Prove that, for each $n\in\mathbb{N}$, $f_n(z)=|z|e^{in\arg(z)}$ is Lipschitz with constant n.

Question

Suppose $n\in\mathbb{N}$ and $f_n:(\mathbb{C},|\cdot|)\rightarrow(\mathbb{C},|\cdot|)$ is defined via the map $z\mapsto |z|e^{in\arg(z)}$. How do you show that $f_n$ is a Lipschitz map with Lipschitz constant n? I know if $f$ maps the unit ball to the unit ball and $n=1$ that $f$ is Lipschitz with constant 1; can be see as an application of the Fundamental Theorem of Calculus. However I am not sure how to extend this. Naturally here $|\cdot|$ denoted the Euclidean norm.

Answer 1

As already observed by Oliver Diaz it suffices to show that $$ \tag{*} |f_n(z) - 1 | \le n^2 |z-1| $$ because of the homogeneity of the function. For $z= re^{i \phi}$ this is equivalent to $$ \begin{align} &r^2 - 2r \cos(n \phi) + 1 \le n^2 \bigl[r^2 - 2r \cos(\phi) + 1 \bigr] \\ \iff &(r-1)^2 + 2r (1-\cos(n\phi)) \le n^2 \left[(r-1)^2 + 2r(1- \cos(\phi))\right] \\ \iff &(r-1)^2 + r \sin^2\left(\frac{n \phi}{2}\right) \le n^2 \left[(r-1)^2 + r\sin^2\left(\frac{ \phi}{2}\right)\right] \, . \end{align} $$ We see that it suffices to prove $$ \begin{align} &\sin^2\left(\frac{n \phi}{2}\right) \le n^2 \sin^2\left( \frac{\phi}{2} \right) \\ \iff & \left|\sin\left(\frac{n \phi}{2}\right) \right|\le n \left| \sin\left( \frac{\phi}{2} \right) \right| \end{align} $$ and that can easily be shown with induction, see for example How to prove by induction that $|\sin(nx)| \leq n|\sin x|$?.

One can also see that equality holds in $(*)$ for $n \ge 2$ if and only if $r=1$ and $\sin\left( \frac{\phi}{2} \right) = 0$, that is if and only if $z=1$.

Answer 2

This is a suggestion. It is too impractical to put this as a comment:

Without loss of generalization, suppose $z=Re^{i\theta}$, $w=re^{i\psi}$ with $R\geq r$ so that $\rho=\frac{r}{R} \leq 1$. Then, the question is whether

$$ |1-\rho e^{in(\psi-\theta)}|\leq n|1-\rho e^{i(\psi-\theta)}| $$ for all $1<\rho\leq1$ and $\psi,\theta$. In other words, if is true that

$$ 1+\rho^2 - 2\rho\cos(n\varphi) \leq n^2(1+\rho^2 -2\rho\cos\varphi)\tag{1}\label{one} $$

for all $0<\rho\leq1$ and $0\leq\varphi\leq2\pi$. Here is a plot ($n=12$) that suggests $$ g_\rho(\varphi)=\frac{n^2\cos\varphi -\cos n\varphi}{n^2-1}\leq 1 \tag{2}\label{two} $$

I had not been worked out the proof of $\eqref{two}$, which would imply $\eqref{one}$ since $1\leq \frac{1+\rho^2}{2\rho}$, but is should be straight forward for you to to do this by showing that $0$, $\pi$, and $2\pi$ are the only critical points of $g_\rho$. This may require a careful analysis of the values of $g'_\rho$ around those critical points in addition to some convexity argument

Answer 3

What the function actually looks like is that it sends $z$ to the point on the circle centered at the origin, with radius $|z|$, to a point with argument $\arg(z^n) = n \arg (z)$. Thus for any $x,y\in\mathbb{C}$, we have a diagram that looks roughly like this (where $x',y'$ are $f_n(x),f_n(y)$):

Thus it suffices to prove the following lemma:

Lemma: In a triangle with angle $\alpha$ in between sides of length $a,b$ and third side $c$, the triangle with angle $n\alpha$ in between sides of lengths $a,b$ and third side $c'$ satisfies $c'\le nc$.

Firstly, we prove this lemma for $n$ odd. If we reflect the initial triangle $n-1$ times, it's easy to see that the lemma is true -- the shortest path between two points is a straight line, as can be seen in this example when $n=5$.:

Next, if the lemma is true for $n=i$ and $n=j$, it is certainly true for $n=ij$, by applying it twice. This follows if we let the side length $c_n=g_n(a,b,\alpha)$ be a function of $a,b$ and the included angle. Then $$c_{ij}\ge ic_j\ge ijc_j,$$ where the first inequality follows from the assumption that it is true for $n=i$, and the latter for $n=j$.

Thus it suffices to prove the case when $n=2$.

In this case, we have:

$$c=\sqrt{a^2+b^2-2ab\cos\alpha}$$ $$c'=\sqrt{a^2+b^2-2ab\cos(2\alpha)}$$

So this is equivalent to proving that $$2^2(a^2+b^2-2ab\cos\alpha)\ge a^2+b^2-2ab\cos(2\alpha).$$ Rearrange this to get $$3\frac{a^2+b^2}{2ab}\ge 4\cos\alpha - \cos(2\alpha).$$

But $a^2+b^2\ge 2ab$, so $\frac{a^2+b^2}{2ab}\ge1$, and thus it suffices to show that $$3\ge 4\cos\alpha-\cos(2\alpha).$$

Using the double angle formula for cos ($\cos(2\alpha)=2\cos^2\alpha-1$), this simplifies directly to $$2(\cos\alpha-1)^2\ge0,$$

and we are done.

