# Pointwise rotation
for homeomorphisms with integrable distortion
and controlled compression

Lauri Hitruhin, Banhirup Sengupta

###### Abstract

We obtain sharp rotation bounds for homeomorphisms $f:\mathbb{C}\to\mathbb{C}$
whose distortion is in $L^{p}_{loc}$, $p\geq 1$, and whose inverse have
controlled modulus of continuity. The interest in this class is partially
motivated by examples arising from fluid mechanics. We also present examples
proving sharpness in a strong sense.

##  1 Introduction

We call $f:\mathbb{C}\to\mathbb{C}$ a mapping of finite distortion if it
belongs to the Sobolev space $f\in W^{1,1}_{loc}(\mathbb{C};\mathbb{C})$, its
Jacobian determinant $\det(Df)=J(\cdot,f)$ is locally integrable, and there
exists a positive measurable function
$\mathbb{K}(\cdot,f):\mathbb{C}\to[1,+\infty]$ such that

$|Df(z)|^{2}\leq\mathbb{K}(z,f)\cdot J(z,f)$

at almost every point $z\in\mathbb{C}$. Here $|Df(z)|$ stands for the operator
norm of the differential matrix $Df(z)$ of $f$ at the point $z$. If
$\mathbb{K}(\cdot,f)\in L^{\infty}$, then $f$ is said to be a mapping of
bounded distortion or $K$-quasiregular, where
$K=\|\mathbb{K}(\cdot,f)\|_{\infty}$. Furthermore, $f$ is called
$K$-quasiconformal if $f$ is a homeomorphism. Partially motivated by questions
in nonlinear elasticity, mappings of finite distortion came into existence as
a generalizaion of quasiregular maps. The interested reader should refer to
the monographs [1] and [11] for a background in quasiregular maps and mappings
of finite distortion.

Lately there has been a significant upsurge of interest in understanding the
pointwise rotational properties of planar homeomorphisms of finite distortion,
along with the spiraling rate of these maps, see [2, 4, 7, 8, 9, 10]. To be
precise, given such a homeomorphism $f:\mathbb{C}\to\mathbb{C}$ normalized by
$f(0)=0$ and $f(1)=1$, one is interested in the maximal growth of
$|\operatorname{arg}(f(r))|$ as $r\to 0$. This growth represents the winding
number of the image $f([r,1])$ around the origin as $r\to 0^{+}$. It is known
that this quantity admits several speeds of growth depending on the class of
maps under study. For example, it was shown in [2] that if $f$ is
$K$-quasiconformal then

$|\operatorname{arg}(f(r))|\leq\frac{1}{2}\left(K-\frac{1}{K}\right)\,\log\left(\frac{1}{r}\right)+c_{K},\hskip
28.45274pt\text{for all }0<r<1.$ (1)

On the other hand, for homeomorphisms of finite distortion situation changes
drastically and the order of spiraling depends on integrability of the
distortion function. Namely, the first named author discovered in [8] that if
$e^{\mathbb{K}(\cdot,f)}\in L^{p}_{loc}$ for some $p>0$ then

$|\operatorname{arg}(f(z))|\leq\frac{c}{p}\,\log^{2}\left(\frac{1}{|z|}\right),\hskip
28.45274pt\text{for small enough }|z|,$

and moreover this is sharp up to the constant $c>0$. In other words, there is
a certain payoff to transit from boundedness to exponential integrability of
$\mathbb{K}(\cdot,f)$. Precisely, it is an increment in the exponent of the
logarithmic term. Further optimal results were obtained later on in [9] for
homeomorphisms with integrable distortion, that is, when
$\mathbb{K}(\cdot,f)\in L^{p}_{loc}$ for some $p>1$,

$|\operatorname{arg}(f(z))|\leq\frac{c}{|z|^{\frac{2}{p}}},\hskip
28.45274pt\text{for small enough }|z|$ (2)

or even when $\mathbb{K}(\cdot,f)\in L^{1}_{loc}$,

$\limsup_{|z|\to 0}|z|^{2}|\operatorname{arg}(f(z))|=0.$ (3)

The above estimates show, as one would expect, that more spiraling can be
allowed by relaxing the degree of integrability of $\mathbb{K}(\cdot,f)$.

One of the most important applications of mappings of finite distortion is
their prominent role in the study of fluid dynamics. Recently it has been
proven that there is an important class of homeomorphisms, arising from fluid
mechanics, all of which have distortion function in some $L^{p}_{loc}$ space
and are bi-Hölder. This class, first found at [6], contains as an example the
flow maps of any Yudovich solution to the planar incompressible Euler system
of equations

$\begin{cases}\omega_{t}+(\mathbf{v}\cdot\nabla)\omega=0\\\
\operatorname{div}(\mathbf{v})=0\\\ \omega(0,\cdot)=\omega_{0}.\end{cases}$
(4)

Here $\omega=\omega(t,z):[0,T]\times\mathbb{C}\to\mathbb{C}$ is the unknown,
$\omega_{0}\in L^{\infty}(\mathbb{C};\mathbb{C})$ is given, and $\mathbf{v}$
is the velocity field. The Biot-Savart law,

$\mathbf{v}=\frac{i}{2\pi\bar{z}}\ast\omega$

provides the precise relation between $v$ and $\omega$. The fundamental result
by Yudovich [15] proved existence and uniqueness of a solution $\omega\in
L^{\infty}([0,T];L^{\infty}(\mathbb{C};\mathbb{C}))$ for any given
$\omega_{0}$. In particular, the corresponding velocity field $\mathbf{v}$
belongs to the Zygmund class, and therefore the classical Cauchy-Lipschitz
theory guarantees for the ODE

$\begin{cases}\frac{d}{dt}X(t,z)=\mathbf{v}(t,X(t,z))\\\ X(0,z)=z\end{cases}$

both existence and uniqueness of a flow map
$X:[0,T]\times\mathbb{C}\to\mathbb{C}$. It was proven in [6] that, for small
enough $t>0$, each of the flow homeomorphisms
$X_{t}=X(t,\cdot):\mathbb{C}\to\mathbb{C}$ is indeed a mapping of finite
distortion. Moreover, for each such small values of $t>0$ there is a number
$p(t)>1$ such that the distortion function $\mathbb{K}(\cdot,X_{t})$ belongs
to $L^{p}_{loc}$ whenever $p<p(t)$. The Hölder continuity of both $X_{t}$ and
$X_{t}^{-1}$ was proven in [14].

Inspired by this connection between mappings of finite distortion and fluid
dynamics the authors reformulated the bound (2) in [5] to fit this situation.
More precisely, it was shown that if $f:\mathbb{C}\to\mathbb{C}$ is a
homeomorphism with finite distortion such that $\mathbb{K}(\cdot,f)\in
L^{p}_{loc}$ for some $p>1$ and $f$ has a $\alpha$-Hölder continuous inverse,
then

$|\operatorname{arg}(f(z))|\leq
C\,\sqrt{\alpha}\,|z|^{-\frac{1}{p}}\,\log^{\frac{1}{2}}\left(\frac{1}{|z|}\right),\hskip
28.45274pt\text{for small enough }|z|,$ (5)

which drastically improves the bound (2) obtained without the Hölder
assumption. This is not surprising as the local rotational properties depend
not just on the integrability of $\mathbb{K}(\cdot,f)$ but also on the local
stretching properties of the mapping, see [2, 8, 9]. In this article, we
extend our previous work from [5] to a more general class of homeomorphisms,
which have $L^{p}_{loc}$ distortion for $p\geq 1$ and the inverse having
predetermined modulus of continuity.

######  Theorem 1.

Let $f:\mathbb{C}\to\mathbb{C}$ be a homeomorphism of finite distortion such
that $f(0)=0$, $f(1)=1$,
and assume that $\mathbb{K}(\cdot,f)\in L^{p}_{loc};$ $p>1.$ Then

$\left|\operatorname{arg}\left(f(z)\right)\right|\leq
C\,|z|^{-\frac{1}{p}}\,\log^{\frac{1}{2}}\left(\frac{1}{\underset{\rm|\omega|=|z|}{\rm\min}|f(\omega)|}\right)\quad\text{when
$|z|$ is small.}$ (6)

Furthermore, if we assume that $\mathbb{K}(\cdot,f)\in L^{1}_{loc},$ then

$\limsup_{|z|\to
0}\frac{|z|}{\sqrt{\log\left(\frac{1}{\underset{\rm|\omega|=|z|}{\rm\min}|f(\omega)|}\right)}}|\operatorname{arg}(f(z))|=0.$
(7)

Towards the optimality of Theorem 1, we can show the following.

######  Theorem 2.

Let $\varphi$ be a radially increasing homeomorphism with $p$-integrable
distortion, $p\geq 1$, such that

$e^{-g_{\varphi,p}(|z|)|z|^{-\frac{2}{p}}}\leq|\varphi(z)|<|z|^{4}\quad\text{when
$|z|$ is small,}$ (8)

where $g_{\varphi,p}:\mathbb{R}\to\mathbb{R}$ is an increasing continuous
function with $g(r)\to 0$ when $r\to 0$. Then we can choose an increasing onto
homeomorphism $h:[0,+\infty)\to[0,+\infty)$, which can converge to zero as
slow as we want, and find a homeomorphism $\bar{f}:\mathbb{C}\to\mathbb{C}$
with the following properties:

  * •

$\bar{f}$ is a homeomorphism of finite distortion, with
$\mathbb{K}(\cdot,\bar{f})\in L^{p}_{loc}$.

  * •

$\bar{f}(0)=0$, $\bar{f}(1)=1$.

