# One-dimensional half-harmonic maps into the circle and their degree

Ali Hyder
Tata Institute of Fundamental Research
<EMAIL_ADDRESS>The author is partially supported by SERB SRG/2022/001291
Luca Martinazzi
Sapienza, Università di Roma
<EMAIL_ADDRESS>The author is partially supported by Fondazione
CARIPLO and Fondazione CDP grant n. 2022-2118, and by the PRIN Project
2022PJ9EFL _Geometric Measure Theory: Structure of Singular Measures,
Regularity Theory and Applications in the Calculus of Variations_.

###### Abstract

Given a half-harmonic map
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ minimizing the
fractional Dirichlet energy under Dirichlet boundary conditions in
$\mathbb{R}\setminus I$, we show the existence of a second half-harmonic map,
minimizing the Dirichlet energy in a different homotopy class. This is a first
step towards addressing a problem raised by Brezis and is based on the study
of the degree of fractional Sobolev maps and a sharp estimate à la Brezis-
Coron. We give examples showing that it is in general not possible to minimize
in every homotopy class and show a contrast with the 2-dimensional case.

##  1 Introduction

Half-harmonic maps were introduced by Da Lio and Rivière [8, 9] as critical
points of the $L^{2}$-norm of the $\tfrac{1}{4}$-Laplacian. In this paper we
will focus on the special case of half-harmonic maps from $\mathbb{R}$ into
the circle $\mathbb{S}^{1}$. More precisely, consider the space
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C})$ of measurable maps $u$ from
$\mathbb{R}$ into the complex plane such that

$[u]_{H^{\frac{1}{2},2}(\mathbb{R})}^{2}:=\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{|u(x)-u(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi<\infty,$

and its non-convex subset

$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})=\left\\{u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C}):|u(x)|=1\text{
for a.e. }x\in\mathbb{R}\right\\},$

endowed with the same seminorm.

For such maps, following [8], we introduce the fractional energy

$E(u):=\int_{\mathbb{R}}\left|(-\Delta)^{\frac{1}{4}}u\right|^{2}dx,\quad
u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1}).$ (1)

This energy can be characterized in different equivalent ways, as we shall
briefly recall. First of all, for
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C})$, its Poisson harmonic
extension is well defined on $\mathbb{R}^{2}_{+}:=\mathbb{R}\times(0,\infty)$
as

$\tilde{u}(x,y):=\frac{1}{\pi}\int_{\mathbb{R}}\frac{yu(\xi)}{y^{2}+(x-\xi)^{2}}d\xi.$

Notice that if $u$ takes values into $\mathbb{S}^{1}$, $\tilde{u}$ takes
values into the closure of the unit disk

$D^{2}:=\\{(x,y)\in\mathbb{R}^{2}:x^{2}+y^{2}<1\\}.$

Then we have (see the appendix)

$\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}(\mathbb{R})}^{2}=\int_{\mathbb{R}^{2}_{+}}|\nabla\tilde{u}|^{2}dxdy=\frac{1}{2\pi}[u]_{H^{\frac{1}{2},2}(\mathbb{R})}^{2},\quad
u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C}),$ (2)

and we can equivalently use any of these formulas to define $E(u)$. While the
characterizations of $E(u)$ in terms of the extension $\tilde{u}$ or in terms
of a double integral are more convenient to do estimates, the original
definition of $E(u)$ as in (1) makes the analogy with harmonic maps more
evident. Indeed, interpreting $(-\Delta)^{\frac{1}{4}}u$ as a half-derivative
of $u$, $E(u)$ can be seen as the fractional analog for functions in
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ of the classical
Dirichlet integral

$E_{2}(u):=\int_{\Omega}|\nabla u|^{2}dxdy,\quad u\in H^{1,2}(\Omega,M),$

where $\Omega\subset\mathbb{R}^{2}$, and $M\subset\mathbb{R}^{N}$ is an
embedded Riemannian manifold. In this analogy, an important property that $E$
shares with $E_{2}$, and that will play an important role is the conformal
invariance:

$E(u)=E(u\circ\phi)\quad\text{for
}u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1}),$

where $\phi:\mathbb{R}\to\mathbb{R}$ is the restriction of a conformal map
sending $\mathbb{R}^{2}_{+}$ into itself and its boundary to itself.

Concerning $E_{2}$, one of the model problems in the calculus of variations is
the study of its minimizers or, more generally, its critical points under
suitable constraints, leading to harmonic maps (harmonic functions, in the
special case $M=\mathbb{R}$). In particular, the following uniqueness question
was raised by Giaquinta and Hildebrandt [14]: Consider $M=\mathbb{S}^{2}$,
$\Omega=D^{2}\subset\mathbb{R}^{2}$, $g\in C^{\infty}(\partial
D^{2},\mathbb{S}^{2})$, and let $u_{0}$ be a minimizer of $E_{2}$ in

$H^{1,2}_{g}(D^{2},\mathbb{S}^{2}):=\\{u\in
H^{1,2}(D^{2},\mathbb{S}^{2}):u|_{\partial D^{2}}=g\\}.$

Assume that $g$ is nonconstant. Does $E_{2}$ necessarily have a second
critical point $\bar{u}\neq u_{0}$? The question was answered in the
affirmative by Brezis and Coron:

######  Theorem A (Brezis-Coron [3])

If $g\in C^{\infty}(\partial D^{2},\mathbb{S}^{2})$ is not constant, then
$E_{2}$ has at least two critical points in
$H^{1,2}_{g}(D^{2},\mathbb{S}^{2})$.

The first aim of this paper is to show that a similar result holds for the
fractional Dirichlet energy $E$. More precisely, let $I=(-1,1)$ and, for a
given $g\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ ,set

$\mathcal{E}_{g}=\left\\{u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1}):u=g\text{
in }\mathbb{R}\setminus I\right\\}.$

By direct methods it is straightforward to see that there exists
$u_{0}\in\mathcal{E}_{g}$ such that

$E(u_{0})=\min_{\mathcal{E}_{g}}E.$

By a first variation argument (see e.g. [23, Sec. 1]), one sees that $u_{0}$
is a half-harmonic map, i.e. it satisfies

$(-\Delta)^{\frac{1}{2}}u_{0}\perp T_{u_{0}}\mathbb{S}^{1}\quad\text{in }I.$
(3)

Moreover, any solution to (3) is smooth in $I$, as shown by Da Lio and Rivière
[8] (see also [18] for a more recent proof and [10], [21], [22], [23] for the
rapidly developing regularity theory of fractional harmonic maps).

We now want to show that such minimizer is never unique, unless it is constant
($g=u_{0}\equiv p\in\mathbb{S}^{1}$).

###### Theorem 1.1

If $g$ is non-constant in $\mathbb{R}\setminus I$, there exists $u^{*}\neq
u_{0}$ critical point of $E$ in $\mathcal{E}_{g}$.

In order to prove Theorem 1.1 we shall use the same strategy of Brezis-Coron
[3] of minimizing in homotopy classes, although this will lead us to a
geometrically different approach based on holomorphic maps.

We start by defining the Brower degree of
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ as

$\deg(u):=\deg(u\circ\Pi_{+}),$ (4)

where

$\Pi_{+}:\mathbb{S}^{1}\setminus\\{i\\}\to\mathbb{R},\quad\Pi_{+}(x+iy)=\frac{x}{1-y}$
(5)

is the stereographic projection of $\mathbb{S}^{1}\subset\mathbb{C}$ from the
North pole $i$ and $\deg(u\circ\Pi_{+})$ is the Brouwer degree of a map in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$, see e.g. [6] and Section
2.111Of course, if we used the stereographic projection from the South pole
$\Pi_{-}(x+iy)=\frac{x}{1+y}$ we would obtain the opposite degree. With our
choice, if $u(x)$ covers $\mathbb{S}^{1}$ once in the counter-clockwise
direction as $x$ goes from $-\infty$ to $+\infty$, then $\deg(u)=1$.

Notice that $\Pi_{+}$ can be extended to a biholomorphic map $\Psi$ from
$\overline{D^{2}}\setminus\\{i\\}$ onto $\bar{\mathbb{R}}^{2}_{+}$ (we
identify $\mathbb{R}^{2}$ with $\mathbb{C}$), with inverse $\Phi$, see
(33)-(34) below. Because of the conformal invariance of the Dirichlet integral
we have

$E(u)=\int_{\mathbb{R}^{2}_{+}}|\nabla\tilde{u}|^{2}dxdy=\int_{D^{2}}|\nabla(\tilde{u}\circ\Psi)|^{2}dxdy=:E(u\circ\Pi_{+}),$

and under these trasformations we can identify
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ and
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. In fact, all the results
of this paper can be stated equivalently for maps on $\mathbb{R}$ or on
$\mathbb{S}^{1}$. With this in mind, (4) is well defined for
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$.

Now define

$\mathcal{E}_{g,k}=\\{u\in\mathcal{E}_{g}:\deg(u)=k\\}.$

Then Theorem 1.1 follows almost immediately from the following:

###### Theorem 1.2

Let $u_{0}$ minimize $E$ in $\mathcal{E}_{g}$ for some
$g\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$, and consider a point
$x_{0}\in I$ such that $u_{0}^{\prime}(x_{0})\neq 0$. Then for
$k=\deg(u_{0})-\mathrm{sign}(u_{0}(x_{0})\wedge u_{0}^{\prime}(x_{0}))$,222for
two vectors $v,w\in\mathbb{R}^{2}\simeq\mathbb{C}$ we define $v\wedge
w:=v_{1}w_{2}-v_{2}w_{1}$. $E$ has a minimizer $u^{*}$ in $\mathcal{E}_{g,k}$.

The main difficulty in Theorem 1.2 is that the weak convergence does not
preserve the degree, so that a minimizing sequence in $\mathcal{E}_{g,k}$ can
concentrate at some points, and weakly converge to a function in
$\mathcal{E}_{g,\ell}$ for some $\ell\neq k$, see also Proposition 3.1.

The choice of $k$ in Theorem 1.2 has the following geometric meaning: If
$u_{0}$ at $x_{0}$ is moving along $\mathbb{S}^{1}$ in a given direction, then
the degree of $u^{*}$ is obtained by turning around $\mathbb{S}^{1}$ once in
the opposite direction. More precisely, in Lemma 4.1 we shall glue near
$x_{0}$ a concentrated circle with orientation opposite to
$u^{\prime}(x_{0})$. This choice of orientation is necessary since, in general
we cannot expect $E$ to have a minimizer in $\mathcal{E}_{g,k}$ for every
$k\in\mathbb{Z}$, as the following example shows.

###### Theorem 1.3

Consider for $\lambda>0$ the map $g_{\lambda}:\mathbb{R}\to\mathbb{S}^{1}$
given by

$g_{\lambda}(x)=\Pi_{+}^{-1}\left(\frac{x}{\lambda}\right),$ (6)

where $\Pi_{+}$ is as in (5). Then:

  1. (i)

for every $\lambda>0$, $u_{0}=g_{\lambda}$ is the only minimizer of $E$ in
$\mathcal{E}_{g_{\lambda},1}$;

  2. (ii)

there exists $\lambda^{*}>1$ such that for $\lambda\geq\lambda^{*}$,
$u_{0}=g_{\lambda}$ is the only minimizer of $E$ in
$\mathcal{E}_{g_{\lambda}}$;

  3. (iii)

for every $\lambda>0$, $E$ has a minimizer $u^{*}$ in
$\mathcal{E}_{g_{\lambda},0}$, and there exists $\lambda_{0}>1$ such that for
$\lambda\in(0,\lambda_{0})$, $u^{*}$ is an absolute mimimizer, i.e. it
mimimizes $E$ in $\mathcal{E}_{g_{\lambda}}$, while $g_{\lambda}$ does not
minimize $E$ in $\mathcal{E}_{g_{\lambda}}$;

  4. (iv)

for every $\lambda>0$, $E$ has no minimizer in $\mathcal{E}_{g_{\lambda},k}$
for $k>1$.

