Ovals and support functions of plane curves

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Let α(s)\alpha (s) be a closed plane curve. Define the diameter dαd_\alpha of α(s)\alpha (s) to be

dα=sup⁡t,s∈R∥α(s)−α(t)∥. d_\alpha = \sup _{t,s \in \mathbb {R}} \left\| \alpha (s) - \alpha (t)\right\| .

Now assume that the curvature k(s)≥1k(s) \geq 1 for all ss.

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(a)

For any N∈Z+N \in \mathbb {Z}^+ sketch an example of such an α(s)\alpha (s) with dα>Nd_\alpha > N.

(b)

Assume further that α\alpha is a simple closed curve. Prove that dα≤2d_\alpha \leq 2 (or some other constant independent of α\alpha; 2 is the best possible such constant).

We assume that in the motion defined in problem 103 the point w=f(z)w=f(z) describes in the positive direction a closed curve without double points. This curve is everywhere convex (seen from outside) if and only if for ∣z∣=r\left|z\right|=r

\rezf′′(z)f′(z)>−1. \re z\frac{f''(z)}{f'(z)} > -1.
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The set of points with respect to which a closed curve is star-shaped forms a convex set. Prove this purely geometric proposition for analytic curves with the help of problem 109.

Let R\mathfrak {R} denote a finite convex domain in the w=u+ivw=u+iv-plane. For a fixed angle φ\varphi the expression

ucos⁡φ+vsin⁡φ=\rew‾eiφ u\cos \varphi + v\sin \varphi = \re \overline{w}e^{i\varphi }

assumes a certain maximum h(φ)h(\varphi ) in R\mathfrak {R}. The function h(φ)h(\varphi ) is periodic with period 2π2\pi and is called the support function of R\mathfrak {R}. The straight line

ucos⁡φ+vsin⁡φ−h(φ)=0 u\cos \varphi + v\sin \varphi - h(\varphi ) = 0

is a line of support; its normal points away from R\mathfrak {R} and forms the angle φ\varphi with the positive uu-axis. If R\mathfrak {R} extends to infinity these definitions are modified insofar as a finite maximum exists only in a sector with an opening ≤π\leq \pi. The two cases: the infinite strip and the half-plane are exceptions; then there exist only two lines of support, or only one, respectively.

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Find the support function of the convex domain that consists of the point a=∣a∣eiαa=\left|a\right|e^{i\alpha }.

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The function w=f(z)w=f(z) establishes a one to one correspondence between the disk ∣z∣≤r\left|z\right|\leq r and the convex domain R\mathfrak {R}. Suppose that f(z)f(z) is regular and f′(z)≠0f'(z)\neq 0 at a certain point zz on the boundary, ∣z∣=r\left|z\right|=r. In this case a definite line of support (tangent) passes through the boundary point w=f(z)w=f(z) of R\mathfrak {R}. Express the corresponding quantities φ\varphi and h(φ)h(\varphi ) in terms of f(z)f(z).

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The function w=log⁡(1+z)w=\log (1+z) maps the disk ∣z∣=1\left|z\right|=1 onto an infinite domain contained in the strip −π2<Im(w)<π2-\frac{\pi }{2}<\mathrm{Im}\left(w\right)<\frac{\pi }{2}. Its support function (defined only for −π2≤φ≤π2-\frac{\pi }{2}\leq \varphi \leq \frac{\pi }{2}) is

h(φ)=cos⁡φ⋅log⁡(2cos⁡φ)+φsin⁡φ. h(\varphi ) = \cos \varphi \cdot \log (2\cos \varphi ) + \varphi \sin \varphi .
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The function w=2iarcsin⁡izw=\frac{2}{i}\arcsin iz maps the disk ∣z∣≤1\left|z\right|\leq 1 onto a finite convex domain which has two corners and lies in the strip −π≤Im(w)≤π-\pi \leq \mathrm{Im}\left(w\right)\leq \pi. Its support function is

h(φ)={cos⁡φlog⁡(cos⁡2φ+2cos⁡φ)2+2sin⁡φarcsin⁡(2sin⁡φ),for 0≤φ≤π4;πsin⁡φ,for π4≤φ≤π2; h(\varphi ) = \begin{cases} \cos \varphi \log (\sqrt{\cos 2\varphi }+\sqrt{2}\cos \varphi )^{2} + 2\sin \varphi \arcsin (\sqrt{2}\sin \varphi ), & \text{for } 0\leq \varphi \leq \frac{\pi }{4}; \\ \pi \sin \varphi , & \text{for } \frac{\pi }{4}\leq \varphi \leq \frac{\pi }{2}; \end{cases} h(φ+π)=h(−φ)=h(φ). h(\varphi +\pi ) = h(-\varphi ) = h(\varphi ).
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The equation we−w+1=zwe^{-w+1}=z defines w=f(z)w=f(z). It maps the disk ∣z∣≤1\left|z\right|\leq 1 onto a finite convex domain which has one corner and lies in the half-plane Re(w)≤1\mathrm{Re}\left(w\right)\leq 1. Its support function is