  * •

There exists a decreasing sequence $\\{r_{n}\\}$, such that

$|\bar{f}(r_{n})|=|\varphi(r_{n})|$ (9)

and

$\left|\operatorname{arg}\left(\bar{f}(r_{n})\right)\right|\geq
r_{n}^{-\frac{1}{p}}\,\log^{\frac{1}{2}}\left(\frac{1}{|\bar{f}(r_{n})|}\right)\,h(r_{n}).$
(10)

Furthermore, this sequence $\\{r_{n}\\}$ can be chosen to be a subsequence for
an arbitrary decreasing sequence $\\{\lambda_{n}\\}\to 0$ of positive numbers.

Note that the homeomorphism $\bar{f}$ in Theorem 2 is radial and hence
$\underset{\rm|\omega|=|z|}{\rm\min}|\bar{f}(\omega)|=|\bar{f}(z)|$. Since $h$
can be chosen to approach zero at any speed and sequence $\lambda_{n}$ can be
chosen freely, Theorem 2 shows that the upper bound provided in Theorem 1 is
essentially sharp when we restrict modulus to satisfy (8).

Let us now briefly give some explanation for the bounds (8). The one on the
right specifies that we are studying mappings that compress stronger than
Hölder maps, which were studied in [5], and thus have faster maximal spiraling
rate than given in (5). On the other hand, the bound on the left is always
satisfied when $p=1$, see [9], and when $p>1$ it is exact up to the gauge
function $g_{\varphi,p}$, see [12]. Studying rotation under extremal
compression leads to the extremal pointwise spiraling as shown in [9]. Thus
Theorem 2, together with examples in [9] proving optimality of the extremal
spiraling rate (2), show that whenever mapping $f$ is compressing we have
essentially sharp spiraling rates.

As a Corollary to Theorem 1 we can extend the case with Hölder bounds and
$p>1$, see [5, Theorem 1], to the borderline situation $p=1$.

######  Corollary 3.

Let $f:\mathbb{C}\to\mathbb{C}$ be a homeomorphism of finite distortion such
that $f(0)=0$ and $f(1)=1$, and assume that $\mathbb{K}(\cdot,f)\in
L^{1}_{loc}$. Moreover, let us suppose that

$|f(x)-f(y)|\geq C\left|x-y\right|^{\alpha}\hskip 28.45274pt\text{if
}|x-y|\text{ is small,}$

for some $\alpha\geq 1$. Then

$\limsup_{|z|\to
0}\frac{|z|}{\sqrt{\log\left(\frac{1}{|z|}\right)}}|\operatorname{arg}(f(z))|=0.$
(11)

Note that in the case $p=1$ we get an improvement in the form of vanishing
limsup compared to the case $p>1$, which is described by the bound (5). This
is analogous to the maximal spiraling bounds (2) and (3), where the exact same
improvement happens.

Finally, we prove the optimality of the above result in a strong sense.

######  Theorem 4.

Given an increasing, onto homeomorphism $h:[0,+\infty)\to[0,+\infty)$, an
arbitrary $\epsilon>0$ and a real number $\beta\geq 1$, there exists a
homeomorphism $\bar{f}:\mathbb{C}\to\mathbb{C}$ with the following properties:

  * (a)

$\bar{f}$ is a mapping of finite distortion, with
$\mathbb{K}(\cdot,\bar{f})\in L^{1}_{loc}$

  * (b)

$\bar{f}(0)=0$, $\bar{f}(1)=1$

  * (c)

If $\alpha\geq 2\left(\beta+2\right)+\epsilon$, then
$|\bar{f}(x)-\bar{f}(y)|\geq C|x-y|^{\alpha}$ whenever $|x-y|<1.$

  * (d)

There exists a decreasing sequence $\\{r_{n}\\}$, with limit $r_{n}\to 0^{+}$
as $n\to\infty$, for which

$|\operatorname{arg}(\bar{f}(r_{n}))|\geq\frac{h(r_{n})}{r_{n}}\,\left(\beta\log\left(\frac{1}{r_{n}}\right)\right)^{\frac{1}{2}}$

and again the sequence $r_{n}$ can be chosen to be a subsequence for an
arbitrary decreasing sequence $\\{\lambda_{n}\\}\to 0$ of positive numbers.

Towards the proof of Theorem 4, we modify the construction from [5, Theorem 3]
giving optimality for the bound (5). However, this construction as written in
[5] does not cover the case $\mathbb{K}\in L^{1}_{loc}$, and thus some changes
in the argument are necessary. Also we note that Corollary 3 is extremely
sharp as the homeomorphism $h$ in Theorem 4 can go to zero as slow as we wish.

The paper is structured as follows. In Section 2 we introduce the basic
preliminaries. In Section 3 we prove the positive results. Finally, we prove
optimality in Section 4.

Acknowledgements. We would like to thank Albert Clop for some interesting
comments and suggestions. B.S. is partially supported by projects
MTM2016-81703-ERC, MTM2016-75390 (Spanish Government) and 2017SGR395 (Catalan
Government). L.H. was partially supported by the ERC grant 834728 Quamap and
by the Finnish Academy of Science projects 13316965 and 1346562.

##  2 Preliminaries

A mapping $f:\mathbb{C}\to\mathbb{C}$ is said to be $\alpha$-Hölder
continuous, or simply Hölder from above, if there exist constants
$C,d,\alpha>0$ such that for any two points $x,y\in\mathbb{C}$, with
$|x-y|<d$, one has

$|f(x)-f(y)|\leq C|x-y|^{\alpha}.$

Similarly, we say $f$ is $\beta$-Hölder from below if there are constants
$\bar{C},d,\beta>0$ such that for any two points $x,y\in\mathbb{C}$, with
$|x-y|<d$, one has

$|f(x)-f(y)|\geq\bar{C}|x-y|^{\beta}.$

A mapping $f:\mathbb{C}\to\mathbb{C}$ is called bi-Hölder if it is both Hölder
from above and from below.

Let us next define pointwise spiraling. Assume $f:\mathbb{C}\to\mathbb{C}$ is
a homeomorphism of finite distortion and fix a point $z_{0}\in\mathbb{C}$. In
order to study the pointwise rotation of $f$ at the point $z_{0}$ we start by
fixing an argument $\theta\in[0,2\pi)$ and looking at how the quantity

$\operatorname{arg}(f(z_{0}+te^{i\theta})-f(z_{0}))$

changes as the parameter $t$ goes from 1 to a small $r$. This can also be
understood as the winding of the path
$f\left([z_{0}+re^{i\theta},z_{0}+e^{i\theta}]\right)$ around the point
$f(z_{0})$. As we are interested in the maximal pointwise spiraling we
naturally need to find the maximum over all directions $\theta$,

$\sup_{\theta\in[0,2\pi)}|\operatorname{arg}(f(z_{0}+re^{i\theta})-f(z_{0}))-\operatorname{arg}(f(z_{0}+e^{i\theta})-f(z_{0}))|.$
(12)

The maximal pointwise rotation is precisely the behavior of the above quantity
(12) when $r\to 0$. That is, we say that the map $f$ _spirals at the point
$z_{0}$ with a rate $g$_, where $g:[0,\infty)\to[0,\infty)$ is a continuous
decreasing function, if

$\limsup_{r\to
0}\frac{\sup_{\theta\in[0,2\pi)}|\operatorname{arg}(f(z_{0}+re^{i\theta})-f(z_{0}))-\operatorname{arg}(f(z_{0}+e^{i\theta})-f(z_{0}))|}{g(r)}=C$
(13)

for some constant $C>0$. Finding maximal pointwise rotation for a given class
of mappings is equivalent to find maximal growth of function $g$ when $r\to 0$
for this class. Note that in (13) we must use limit superior as the limit
itself might not exist. Furthermore, for a given mapping $f$ there might be
many different sequences $r_{n}\to 0$ along which it has profoundly different
rotational behavior.

Our proof of Theorem 1 relies heavily on the modulus of path families. We give
here the main definitions, see, for example, [13] for a closer look at the
topic. An image of a line segment $I$ under a continuous mapping is called a
_path_ , and we denote by $\Gamma$ a family of paths. Given a path family
$\Gamma$, we say that a Borel measurable function $\rho$ is _admissible for
$\Gamma$_ if any rectifiable $\gamma\in\Gamma$ satisfies

$\int_{\gamma}\rho(z)dz\geq 1.$

The _modulus of the path family $\Gamma$_ is then defined by

$M(\Gamma)=\inf_{\rho\text{ admissible}}\int_{\mathbb{C}}\rho^{2}(z)\,dA(z),$

where $dA(z)$ denotes the Lebesgue measure on $\mathbb{C}$. Intuitively, the
modulus is big if the family $\Gamma$ has _lots_ of short paths, and it is
small if the paths are long and there are _not many_ of them.

We will also need a weighted version of the modulus. Any measurable, locally
integrable function $\omega:\mathbb{C}\to[0,\infty)$ will be called _weight
function_. In our case the weight function $\omega$ will always be the
distortion function $\mathbb{K}(\cdot,f)$. Then, we define the weighted
modulus $M_{\omega}(\Gamma)$ by

$M_{\omega}(\Gamma)=\inf_{\rho\text{
admissible}}\int_{\mathbb{C}}\rho^{2}(z)\,\omega(z)\,dA(z).$

Finally, we need the modulus inequality

$M(f(\Gamma))\leq M_{\mathbb{K}(\cdot,f)}(\Gamma)$ (14)

which holds for any homeomorphism $f$ of finite distortion for which the
distortion $\mathbb{K}(\cdot,f)$ is locally integrable, proven by the first
named author in [9].