Theorem 1.3 can be seen as a $1$-dimensional counterpart of Theorem 2 in [3],
where for the boundary datum

$g_{R}:\partial D^{2}\to\mathbb{S}^{2},\quad
g_{R}(x,y):=(Rx,Ry,\sqrt{1-R^{2}}),\quad 0<R<1,$ (7)

it was shown that:

  1. 1.

the absolute minimizer of $E_{2}(u)=\int_{D^{2}}|\nabla u|^{2}dxdy$ in
$H^{1,2}_{g_{R}}(D^{2},\mathbb{S}^{2})$ is

$\underline{u}(x,y)=\frac{2\lambda}{\lambda^{2}+x^{2}+y^{2}}(x,y,\lambda)+(0,0,-1),\quad\lambda:=\frac{1}{R}+\sqrt{\frac{1}{R^{2}}-1};$

  2. 2.

the function

$\overline{u}(x,y)=\frac{2\mu}{\mu^{2}+x^{2}+y^{2}}(x,y,-\mu)+(0,0,1),\quad\mu:=\frac{1}{R}-\sqrt{\frac{1}{R^{2}}-1}$

minimizes in the homotopy class obtained by lowering the Brouwer degree
(suitably definied) by $1$, hence it is a second harmonic map in
$H^{1,2}_{g_{R}}(D^{2},\mathbb{S}^{2})$;

  3. 3.

there is no minimizer in _any_ other homotopy class.

The analogy between Theorem 1.3 and Theorem 2 of [3] is very strong if one
considers that the functions $\overline{u}$ and $\underline{u}$ above (and
their restriction $g_{R}$ in (7)) are rescalings the inverse stereographic
projection from the South and North pole respectively.

On the other hand, in contrast with [3, Thm. 2], in Theorem 1.3 (iii), for
$\lambda$ slightly bigger than $1$ (which would be the analog of $R$ slightly
smaller than $1$ in (7)) we do not have that $u_{0}=g_{\lambda}$ is the
absolute minimizer of $E$ in $\mathcal{E}_{g_{\lambda}}$, as

$\inf_{\mathcal{E}_{g_{\lambda},0}}E<E(g_{\lambda})=\min_{\mathcal{E}_{g_{\lambda},1}}E,$

see Proposition 5.2, so that in part (ii) of Theorem 1.3 one cannot take
$\lambda^{*}=1$. Moreover, it is open whether in part (iv) of Theorem 1.3, $E$
has minimizers in $\mathcal{E}_{g_{\lambda},k}$ for $k<0$.

The first of these facts might sound a bit surprising at first, since for
$\lambda>1$, $g_{\lambda}|_{\mathbb{R}\setminus I}$ covers more than half of
the circle, hence any map in $\mathcal{E}_{g_{\lambda},0}$ must cover more
than half the circle at least _twice_ (in opposite directions), while maps in
$\mathcal{E}_{g_{\lambda},1}$ can cover the whole circle only once. The non-
locality of the energy, when $\lambda$ is not much bigger than $1$ makes the
first option more convenient. Of course this is not the case with the local
energy $\int_{\mathbb{R}}|u^{\prime}|dx$, as in this case we have

$\inf_{u\in\mathcal{E}_{g_{\lambda},0}}\int_{\mathbb{R}}|u^{\prime}|dx>2\pi=\inf_{u\in\mathcal{E}_{g_{\lambda},1}}\int_{\mathbb{R}}|u^{\prime}|dx=\int_{\mathbb{R}}|g_{\lambda}^{\prime}|dx\quad\text{for
}\lambda>1.$

Also the second contrast between Theorem 2 of [3] and Theorem 1.3 (part iv)
above arises from the nonlocal nature of the fractional energy $E$. Indeed the
method used by Brezis and Coron of gluing functions $u_{1}$ and $u_{2}$ via a
Kelvin transformation and expressing $E_{2}$ of the glued functions as
$E_{2}(u_{1})+E_{2}(u_{2})$, does not apply to the energy $E$.

Acknowledgements The second author first learnt about degree theory in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ (more generally in
$VMO(\mathbb{S}^{1},\mathbb{S}^{1})$) and the problem of minimizing in
different homotopy classes during a very inspiring talk of H. Brezis in honour
of G. Stampacchia held in Pisa in November 2010. He was later made aware of
the potential applications in the study of the regularity theory of
$2$-dimensional harmonic maps by T. Rivière. This will be developed in future
works.

##  2 Degree for maps in $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$
and $\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$

We first recall some facts about the degree of maps in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ and
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$. For a classical
introduction to the topic we refer to [6].

Given a function $u\in C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$, one typically
defines its Brouwer degree via a lifting, i.e. a map $\varphi\in
C^{0}(\mathbb{R},\mathbb{R})$ such that

$u(e^{i\theta})=e^{i\varphi(\theta)}.$ (8)

Since $u(e^{i\theta})=u(e^{i(\theta+2\pi)})$, it follows that
$\varphi(2\pi)-\varphi(0)\in 2\pi\mathbb{Z}$, and one defines

$\deg(u):=\frac{\varphi(2\pi)-\varphi(0)}{2\pi}\in\mathbb{Z}.$ (9)

On the other hand, due to the nature of our problems, it will be more natural
to work with integro-differential definitions of the Brouwer degree. We start
recalling that for $u\in C^{\infty}(\mathbb{S}^{1},\mathbb{S}^{1})$ its
Brouwer degree can be computed via the de Rahm cohomology as

$\deg(u)=\frac{1}{2\pi}\int_{\mathbb{S}^{1}}u^{*}(d\theta)=\frac{1}{2\pi}\int_{\mathbb{S}^{1}}u\wedge\dot{u}\,d\theta=\frac{1}{2\pi
i}\int_{\mathbb{S}^{1}}\bar{u}\dot{u}\,d\theta.$ (10)

Here we used that for vectors $v,w\in\mathbb{R}^{2}\simeq\mathbb{C}$,
$\bar{v}w=v\cdot w+i(v\wedge w)$, where
$\bar{v}w=(v_{1}-iv_{2})(w_{1}+iw_{2})$ is the usual product in $\mathbb{C}$,
while $v\wedge w:=(v_{1}w_{2}-v_{2}w_{1})$ and $v\cdot
w:=v_{1}w_{1}+v_{2}w_{2}$, and we used that $u\cdot\dot{u}=0$.

This definition is equivalent to the previous one, since for $\varphi\in
C^{\infty}(\mathbb{R})$ lifting $u$ in the sense of (8), we have

$\frac{1}{i}\int_{\mathbb{S}^{1}}\bar{u}\dot{u}\,d\theta=\int_{0}^{2\pi}\varphi^{\prime}(x)\,dx={\varphi(2\pi)-\varphi(0)}.$

Further, writing $u$ as Fourier series

$u(e^{i\theta})=\frac{1}{\sqrt{2\pi}}\sum_{k\in\mathbb{Z}}a_{k}e^{ik\theta},\quad
a_{k}:=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{S}^{1}}u(e^{i\theta})e^{-ik\theta}d\theta,$

with a simple computation we obtain (still assuming $u$ to be smooth)

$\frac{1}{i}\int_{\mathbb{S}^{1}}\bar{u}\dot{u}\,d\theta=\sum_{k\in\mathbb{Z}}k|a_{k}|^{2}.$
(11)

While the integral in (11) is naturally defined for $u\in
C^{1}(\mathbb{S}^{1},\mathbb{S}^{1})$, the sum on the right-hand side is
defined for any $u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$,555Actually, also the
integral in (11) can be defined for $u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ using the duality between
$H^{-\frac{1}{2},2}$ and $H^{\frac{1}{2},2}$, as observed by Boutet de Monvel
and Gabber (see [6, p. 224] and [2, Sec. 3]). (see the Appendix for the
definitions of fractional Sobolev spaces and their seminorms), so that we can
also define

$\deg(u):=\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}k|a_{k}|^{2},\quad u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1}).$ (12)

It follows at once that

$|\deg(u)|\leq\frac{1}{2\pi}\sum_{k\in\mathbb{Z}}|k||a_{k}|^{2}=\frac{1}{2\pi}[u]_{H^{\frac{1}{2},2}}^{2}.$
(13)

Moreover, with the Hilbert transform

$\mathcal{H}(u)=\sum_{k\in\mathbb{Z}\setminus\\{0\\}}\frac{k}{|k|}a_{k}e^{ik\theta},\quad
u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1}),$

we can write

$\begin{split}\sum_{k\in\mathbb{Z}}k|a_{k}|^{2}&=\frac{1}{2\pi}\int_{\mathbb{S}^{1}}\sum_{k\in\mathbb{Z}\setminus\\{0\\}}\frac{k}{|k|}\sqrt{|k|}\bar{a}_{k}e^{-ik\theta}\sum_{\ell\in\mathbb{Z}\setminus\\{0\\}}\sqrt{|l|}a_{\ell}e^{i\ell\theta}d\theta\\\
&={-}\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{u}](-\Delta)^{\frac{1}{4}}ud\theta,\end{split}$
(14)

(see (61) for the definition of $(-\Delta)^{\frac{1}{4}}$) hence, from (12),
we have the equivalent definition

$\deg(u):={-}\frac{1}{2\pi}\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{u}](-\Delta)^{\frac{1}{4}}ud\theta,\quad
u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1}).$ (15)

Going back to (10), for $u\in C^{\infty}(\mathbb{S}^{1},\mathbb{S}^{1})$,
which we write as $u=u_{1}+iu_{2}$, considering a smooth extension
$U=U_{1}+iU_{2}$ to the unit disk $\overline{D^{2}}$, and using Stokes’
theorem we get

$\begin{split}\frac{1}{i}\int_{\mathbb{S}^{1}}\bar{u}\dot{u}d\theta&=\frac{1}{i}\int_{\mathbb{S}^{1}}\bar{u}du=\frac{1}{i}\int_{D^{2}}d(\bar{U}dU)=\frac{1}{i}\int_{D^{2}}d[(U_{1}-iU_{2})(dU_{1}+idU_{2})]\\\
&=2\int_{D^{2}}dU_{1}\wedge dU_{2}=2\int_{D^{2}}\left(\frac{\partial
U_{1}}{\partial x}\frac{\partial U_{2}}{\partial y}-\frac{\partial
U_{1}}{\partial y}\frac{\partial U_{2}}{\partial
x}\right)dxdy=2\int_{D^{2}}JUdxdy.\end{split}$

Considering that $|JU|\leq\tfrac{1}{2}|\nabla U|^{2}$, the right-hand side is
well defined for any $u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$
choosing as $U$ any $H^{1}$ extension, for instance the harmonic extension
$\tilde{u}\in H^{1,2}(D^{2},\mathbb{C})$. Hence we can also define

