{\overline{z}}}$ , which maps $\mathcal{A}_{r,R}$ to itself. Under the same notation as in Proposition 2.3, we have

\begin{equation*}\begin{alignedat}{1}\Omega & \,:\!=\big\{\sqrt{rR}<|z|<R\big\}{,}\\\Lambda & \,:\!=\{|z|=\sqrt{rR}\}\end{alignedat}\end{equation*}

and we can check easily that f satisfies the requirements of Proposition 2.3). Therefore we can eliminate $U\setminus\Omega$ from consideration, and we obtain the lower bound $\sqrt{{\scriptstyle rR}}$ . In order to get the upper bound ${{(R+r)}/{2}}$ we can see, as illustrated by Figure 5, that the line $\Delta$ is a partial symmetry axis. The result follows.

Figure 5: Bounds for the annulus.

Example 3.4. Let $\mathscr{H}$ be the region $\{|x|>|y|, x^2-y^2<1\}$ ; this is the region bounded by the lines $y = \pm x$ and the hyperbola $x^2-y^2=1$ ; see Figure 6.

Figure 6: Bounds for the hyperbolic region.

It is perhaps not obvious for which p we have $\mathbb{E}_{w}[\tau_{\mathscr{H}}^p]< \infty$ , but we can show that $\mathbb{E}_{w}[\tau_{\mathscr{H}}^p]< \infty$ for any $p>0$ and $w \in \mathscr{H}$ , as follows. $\mathscr{H}$ is contained in the union of two infinite strips which are orthogonal. Any strip has all moments of its exit time finite: Brownian motion is rotation-invariant, so the moments are the same as for a horizontal strip, and these moments in turn are the same as for a one-dimensional Brownian motion from a bounded interval, since that is what we obtain when we project the Brownian motion onto the imaginary axis; these moments are well known to be finite for all p, and in fact they can be calculated explicitly for integer p using the Hermite polynomials. We would like to conclude that the union of these two strips must then have finite pth moment, but easy examples show that it is not necessarily the case that the union of two domains with finite pth moment must itself have finite pth moment. A method does exist for reaching the desired conclusion, however, and it is contained in Theorem 3 and Lemmas 1 and 2 of [Reference Markowsky13]. It is straightforward to verify that our infinite strips satisfy the required conditions: their intersection is bounded, and boundary arcs intersect at non-zero angles. Therefore the exit time for their union has finite pth moments for all p, and thus so does $\mathscr{H}$ . See [Reference Markowsky13] for details.

Now let us see how our methods can be used to localize the pth centers. The real axis is an axis of symmetry, but the domain is not $\Delta$ -convex, so we may not apply Proposition 2.1. However, all p-centers lie on the real axis, and we may prove this as follows. The map $f(z) = \sqrt{z}$ maps the strip $\{0<\textrm{Re}(z)<1\}$ conformally onto $\mathscr{H}$ . Any horizontal line can be used in Proposition 2.2, and we note that

\begin{equation*}|f'(z)| = \frac{1}{2\sqrt{|z|}}\end{equation*}

is monotone decreasing in $|z|$ and therefore in $|\textrm{Im}(z)|$ . Thus all pth centers must lie on the image of the real axis under f, which is again the real axis. So we need only consider points on ${\mathbb R}$ . The line $\{\textrm{Re}(z)={{1}/{2}}\}$ is a partial symmetry axis, which gives a lower bound of ${{1}/{2}}$ for all pth centers. For an upper bound, note that $\{\textrm{Re}(z) = {{1}/{2}}\}$ is another axis of symmetry of $\{0<\textrm{Re}(z)<1\}$ , and the monotonicity of the derivative shows again via Proposition 2.2 that we only need to look in the region

\begin{equation*}f\biggl(\biggl\{0<\textrm{Re}(z)<\frac{1}{2}\biggr\}\biggr),\end{equation*}

which is the region $\{x^2-y^2< {{1}/{2}}\}$ inside U. This gives an upper bound of ${{1}/{\sqrt{2}}}$ on the real axis. Thus all p-centers lie on

\begin{equation*}\biggl\{\textrm{Im}(z) = 0,\ \frac{1}{2} < \textrm{Re}(z) < \frac{1}{\sqrt{2}}\biggr\}.\end{equation*}

This set is in bold in Figure 6.