##  3 Spiraling bounds

In this section, we prove Theorems 1 and 3 following closely the proof for
extremal spiraling rate (2) in [9].

PROOF OF _THEOREM 1_:

###### Proof.

Let $f$ satisfy the hypothesis of Theorem 1, and let
$z\in\mathbb{C}\setminus\\{0\\}$ be such that $|z|<1.$ Our goal is to estimate
the _winding number_ $n(z)$ of the image set
$f\left(\left[z,\frac{z}{|z|}\right]\right)$ around the origin (recall that
$f(0)=0$). We will bound $n(z)$ using the modulus inequality (14). More
precisely, we will prove that

$\displaystyle n(z)\leq
C\,|z|^{-\frac{1}{p}}\,\log^{\frac{1}{2}}\left(\frac{1}{\underset{\rm|z_{0}|=|z|}{\rm\min}|f(z_{0})|}\right)$

which is equivalent to Theorem 1 when $p>1$, and

$\displaystyle\limsup_{|z|\to
0}\frac{|z|}{\sqrt{\log\left(\frac{1}{\underset{\rm|z_{0}|=|z|}{\rm\min}|f(z_{0})|}\right)}}n(z)=0$

for $p=1$.

Let us first prove $p>1$ case. To this end, choose an arbitrary point
$z_{0}\in\mathbb{C}\setminus\\{0\\}$ such that $|z_{0}|<1$. Without loss of
generality we can assume that $z_{0}$ is real and positive. Next, fix line
segments $E=[z_{0},1]$ and $F=(-\infty,0]$, and let $\Gamma$ be the family of
paths connecting them. Then we estimate the modulus term
$M_{\mathbb{K}(\cdot,f)}(\Gamma)$ from above. Fix balls
$B_{j}=(2^{j}z_{0},2^{j}z_{0})$, $j\in\\{0,1,...,n\\}$ and let $n$ be the
smallest integer such that $2^{n}z_{0}\geq 1$. Let us define

$\displaystyle\rho_{0}(z)=\begin{cases}\frac{2}{r(B_{0})}\hskip
5.69046pt\mbox{if}\hskip 5.69046ptz\in B_{0}\\\ \frac{2}{r(B_{1})}\hskip
5.69046pt\mbox{if}\hskip 5.69046ptz\in B_{1}\setminus B_{0}\\\ \vdots\\\
\frac{2}{r(B_{n})}\hskip 5.69046pt\mbox{if}\hskip 5.69046ptz\in B_{n}\setminus
B_{n-1}\\\ 0\hskip 17.07182pt\mbox{otherwise}\end{cases}$

It is clear that any $z\in E$ belongs to some ball $\frac{1}{2}B_{j}$ and that
$\rho_{0}(z)\geq\frac{2}{r(B_{j})}$, whenever $z\in B_{j}$. Moreover, since
$B_{j}\cap F=\phi$ for every $j$, this implies that $\rho_{0}(z)$ is
admissible with respect to $\Gamma$. Therefore, one can estimate the modulus
from above as follows:

$\displaystyle M_{\mathbb{K}(\cdot,f)}(\Gamma)$
$\displaystyle=\inf_{\rho\hskip
2.84544ptadmissible}\int_{\mathbb{C}}\mathbb{K}(\cdot,f)\rho^{2}(z)\,dA(z)$
$\displaystyle\leq\int_{\mathbb{C}}\mathbb{K}(\cdot,f)\rho_{0}^{2}(z)\,dA(z)$
$\displaystyle\leq\|\mathbb{K}(\cdot,f)\|_{L^{p}\left(B(0,4)\right)}\left(\int_{B(0,4)}\rho_{0}^{\frac{2p}{p-1}}(z)\,dA(z)\right)^{\frac{p-1}{p}}$
$\displaystyle\leq
c_{f,p}\left(\int_{B(0,4)}\rho_{0}^{\frac{2p}{p-1}}(z)\,dA(z)\right)^{\frac{p-1}{p}}$

Let us now estimate the integral term by using the definition of $\rho_{0}$.

$\displaystyle\int_{B(0,4)}\rho_{0}^{\frac{2p}{p-1}}(z)\,dA(z)$
$\displaystyle\leq\sum_{j=0}^{n}\int_{B_{j}}\left(\frac{2}{r(B_{j})}\right)^{\frac{2p}{p-1}}\,dA(z)$
$\displaystyle=\sum_{j=0}^{n}|B_{j}|\left(\frac{2}{r(B_{j})}\right)^{\frac{2p}{p-1}}$
$\displaystyle=c_{p}\sum_{j=0}^{n}\frac{\left(r(B_{j})\right)^{2}}{\left(r(B_{j})\right)^{\frac{2p}{p-1}}}$
$\displaystyle=c_{p}\sum_{j=0}^{n}\frac{1}{z_{0}^{\frac{2}{p-1}}}\frac{1}{2^{\frac{2j}{p-1}}}$
$\displaystyle=c_{p}z_{0}^{-\frac{2}{p-1}}\sum_{j=0}^{n}\frac{1}{2^{\frac{2j}{p-1}}}$

The series $\sum_{j=0}^{n}\frac{1}{2^{\frac{2j}{p-1}}}$ converges to a
constant depending on p for any fixed $p>1$, and hence

$M_{\mathbb{K}(\cdot,f)}(\Gamma)\leq c_{f,p}z_{0}^{-\frac{2}{p}}.$ (15)

Next, we would like to estimate the modulus term $M\left(f(\Gamma)\right)$
from below for $p\geq 1$. Let us recall that $f(0)=0$ and define
$M\left(f(\Gamma)\right)$ in polar coordinates as follows:

$\displaystyle M\left(f(\Gamma)\right)$ $\displaystyle=\inf_{\rho\hskip
2.84544ptadmissible}\int_{\mathbb{C}}\rho^{2}(z)\,dA(z)$
$\displaystyle=\inf_{\rho\hskip
2.84544ptadmissible}\int_{0}^{2\pi}\int_{0}^{\infty}\rho^{2}(r,\theta)r\hskip
2.84544ptdrd\theta$

and provide a lower bound for

$\displaystyle\int_{0}^{\infty}\rho^{2}(r,\theta)rdr$

for an arbitrary direction $\theta\in[0,2\pi)$ and an arbitrary admissible
$\rho$. To this end, let us fix a direction $\theta$ and consider the half
line $L_{\theta}$ starting from the origin to the direction $\theta$. Assume
that the image set $f(E)$ winds once around the origin when $z$ moves from a
point $t_{0}$ to a point $t_{2}$ along $E$ and $f(t_{0}),f(t_{2})\in
L_{\theta}$. Since $f$ is a homeomorphism and the image $f(F)$ contains the
origin and points with big modulus, we can deduce that $f(F)$ must intersect
the line segment $(f(t_{2}),f(t_{0}))$ at least once, say at a point
$f(t_{1})$, with $t_{1}\in F$. Moreover, we can choose $t_{1}$ in such a way
that either the line segment $(f(t_{1}),f(t_{0}))$ or the line segment
$(f(t_{2}),f(t_{1}))$ belongs to the path family $f(\Gamma)$. It is evident
that $f(E)$ cycles around the origin
$n(z_{0})=\left\lfloor{\frac{|\operatorname{arg}\left(f(z_{0})\right)-\operatorname{arg}\left(f(1)\right)|}{2\pi}}\right\rfloor$
times. So, it is possible to find at least

$\displaystyle
n(z_{0})=\left\lfloor{\frac{|\operatorname{arg}\left(f(z_{0})\right)-\operatorname{arg}\left(f(1)\right)|}{2\pi}}\right\rfloor-1$

disjoint line segments belonging to the path family $f(\Gamma)$ using this
argument. Note that $n(z_{0})$ does not depend on the direction $\theta$.
Since we are interested in extremal rotation, it can be assumed that $f(E)$
winds around the origin at least once, which makes it clear that $n(z_{0})$ is
non-negative. Now, the $n(z_{0})$ disjoint line segments can be written in the
form $\left(x_{j}e^{i\theta},y_{j}e^{i\theta}\right)\subset L_{\theta}$, where
$j\in\\{1,2,...,n(z_{0})\\}$ and $x_{j}$,$y_{j}$ are positive real numbers
satisfying

$\displaystyle 0<r_{f}\leq x_{1}<y_{1}<...<x_{n(z_{0})}<y_{n(z_{0})}\leq
c_{f}$

where $c_{f}=\sup_{z\in E}|f(z)|$ and $r_{f}=\inf_{z\in E}|f(z)|$. Hence we
can write

$\displaystyle\int_{0}^{\infty}\rho^{2}(r,\theta)rdr\geq\sum_{j=1}^{n(z_{0})}\int_{x_{j}}^{y_{j}}\rho^{2}(r,\theta)rdr.$

Next, let us consider the Hölder inequality with the functions
$f(r)=\rho\sqrt{r}$ and $g(r)=\frac{1}{\sqrt{r}}$, which after squaring both
sides gives

$\displaystyle\int_{x_{j}}^{y_{j}}\rho^{2}(r,\theta)rdr\geq\left(\int_{x_{j}}^{y_{j}}\rho(r,\theta)dr\right)^{2}\left(\int_{x_{j}}^{y_{j}}\frac{1}{r}dr\right)^{-1}\geq\frac{1}{\log\left(\frac{y_{j}}{x_{j}}\right)}.$

The last inequality holds true as $\rho$ is admissible with respect to
$f(\Gamma)$ and the line segments $(x_{j}e^{i\theta},y_{j}e^{i\theta})$ belong
to the path family $f(\Gamma)$. Therefore,