$\deg(u):=\frac{1}{\pi}\int_{D}JUdxdy,\quad u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1}),\,U\in
H^{1,2}(D^{2},\mathbb{C}),\,U|_{\partial D^{2}}=u.$ (16)

In a similar fashion, if $u\in C^{\infty}(\mathbb{R},\mathbb{S}^{1})$ is
constant outside a bounded interval, say $u\equiv a$ in
$\mathbb{R}\setminus[-R,R]$, we can write

$\displaystyle\deg(u)=\frac{1}{2\pi}\int_{\mathbb{R}}u^{*}(d\theta)=\frac{1}{2\pi}\int_{\mathbb{R}}u\wedge
u^{\prime}dx=\frac{1}{2\pi i}\int_{\mathbb{R}}\bar{u}u^{\prime}dx.$ (17)

Now, writing $w=u-a$, we have that $w$ is compactly supported and
$\int_{\mathbb{R}}\bar{u}u^{\prime}dx=\int_{\mathbb{R}}\bar{w}w^{\prime}dx$.
Then, we can write $w^{\prime}$ as Fourier integral and compute

$\displaystyle\deg(u)$ $\displaystyle=\frac{1}{2\pi
i}\int_{\mathbb{R}}\bar{w}w^{\prime}dx=\frac{1}{2\pi
i}\int_{\mathbb{R}}\bar{w}(x)\left(\frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{i\xi
x}\xi\hat{w}(\xi)d\xi\right)dx$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}}\xi\hat{w}(\xi)\left(\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{i\xi
x}\bar{w}(x)dx\right)d\xi=\frac{1}{2\pi}\int_{\mathbb{R}}\xi\hat{w}(\xi)\check{\bar{w}}(\xi)d\xi$
$\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}}\xi|\hat{w}(\xi)|^{2}d\xi,$ (18)

where in the last equality we have used that $\bar{\hat{w}}=\check{\bar{w}}$.
On the other hand

$\displaystyle\int_{\mathbb{R}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{w}]$
$\displaystyle(-\Delta)^{\frac{1}{4}}wdx=\int_{\mathbb{R}}\mathcal{F}^{-1}\left(\frac{\xi}{|\xi|}\mathcal{F}((-\Delta)^{\frac{1}{4}}\bar{w})\right)(-\Delta)^{\frac{1}{4}}wdx$
$\displaystyle=\int_{\mathbb{R}}\frac{\xi}{|\xi|}\mathcal{F}((-\Delta)^{\frac{1}{4}}\bar{w})\mathcal{F}^{-1}\left((-\Delta)^{\frac{1}{4}}w\right)d\xi=\int_{\mathbb{R}}\frac{\xi}{|\xi|}|\xi|^{\frac{1}{2}}\hat{\bar{w}}(\xi)|\xi|^{\frac{1}{2}}\check{w}(\xi)d\xi$
$\displaystyle=\int_{\mathbb{R}}\xi\bar{\check{w}}(\xi)\check{w}(\xi)d\xi=\int_{\mathbb{R}}\xi|\check{w}(\xi)|^{2}d\xi=-\int_{\mathbb{R}}\xi|\hat{w}(\xi)|^{2}d\xi,$
(19)

where $(-\Delta)^{\frac{1}{4}}$ is defined in (53) and $\mathcal{H}$ denotes
the Hilbert transform:

$\displaystyle\mathcal{H}({v}):=\mathcal{F}^{-1}\left(\frac{\xi}{|\xi|}\hat{{v}}\right).$
(20)

Putting (2) and (2) together yields

$\deg(u)={-}\frac{1}{2\pi}\int_{\mathbb{R}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}{\bar{w}}](-\Delta)^{\frac{1}{4}}{w}dx={-}\frac{1}{2\pi}\int_{\mathbb{R}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{u}](-\Delta)^{\frac{1}{4}}udx,$
(21)

which can be extended by density to every
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$. It follows from (21)
and Hölder’s inequality

$|\deg(u)|\leq\frac{1}{2\pi}\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}}^{2}=\frac{1}{2\pi}E(u).$
(22)

Finally we remark that by composing with the conformal map
$\Psi(z)=-i\frac{z+i}{z-i}$, sending $D^{2}$ onto $\mathbb{R}^{2}_{+}$ (see
(34)), one infers from (16) also

$\deg(u)=\frac{1}{\pi}\int_{\mathbb{R}^{2}_{+}}JUdxdy,\quad U\in
H^{1,2}(\mathbb{R}^{2}_{+}),\,U|_{\mathbb{R}\times\\{0\\}}=u.$ (23)

###### Proposition 2.1

Given $u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ (resp.
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$), the definitions of
Brouwer degree given by (12), (15) and (16) (resp. (2), (21) and (23)) are
equivalent and take integer values. Moreover, given
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$,
$\deg(u)=\deg(u\circ\Pi_{+})$

###### Proof.

The equivalence of the above formulas for $u\in
C^{1}(\mathbb{S}^{1},\mathbb{S}^{1})$ or for $u\in
C^{1}(\mathbb{R},\mathbb{S}^{1})$ constant outside an interval follows from
the above computations and is an integer by classical theory. When $u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ or
$u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$, this follows by
approximation using Propositions A.2, A.5 and A.6. ∎

Finally, one might wonder whether for $u\in C^{0}\cap
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ the degree defined in (9)
coincides with the one defined in (12), (15) and (16). That this is the case
follows essentially from Lemma A.4, since the averaging functions
$u_{\operatorname{av},\varepsilon}$ converge uniformly to $u$ as
$\varepsilon\to 0$ for $u\in C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$, hence
$\deg(u_{\operatorname{av},\varepsilon})=\deg(u)$ for $\varepsilon$ small.

##  3 Degree jumps and surgeries

###  3.1 Degree jumps under weak convergence

The first type of degree jump that we will consider is due to the fact that
the degree is not continuous with respect to the weak convergence in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$, or in
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$.

We shall say that a sequence
$(u_{n})\subset\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ weakly
converges to $u$ in $\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ if

$\lim_{n\to\infty}\int_{\mathbb{R}}(-\Delta)^{\frac{1}{4}}u_{n}(-\Delta)^{\frac{1}{4}}vdx=\int_{\mathbb{R}}(-\Delta)^{\frac{1}{4}}u(-\Delta)^{\frac{1}{4}}vdx,\quad\text{for
}v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C}).$

As already mentioned, weakly converging sequences might jump from a homotopy
class to another. The next proposition, which is the analog [16, Lemma 3], and
(15) in [3] (see also [15] and [4, Lemma 1]), quantifies the minimal energy
cost of such jumps.

###### Proposition 3.1

Let
$(u_{n})\subset\mathcal{E}_{k}=\mathcal{E}_{k}(\mathbb{R},\mathbb{S}^{1}):=\\{v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1}):\deg(v)=k\\}$,
converge weakly in $\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ to
$u\in\mathcal{E}_{\ell}$. Then

$E(u)\leq\liminf_{n\to\infty}E(u_{n})-2\pi|k-\ell|.$

Equivalently, if
$(u_{n})\subset\mathcal{E}_{k}=\mathcal{E}_{k}(\mathbb{S}^{1},\mathbb{S}^{1}):=\\{v\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1}):\deg(v)=k\\}$ converge weakly
in $\dot{H}^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ to
$u\in\mathcal{E}_{\ell}$, then

$E(u)\leq\liminf_{n\to\infty}E(u_{n})-2\pi|k-\ell|.$

In the proof of Proposition 3.1 we will use the following:

###### Lemma 3.2

For $v,w\in H^{\frac{1}{2},2}(\mathbb{R},\mathbb{C})$, we have
$\mathcal{H}((-\Delta)^{\frac{1}{4}}w)=(-\Delta)^{\frac{1}{4}}(\mathcal{H}(w))$
and

$\int_{\mathbb{S}^{1}}\mathcal{H}((-\Delta)^{\frac{1}{4}}\bar{v})(-\Delta)^{\frac{1}{4}}wd\theta=\overline{\int_{\mathbb{S}^{1}}\mathcal{(}-\Delta)^{\frac{1}{4}}v(-\Delta)^{\frac{1}{4}}(\mathcal{H}(\bar{w}))d\theta}.$

###### Proof.

The proof follows easily from the definition of the Hilbert transform and
$(-\Delta)^{\frac{1}{4}}$ on $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$. ∎

_Proof of Proposition 3.1_ We prove the proposition for
$(u_{n})\subset\mathcal{E}_{k}(\mathbb{S}^{1},\mathbb{S}^{1})$. We define

$F_{\pm}(u)=E(u)\pm 2\pi\deg(u),$

and prove that $F_{\pm}$ is weakly lower semicontinuous in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. It will be convenient to
use the degree formula (15) Set $w_{n}:=u_{n}-u\rightharpoonup 0$. Using (15),
and Lemma 3.2 we get

$\displaystyle F_{\pm}(u_{n})$
$\displaystyle=E(u)+E(w_{n})+o(1){\mp}\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}(\bar{u}+\bar{w}_{n})](-\Delta)^{\frac{1}{4}}(u+w_{n})d\theta$
$\displaystyle=F_{\pm}(u)+E(w_{n}){\mp}\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{w}_{n}](-\Delta)^{\frac{1}{4}}w_{n}d\theta+o(1),$

provided

$\displaystyle\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{w}_{n}](-\Delta)^{\frac{1}{4}}ud\theta\to
0\quad\text{and
}\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{u}](-\Delta)^{\frac{1}{4}}w_{n}d\theta\to
0.$ (24)

Then, using that

$\left|\int_{\mathbb{S}^{1}}\mathcal{H}[(-\Delta)^{\frac{1}{4}}\bar{w}_{n}](-\Delta)^{\frac{1}{4}}w_{n}d\theta\right|\leq\|(-\Delta)^{\frac{1}{4}}\bar{w}_{n}\|_{L^{2}}^{2}=E(w_{n})$

we obtain $F_{\pm}(u)\leq F_{\pm}(u_{n})+o(1)$ as $n\to\infty$, and we
conclude that $F_{\pm}$ is weakly lower-semicontinuous. Then the lemma follows
from

$E(u)=F_{\pm}(u)\mp 2\pi\ell\leq\liminf_{n\to\infty}F_{\pm}(u_{n})\mp
2\pi\ell=\liminf_{n\to\infty}E(u_{n})\pm 2\pi(k-\ell).$

It follows from the definition of weak convergence and Lemma 3.2 that (24)
hold true, and we conclude the proof of the proposition. $\square$

_Remark._ The constant $2\pi$ is sharp, as can be proven by considering a
concentrating sequence $(u_{n})$ of (restrictions of) Möbius functions with
$\deg(u_{n})=1$, $E(u_{n})=2\pi$, $u_{n}\rightharpoonup u\equiv C$ as
$n\to\infty$. More generally, using (restriction of) Blaschke products as in
(45), one can construct a sequence $(u_{n})$ with $\deg(u_{n})=k>0$,
$E(u_{n})=2\pi k$ (by Lemma 5.1), $u_{n}\rightharpoonup u$,
$\deg(u)=\ell\in[0,k)$, $E(u)=2\pi\ell$.