Example 3.5. Let $\mathscr{C}_{R}$ be the crescent-like shape limited by the two circles

\begin{equation*}\biggl\{\biggl|z-\frac{1}{2}\biggr| = \frac{1}{2}\biggr\}\quad\text{and}\quad \biggl\{\biggl|z-\frac{R}{2}\biggr| = \frac{R}{2}\biggr\}\end{equation*}

(see Figure 7). $\mathscr{C}_{R}$ is the image of the region

\begin{equation*}\biggl\{\frac{1}{R}< \textrm{Re}(z) < 1\biggr\}\end{equation*}

under the conformal map $f(z) = {{1}/{z}}$ . Note that $|f'(z)| = {{1}/{|z|^2}}$ is monotone decreasing in $|z|$ , so by the same argument as in Example 3.4, all pth centers lie on the real axis. Furthermore,

\begin{equation*}\biggl\{\textrm{Re}(z) = \frac{R+1}{2}\biggr\}\end{equation*}

is a partial symmetry axis for $\mathscr{C}_{R}$ , and this allows us to eliminate the region

\begin{equation*}\biggl\{\textrm{Re}(z) > \frac{R+1}{2}\biggr\}\end{equation*}

from consideration. We may also use an axis of symmetry

\begin{equation*}\biggl\{\textrm{Re}(z) = \frac{1+{{1}/{R}}}{2}\biggr\}\end{equation*}

to conclude via Proposition 2.2 that we can exclude the region

\begin{equation*}f\biggl(\biggl\{\frac{1}{R}<\textrm{Re}(z)<\frac{1+ {{1}/{R}}}{2}\biggr\}\biggr),\end{equation*}

which is the region

Figure 7: Bounds for the crescent region.

\begin{equation*}\mathscr{C}_{R} \cap \biggl\{\biggl|z-\frac{R}{R+1}\biggr| < \frac{R}{R+1}\biggr\},\end{equation*}

in the search for pth centers. We see that all p-centers lie on the interval

\begin{equation*}\biggl\{\textrm{Im}(z)=0, \frac{2R}{R+1} < \textrm{Re}(z) < \frac{R+1}{2}\biggr\},\end{equation*}

which is in bold for $R=2$ in Figure 7.

4. Concluding remarks

As remarked earlier, in Figure 1 the regions $\mathcal{A}$ and $\mathcal{B}$ do not fill all of U, and it is natural to search for a better bound by finding a conformal map that fills the entire domain. Let us consider the Schwarz–Christoffel transformation sending the unit disk to $U({\theta})$ given by (see [Reference Driscoll and Trefethen8, Chapter 2])

\begin{equation*}f(z) = A + C \int_{0}^{z} (1-w)^{{{\theta}/{\pi}}-1}(1+w)^{{{\theta}/{\pi}}-1} (1+{\textrm{i}} w)^{{{-2{\theta}}/{\pi}}}\,{\textrm{d}} w\end{equation*}

for appropriate choices of constants A and C; note that this is chosen so that 1 and $-1$ are mapped to the base angles, and i is mapped to the top angle. We have

\begin{equation*}|f'(z)| = |C||1-z|^{{{\theta}/{\pi}}-1}|1+z|^{{{\theta}/{\pi}}-1} |1+{\textrm{i}} z|^{{{-2{\theta}}/{\pi}}}.\end{equation*}

It can then be checked that $|f'(z)| > |f'(\bar z)|$ whenever $\textrm{Re}(z)>0$ , so Proposition 2.2 implies that no pth centers can be found in the image of ${\mathbb D} \cap \{\textrm{Re}(z)<0\}$ . From this point a numerical method can be employed, and the resulting bound should improve the one we found, if desired.

As was mentioned in connection with $\Delta$ -convexity, there are some purely geometrical consequences of our results. In that context, the following may be proved.

As a corollary of this, and of our probabilistic results above, we obtain the following.