Answer 4

Let$$ f(z)=|z|e^{in\arg(z)}\tag1 $$ For $n=1$, $f(z)=z$, so assume $n\ge2$.

Define $r=\frac{2|z||w|}{|z|^2+|w|^2}\le1$ and $\theta=\arg(z)-\arg(w)\in[-\pi,\pi]$. Then $$ \begin{align} \frac{|f(z)-f(w)|^2}{|z-w|^2} &=\frac{|z|^2+|w|^2-2|z||w|\cos(n(\arg(z)-\arg(w)))}{|z|^2+|w|^2-2|z||w|\cos(\arg(z)-\arg(w))}\tag{2a}\\[6pt] &=\frac{1-r\cos(n\theta)}{1-r\cos(\theta)}\tag{2b}\\ &=1+r\frac{\cos(\theta)-\cos(n\theta)}{1-r\cos(\theta)}\tag{2c}\\ &\le1+r\left|\frac{\cos(\theta)-\cos(n\theta)}{1-r\cos(\theta)}\right|\tag{2d}\\ &\le1+\left|\frac{\cos(\theta)-\cos(n\theta)}{1-\cos(\theta)}\right|\tag{2e}\\ &=1+\left|\frac{1-\cos(n\theta)}{1-\cos(\theta)}-1\right|\tag{2f}\\[3pt] &\le n^2\tag{2g} \end{align} $$ Explanation:
$\text{(2a)}$: Law of Cosines
$\text{(2b)}$: apply the definitions of $r$ and $\theta$
$\text{(2c)}$: algebra
$\text{(2d)}$: Triangle Inequality
$\text{(2e)}$: $r\le1$
$\text{(2f)}$: algebra
$\text{(2g)}$: $0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le n^2$

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$\boldsymbol{0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le n^2}$

Since $\cos(x)$ is even, we need only consider $0\lt\theta\le\pi$. $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}\theta}\frac{1-\cos(n\theta)}{1-\cos(\theta)} &=\frac{n(1-\cos(\theta))\sin(n\theta)-(1-\cos(n\theta))\sin(\theta)}{(1-\cos(\theta))^2}\tag{3a}\\ &=\frac{\sin(\theta)\sin(n\theta)}{(1-\cos(\theta))^2}(n\tan(\theta/2)-\tan(n\theta/2))\tag{3b} \end{align} $$ Since $\tan(x)$ is convex, for $0\lt\theta\le\frac\pi{n}$, $$ \overbrace{\ \sin(n\theta)\ }^{\ge0}\overbrace{(n\tan(\theta/2)-\tan(n\theta/2))}^{\le0}\le0\tag4 $$ For $\frac\pi{n}\lt\theta\le\frac{2\pi}n$, $$ \overbrace{\ \sin(n\theta)\ }^{\le0}(\overbrace{n\tan(\theta/2)}^{\ge0}-\overbrace{\tan(n\theta/2)}^{\le0})\le0\tag5 $$ Thus, for $0\lt\theta\le\frac{2\pi}n$, inequalities $(3)$, $(4)$, and $(5)$ show that $\frac{1-\cos(n\theta)}{1-\cos(\theta)}$ is decreasing and the limit as $\theta\to0$ is $n^2$. Therefore, $$ 0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le n^2\tag6 $$ For $\frac{2\pi}n\lt\theta\le\pi$, we have $\frac8{n^2}\le\frac2{\pi^2}\theta^2\le1-\cos(\theta)\le\frac12\theta^2$. Therefore, $$ 0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le\frac{n^2}4\tag7 $$ Inequalities $(6)$ and $(7)$ show that $0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le n^2$.

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Another approach to $\boldsymbol{0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le n^2}$

Martin R mentions that we can use the relation $1-\cos(x)=2\sin^2\left(\frac x2\right)$.

As he mentions, induction can be used to prove $$ |\sin(nx)|\le n\,|\sin(x)|\tag8 $$ $(8)$ obviously holds for $n=0$, so assume it holds for some $n$. Then $$ \begin{align} |\sin((n+1)x)| &=|\sin(nx)\cos(x)+\cos(nx)\sin(x)|\tag{9a}\\ &\le|\sin(nx)\cos(x)|+|\cos(nx)\sin(x)|\tag{9b}\\ &\le|\sin(nx)|+|\sin(x)|\tag{9c}\\ &\le n|\sin(x)|+|\sin(x)|\tag{9d}\\ &=(n+1)|\sin(x)|\tag{9e} \end{align} $$ Thus, $(8)$ holds for all $n\ge0$.

$(8)$ implies that $$ 0\le\frac{\sin^2\left(\frac{n\theta}2\right)}{\sin^2\left(\frac{\theta}2\right)}\le n^2\tag{10} $$ which is equivalent to $$ 0\le\frac{1-\cos(n\theta)}{1-\cos(\theta)}\le n^2\tag{11} $$