$\displaystyle\int_{0}^{\infty}\rho^{2}(r,\theta)rdr\geq\sum_{j=1}^{n(z_{0})}\frac{1}{\log\left(\frac{y_{j}}{x_{j}}\right)}.$

It is quite clear from the definition of $c_{f}$ that

$\displaystyle\sum_{j=1}^{n(z_{0})}\frac{1}{\log\left(\frac{y_{j}}{x_{j}}\right)}\geq\sum_{j=1}^{n(z_{0})-1}\frac{1}{\log\left(\frac{x_{j+1}}{x_{j}}\right)}+\frac{1}{\log\left(\frac{c_{f}}{x_{n(z_{0})}}\right)}.$

Next, let us consider the AM-HM inequality. For every positive integer
$a_{j}$,

$\displaystyle\sum_{j=1}^{n}a_{j}\geq\frac{n^{2}}{\sum_{j=1}^{n}\frac{1}{a_{j}}}.$

At this point, we would like to use AM-HM with the precise choices

$a_{j}=\frac{1}{\log\left(\frac{x_{j+1}}{x_{j}}\right)}\hskip 14.22636pt\text{
if }j\in\\{1,2,...,n(z_{0})-1\\},\text{ and }\hskip
28.45274pta_{n(z_{0})}=\frac{1}{\log\left(\frac{c_{f}}{x_{n(z_{0})}}\right)},$

which gives

$\displaystyle\sum_{j=1}^{n(z_{0})}\frac{1}{\log\left(\frac{y_{j}}{x_{j}}\right)}\geq\frac{n^{2}(z_{0})}{\log\left(\frac{c_{f}}{x_{1}}\right)}\geq\frac{n^{2}(z_{0})}{\log\left(\frac{c_{f}}{r_{f}}\right)}.$

Therefore,

$\int_{0}^{\infty}\rho^{2}(r,\theta)rdr\geq\frac{n^{2}(z_{0})}{\log\left(\frac{c_{f}}{r_{f}}\right)}.$
(16)

The constant $c_{f}$ is finite and does not depend on either $\theta$ or
$z_{0}$, at least for small $z_{0}$. So, it is irrelevant at the limit
$z_{0}\to 0$. Hence the estimate (16) implies that

$M\left(f(\Gamma)\right)\geq\frac{cn^{2}(z_{0})}{\log\left(\frac{1}{\underset{\rm|z|=|z_{0}|}{\rm\min}|f(z)|}\right)}.$
(17)

Next, use the modulus inequality (14) to get

$\displaystyle\frac{n^{2}(z_{0})}{\log\left(\frac{1}{\underset{\rm|z|=|z_{0}|}{\rm\min}|f(z)|}\right)}\leq
c_{f,p}z_{0}^{-\frac{2}{p}},$

which implies the desired estimate (6).

To prove $p=1$ case, we will again use the modulus inequality (14). Note that
we have already lower bound for $M\left(f(\Gamma)\right)$ from (17) for any
$p\geq 1$. Therefore, we just need to estimate modulus term
$M_{\mathbb{K}(\cdot,f)}(\Gamma)$ from above. To this end, let us define the
function

$\displaystyle\rho_{0}(z)=\begin{cases}\frac{1}{z_{0}}\hskip
11.38092pt\mbox{if}\hskip 5.69046ptdist(z,E)<z_{0}\\\ 0\hskip
17.07182ptotherwise\\\ \end{cases}$

Note that $\rho_{0}$ is admissible with respect to the path family $\Gamma$.
Therefore,

$\displaystyle M_{\mathbb{K}(.,f)}(\Gamma)$
$\displaystyle\leq\int_{\mathbb{C}}\mathbb{K}(.,f)\rho_{0}^{2}(z)dA(z)$
$\displaystyle=\frac{1}{z_{0}^{2}}\int_{\\{z:dist(z,E)<z_{0}\\}}\mathbb{K}(.,f)(z)dA(z).$

Denote

$\int_{\\{z:dist(z,E)<z_{0}\\}}\mathbb{K}(.,f)(z)dA(z)=C_{f}(z_{0})$

and note that since $\mathbb{K}(.,f)(z)\in L^{1}_{loc}(\mathbb{C})$ and

$\left|\\{z:dist(z,E)<z_{0}\\}\right|\to 0$

it follows that $C_{f}(z_{0})\to 0$ as $z_{0}\to 0$, and thus

$M_{\mathbb{K}(.,f)}(\Gamma)\leq\frac{C_{f}(z_{0})}{z_{0}^{2}}.$ (18)

Next, we use the modulus inequality (14), bounds (17) and (18) to get

$\displaystyle\frac{n^{2}(z_{0})}{\log\left(\frac{1}{\underset{\rm|z|=|z_{0}|}{\rm\min}|f(z)|}\right)}\leq\frac{C_{f}(z_{0})}{z_{0}^{2}}$

which implies the desired estimate (7). Hence, Theorem 1 is proved. ∎

PROOF OF _THEOREM 3_:

###### Proof.

As in Theorem 1, our aim is to estimate the _winding number_ $n(z)$ of the
image set $f\left(\left[z,\frac{z}{|z|}\right]\right)$ around the origin
(recall that $f(0)=0$), when $f$ satisfies the hypothesis of Theorem 3 and
$z\in\mathbb{C}\setminus\\{0\\}$ such that $|z|<1$. We will estimate $n(z)$
using the modulus inequality for homeomorphisms with integrable distortion
(14). More precisely, we will show that

$\displaystyle\limsup_{|z|\to
0}\frac{|z|}{\sqrt{\log\left(\frac{1}{|z|}\right)}}n(z)=0,$

which is equivalent to Theorem 3.

To this end, let us choose an arbitrary point
$z_{0}\in\mathbb{C}\setminus\\{0\\}$ such that $|z_{0}|<1$ and which we can
again assume to be positive and real. Next, as in the proof of Theorem 1, we
fix line segments $E=[z_{0},1]$ and $F=(-\infty,0]$, and let $\Gamma$ be the
family of paths connecting them. We have already estimated modulus term
$M_{\mathbb{K}(\cdot,f)}(\Gamma)$ in (18) and thus we can concentrate on
estimating $M\left(f(\Gamma)\right)$ from below.

To this end we use the exact same steps as in the proof of Theorem 1 until the
lower bound (16), where we now estimate the constant $r_{f}$ using the Hölder
modulus of continuity assumption on the inverse of our map $f$. That is, we
estimate

$|f(z_{0})|\geq C\left|z_{0}\right|^{\alpha}$

for sufficiently small $z_{0}$, and obtain

$M\left(f(\Gamma)\right)\geq\frac{n^{2}(z_{0})}{\alpha\,\log\left(\frac{1}{|z_{0}|}\right)}.$

The estimates for moduli combined with the modulus inequality (14) results in

$\displaystyle\frac{n^{2}(z_{0})}{\alpha\,\log\left(\frac{1}{|z_{0}|}\right)}\leq\frac{C_{f}(z_{0})}{z_{0}^{2}},$

which provides the desired estimate (11). Hence Theorem 3 is proved. ∎

##  4 Optimal Results

We are going to prove Theorems 2 and 4 in this section.

PROOF OF _THEOREM 2_:

###### Proof.

We prove Theorem 2 in two steps. In the first step, we construct a map which
_only rotates_. This map will have the correct spiraling rate but the
distortion of the map will not belong to the desired space. To overcome this
barrier we compose it with a radial stretching map, which gives us better
control over distortion.

Given an arbitrary annulus $A=B(0,R)\setminus B(0,r)$ let us define the
corresponding rotation map as

$\phi_{A}(z)=\begin{cases}z&|z|>R\\\
z\,e^{i\alpha\log\left|\frac{z}{R}\right|}&r\leq|z|\leq R\\\
z\,e^{i\alpha\,\log\frac{r}{R}}&|z|<r\end{cases}$ (19)

Here $0<r<R$, and $\alpha\in\mathbb{R}$. It is clear that
$\phi_{A}:\mathbb{C}\to\mathbb{C}$ is bilipschitz, hence quasiconformal (its
quasiconformality constant depends only on $\alpha$), and moreover it is
conformal outside the annulus $A$. Moreover, $|\phi_{A}(te^{i\theta})|=t$ for
each $t>0$ and $\theta\in\mathbb{R}$. This means that $\phi_{A}$ leaves fixed
all circles centered at $0$. It is easy to check that the jacobian determinant
$J(z,\phi_{A})=1$ for each $z$.