_Remark._ One can give an analogous proof of Proposition 3.1 by working with
$\tilde{u}_{k}\rightharpoonup\tilde{u}$ in $H^{1,2}(\mathbb{R}^{2}_{+})$ and
using (23), together with $|J\tilde{u}|\leq\frac{1}{2}|\nabla\tilde{u}|^{2}$.

_Remark._ In attempting to formulate a minimization problem in
$\dot{W}^{s,\frac{1}{s}}(\mathbb{R},\mathbb{S}^{1})$ or in
$W^{s,\frac{1}{s}}(\mathbb{S}^{1},\mathbb{S}^{1})$ with $s\neq\frac{1}{2}$, in
the spirit of [19, 20], one of the main obstacles would be finding the analog
of the sharp constant $2\pi$.

###  3.2 Surgeries

Next we will consider a degree change due to surgery, namely, given a closed
arc $A\subset\mathbb{S}^{1}$ we will replace a function $u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ in $A$ with another map
$v:A\to\mathbb{S}^{1}$ such that the resulting map is still in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. Up to a Möbius
transformation, we can assume
$A=\mathbb{S}^{1}_{+}:=\mathbb{S}^{1}\cap\\{(x,y)\in\mathbb{R}^{2}:y\geq
0\\}$.

Set the gluing of $v$ to $u$ as

$v\&u(e^{i\theta}):=\begin{cases}v(e^{i\theta})&\text{for
}0\leq\theta\leq\pi\\\ u(e^{i\theta})&\text{for }\pi<\theta<2\pi,\end{cases}$
(25)

and the gluing of $v$ to the reflection of $u$ as

$v\\#u(e^{i\theta}):=\begin{cases}v(e^{i\theta})&\text{for
}0\leq\theta\leq\pi\\\ u(e^{-i\theta})&\text{for }\pi<\theta<2\pi.\end{cases}$
(26)

If $u\in C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$, $v\in
C^{0}(\mathbb{S}^{1}_{+},\mathbb{S}^{1})$ and $u(\pm 1)=v(\pm 1)$, so that
$v\&u,v\\#u\in C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$, it is fairly easy to see
that

$\deg(v\&u)=\deg(u)+\deg(v\\#u).$

In particular the degree change only depends on the values of $u$ and its
replacement in $\mathbb{S}^{1}_{+}$, as it is well known.

We want to prove the same formula for $u,v\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. Two main difficulties
arise. First of all we do not have an analog of $u(\pm 1)=v(\pm 1)$ in terms
of traces of functions in $H^{\frac{1}{2},2}$. As Proposition A.10 shows, and
contrary to what happens for maps in $H^{1,2}(\Omega,\mathbb{S}^{2})$ with
$\Omega\Subset\mathbb{R}^{2}$, a map $u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ might not have a trace on
the boundary of an arc $A$ of $\mathbb{S}^{1}$, and in fact, for some point
$e^{i\theta_{0}}\in\mathbb{S}^{1}$, $u(e^{i\theta})$ can wind infinitely many
times around $\mathbb{S}^{1}$ as $\theta\to\theta_{0}^{+}$ and infinitely many
times, but with opposite orientation as $\theta\to\theta_{0}^{-}$. The heavy
cancelation due to the opposite orientation of the windings allow the degree
of $u$ to still be finite, but this suggests extra care in cutting, gluing and
counting the degree of functions in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

This might appear to be only a problem of trace (in particular of the
unboundedness of functions in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$), but problems arise even
if one considers functions in $H^{\frac{1}{2},2}\cap
C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$, as Proposition A.8 shows. In fact, one
can take a map $H^{\frac{1}{2},2}\cap C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$
and replace it locally by a constant map obtaining a new map that, in spite of
being continuous, does not belong to
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. Both phenomena are
essentially due to the nonlocal nature of the $H^{\frac{1}{2},2}$-seminorm,
and in particular to cancelations in the double integral defining it. Because
of them, it will be necessary to assume that $v\&u,v\\#u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$:

###### Proposition 3.3

Consider $u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ and
$v:\mathbb{S}^{1}_{+}\to\mathbb{S}^{1}$. Set $v\&u$ and $v\\#u$ as in (25) and
(26) and assume that $v\&u,v\\#u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. Then

$\deg(v\&u)=\deg(u)+\deg(v\\#u).$ (27)

###### Proof.

Consider $\gamma:[-1,1]\times\\{0\\}\to\mathbb{S}^{1}_{-}$ given by
$\gamma(x,0)=e^{i\frac{\pi}{2}(x-1)}$. Clearly $\gamma$ is a diffeomorphism
and

$|\gamma^{\prime}|\equiv\frac{\pi}{2},\quad|(\gamma^{-1})^{\prime}|\equiv\frac{2}{\pi}.$
(28)

On $[-1,1]\times\\{0\\}$ define the function

$u^{*}(x,0)=u(\gamma(x,0))=u(e^{i\frac{\pi}{2}(x-1)}).$

Now define the maps $u^{*}_{\pm}:\partial D^{2}_{\pm}\to\mathbb{S}^{1}$

$\begin{cases}u^{*}_{\pm}:=u&\text{on }\mathbb{S}^{1}_{\pm}\\\
u^{*}_{\pm}=u^{*}&\text{on }(-1,1)\times\\{0\\},\end{cases}$

and $v^{*}:\partial D^{2}_{+}\to\mathbb{S}^{1}$

$\begin{cases}v^{*}=v&\text{on }\mathbb{S}^{1}_{+}\\\ v^{*}=u^{*}&\text{on
}(-1,1)\times\\{0\\}.\end{cases}$

Since $|(x,0)-(\xi,0)|\geq\frac{2}{\pi}|\gamma(x,0)-\gamma(\xi,0)|$ for
$x,\xi\in[-1,1]$, also using (28), we get

$\begin{split}\int_{-1}^{1}\int_{-1}^{1}\frac{|u^{*}(x,0)-u^{*}(\xi,0)|^{2}}{|(x,0)-(\xi,0)|^{2}}dxd\xi&\leq\frac{\pi^{2}}{4}\int_{-1}^{1}\int_{-1}^{1}\frac{|u(\gamma(x,0))-u(\gamma(\xi,0))|^{2}}{|\gamma(x,0)-\gamma(\xi,0)|^{2}}dxd\xi\\\
&=\int_{\mathbb{S}^{1}_{-}}\int_{\mathbb{S}^{1}_{-}}\frac{|u(e^{is})-u(e^{it})|^{2}}{|e^{is}-e^{it}|^{2}}dsdt.\end{split}$
(29)

Moreover

$|e^{is}-(x,0)|\geq\frac{|e^{is}-\gamma(x,0)|}{2}\quad\text{for
}e^{is}\in\mathbb{S}^{1}_{-},\,x\in[-1,1],$

hence with (28)

$\begin{split}\int_{[-1,1]\times\\{0\\}}\int_{\mathbb{S}^{1}_{-}}\frac{|u^{*}_{-}(e^{is})-u^{*}_{-}(x,0)|^{2}}{|e^{is}-(x,0)|^{2}}dsdx&\leq{4}\int_{[-1,1]\times\\{0\\}}\int_{\mathbb{S}^{1}_{-}}\frac{|u(e^{is})-u(\gamma(x,0))|^{2}}{|e^{is}-\gamma(x,0)|^{2}}dsdx\\\
&=\frac{8}{\pi}\int_{\mathbb{S}^{1}_{-}}\int_{\mathbb{S}^{1}_{-}}\frac{|u(e^{is})-u(e^{it})|^{2}}{|e^{is}-e^{it}|^{2}}dsdt.\end{split}$

Together with (29), it follows that

$[u^{*}_{-}]_{H^{\frac{1}{2},2}(\partial D^{2}_{-})}\leq
C[u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1}_{-})}.$

Considering that for $e^{is}\in\mathbb{S}^{1}_{+}$ and $x\in[-1,1]$ we have

$|e^{is}-(x,0)|\geq\frac{|e^{is}-\gamma(x,0)|}{2},$

we also infer

$\begin{split}\int_{[-1,1]\times\\{0\\}}\int_{\mathbb{S}^{1}_{+}}\frac{|u^{*}_{+}(e^{is})-u^{*}_{+}(x,0)|^{2}}{|e^{is}-(x,0)|^{2}}dsdx&\leq
4\int_{[-1,1]\times\\{0\\}}\int_{\mathbb{S}^{1}_{+}}\frac{|u(e^{is})-u(\gamma(x,0))|^{2}}{|e^{is}-\gamma(x,0)|^{2}}dsdx\\\
&=\frac{8}{\pi}\int_{\mathbb{S}^{1}_{-}}\int_{\mathbb{S}^{1}_{+}}\frac{|u(e^{is})-u(e^{it})|^{2}}{|e^{is}-e^{it}|^{2}}dsdt.\end{split}$

Similarly

$\int_{[-1,1]\times\\{0\\}}\int_{\mathbb{S}^{1}_{+}}\frac{|v^{*}(e^{is})-v^{*}(x,0)|^{2}}{|e^{is}-(x,0)|^{2}}dsdx\leq\frac{8}{\pi}\int_{\mathbb{S}^{1}_{-}}\int_{\mathbb{S}^{1}_{+}}\frac{|v(e^{is})-u(e^{it})|^{2}}{|e^{is}-e^{it}|^{2}}dsdt.$

Together with (29), it follows that

$[u^{*}_{+}]_{H^{\frac{1}{2},2}(\partial D^{2}_{-})}\leq
C[u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1})},\qquad[v^{*}]_{H^{\frac{1}{2},2}(\partial
D^{2}_{-})}\leq C[v\&u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1})}.$

Let us now call $U_{+}\in H^{1,2}(D^{2}_{+},\mathbb{C})$, $U_{-}\in
H^{1,2}(D^{2}_{-},\mathbb{C})$, $V\in H^{1,2}(D^{2}_{+},\mathbb{C})$ any
extensions of $u^{*}_{+}$, $u^{+}_{-}$ and $v^{*}$, respectively, which exist
thanks to Proposition A.1. Since $U_{+}$, $U_{-}$ and $V$ have the same trace
$u^{*}$ on $(-1,1)\times\\{0\\}$ they can be patched together to obtain
extensions $U$, $V\&U$ and $V\\#U$ of $u$, $v\&u$ and $v\\#u$, respectively,
namely

$U=\begin{cases}U_{+}&\text{in }D^{2}_{+}\\\ U_{-}&\text{in
}D^{2}_{-},\end{cases}\qquad V\&U=\begin{cases}V&\text{in }D^{2}_{+}\\\
U_{-}&\text{in }D^{2}_{-},\end{cases}$

and

$V\\#U(z)=\begin{cases}V(z)&\text{for }z\in D^{2}_{+}\\\
U_{+}(\bar{z})&\text{for }z\in D^{2}_{-}\end{cases}$

Then, thanks to (16)

$\begin{split}\deg(u\&v)&=\frac{1}{\pi}\int_{D^{2}}J(V\&U)\,dxdy\\\
&=\frac{1}{\pi}\left(\int_{D^{2}_{+}}JV\,dxdy-\int_{D^{2}_{+}}JU_{+}\,dxdy\right)+\frac{1}{\pi}\left(\int_{D^{2}_{+}}JU_{+}\,dxdy+\int_{D^{2}_{-}}JU_{-}\,dxdy\right)\\\
&=\frac{1}{\pi}\int_{D^{2}}J(V\\#U)\,dxdy+\frac{1}{\pi}\int_{D^{2}}JU\,dxdy\\\
&=\deg(v\\#u)+\deg(u).\end{split}$

∎

Composing with a suitable conformal map from $D^{2}$ to $\mathbb{R}^{2}_{+}$
Proposition 3.3 implies:

###### Proposition 3.4

Consider $u,v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ and
$\varepsilon>0$ such that for

$v\&u(x):=\begin{cases}v(x)&\text{for }|x|\leq\varepsilon\\\ u(x)&\text{for
}|x|>\varepsilon\end{cases}$

and

$v\\#u(x):=\begin{cases}v(x)&\text{for }|x|\leq\varepsilon\\\
u\big{(}\frac{\varepsilon^{2}}{x}\big{)}&\text{for
}|x|>\varepsilon\end{cases}$

one has $v\&u,v\\#u\in H^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$. Then

$\deg(v\&u)=\deg(u)+\deg(v\\#u).$ (30)

##  4 Proof of Theorems 1.1 and 1.2

The following lemma is the crucial energy estimate.