h(φ)=cos⁡φfor−π4≤φ≤π4; h(\varphi ) = \cos \varphi \quad \text{for} \quad -\frac{\pi }{4}\leq \varphi \leq \frac{\pi }{4};

in the sector π4<φ<7π4\frac{\pi }{4}<\varphi <\frac{7\pi }{4} it is given in parametric form

h(φ)eiφ=w1−w\re(1−w) h(\varphi )e^{i\varphi } = \frac{w}{1-w}\re (1-w)

whereby ww describes the boundary, ∣we−w+1∣=1\left|we^{-w+1}\right|=1.

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Let h(φ)h(\varphi ) denote the support function of the infinite convex domain considered in problem 114. The sequence f1(z),f2(z),…,fn(z),…f_{1}(z),f_{2}(z),\ldots ,f_{n}(z),\ldots

fn(z)=z(z−1)(z−2)⋯(z−n+1)(z−n)n!e−rh(φ),n=1,2,3,…, f_{n}(z) = \frac{\sqrt{z}(z-1)(z-2)\cdots (z-n+1)(z-n)}{n!}e^{-rh(\varphi )}, \quad n=1,2,3,\ldots ,

where z=reiφz=re^{i\varphi }, is bounded in the entire half-plane Re(z)≥0\mathrm{Re}\left(z\right)\geq 0 [problems 12, and Part II, problem 220]; h(φ)h(\varphi ) is the smallest function of the angle φ\varphi that keeps the sequence bounded.

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Let h(φ)h(\varphi ) denote the support function of the convex domain considered in problem 115. The sequence f1(z),f2(z),…,fn(z),…f_{1}(z),f_{2}(z),\ldots ,f_{n}(z),\ldots

fn(z)=z(1−z212)(1−z222)⋯(1−z2n2)e−rh(φ),n=1,2,3,…, f_{n}(z) = z\left(1-\frac{z^{2}}{1^{2}}\right)\left(1-\frac{z^{2}}{2^{2}}\right)\cdots \left(1-\frac{z^{2}}{n^{2}}\right)e^{-rh(\varphi )}, \quad n=1,2,3,\ldots ,

where z=reiφz=re^{i\varphi }, is bounded in the entire zz-plane [problems 13, and Part II, problem 221]; h(φ)h(\varphi ) is the smallest function of the angle φ\varphi that keeps the sequence bounded.

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Let h(φ)h(\varphi ) denote the support function of the convex domain considered in problem 116. Then

∣1+zn∣ne−rh(φ)≤1,n=1,2,3,…, \left|1+\frac{z}{n}\right|^{n}e^{-rh(\varphi )} \leq 1, \quad n=1,2,3,\ldots ,

z=reiφz=re^{i\varphi }, in the entire plane.

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On a closed convex curve with continuous curvature there are always at least three different pairs of points with the following property: The tangents at the two points of a pair are parallel and the curvature of the curve at the two points is the same.

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Remark.
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W. Blaschke, problem: Arch. Math. Phys. Ser. 3, 26, p. 65 (1917). Solution by G. Szegö: Arch. Math. Phys. Ser. 3, 28, pp. 183--184 (1920). Cf. also W. Süss: Jber. deutsch. Math. Verein. 33, Section 2, pp. 32--33 (1924).

Let the boundary ∂D\partial D of a domain DD consist of one smooth closed curve with parametric equation z=z(t),a⩽t⩽bz = z(t), a \leqslant t \leqslant b, where the function z(t)z(t) has, on the segment [a,b][a, b], a continuous and nonzero derivative. Prove that:

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(1)

the domain DD is convex if and only if

Im(z′′(t)z′(t))⩾0,a⩽t⩽b \mathrm{Im}\left(\frac{z^{\prime \prime }(t)}{z^{\prime }(t)}\right) \geqslant 0, \quad a \leqslant t \leqslant b
(2)

the domain DD is star-shaped with respect to the point z0∈Dz_{0} \in D if and only if

Im(z′(t)z(t)−z0)⩾0,a⩽t⩽b \mathrm{Im}\left(\frac{z^{\prime }(t)}{z(t)-z_{0}}\right) \geqslant 0, \quad a \leqslant t \leqslant b

References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.