Next, let us consider sequence $\\{r_{n}\\}$ such that
$0<r_{n+1}<\frac{r_{n}}{2e}$ and $r_{1}<\frac{1}{e}$. Also, let
$R_{n}=er_{n}$, which ensures that $2r_{n+1}<R_{n+1}<\frac{r_{n}}{2}$. Let us
now construct disjoint annuli $A_{n}=B(0,R_{n})\setminus B(0,r_{n})$, and set
$\\{f_{n}\\}_{n}$ to be a sequence of maps, constructed in an iterative way as
follows. For $n=1$, we set

$f_{1}(z)=\phi_{A_{1}}(z)=\begin{cases}z&|z|>R_{1}\\\
z\,e^{i\alpha_{1}\,\log\frac{|z|}{R_{1}}}&r_{1}\leq|z|\leq R_{1}\\\
z\,e^{-i\alpha_{1}}&|z|<r_{1}\\\ \end{cases}$ (20)

where $\alpha_{1}\in\mathbb{R}$, $\alpha_{1}\geq 1$, is to be determined
later. We then define $f_{n}$ for $n\geq 2$ as

$\displaystyle f_{n}(z)=\phi_{f_{n-1}(A_{n})}\circ f_{n-1}(z)$

again for some values $\alpha_{n}\in\mathbb{R}$, $\alpha_{n}\geq 1$, to be
determined later. Clearly, each $f_{n}:\mathbb{C}\to\mathbb{C}$ is
quasiconformal, and conformal outside the annuli $A_{i}$, $i=1,\dots,n$. It is
also clear that $f_{n}(z)=f_{n-1}(z)$ on the unbounded component of
$\mathbb{C}\setminus f_{n-1}(A_{n})$ (i.e. outside of $B(0,R_{n})$). This
proves that the sequence $f_{n}$ is uniformly Cauchy and hence it converges to
a map $f$, that is,

$\displaystyle f=\lim_{n\to\infty}f_{n}$

which is again a homeomorphism by construction. Now, since $f_{n}$ is
quasiconformal for every $n$ and $f_{n}(z)=f_{n-1}(z)$ everywhere except
inside the ball $B(0,R_{n})$, where $R_{n}\to 0$ as $n\to\infty$, the limit
map $f$ is absolutely continuous on almost every line parallel to the
coordinate axes and differentiable almost everywhere.

It is helpful to note that each $f_{n}$ leaves fixed all circles centered at
the origin, so in particular we have $f_{n}(A_{j})=A_{j}$ for each $j$, and
therefore $\phi_{f_{n-1}(A_{n})}=\phi_{A_{n}}$. Direct calculation shows that

$|D\phi_{A_{n}}(z)|=|\partial\phi_{A_{n}}(z)|+|\overline{\partial}\phi_{A_{n}}(z)|=\begin{cases}1&|z|>R_{n}\\\
\frac{|2+i\alpha_{n}|+|\alpha_{n}|}{2}&r_{n}\leq|z|\leq R_{n}\\\
1&|z|<r_{n}\end{cases}$

which allows us to estimate that

$\displaystyle|\partial f(z)|+|\bar{\partial}f(z)|\leq 2\alpha_{n}\hskip
28.45274pt\text{whenever }z\in A_{n},$

and $|Df(z)|\leq 1$ otherwise. Therefore, in order to have $Df(z)\in
L^{1}_{loc}(\mathbb{C})$ it suffices that

$\sum_{n}\alpha_{n}\,r_{n}^{2}<+\infty.$ (21)

This, together with the absolute continuity, guarantees $f\in
W^{1,1}_{loc}(\mathbb{C})$. Also, since $f$ is a homeomorphism, we have that
$J_{f}(z)\in L^{1}_{loc}(\mathbb{C})$, and in fact $J(z,f)=1$ at almost every
$z\in\mathbb{C}$. Therefore, $f$ is a homeomorphism of finite distortion, with
distortion function

$\displaystyle\mathbb{K}(z,f)=\frac{|Df(z)|^{2}}{J(z,f)}\leq\begin{cases}4\alpha_{n}^{2}&z\in
A_{n},\\\ 1&\text{otherwise.}\end{cases}$

Especially, in order to have $\mathbb{K}(\cdot,f)\in L^{p}_{loc}$, it suffices
to ensure the convergence of the series

$\sum_{n=1}^{\infty}|A_{n}|(4\alpha_{n}^{2})^{p}\simeq\sum_{n=1}^{\infty}\alpha_{n}^{2p}\,r_{n}^{2}$
(22)

which can be done by choosing $\alpha_{n}$ properly. Note that if (22) holds,
then also (21) holds, because our choice of $\alpha_{n}$ will guarantee
$\alpha_{n}\geq 1$. The last restriction to choose our $\alpha_{n}$ comes from
rotational behavior of $f$. It is clear from the above construction that
$f(0)=0$, $f(1)=1$ and

$\left|\operatorname{arg}\left(f(r_{n})\right)\right|\geq\left|\operatorname{arg}\left(\left(\frac{1}{e}\right)^{1+i\alpha_{n}}\right)\right|=\alpha_{n}$

for every $r_{n}$. Let us choose
$\alpha_{n}=r_{n}^{-1/p}\,\log^{1/2}(1/\varphi(r_{n}))$. This implies that

$\begin{aligned} \left|\operatorname{arg}\left(f(r_{n})\right)\right|\geq
r_{n}^{-1/p}\,\log^{1/2}(1/\varphi(r_{n}))\end{aligned},$

which shows that this map would be optimal for Theorem 1.
However, with this particular choice of $\alpha_{n}$,

$\displaystyle\sum_{n=1}^{\infty}\,\alpha_{n}^{2p}\,r_{n}^{2}=\sum_{n=1}^{\infty}\,\log^{p}(1/\varphi(r_{n}))$

which is certainly not finite. This means that $\mathbb{K}(\cdot,f)\notin
L^{p}_{loc}$.

Hence we need to modify the construction by adding a stretching factor to our
building blocks, which lets us reduce the distortion while preserving
spiraling rate. This is precisely done by substituting the logarithmic spiral
map $z|z|^{i\alpha}=ze^{i\alpha\log|z|}$ by a complex power
$z|z|^{q+i\alpha}=z|z|^{q}\,e^{i\alpha\log|z|}$ at each iterate. Let us
explain this process in detail.

Similarly as in the previous construction, we consider a rapidly decreasing
sequence $\\{r_{n}\\}$ such that $r_{n+1}<\frac{r_{n}}{2e}$,
$r_{1}<\frac{1}{e}$ and set $R_{n}=er_{n}$. Given an arbitrary annulus
$A=B(0,R)\setminus B(0,r)$ we define the corresponding composition map as
follows:

$\displaystyle\phi_{A}(z)=\begin{cases}z&|z|>R\\\
z\,\left|\frac{z}{R}\right|^{q-1}\,e^{i\alpha\log\frac{|z|}{R}}&r\leq|z|\leq
R\\\
z\left(\frac{r}{R}\right)^{q-1}\,e^{i\alpha\log\frac{r}{R}}&|z|<r\end{cases}$
(23)

Note that we will always choose $q\geq 1$. Direct calculation shows that

$|\partial\phi_{A}(z)|+|\overline{\partial}\phi_{A}(z)|=\begin{cases}1&|z|>R\\\
R^{1-q}|z|^{q-1}\frac{|q+1+i\alpha|+|q-1+i\alpha|}{2}&r\leq|z|\leq R\\\
R^{1-q}r^{q-1}&|z|<r\end{cases}$ (24)

and also that

$J(z,\phi_{A})=\begin{cases}1&|z|>R\\\
q\,\left(\frac{|z|}{R}\right)^{2(q-1)}&r\leq|z|\leq R\\\
\left(\frac{r}{R}\right)^{2(q-1)}&|z|<r\end{cases}$ (25)

whence

$\mathbb{K}(z,\phi_{A})=\begin{cases}1&|z|>R\\\
\frac{(|q+1+i\alpha|+|q-1+i\alpha|)^{2}}{4q}&r\leq|z|\leq R\\\
1&|z|<r\end{cases}$ (26)

In particular, if $2\leq q+1\leq\alpha$, which will be satisfied for our
choices of $\alpha$ and $q$, then one may estimate
$\|\mathbb{K}(\cdot,\phi_{A})\|_{\infty}\leq\frac{4\alpha^{2}}{q}$. Next, let
us construct the sequence of maps $f_{n}$ in an iterative way as follows. For
$n=1$, we set

$f_{1}(z)=\phi_{A_{1}}(z)=\begin{cases}z&|z|<R_{1}\\\
z\,\left|\frac{z}{R_{1}}\right|^{q_{1}-1}\,e^{i\alpha_{1}\log\frac{|z|}{R_{1}}}&r_{1}\leq|z|\leq
R_{1}\\\ z\left(\frac{1}{e}\right)^{q_{1}-1}\,e^{-i\alpha_{1}}&|z|<r_{1}\\\
\end{cases}$ (27)

where $q_{1}$ and $\alpha_{1}$ are to be determined later. Next, assuming we
have $f_{1},\dots,f_{n-1}$, we define $f_{n}$ for $n\geq 2$ as:

$\displaystyle f_{n}(z)=\phi_{f_{n-1}(A_{n})}\circ f_{n-1}(z)$

Note that $\phi_{f_{n-1}(A_{n})}$ is determined by the inner and outer radii
of $\phi_{f_{n-1}(A_{n})}$ (which are already available since
$f_{1},\dots,f_{n-1}$ are known) as well as for the parameters $q_{n}$ and
$\alpha_{n}$, which will be determined later. Clearly, each
$f_{n}:\mathbb{C}\to\mathbb{C}$ is quasiconformal, and conformal outside the
annuli $A_{i}$, $i\in\\{1,...,n\\}$. Moreover, one can easily show that

$\mathbb{K}(\cdot,f_{n})=\prod_{j=1}^{n}\mathbb{K}(\cdot,f_{n-j}\circ\phi_{f_{n-j}(A_{n-j+1})})\\\
=\prod_{j=1}^{n}\mathbb{K}(\cdot,\phi_{A_{n-j+1}})$

so that $\mathbb{K}(z,f_{n})\leq C\frac{\alpha_{j}^{2}}{q_{j}}$ whenever $z\in
A_{j}$, $j=1\dots n$ while $\mathbb{K}(\cdot,f_{n})=1$ otherwise. In a similar
way, we can use that $|D\phi_{A}(z)|\leq C\alpha$ when $z\in A$ (and
$|D\phi_{A}(z)|\leq 1$ at all other points) to obtain that $|Df_{n}|\leq
C\alpha_{j}$ on $A_{j}$, $j=1\dots n$, and $|Df_{n}|\leq 1$ otherwise.