###### Lemma 4.1

Let $u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ be such that for
some $x_{0}\in\mathbb{R}$, $u$ is smooth near $x_{0}$, $u^{\prime}(x_{0})\neq
0$ and

$\partial_{y}\tilde{u}(x_{0},0)\perp T_{u(x_{0})}\mathbb{S}^{1},$ (31)

where $\tilde{u}$ is the Poisson harmonic extension of $u$. Then for every
$\varepsilon>0$ sufficiently small there exists $v\in
H^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ such that $v=u$ in
$\mathbb{R}\setminus[x_{0}-2\varepsilon,x_{0}+2\varepsilon]$, $\deg v=\deg
u\pm 1$ and

$E(v)<E(u)+2\pi.$

More precisely, $\deg v=\deg u-\mathrm{sign}(u(x_{0})\wedge
u^{\prime}(x_{0}))$ and we can choose $v$ (depending on $\varepsilon>0$) such
that

$E(v)\leq E(u)+2\pi-(1-\ln
2)|u^{\prime}(x_{0})|^{2}\varepsilon^{2}+o(\varepsilon^{2}),\quad\text{as
}\varepsilon\to 0.$

###### Proof.

Using the harmonic extension, we are able to closely follow [3, pp. 206-207].
Up to a translation and a rotation we can set $x_{0}=0$, $u(x_{0})=-i=(0,-1)$.
By assumption

$\nabla\tilde{u}(0,0)=\begin{pmatrix}a&0\\\ 0&b\end{pmatrix}$ (32)

where $u^{\prime}(0)=(a,0)$, $\partial_{y}\tilde{u}(0,0)=(0,b)$. Up to a
reflection, we can also assume $a>0$.

We now consider the inverse stereographic projection from the North pole

$\Pi^{-1}_{+}:\mathbb{R}\to\mathbb{S}^{1},\quad\Pi^{-1}_{+}(x)=\left(\frac{2x}{1+x^{2}},\frac{-1+x^{2}}{1+x^{2}}\right),$

which can be extended to the biholomorphic map
$\Phi:\overline{\mathbb{R}^{2}_{+}}\to\overline{D^{2}}\setminus\\{i\\},$
which, using the complex notation $z=x+iy$, is

$\Phi(z)=\frac{z-i}{1-iz}=\left(\frac{2x}{(1+y)^{2}+x^{2}},\frac{-1+x^{2}+y^{2}}{(1+y)^{2}+x^{2}}\right).$
(33)

Notice that
$\Phi(\overline{\mathbb{R}^{2}_{+}})=\overline{D^{2}}\setminus\\{i\\}$. Its
inverse is

$\Psi:\overline{D^{2}}\setminus\\{i\\}\to\overline{\mathbb{R}^{2}_{+}},\quad\Psi(z)=\Phi^{-1}(z)=-i\frac{z+i}{z-i}.$
(34)

Similarly, the inverse stereographic projection from the South pole

$\Pi_{-}^{-1}(x)=\left(\frac{2x}{1+x^{2}},\frac{1-x^{2}}{1+x^{2}}\right)$

can be extended to the anti-biholomorphic map
$\bar{\Phi}:\overline{\mathbb{R}^{2}_{+}}\to\overline{D^{2}}\setminus\\{-i\\}$,

$\bar{\Phi}(z)=\frac{\bar{z}+i}{1+i\bar{z}}=\left(\frac{2x}{(1+y)^{2}+x^{2}},\frac{1-x^{2}-y^{2}}{(1+y)^{2}+x^{2}}\right).$

We now set for $\lambda=\mu\varepsilon^{2}$, with $\varepsilon>0$ sufficiently
small and $\mu>0$ to be fixed depending on $a$ and $b$ only,

$U_{\varepsilon}(z)=\begin{cases}\bar{\Phi}_{\lambda}(z):=\bar{\Phi}(\tfrac{z}{\lambda})=\frac{\frac{\bar{z}}{\lambda}+i}{1+i\frac{\bar{z}}{\lambda}}&\text{in
}B_{\varepsilon}^{+}(0)\\\ \Phi\begin{pmatrix}A_{1}r+B_{1}\\\
A_{2}r+B_{2}\end{pmatrix}&\text{in }B_{2\varepsilon}^{+}(0)\setminus
B_{\varepsilon}^{+}(0)\\\ \tilde{u}(z)&\text{in }\mathbb{R}^{2}_{+}\setminus
B_{2\varepsilon}^{+}(0)\end{cases}$

with $A_{1},A_{2},B_{1},B_{2}$ to be determined as functions of
$\varepsilon,\theta$ to make $U_{\varepsilon}$ continuous. In particular,
since $\Psi=\Phi^{-1}$, we want

$\begin{pmatrix}A_{1}(\theta)r+B_{1}(\theta)\\\
A_{2}(\theta)r+B_{2}(\theta)\end{pmatrix}=\Psi(\tilde{u}(re^{i\theta}))\quad\text{for
}r=2\varepsilon,$ (35)

and

$\begin{pmatrix}A_{1}(\theta)r+B_{1}(\theta)\\\
A_{2}(\theta)r+B_{2}(\theta)\end{pmatrix}=\Psi(\bar{\Phi}_{\lambda}(re^{i\theta}))\quad\text{for
}r=\varepsilon.$ (36)

Notice that this implies that $A_{2}(0)=A_{2}(\pi)=B_{2}(0)=B_{2}(\pi)=0$,
since $\Psi\circ\tilde{u}$ and $\Psi\circ\bar{\Phi}_{\lambda}$ send
$\mathbb{R}\times\\{0\\}$ into itself. It follows in particular that
$U_{\varepsilon}$ sends $\mathbb{R}\times\\{0\\}$ into $\mathbb{S}^{1}$.

Using polar coordinates at $(0,0)$, from (32) we get (identifying $i$ with
$(0,-1)$, with a little abuse of notation) the Taylor expansion

$\displaystyle\tilde{u}(re^{i\theta})$ $\displaystyle=-i+\begin{pmatrix}a&0\\\
0&b\end{pmatrix}\begin{pmatrix}r\cos\theta\\\
r\sin\theta\end{pmatrix}+O(r^{2})=-i+\begin{pmatrix}ar\cos\theta\\\
br\sin\theta\end{pmatrix}+O(r^{2}),$

We then compute from (34)

$\Psi(z)=\frac{1}{2}(z-(-i))+O((z-(-i))^{2}),\quad\text{as }z\to-i,$

so that

$\Psi(\tilde{u}(re^{i\theta}))=\frac{1}{2}\begin{pmatrix}ar\cos\theta\\\
br\sin\theta\end{pmatrix}+O(r^{2})\quad\text{as }r\to 0.$

This and (35) give

$\begin{cases}2\varepsilon
A_{1}(\theta)+B_{1}(\theta)=a\varepsilon\cos\theta+O(\varepsilon^{2})\\\
2\varepsilon
A_{2}(\theta)+B_{2}(\theta)=b\varepsilon\sin\theta+O(\varepsilon^{2}).\end{cases}$
(37)

Moreover

$\bar{\Phi}_{\lambda}(z)=-i+2\frac{\lambda}{\bar{z}}+O(\varepsilon^{2}),\quad\text{for
}|z|=\varepsilon,$ (38)

hence

$\Psi(\bar{\Phi}_{\lambda}(\varepsilon
e^{i\theta}))=\frac{\lambda}{\varepsilon
e^{-i\theta}}+O(\varepsilon^{2})=\mu\varepsilon\begin{pmatrix}\cos\theta\\\
\sin\theta\end{pmatrix}+O(\varepsilon^{2}),$

and together with (36) we get

$\begin{cases}\varepsilon
A_{1}(\theta)+B_{1}(\theta)=\mu\varepsilon\cos\theta+O(\varepsilon^{2})\\\
\varepsilon
A_{2}(\theta)+B_{2}(\theta)=\mu\varepsilon\sin\theta+O(\varepsilon^{2}).\end{cases}$
(39)

Solving (37)-(39) yields

$\begin{cases}A_{1}(\theta)=(a-\mu)\cos\theta+O(\varepsilon)\\\
A_{2}(\theta)=(b-\mu)\sin\theta+O(\varepsilon)\\\
B_{1}(\theta)=\varepsilon(2\mu-a)\cos\theta+O(\varepsilon^{2})\\\
B_{2}(\theta)=\varepsilon(2\mu-b)\sin\theta+O(\varepsilon^{2}).\end{cases}$
(40)

We now easily estimate

$\begin{split}\int_{\mathbb{R}^{2}_{+}\setminus
B_{2\varepsilon}^{+}(0)}|\nabla
U_{\varepsilon}|^{2}dxdy&=E(u)-\int_{B_{2\varepsilon}^{+}(0)}|\nabla\tilde{u}|^{2}dxdy=E(u)-2\pi\varepsilon^{2}(a^{2}+b^{2})+o(\varepsilon^{2}).\end{split}$
(41)

Observing that $\bar{\Phi}_{\lambda}$ is anticonformal, so that
$|\nabla\bar{\Phi}_{\lambda}|^{2}=2|J\bar{\Phi}_{\lambda}|$, considering that
$\bar{\Phi}_{\lambda}$ is a diffeomorphism of $\mathbb{R}^{2}_{+}$ onto
$D^{2}$, and taking (38) into account, we see that
$\bar{\Phi}_{\lambda}(B_{\varepsilon}^{+}(0))$ is the unit disk minus a region
of area $2\pi\mu^{2}\varepsilon^{2}+o(\varepsilon^{2})$, so that

$\begin{split}\int_{B_{\varepsilon}^{+}(0)}|\nabla\bar{\Phi}_{\lambda}|^{2}dxdy&=2\mathrm{Area}\left(\bar{\Phi}_{\lambda}(B_{\varepsilon}^{+}(0))\right)=2\pi-4\pi\mu^{2}\varepsilon^{2}+o(\varepsilon^{2}).\end{split}$
(42)

From (33) we obtain $\nabla\Phi(z)=2Id+o(1)$ as $z\to 0$, and by the chain
rule we finally obtain

$\begin{split}\int_{B_{2\varepsilon}^{+}(0)\setminus
B_{\varepsilon}^{+}(0)}|\nabla
U_{\varepsilon}|^{2}dxdy&=4(1+o(1))\int_{B_{2\varepsilon}^{+}(0)\setminus
B_{\varepsilon}^{+}(0)}\left|\nabla\begin{pmatrix}A_{1}(\theta)r+B_{1}(\theta)\\\
A_{2}(\theta)r+B_{2}(\theta)\end{pmatrix}\right|^{2}dxdy\\\
&=4(1+o(1))\int_{0}^{\pi}\int_{\varepsilon}^{2\varepsilon}\sum_{\ell=1}^{2}\left[A_{\ell}^{2}+\left(A_{\ell}^{\prime}+\frac{B_{\ell}^{\prime}}{r}\right)^{2}\right]rdrd\theta\\\
&=2\pi\varepsilon^{2}[a^{2}+b^{2}-2\mu^{2}+(a^{2}+b^{2}+8\mu^{2}-4a\mu-4b\mu)\ln
2]+o(\varepsilon^{2}).\end{split}$ (43)

Summing (41), (42) and (43) we get

$\int_{\mathbb{R}^{2}_{+}}|\nabla
U_{\varepsilon}|^{2}dxdy=E(u)+2\pi-2\pi\varepsilon^{2}(4\mu^{2}-(8\mu^{2}+a^{2}+b^{2}-4a\mu-4b\mu)\ln
2)+o(\varepsilon^{2}).$

Then, as in [3], we can choose $\mu=\max\\{a/2,b/2\\}$ to get

$\begin{split}4\mu^{2}-(8\mu^{2}+a^{2}+b^{2}-4a\mu-4b\mu)\ln
2&=\max\\{a^{2},b^{2}\\}-(a-b)^{2}\ln 2\\\ &\geq(1-\ln
2)\max\\{a^{2},b^{2}\\}\geq(1-\ln 2)a^{2}>0.\end{split}$

Now, setting $v=u_{\varepsilon}=U_{\varepsilon}|_{\mathbb{R}\times\\{0\\}}$,
we conclude

$E(v)\leq\int_{\mathbb{R}^{2}_{+}}|\nabla U_{\varepsilon}|^{2}dxdy\leq
E(u)+2\pi-(1-\ln 2)a^{2}\varepsilon^{2}+o(\varepsilon^{2})<E(u)+2\pi,$

for some $\varepsilon>0$ small enough. The finiteness of $E(v)$ implies that
$v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C})$ and we have already
noticed that $U_{\varepsilon}|_{\mathbb{R}\times\\{0\\}}$ takes values into
$\mathbb{S}^{1}$, so that
$v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$.

Let us now set

$\displaystyle w\&u(x):=\left\\{\begin{array}[]{ll}w(x)&\quad\text{for
}|x|\leq 2\varepsilon\\\ u(x)&\quad\text{for }|x|\geq
2\varepsilon,\end{array}\right.\quad
w\\#u(x):=\left\\{\begin{array}[]{ll}w(x)&\quad\text{for }|x|\leq
2\varepsilon\\\ u(\frac{4\varepsilon^{2}}{x})&\quad\text{for }|x|\geq
2\varepsilon,\end{array}\right.$

so that $v=w\&u$, where $w(x):=U_{\varepsilon}(x)$ for $|x|\leq 2\varepsilon.$
Since $u$ is smooth near the origin, it follows from the estimates on $A_{1}$
and $B_{1}$ that $\mathbb{S}^{1}_{+}$ is covered by the function $w\\#u$ only
once in clockwise direction, and hence $\deg(w\\#u)=-1$. Therefore, by
Proposition 3.4 $\deg(v)=\deg(u)-1.$ ∎

###### Corollary 4.2

Let $u_{0}$ minimize $E$ in $\mathcal{E}_{g}$, take $x_{0}\in I$ such that
$u^{\prime}(x_{0})\neq 0$ and set $k=\deg(u_{0})-\mathrm{sign}(u(x_{0})\wedge
u^{\prime}(x_{0}))$. Then

$\inf_{\mathcal{E}_{g,k}}E<\inf_{\mathcal{E}_{g}}E+2\pi.$

###### Proof.

Since $u_{0}$ is half-harmonic it is smooth in $I$ by [8], and by [23, Sec. 1]
it satisfies condition (31) at every $x_{0}\in I$, hence Lemma 4.1 applies. ∎

_Proof of Theorem 1.2 (completed)._ Given $u_{0}$ minimizing $E$ in
$\mathcal{E}_{g}$, take $x_{0}\in I$ such that $u_{0}^{\prime}(x_{0})\neq 0$
(such $x_{0}$ exists, thanks to Lemma 4.3) and set
$k=-\mathrm{sign}(u(x_{0})\wedge u^{\prime}(x_{0}))$. Consider a minimizing
sequence $(u_{n})\subset\mathcal{E}_{g,k}$. Up to a subsequence
$u_{n}\rightharpoonup u^{*}$ weakly in
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ where
$u^{*}\in\mathcal{E}_{g,\ell}$ for some $\ell\in\mathbb{Z}$. According to
Proposition 3.1

$\inf_{\mathcal{E}_{g}}E\leq
E(u^{*})\leq\liminf_{n\to\infty}E(u_{n})-2\pi|k-\ell|=\inf_{\mathcal{E}_{g,k}}E-2\pi|k-\ell|.$

If $k\neq\ell$ this contradicts Corollary 4.2, hence $\ell=k$ and
$u^{*}\in\mathcal{E}_{g,k}$ is the desired minimizer of $E$ in
$\mathcal{E}_{g,k}$. $\square$

_Proof of Theorem 1.1._ If $g\not\equiv p$ in $\mathbb{R}\setminus I$, then
$u_{0}$ cannot be constant in $I$ by Lemma 4.3 below, so Theorem 1.2 can be
applied. $\square$

###### Lemma 4.3

Let $u\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ be both half-
harmonic and constant in $I=(-1,1)$. Then $u$ is constant in all of
$\mathbb{R}$.

###### Proof.

Up to a rotation, we may assume that $u\equiv i=(0,1)$ in $I$. Writing
$u=v+iw$, with $v$ and $w$ real valued, we obtain from (3) that

$v=(-\Delta)^{\frac{1}{2}}v=0\quad\text{in }I.$ (44)

This implies that $v\equiv 0$ everywhere in $\mathbb{R}$ by the unique
continuation property of fractional Laplacians, see e.g. [13, Thm. 1.2]. An
elementary proof of this fact, in this simple situation, can also be obtained
by considering the Poisson harmonic extension $\tilde{v}$ of $v$ (which is
smooth up to $I\times\\{0\\}$ by elliptic regularity). Then by (44)

$\frac{\partial\tilde{v}(x,y)}{\partial y}\bigg{|}_{y=0}=0,\quad\text{for
}x\in I.$

Considering $B_{-}:=\\{(x,y)\in\mathbb{R}^{2}:x^{2}+y^{2}<1,\,y<0\\}$ and

$\tilde{\tilde{v}}=\begin{cases}\tilde{v}&\text{in }\mathbb{R}^{2}_{+}\\\
0&\text{in }B_{-},\end{cases}$

we obtain that $\tilde{\tilde{v}}$ is harmonic, hence it vanishes identically
in $\mathbb{R}^{2}_{+}\cup B_{-}$ by the unique continuation property of
harmonic maps. Then $v\equiv 0$ in $\mathbb{R}$.

Since $v^{2}+w^{2}=1$ a.e., it follows that $w(x)\in\\{-1,1\\}$ for a.e. $x$,
and since $w\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{R})$, $w$ must be
constant, hence $w\equiv 1$ and $u\equiv(0,1)$, see [5, Corollary 6.2]. ∎

##  5 Proof of Theorem 1.3

Let us first recall a few facts about Blaschke products. Blaschke products are
holomorphic maps
$H:\overline{D^{2}}\subset\mathbb{C}\to\overline{D^{2}}\subset\mathbb{C}$ of
the form

$H(z)=e^{i\theta_{0}}\prod_{j=1}^{k}\frac{z-a_{j}}{1-\bar{a}_{j}z},\quad
a_{1},\dots,a_{k}\in D^{2},\,\theta_{0}\in\mathbb{R}.$

They are the only holomorphic maps of $\overline{D^{2}}$ into itself sending
$\mathbb{S}^{1}$ into itself. Then, by conformality, as in the degree formula
(16)

$\int_{D}|\nabla H|^{2}dxdy=2\int_{D}JHdxdy=2\pi\deg H|_{\mathbb{S}^{1}}=2\pi
k.$

By invariance of the Dirichlet energy under conformal trasformations, we also
have for $h(x):=H\circ\Pi_{+}^{-1}$ and its Poisson extension
$\tilde{h}=H\circ\Phi$

$E(h)=\int_{\mathbb{R}^{2}_{+}}|\nabla\tilde{h}|^{2}dxdy=\int_{D^{2}}|\nabla
H|^{2}dxdy=2\pi k.$

Every such map has the form

$\tilde{h}(z)=e^{i\theta_{0}}\prod_{j=1}^{k}\frac{z-\beta_{j}}{z-\bar{\beta}_{j}},\quad\beta_{1},\dots\beta_{k}\in\mathbb{R}^{2}_{+},\,\theta_{0}\in\mathbb{R}.$
(45)

We define for $k>0$ the set of Blaschke products

$\mathcal{B}_{k}=\left\\{\tilde{h}\text{ as in }\eqref{Blas}\right\\}$

and the set of their conjugates

$\mathcal{B}_{-k}=\left\\{\bar{\tilde{h}}:\tilde{h}\in\mathcal{B}_{k}\right\\}.$

For $k=0$, $\mathcal{B}_{0}$ are just constant maps into $\mathbb{S}^{1}$.

Notice that for $\tilde{h}\in\mathcal{B}_{\pm k}$ and $h(x)=\tilde{h}(x,0)$,
we have $\deg(h)=\pm k$, $E(h)=2\pi|k|$.

The following lemma is probably well known. We state it for completeness.

###### Lemma 5.1

Consider for $k\in\mathbb{Z}$

$\mathcal{E}_{k}:=\\{v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1}):\deg(v)=k\\}.$

Then

$\inf_{\mathcal{E}_{k}}E=2\pi|k|$

and the infimum is attained only by the restrictions of maps
$\tilde{h}\in\mathcal{B}_{k}$.

###### Proof.

(Sketch) By conformal invariance of $E$, we can prove the lemma for functions
in $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. By the degree formula
(16) we have for $v\in\mathcal{E}_{k}$

$2\pi|k|=2\left|\int_{D}J\tilde{v}dxdy\right|\leq\int_{D}|\nabla\tilde{v}|^{2}dxdy=:E(v),$

where $\tilde{v}$ is the harmonic extension of $v$ to the whole disk, and the
inequality is strict unless $\tilde{v}$ is everywhere conformal or
anticonformal, which corresponds to $v\in\mathcal{B}_{k}$ or
$v\in\mathcal{B}_{-k}$, respectively. ∎

_Proof of Theorem 1.3 \- Part (i)._ Since $g_{\lambda}$ is the restriction to
$\mathbb{R}\times\\{0\\}$ of
$\Phi_{\lambda}:=\Phi\left(\tfrac{\cdot}{\lambda}\right)\in\mathcal{B}_{1}$,
the minimality of $g_{\lambda}$ and its uniqueness follow at once from Lemma
5.1.