By construction, we have $f_{n}(z)=f_{n-1}(z)$ whenever $z\notin B(0,R_{n})$.
Thus $\\{f_{n}\\}_{n}$ converges uniformly to a map $\bar{f}$, that is,

$\displaystyle\bar{f}=\lim_{n\to\infty}f_{n}$

which is again a homeomorphism by construction. A similar argument to the one
before shows that $\bar{f}$ is absolutely continuous on almost every line
parallel to the coordinate axis. For almost every fixed $z_{0}$ there is a
neighbourhood of $z_{0}$ such that the sequence $\\{f_{n}(z)\\}_{n}$ remains
constant for $n$ very large and $z$ in that neighbourhood. Therefore the same
happens to the sequences $Df_{n}(z)$, $J(z,f_{n})$ and $\mathbb{K}(z,f_{n})$,
and so their limits are precisely $D\bar{f}(z)$, $J(z,\bar{f})$ and
$\mathbb{K}(z,\bar{f})$. Especially, in order to have $D\bar{f}\in
L^{1}_{loc}$ it suffices that

$\sum_{n=1}^{\infty}|A_{n}|\,\alpha_{n}<+\infty.$ (28)

In case this holds true, then $\bar{f}$ is a homeomorphism in $W^{1,1}_{loc}$,
and as a consequence its jacobian determinant $J(\cdot,\bar{f})\in
L^{1}_{loc}$. Moreover, in order to have $\mathbb{K}(\cdot,\bar{f})\in
L^{p}_{loc}$; ($p\geq 1$) one needs to require that

$\sum_{n=1}^{\infty}|A_{n}|\,\frac{\alpha_{n}^{2p}}{q_{n}^{p}}<+\infty.$ (29)

Again, as it was the case for the pure rotation example, when $p>1$ condition
(29) implies (28) if $q_{n}^{\frac{p}{2p-1}}\leq\alpha_{n}$, and for $p=1$
case we must verify $q_{n}\leq\alpha_{n}$. So, our parameters $\alpha_{n}$ and
$q_{n}$ need to be chosen according to these constrains as well as the purpose
of $\bar{f}$ to be optimal for Theorem 1. To this end, note that
$\bar{f}(0)=0$, $\bar{f}(1)=1$ and

$\left|\operatorname{arg}\left(\bar{f}(r_{n})\right)\right|\geq\,\left|\operatorname{arg}\left(\left(\frac{1}{e}\right)^{q_{n}+i\alpha_{n}}\right)\right|=|\alpha_{n}|,$
(30)

which motivates us to choose

$q_{n}=\begin{cases}\log\frac{e\,r_{1}}{|\varphi(r_{1})|}&n=1\\\
\log\left(\frac{e\,\cdot\,r_{n}\cdot\,\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}}{|\varphi(r_{n})|}\right)&n\geq
2\end{cases}$ (31)

and

$\alpha_{n}=h(r_{n})\,\left(\log\frac{1}{|\varphi(r_{n})|}\right)^{1/2}\,r_{n}^{-\frac{1}{p}}$
(32)

where $h$ is a monotone non-increasing gauge function such that $h(r)\to 0$ as
$r\to 0$ which we specify later.

Next, we show that $q_{n}\leq\alpha_{n}$ for all $p\geq 1$, from which
$q_{n}^{\frac{p}{2p-1}}\leq\alpha_{n}$ then also follows. At this point, we
impose an ansatz on $r_{n}$:

$r_{n}<\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)},$ (33)

which is feasible as the radii $r_{n}$ can be assumed to decrease as fast as
we want. Let us then recall our assumption on $\varphi$ to satisfy compression
bound:

$|\varphi(z)|\geq e^{-g_{\varphi,p}(|z|)|z|^{-\frac{2}{p}}},$ (34)

where $g_{\varphi,p}:\mathbb{R}\to\mathbb{R}$ is some increasing gauge
function such that $|g_{\varphi,p}|\to 0$ as $|z|\to 0$. Now, let us proceed
with the calculations.

$\displaystyle q_{n}$ $\displaystyle=\log\,\left(\frac{e\,\cdot
r_{n}\,\cdot\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+....+q_{1}-(n-1)}}{|\varphi(r_{n})|}\right)$
$\displaystyle\leq\log\,\frac{1}{|\varphi(r_{n})|}$
$\displaystyle=\log^{\frac{1}{2}}\frac{1}{|\varphi(r_{n})|}\cdot\,\log^{\frac{1}{2}}\frac{1}{|\varphi(r_{n})|}$
$\displaystyle\leq\log^{\frac{1}{2}}\frac{1}{|\varphi(r_{n})|}\cdot\,\sqrt{g_{\varphi,p}(r_{n})}\frac{1}{r_{n}^{\frac{1}{p}}}$
$\displaystyle\leq\log^{\frac{1}{2}}\frac{1}{|\varphi(r_{n})|}\cdot\,\frac{h(r_{n})}{r_{n}^{\frac{1}{p}}}=\alpha_{n}$

where the last inequality holds for $h$ converging to zero slowly enough. Note
that, from [9, Theorem 1.6] we see that if $p=1$ then the compression bound
(34) is always satisfied with some $g_{\varphi}$. Thus our choices for $q_{n}$
and $\alpha_{n}$ satisfy technical constrains.

Next, we show that estimate (29) governing integrability of the distortion
holds true for $p\geq 1$. We start by estimating

$\displaystyle|A_{n}|\frac{\alpha_{n}^{2p}}{q_{n}^{p}}$
$\displaystyle=C\,r_{n}^{2}\,\frac{h^{2p}\,r_{n}^{-2}\,\log^{p}\left(\frac{1}{|\varphi(r_{n})|}\right)}{\log^{p}\,\left(\frac{e\,\cdot
r_{n}\,\cdot\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+....+q_{1}-(n-1)}}{|\varphi(r_{n})|}\right)}$
$\displaystyle\leq
C\,h^{2p}\,\frac{\log^{p}\left(\frac{1}{|\varphi(r_{n})|}\right)}{\log^{p}\,\left(\frac{r_{n}^{2}}{|\varphi(r_{n})|}\right)}.$

It is easy to check that
$1<\frac{\log^{p}\left(\frac{1}{|\varphi(r_{n})|}\right)}{\log^{p}\,\left(\frac{r_{n}^{2}}{|\varphi(r_{n})|}\right)}\leq
2^{p}$, using the condition (8), and therefore, up to constants, condition
(29) is equivalent to

$\sum_{n}h(r_{n})^{2p}<+\infty$

which we can always satisfy by choosing $r_{n}$ small enough. Having (29)
fulfilled, our map $\bar{f}$ is a mapping of finite distortion with
$\mathbb{K}(\cdot,\bar{f})\in L^{p}_{loc}$.

Next we must show that our mapping $f$ has right compression and spiraling
behaviour. Let us start with modulus and show that
$|\bar{f}(r_{n})|=|\varphi(r_{n})|$ by calculating

$\displaystyle|\bar{f}(r_{n})|$
$\displaystyle=\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}\cdot
r_{n}\,\left(\frac{r_{n}}{R_{n}}\right)^{q_{n}-1}$
$\displaystyle=\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}\cdot
r_{n}\,\left(\frac{1}{e}\right)^{q_{n}-1}$ $\displaystyle=|\varphi(r_{n})|,$

where the last line follows from the penultimate due to our choice of $q_{n}$.

For the spiraling part we must show that the rotation bound (10) holds true.
But this follows directly from (30), (32) and from the above modulus equation.
Finally we note that in the proof we only need to assure that the sequence
$r_{n}$ decreases fast enough, and thus we can choose it to be a subsequence
for an arbitrary predefined sequence $\\{\lambda_{n}\\}$. This concludes the
proof of Theorem 2. ∎

PROOF OF _THEOREM 4_:

###### Proof.

We prove Theorem 4 in two steps similarly to Theorem 2. In the first step we
construct a map which _only rotates_. This map already provides the optimal
result in the exponent scale. Then, as in the previous construction, we
compose this map with radial stretching mapping and finish the proof.

Given an arbitrary annulus $A=B(0,R)\setminus B(0,r)$ we define the
corresponding rotation map $\phi_{A}$ as in (19). It is clear that
$\phi_{A}:\mathbb{C}\to\mathbb{C}$ is quasiconformal, and moreover it is
conformal outside the annulus $A$. Furthermore, $\phi_{A}$ leaves fixed all
circles centered at $0$, and the Jacobian determinant $J(z,\phi_{A})=1$ for
each $z$.