_Part (ii)._ We have $E(u_{0})=E(g_{\lambda})=2\pi$. Except when $k=1$, none
of the Blaschke products in $\mathcal{B}_{k}$ matches $g_{\lambda}$ on
$\mathbb{R}\setminus I$, hence, by Lemma 5.1, for every
$u\in\mathcal{E}_{g_{\lambda},k}$, $k\in\mathbb{Z}\setminus\\{0,1\\}$

$E(u)>2\pi|k|\geq 2\pi=E(u_{0}).$

Still by Lemma 5.1, $E(u)>2\pi=E(u_{0})$ for every
$u\in\mathcal{E}_{g_{\lambda},1}\setminus\\{u_{0}\\}$. It remains to exclude
for $\lambda$ sufficiently large the existence of a function
$u\in\mathcal{E}_{g_{\lambda},0}$ such that $E(u)\leq E(g_{\lambda})$.

Consider $\lambda_{n}\to\infty$. We have by dominated convergence

$\begin{split}\lim_{n\to\infty}\frac{1}{2\pi}\int_{\mathbb{R}\setminus
I}\int_{\mathbb{R}\setminus
I}\frac{|g_{\lambda_{n}}(x)-g_{\lambda_{n}}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi&=\lim_{n\to\infty}\frac{1}{2\pi}\int_{\mathbb{R}\setminus\left[-\tfrac{1}{\lambda_{n}},\tfrac{1}{\lambda_{n}}\right]}\int_{\mathbb{R}\setminus\left[-\tfrac{1}{\lambda_{n}},\tfrac{1}{\lambda_{n}}\right]}\frac{|g_{1}(x)-g_{1}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi\\\
&=\frac{1}{2\pi}[g_{1}]_{H^{\frac{1}{2},2}(\mathbb{R})}^{2}=E(g_{1})=2\pi,\end{split}$

while, for any sequence of functions
$(u_{n})\subset\mathcal{E}_{g_{\lambda_{n}},0}$ we have

$\liminf_{n\to\infty}\frac{1}{2\pi}\left(\int_{\mathbb{R}}\int_{I}\frac{|u_{n}(x)-u_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi+\int_{I}\int_{\mathbb{R}\setminus
I}\frac{|u_{n}(x)-u_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi\right)\geq 2\pi.$ (46)

In order to prove (46), consider the function

$v_{n}(x)=\begin{cases}u_{n}(x)&\text{for }|x|\leq 1\\\\[8.53581pt]
g_{\lambda_{n}}\left(x\right)&\text{for }1<|x|\leq R\\\\[8.53581pt]
g_{\lambda_{n}}\left(\tfrac{R^{2}}{x}\right)&\text{for }|x|>R.\end{cases}$
(47)

For $R\geq 2$ we rewrite the functions $u_{n}$ and $v_{n}$ as

$u_{n}=w_{1}\&w_{2},\quad v_{n}=w_{1}\\#w_{2},\quad\text{where }\quad
w_{1}:=u_{n}|_{[-R,R]},\quad w_{2}:=g_{\lambda_{n}},$

and hence, by Proposition 3.4

$\deg(v_{n})=\deg(u_{n})-\deg(w_{2})=-1.$

By Lemma 5.1 we have

$E(v_{n})\geq 2\pi.$

On the other hand, from (47) we estimate

$\int_{\mathbb{R}\setminus I}\int_{\mathbb{R}\setminus
I}\frac{|v_{n}(x)-v_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi=o(1),\quad\text{as
}n\to\infty.$ (48)

Indeed, for $R\geq 2$ writing

$\displaystyle\int_{\mathbb{R}\setminus I}\int_{\mathbb{R}\setminus
I}\frac{|v_{n}(x)-v_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi$
$\displaystyle=\left(\int_{\mathbb{R}\setminus[-R,R]}\int_{\mathbb{R}\setminus[-R,R]}+2\int_{\mathbb{R}\setminus[-R,R]}\int_{[-R,R]\setminus
I}+\int_{[-R,R]\setminus I}\int_{[-R,R]\setminus
I}\right)\frac{|v_{n}(x)-v_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi$
$\displaystyle=:(I)+2(II)+(III),$

one can bound with a change of variables in $x$ and in $y$

$\displaystyle(I)$
$\displaystyle=\int_{[-R,R]}\int_{[-R,R]}\frac{|g_{\lambda_{n}}(x)-g_{\lambda_{n}}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi$
$\displaystyle\leq\|g_{1}^{\prime}\|_{L^{\infty}}\frac{1}{\lambda_{n}^{2}}\int_{[-R,R]}\int_{[-R,R]}dxd\xi\leq\frac{C(R)}{\lambda_{n}^{2}},$

with a change of variables in $\xi$

$\displaystyle(II)$ $\displaystyle=\int_{[-R,R]}\int_{[-R,R]\setminus
I}\frac{|g_{\lambda_{n}}(x)-g_{\lambda_{n}}(\xi)|^{2}}{(x-\frac{R^{2}}{\xi})^{2}}\frac{R^{2}}{\xi^{2}}dxd\xi$
$\displaystyle\leq
R^{2}\frac{\|g_{1}^{\prime}\|}{\lambda_{n}^{2}}\int_{[-R,R]}\int_{[-R,R]\setminus
I}\frac{(x-\xi)^{2}}{(x\xi-R^{2})^{2}}dxd\xi\leq\frac{C(R)}{\lambda_{n}^{2}},$

and finally

$\displaystyle(III)=\int_{[-R,R]\setminus I}\int_{[-R,R]\setminus
I}\frac{|g_{\lambda_{n}}(x)-g_{\lambda_{n}}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi\leq(I)\leq\frac{C(R)}{\lambda_{n}^{2}},$

so that $\eqref{vno1}$ is proven.

Moreover, since $R\geq 2$, we have $|x-y|\geq\frac{|y|}{2}$ for $x\in I$,
$|y|\geq R$, so that

$\begin{split}\int_{\mathbb{R}\setminus[-R,R]}\int_{I}\frac{|v_{n}(x)-v_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi&\leq\int_{\mathbb{R}\setminus[-R,R]}\int_{I}\frac{16}{\xi^{2}}dxd\xi=\frac{64}{R}.\end{split}$

It then follows

$\displaystyle 2\pi\leq E(v_{n})\leq$
$\displaystyle\frac{1}{2\pi}\int_{[-R,R]}\int_{I}\frac{|v_{n}(x)-v_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi+\frac{1}{2\pi}\int_{I}\int_{[-R,R]\setminus
I}\frac{|v_{n}(x)-v_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxdy$
$\displaystyle\quad+\frac{C}{R}+o(1),\quad\text{as }n\to\infty$

and, since $u_{n}=v_{n}$ in $[-R,R]$,

$\begin{split}\frac{1}{2\pi}\int_{[-R,R]}\int_{I}\frac{|u_{n}(x)-u_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxdy+\frac{1}{2\pi}\int_{I}\int_{[-R,R]\setminus
I}\frac{|u_{n}(x)-u_{n}(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi\geq
2\pi-\frac{C}{R}+o(1)\end{split}$

as $n\to\infty$. Since $R>1$ can be taken arbitrarily large, (46) is proven.

Since

$\displaystyle E(u)$ $\displaystyle=\frac{1}{2\pi}\int_{\mathbb{R}\setminus
I}\int_{\mathbb{R}\setminus
I}\frac{|u(x)-u(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi+\frac{1}{2\pi}\int_{\mathbb{R}}\int_{I}\frac{|u(x)-u(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi$
$\displaystyle\qquad+\frac{1}{2\pi}\int_{I}\int_{\mathbb{R}\setminus
I}\frac{|u(x)-u(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi,$

we conclude that

$\liminf_{n\to\infty}E(u_{n})\geq 4\pi,$

hence for $\lambda$ sufficiently large we have $E(u)>2\pi$ for every
$u\in\mathcal{E}_{g_{\lambda},0}$.

_Part (iii)._ Let $u_{0}$ be an absolute minimizer of $E$ in
$\mathcal{E}_{g_{\lambda}}$, and set $k=\deg(u_{0})$. If $k=0$ then we are
done. For $|k|\geq 1$, it follows from Lemma 5.1 that

$2\pi\leq 2\pi|k|=\inf_{\mathcal{E}_{k}}E\leq E(u_{0})\leq
E(g_{\lambda})=2\pi.$

Thus, we can simply take $u_{0}=g_{\lambda}$, and the existence of a minimizer
$u^{*}$ in $\mathcal{E}_{g_{\lambda},0}$ follows at once from Theorem 1.2.

On the other hand, by Proposition 5.2 below, there exists $\lambda_{0}>1$ such
that for $\lambda\in(0,\lambda_{0})$ there exists
$u_{0}\in\mathcal{E}_{g_{\lambda},0}$ absolute mimimizer of $E$ in
$\mathcal{E}_{g_{\lambda}}$ with $E(u_{0})<2\pi=E(g_{\lambda})$.

_Part (iv)._ We claim that for any $k>1$ and $\lambda>0$

$\inf_{\mathcal{E}_{g_{\lambda},k}}E\leq
E(g_{\lambda})+2\pi|k-\deg(g_{\lambda})|=2\pi k.$ (49)

In order to prove (49), take $\delta>0$ sufficiently small, $k-1$ distinct
points $x_{1},\dots,x_{k-1}\in I$ and
$v\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$ such that

  1. 1.

$|x_{j}|<1-\delta$ for $1\leq j\leq k-1$ and $|x_{j}-x_{\ell}|>2\delta$ for
$1\leq j<\ell\leq k-1$,

  2. 2.

$v=g_{\lambda}$ in
$\mathbb{R}\setminus\cup_{j=1}^{k-1}(x_{j}-\delta,x_{j}+\delta)$,

  3. 3.

$v(x_{j})\wedge v^{\prime}(x_{j})<0$,

  4. 4.

$E(v)\leq E(g_{\lambda})+\delta.$

To prove that such $v$ exists, consider the map $\Pi_{+}\circ
g_{\lambda}(x)=\frac{x}{\lambda}$ for $x\in\mathbb{R}$. For a function
$\eta\in C^{\infty}_{c}((-1,1))$ with $\eta^{\prime}(0)=-\frac{2}{\lambda}$
and $\rho>0$ set $\eta_{\rho}(x):=\rho\eta(x/\rho)$ and define

$v_{\rho}:=\Pi_{+}^{-1}\left(\Pi_{+}\circ
g_{\lambda}+\sum_{j=1}^{k-1}\eta_{\rho}(x-x_{j})\right).$

Since for $\rho$ small it holds

$(\Pi_{+}\circ
v_{\rho})^{\prime}(x_{j})=\frac{1}{\lambda}-\eta^{\prime}(0)=-\frac{1}{\lambda}<0,$

we have $v_{\rho}(x_{j})\wedge v^{\prime}_{\rho}(x_{j})<0$ for
$j=1,\dots,k-1$. Moreover, by scale invariance,

$[\eta_{\rho}]_{H^{\frac{1}{2},2}(\mathbb{R})}=\rho[\eta]_{H^{\frac{1}{2},2}(\mathbb{R})}=O(\rho),\quad\text{as
}\rho\to 0,$

hence $E(v_{\rho})\to E(g_{\lambda})$ as $\rho\to 0$, and it suffices to
choose $\rho>0$ small enough.

Then, applying Lemma 4.1 to $v$ at $x_{1},\dots,x_{k-1}$ we obtain a new
function $w$ with $\deg(w)=\deg(v)+k-1=k$ and

$E(w)<E(v)+2\pi(k-1)\leq E(g_{\lambda})+\delta+2\pi(k-1)=2\pi k+\delta.$

Since $\delta>0$ is arbitrary, (49) follows.