Next, we again consider sequence $\\{r_{n}\\}$ such that
$0<r_{n+1}<\frac{r_{n}}{2e}$, $r_{1}<\frac{1}{e}$, and fix $R_{n}=er_{n}$. We
then construct disjoint annuli $A_{n}=B(0,R_{n})\setminus B(0,r_{n})$ and a
sequence of maps $\\{f_{n}\\}_{n}$ iteratively as before. That is, set $f_{1}$
as in (20) and define $f_{n}$ for $n\geq 2$ as

$\displaystyle f_{n}(z)=\phi_{f_{n-1}(A_{n})}\circ f_{n-1}(z)$

for some $\alpha_{n}\geq 1$, to be determined later. We can use word by word
the same arguments as before to deduce that the limit

$\displaystyle f=\lim_{n\to\infty}f_{n}$

is a homeomorphism with integrable distortion if

$\sum_{n}\alpha_{n}\,r_{n}^{2}<+\infty$ (35)

and

$\sum_{n=1}^{\infty}|A_{n}|4\alpha_{n}^{2}\simeq\sum_{n=1}^{\infty}\alpha_{n}^{2}\,r_{n}^{2}<+\infty.$
(36)

Moreover, as we will choose $\alpha_{n}>1$, we see that in fact (36) implies
(35). Hence we only need to choose $\alpha_{n}$ so that (36) is satisfied.
Furthermore, it is clear from the above construction that $f(0)=0$, $f(1)=1$
and

$\left|\operatorname{arg}\left(f(r_{n})\right)\right|\geq\left|\operatorname{arg}\left(\left(\frac{1}{e}\right)^{1+i\alpha_{n}}\right)\right|=\alpha_{n}$
(37)

for every $r_{n}$. Since we want our map to be optimal for Theorem 3, we may
be tempted to choose $\alpha_{n}=\frac{\log^{1/2}(1/r_{n})}{r_{n}}$.
Unfortunately such a choice does not meet the requirement (36) and instead we
are forced to choose

$\alpha_{n}=\frac{h(r_{n})}{r_{n}},$

where $h:[0,\infty)\to[0,\infty)$ is a monotonically decreasing gauge function
such that $\lim_{r\to 0^{+}}h(r)=0$. With this choice, (36) is fulfilled if

$\sum_{n=1}^{\infty}\left(h(r_{n})\right)^{2}<+\infty,$

which we can ensure by choosing small enough $r_{n}$. Note that this does not
provide optimality for Theorem 3 in full generality, but it already gives the
right order in the exponent scale.

Finally, we show that $f^{-1}$ is Hölder continuous with exponent
$\frac{1}{2}$. To this end, let us recall that our map $f$ is actually a limit
of iterates of logarithmic spiral maps inside the annuli
$A_{n}=B(0,R_{n})\setminus B(0,r_{n})$. In particular, as shown in [2], if
$\gamma\in\mathbb{R}$ then the basic logarithmic spiral map
$g(z)=ze^{i\gamma\log|z|}$ is $L$-bilipschitz for a constant $L$ such that
$|\gamma|=L-\frac{1}{L}$. And thus for large $|\gamma|$ one roughly has
$|\gamma|\simeq L$. Since our $f_{n}$ behaves in the annulus $A_{n}$ as a
spiral map with $|\gamma|=\alpha_{n}$, we deduce that the bilipschitz constant
of $f_{n}$ on $A_{n}$ is

$L\simeq|\gamma|=\alpha_{n}=\frac{h(r_{n})}{r_{n}}.$

Let us now start the proof. We first consider the case where $x,y\in A_{n}$,
and hence $f(x)=f_{n}(x)$, $f(y)=f_{n}(y)$. Since $r_{n}>C|x-y|$, we have

$\displaystyle|f(x)-f(y)|=|f_{n}(x)-f_{n}(y)|$
$\displaystyle\gtrsim\frac{r_{n}}{h(r_{n})}|x-y|$
$\displaystyle\geq\frac{C}{h(r_{n})}|x-y|^{2}$ $\displaystyle\geq C|x-y|^{2}$

where we have used the bilipschitz nature of $f_{n}$ on $A_{n}$. The fact that
$f$ is Hölder from below inside the annuli $A_{n}$ with exponent $2$ implies
that in these sets $f^{-1}$ is Hölder continuous with exponent $\frac{1}{2}.$
Here we note, that $f$ and $f^{-1}$ are essentially the same mapping, only the
direction of rotation is changed, and hence $f$ is also Hölder continuous with
exponent $\frac{1}{2}$ inside $A_{n}.$

Then we assume that $x,y\in D_{n}=B(0,r_{n})\setminus B(0,R_{n+1})$. In this
case $f$ is of the form $ze^{i\beta}$, where
$\beta\in\mathbb{R}\setminus\\{0\\}$, which is clearly an isometry and hence
Hölder estimate inside $D_{n}$ is trivial.

Next, we take $x\in A_{n}$ and $y\in D_{n}$. In particular, $|x|\geq|y|$. Then
let $w$ be the point on the outer boundary of $D_{n}$ joining $x$ and $y$. We
have

$\displaystyle|f(x)-f(y)|$ $\displaystyle\leq|f(x)-f(w)|+|f(w)-f(y)|$
$\displaystyle\leq C|x-w|^{\frac{1}{2}}+|w-y|$ $\displaystyle\leq
2C|x-y|^{\frac{1}{2}}$

The same happens if $x\in D_{n-1}$ and $y\in A_{n}$.

So it just remains to see what happens when points are further apart from each
other. Let us first cover the case $x\in A_{n}=B(0,R_{n})\setminus B(0,r_{n})$
and $y\in B(0,R_{n+1})$. Let $L$ be the line joining $x$ and $y$. We divide it
into three parts, viz., $L_{1}$, $L_{2}$ and $L_{3}$. Fix $L_{1}$ so that it
connects $x$ to a point $a$ on the inner boundary of $A_{n}$, giving estimate

$|f(x)-f(a)|=|f_{n}(x)-f_{n}(a)|\leq C|x-a|^{\frac{1}{2}}$

Next, $L_{2}$ connects $a$ to the crossing point of the line $L$ and the inner
boundary of $D_{n}$, which we denote by $b$. And since $f$ is an isometry in
$D_{n}$ an estimate for line segment $L_{2}$ is trivial.

For $L_{3}$ part we note that from $2R_{n+1}<r_{n}<\frac{R_{n}}{2}$ we get
that $|f(a)|>2|f(b)|$ and hence

$|f(b)-f(y)|\leq 2|f(b)|\leq 2|f(b)-f(a)|=2|b-a|.$

Combining these estimates we get

$\displaystyle|f(x)-f(y)|$
$\displaystyle\leq|f(x)-f(a)|+|f(a)-f(b)|+|f(b)-f(y)|$ $\displaystyle\leq
C|x-a|^{\frac{1}{2}}+|a-b|+2|a-b|$ $\displaystyle\leq C|x-y|^{\frac{1}{2}}$

The case $x\in D_{n}$ and $y\in B(0,r_{n+1})$ can be proved in a similar
manner. Thus $f$ is Hölder continuous with exponent $\frac{1}{2}$.

Here we again note that $f$ and $f^{-1}$ are essentially the same mapping
modulo the direction of rotation, and hence $f^{-1}$ is also Hölder continuous
with the exponent $\frac{1}{2}$. Thus $f$ is Hölder from below with the
exponent $2.$

As we discussed before, the above example approaches the borderline stated in
Theorem 3, but it does not attain full optimality yet. To this end, we need to
modify it by adding a stretching factor to our building blocks, which lets us
increase rotation without increasing the distortion. This is done by
replacing, at each iterate, the logarithmic spiral map
$z|z|^{i\alpha}=ze^{i\alpha\log|z|}$ by a complex power
$z|z|^{q+i\alpha}=z|z|^{q}\,e^{i\alpha\log|z|}$. Let us proceed with the
details.

So, similarly as in the previous construction, we consider a rapidly
decreasing sequence $\\{r_{n}\\}$ such that $r_{n+1}<\frac{r_{n}}{2e}$,
$r_{1}<\frac{1}{e}$ and fix $R_{n}=er_{n}$. Given an arbitrary annulus
$A=B(0,R)\setminus B(0,r)$ we define the corresponding radial stretching
combined with rotation map as in (23). As before we will choose $q\geq 1$.

The values of the differential, Jacobian and distortion of $\phi_{A}$ are
already known from (24), (25) and (26). In particular, if $2\leq q+1<\alpha$
then one may estimate
$\|\mathbb{K}(\cdot,\phi_{A})\|_{\infty}\leq\frac{4\alpha^{2}}{q}$. Next, we
construct the sequence of maps $f_{n}$ in an iterative way as before. Let us
set $f_{1}$ as in (27) and $f_{n}$ for $n\geq 2$ as:

$\displaystyle f_{n}(z)=\phi_{f_{n-1}(A_{n})}\circ f_{n-1}(z).$

Each $f_{n}:\mathbb{C}\to\mathbb{C}$ is quasiconformal, and conformal outside
the annuli $A_{i}$, $i\in\\{1,...,n\\}$. Moreover, we still calculate
distortion by

$\mathbb{K}(\cdot,f_{n})=\prod_{j=1}^{n}\mathbb{K}(\cdot,f_{n-j}\circ\phi_{f_{n-j}(A_{n-j+1})})\\\
=\prod_{j=1}^{n}\mathbb{K}(\cdot,\phi_{A_{n-j+1}})$

so that $\mathbb{K}(z,f_{n})\leq C\frac{\alpha_{j}^{2}}{q_{j}}$ whenever $z\in
A_{j}$, $j=1\dots n$ while $\mathbb{K}(\cdot,f_{n})=1$ otherwise. As before we
also use $|D\phi_{A}(z)|\leq C\alpha$ when $z\in A$ (and $|D\phi_{A}(z)|\leq
1$ at all other points) to obtain that $|Df_{n}|\leq C\alpha_{j}$ on $A_{j}$,
$j=1\dots n$, and $|Df_{n}|\leq 1$ otherwise.

Using the exact same arguments as before we see that for the limit

$\displaystyle\bar{f}=\lim_{n\to\infty}f_{n}$

to be a homeomorphism of integrable distortion it is enough to check that

$\sum_{n=1}^{\infty}|A_{n}|\,\alpha_{n}<+\infty$ (38)

and

$\sum_{n=1}^{\infty}|A_{n}|\,\frac{\alpha_{n}^{2}}{q_{n}}<+\infty.$ (39)

Note that as in the case of $f$ (39) implies (38) when $q_{n}<\alpha_{n}$ and
so our parameters $\alpha_{n}$ and $q_{n}$ need to be chosen such that (39) is
satisfied as well as the purpose of $\bar{f}$ to be optimal for Theorem 3.
Thus we choose

$\alpha_{n}=\frac{h(r_{n})}{r_{n}}\,\left(\beta\log\frac{1}{r_{n}}\right)^{1/2}\hskip
56.9055ptq_{n}=\beta\log\frac{1}{r_{n}},\hskip 14.22636pt\beta\geq 1$ (40)

where $h$ is any gauge function such that $h(r)\to 0$ as $r\to 0$ and the
condition $q_{n}<\alpha_{n}$ is satisfied. Indeed, with these choices (39)
becomes

$\sum_{n}\left(h(r_{n})\right)^{2}<+\infty$

which, as before, may always be satisfied by choosing small enough $r_{n}$.
Having (39) fulfilled, our map $\bar{f}$ is a mapping of finite distortion
with $\mathbb{K}(\cdot,\bar{f})\in L^{1}_{loc}$. Furthermore, since we can
bound spiraling from below by $\alpha_{n}$ at the points $r_{n}$ using the
same estimate (37) as before, the resulting map $\bar{f}$ attains the optimal
rotational behavior stated at Theorem 3 modulo the gauge function $h$ which
can be chosen to converge to $0$ as slowly as desired.

Therefore, Theorem 4 will be proven once we show that $\bar{f}$ is Hölder from
below.

To this end, we first observe that the composition of $z\mapsto
ze^{i\alpha\log|z|}$ followed by $z\mapsto z|z|^{q-1}$ is precisely $z\mapsto
z|z|^{q-1}e^{i\alpha\log|z|}$. This observation suggests us to decompose
$\bar{f}=g\circ f$, where $f$ is essentially the first example in this section
(with slightly different choices for the constants $\alpha_{n}$) and $g$ is
constructed by building blocks (23) with $\alpha=0$ at each step. Morally, $f$
leaves fixed all circles centered at $0$ and only rotates inside the annuli
$A_{n}$, while $g$ conveniently stretches each $A_{n}$.

The Hölder nature of $f^{-1}$ has already been proven when
$\alpha_{n}=\frac{h(r_{n})}{r_{n}}$. We need to show that our map $f^{-1}$ is
still Hölder continuous with our new choices for $\alpha_{n}$, which we can
estimate by

$\alpha_{n}=\frac{h(r_{n})}{r_{n}}\,\left(\beta\log\frac{1}{r_{n}}\right)^{1/2}\leq\sqrt{\beta}h(r_{n})\,r_{n}^{-1/(1-\epsilon)}$
(41)

for an arbitrary $\epsilon>0$ and small enough $r_{n}$. This can be done by
exactly the same proof as before once we check that $f$ is Hölder from below
inside the annuli $A_{n}.$ To this end, let us consider two points $x,y\in
A_{n}$ and note that $f(x)=f_{n}(x)$ and $f(y)=f_{n}(y)$. Since
$r_{n}>C|x-y|,$ using the estimate (41) gives

$\displaystyle|f(x)-f(y)|=|f_{n}(x)-f_{n}(y)|$
$\displaystyle\gtrsim\frac{r_{n}^{1/(1-\epsilon)}}{h(r_{n})}|x-y|$
$\displaystyle\geq\frac{C}{h(r_{n})}|x-y|^{1+\frac{1}{1-\epsilon}}$
$\displaystyle\geq C|x-y|^{2+\epsilon}$

where we are using the bilipschitz nature of $f_{n}$ in $A_{n}.$ Therefore, in
order to prove Theorem 4 it remains to prove that $g$ is Hölder from below.

To this end, given any two points $x,y\in B(0,1)$, we can without loss of
generality assume that $|y|\geq|x|$ and let $w$ be the point for which
$|w|=|x|$ and $\operatorname{arg}(w)=\operatorname{arg}(y)$. Now, as $g$ is a
radial stretching map, it follows that

$\displaystyle|g(x)-g(y)|\geq\max\\{|g(x)-g(w)|,|g(y)-g(w)|\\}.$

Moreover,

$\displaystyle\max\\{|x-w|,|y-w|\\}\geq\frac{1}{2}|x-y|.$

Therefore, it is enough to show that both $|g(x)-g(w)|$ and $|g(y)-g(w)|$
satisfy Hölder bounds from below. Note that if $x=0$ then clearly $w=0$ and we
have only the radial part $|g(y)-g(w)|$.

Let us first check the term $|g(x)-g(w)|$. Since $g$ maps radially circles
centered at the origin to similar circles we see that $|g(x)-g(w)|$ gets
contracted the same amount as the modulus $|g(x)|$ is contracted under $g$.
Now we must consider two possibilities, either $x,w\in A_{n}$ or $x,w\in
D_{n}$ for some $n$. Let us first assume $x,w\in A_{n}=B(0,R_{n})\setminus
B(0,r_{n})$ for some $n$. Here we recall the ansatz (33) on $r_{n}$. Then we
can estimate

$\displaystyle|g(x)|$
$\displaystyle=\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}\cdot|x|\,\left(\frac{|x|}{R_{n}}\right)^{q_{n}-1}$
$\displaystyle\geq r_{n}\cdot|x|\,\left(\frac{|x|}{R_{n}}\right)^{q_{n}-1}$
$\displaystyle\geq r_{n}\cdot|x|\,\left(\frac{1}{e}\right)^{q_{n}-1}=e\cdot
r_{n}^{1+\beta}\cdot|x|$

for any $x\in A_{n}$, where in the last step we use (40). Therefore,

$\displaystyle|g(x)-g(w)|\geq e\cdot r_{n}^{1+\beta}\cdot|x-w|$
$\displaystyle\geq C\cdot|x-w|^{2+\beta}$

since $|x-w|<C\cdot r_{n}$ for some fixed constant $C>0$ when $x,w\in A_{n}$.

Next, let $x,w\in D_{n}=B(0,r_{n})\setminus B(0,R_{n+1})$ for some $n$. Using
(33) we get

$\displaystyle|g(x)|$ $\displaystyle\geq
c\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}\cdot
r_{n}^{\beta}\cdot|x|$ $\displaystyle\geq c\cdot r_{n}^{1+\beta}\cdot|x|.$

Thus we can use similar argument as in the previous case to estimate

$\displaystyle|g(x)-g(w)|$ $\displaystyle\geq c\cdot
r_{n}^{1+\beta}\cdot|x-w|$ $\displaystyle\geq c\cdot|x-w|^{2+\beta}$

since $|x-w|<c\cdot r_{n}$ for some fixed constant $c>0$ when $x,w\in D_{n}$.

Since the set $\mathbb{D}\setminus\\{0\\}$ is partitioned by separated annuli
$A_{n}$ and $D_{n}$ we have thus proven that $|g(x)-g(w)|$ satisfies Hölder
estimates from below.

Finally, let us prove the Hölder estimates from below for the term
$|g(y)-g(w)|$. As the mapping $g$ is radial, we can assume that $y$ and $w$
are real. We intend to use the Fundamental Theorem of Calculus, and thus have
to estimate the differential from below. Using (33), as well as the facts that
$q_{n}>1$ and $R_{n}=er_{n}$, we can estimate for any real number
$t\in[r_{n},R_{n}]$ that

$\displaystyle g^{\prime}(t)$
$\displaystyle=\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}\cdot
q_{n}\cdot\left(\frac{t}{R_{n}}\right)^{q_{n}-1}$ $\displaystyle\geq
r_{n}\,q_{n}\cdot\left(\frac{r_{n}}{R_{n}}\right)^{q_{n}-1}$
$\displaystyle=e\,q_{n}\,r_{n}^{1+\beta}$ $\displaystyle\geq c\beta\cdot
t^{1+\beta}\,\log\left(e+\frac{1}{t}\right).$

Next, if $t\in[R_{n+1},r_{n}]$, we have

$\displaystyle g^{\prime}(t)$
$\displaystyle=\left(\frac{1}{e}\right)^{q_{n-1}+q_{n-2}+...+q_{1}-(n-1)}\cdot\left(\frac{1}{e}\right)^{q_{n}-1}$
$\displaystyle\geq e\cdot r_{n}^{1+\beta}$ $\displaystyle\geq c\cdot
t^{1+\beta}$

Thus, as before, since $(0,1)$ is partitioned by the intervals
$[r_{n},R_{n}]$, $[R_{n+1},r_{n}]$ and $[R_{1},1)$, we end up getting that

$\displaystyle g^{\prime}(t)\geq c\cdot t^{1+\beta}$

for every $t\in(0,1)$. Now, we use the fundamental theorem of calculus to get

$\displaystyle|g(y)-g(w)|$ $\displaystyle=\int_{w}^{y}g^{\prime}(t)dt$
$\displaystyle\geq\int_{w}^{y}c\cdot t^{1+\beta}dt$
$\displaystyle=C(\beta)\left(y^{2+\beta}-w^{2+\beta}\right)$
$\displaystyle\geq C|y-w|^{2+\beta}$

This proves that the second term is Hölder from below as well, which in turn
proves that $g$ is Hölder from below with exponent $\left(2+\beta\right).$
This finishes the proof of Theorem 4. ∎

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Lauri Hitruhin
University of Helsinki, Department of Mathematics and Statistics, P.O. Box 68
(Pietari Kalmin katu 5), 00014 University of Helsinki, Helsinki, Finland.
E-mail address<EMAIL_ADDRESS>

Banhirup Sengupta
Departament de Matemàtiques, Universitat Autònoma de Barcelona,
08193-Bellaterra, Catalonia
E-mail address<EMAIL_ADDRESS>