On the other hand, by Lemma 5.1

$\inf_{\mathcal{E}_{g_{\lambda},k}}E\geq\inf_{\mathcal{E}_{k}}E=2\pi k$

and can only be attained by Blaschke products in $\mathcal{B}_{k}$. Since
$\mathcal{B}_{k}\cap\mathcal{E}_{g_{\lambda}}=\emptyset$ for $k>1$, the
infimum of $E$ in $\mathcal{E}_{g_{\lambda},k}$ is not attained. $\square$

###### Proposition 5.2

Consider $g_{\lambda}(x)=\Pi_{+}^{-1}(x/\lambda)$. There exists
$\lambda_{0}>1$ such that for $\lambda\in(0,\lambda_{0})$

$2\pi=E(g_{\lambda})>\inf_{\mathcal{E}_{g_{\lambda}}}E=\inf_{\mathcal{E}_{g_{\lambda},0}}E.$

###### Proof.

As in the proof of Theorem 1.3, Part (i), we have

$E(g_{\lambda})=2\pi=\min_{\mathcal{E}_{g_{\lambda},1}}E.$

It suffices to prove that for $\lambda_{0}>1$ sufficiently close to $1$, and
for $\lambda\in(0,\lambda_{0}]$, we have

$\inf_{\mathcal{E}_{g_{\lambda},0}}E<2\pi,$ (50)

i.e. there exists $u\in\mathcal{E}_{g_{\lambda}}$ with $\deg(u)=0$ and
$E(u)<2\pi$. Then, since a minimizer $u_{0}$ of $E$ in
$\mathcal{E}_{g_{\lambda}}$ exists by direct methods, we must have
$u_{0}\in\mathcal{E}_{g_{\lambda},0}$ by (50) and Lemma 5.1.

We start with the case $\lambda=1$. It is convenient to work on
$\mathbb{S}^{1}$ via stereographic projection. Clearly
$g_{1}\circ\Pi_{+}(e^{i\theta})=e^{i\theta}$. Now we define

$v(e^{i\theta})=\begin{cases}e^{i\theta}&\text{for }\theta\in[0,\pi]\\\
e^{-i\theta}&\text{for }\theta\in[\pi,2\pi].\end{cases}$

Clearly $\deg(v)=0$. We extend $v$ to the function on $\overline{D^{2}}$

$V(z)=\begin{cases}z&\text{for }\mathrm{Im}z\geq 0\\\ \bar{z}&\text{for
}\mathrm{Im}z<0\end{cases}$ (51)

Let $\tilde{v}$ be the harmonic extension of $v$ to $D^{2}$. Then, since $V$
is not harmonic, we have

$E(v)=\int_{D^{2}}|\nabla\tilde{v}|^{2}dxdy<\int_{D^{2}}|\nabla
V|^{2}dxdy=2\pi.$

Then, by conformal invariance, $E(v\circ\Pi_{+}^{-1})<2\pi$, and since
$v\circ\Pi_{+}^{-1}\in\mathcal{E}_{g_{1},0}$, (50) is proven for $\lambda=1$.

By continuity, (50) continues to hold for $\lambda$ in a neighborhood of $1$,
in particular in $[1,\lambda_{0}]$ for some $\lambda_{0}>1$.

For $\lambda\in(0,1)$ we still have (50). Indeed, similar to (51) we can
construct

$V_{\lambda}(z)=\begin{cases}\Phi_{\lambda}\circ\Psi(z)&\text{for
}\mathrm{Im}z\geq 0\\\ \Phi_{\lambda}\circ\Psi(\bar{z})&\text{for
}\mathrm{Im}z<0\end{cases}$

By symmetry and conformality we have, for $D^{2}_{+}=\\{z\in
D^{2}:\mathrm{Im}z>0\\}$,

$\int_{D^{2}}|\nabla V_{\lambda}|^{2}dxdy=2\int_{D^{2}_{+}}|\nabla
V_{\lambda}|^{2}dxdy=2\mathrm{Area}(V_{\lambda}(D^{2}_{+}))<2\pi,$

since $V_{\lambda}$ sends the half disk $D^{2}_{+}$ into a strict subset of
itself666To verify this, observe that $\Phi\circ\Psi=Id$, which of course
sends $D^{2}_{+}$ onto itself, while $\Phi_{\lambda}\circ\Psi=\Phi\circ
M_{\lambda}\circ\Psi$, where $M_{\lambda}(z)=\tfrac{z}{\lambda}$. On the other
hand, $\Psi(D^{2}_{+})=\mathbb{R}^{2}_{+}\setminus D^{2}_{+}$, and
$M_{\lambda}$ for $\lambda\in(0,1)$, being a dilation, sends
$\mathbb{R}^{2}_{+}\setminus D^{2}_{+}$ into a strict subset of itself, which
is sent by $\Phi$ into a strict subset of $D^{2}_{+}$. Setting
$v_{\lambda}=V_{\lambda}|_{\mathbb{S}^{1}}$, as before we have

$E(v_{\lambda}\circ\Pi_{+}^{-1})\leq\int_{D^{2}}|\nabla
V_{\lambda}|^{2}dxdy<2\pi,$

and since $\deg(v_{\lambda})=0$ and $v_{\lambda}\circ\Pi_{+}^{-1}=g_{\lambda}$
on $\mathbb{R}\setminus[-1,1]$, we obtain (50). ∎

##  Appendix A Appendix

###  A.1 Fractional Laplacians and the spaces
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C})$,
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$

For the proofs of the following statements, which are all classic, we refer to
the appendix of [17] and the references therein.

We start working on the real line. Given
$u\in\mathcal{S}(\mathbb{R},\mathbb{C})$ (the Schwartz space of rapidly
decreasing complex-valued functions) and $s>0$, one can define
$(-\Delta)^{s}u$ via Fourier transform as

$(-\Delta)^{s}u=\mathcal{F}^{-1}\left(|\xi|^{2s}\hat{u}\right),$ (52)

where

$\mathcal{F}(u)(\xi)=\hat{u}(\xi):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}u(x)e^{-ix\xi}dx,$

so that

$u(x)=\mathcal{F}^{-1}(\hat{u})(x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{u}(\xi)e^{ix\xi}d\xi.$

This definition can be extended by duality to every function in

$L_{s}(\mathbb{R},\mathbb{C}):=\left\\{u\in
L^{1}_{\operatorname{loc}}(\mathbb{R},\mathbb{C}):\|u\|_{L_{s}}:=\int_{\mathbb{R}}\frac{|u(x)|}{1+|x|^{1+2s}}dx<\infty\right\\},$

in the sense of tempered distributions:

$\langle(-\Delta)^{s}u,\varphi\rangle:=\int_{\mathbb{R}}u\overline{(-\Delta)^{s}\varphi}dx=\int_{\mathbb{R}}u\,\overline{\mathcal{F}^{-1}(|\xi|\hat{\varphi}(\xi))}\,dx,\quad\text{for
every }\varphi\in\mathcal{S}.$ (53)

Then we can define

$\begin{split}\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{C})&=\\{u\in
L_{\frac{1}{4}}(\mathbb{R},\mathbb{C}):(-\Delta)^{\frac{1}{4}}u\in
L^{2}(\mathbb{R},\mathbb{C})\\}\\\ &=\\{u\in
L_{\frac{1}{4}}(\mathbb{R},\mathbb{C}):|\xi|^{\frac{1}{2}}\hat{u}\in
L^{2}(\mathbb{R},\mathbb{C})\\}.\end{split}$ (54)

For $u\in\mathcal{S}(\mathbb{R},\mathbb{C})$ one also has (see e.g. [11,
Section 3]) that

$(-\Delta)^{s}u(x)=K_{s}P.V.\int_{\mathbb{R}}\frac{u(x)-u(\xi)}{|x-\xi|^{1+2s}}d\xi:=K_{s}\lim_{\varepsilon\to
0}\int_{\mathbb{R}\setminus[-\varepsilon,\varepsilon]}\frac{u(x)-u(\xi)}{|x-\xi|^{1+2s}}d\xi,$
(55)

with $K_{\frac{1}{2}}=\frac{1}{\pi}$. In particular

$(-\Delta)^{\frac{1}{2}}u(x)=\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{u(x)-u(\xi)}{(x-\xi)^{2}}dy.$
(56)

Considering now the Poisson extension $\tilde{u}$ of $u$ given by

$\tilde{u}(x,y)=\frac{1}{\pi}\int_{\mathbb{R}}\frac{yu(\xi)}{y^{2}+(x-\xi)^{2}}d\xi,\quad
x\in\mathbb{R},\,y>0,$

we have (see e.g. [17, Prop. A.1])

$\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}}=\|\nabla\tilde{u}\|_{L^{2}}.$ (57)

Indeed, for $u\in\mathcal{S}(\mathbb{R},\mathbb{C})$) we have

$(-\Delta)^{\frac{1}{2}}u=-\frac{\partial\tilde{u}}{\partial
y}\bigg{|}_{y=0},$

hence with (52) and integration by parts

$\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}}^{2}=\int_{\mathbb{R}}\bar{u}(-\Delta)^{\frac{1}{2}}udx=-\int_{\mathbb{R}}\bar{u}\frac{\partial\tilde{u}}{\partial
y}\bigg{|}_{y=0}dx=\int_{\mathbb{R}^{2}_{+}}|\nabla\tilde{u}|^{2}dxdy.$ (58)

Then (57) follows by density.

We also have

$\|(-\Delta)^{\frac{1}{4}}u\|_{L^{2}}^{2}=\frac{1}{2\pi}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{(u(x)-u(\xi))^{2}}{(x-\xi)^{2}}dxd\xi,$
(59)

which can be proven using (56) to get

$\int_{\mathbb{R}}\bar{u}(-\Delta)^{\frac{1}{2}}udx=\frac{1}{\pi}\int_{\mathbb{R}}\bar{u}(x)P.V.\int_{\mathbb{R}}\frac{u(x)-u(\xi)}{(x-\xi)^{2}}d\xi
dx$

and exchanging the roles of $x$ and $\xi$ and summing to obtain

$\displaystyle 2\int_{\mathbb{R}}\bar{u}(-\Delta)^{\frac{1}{2}}udx$
$\displaystyle=\frac{1}{\pi}\int_{\mathbb{R}}\bar{u}(x)P.V.\int_{\mathbb{R}}\frac{u(x)-u(\xi)}{(x-\xi)^{2}}d\xi
dx+\frac{1}{\pi}\int_{\mathbb{R}}\bar{u}(y)P.V.\int_{\mathbb{R}}\frac{u(\xi)-u(x)}{(x-\xi)^{2}}dxd\xi$
$\displaystyle=\frac{1}{\pi}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{|u(x)-u(\xi)|^{2}}{(x-\xi)^{2}}dxd\xi.$

Then, using the first identity in (58), (59) follows.

We can do similar computations for functions defined on $\mathbb{S}^{1}$. We
start by recalling that any map $u\in L^{2}(\mathbb{S}^{1},\mathbb{S}^{1})$
can be written in terms of its Fourier series

$u(e^{i\theta})=\frac{1}{\sqrt{2\pi}}\sum_{k\in\mathbb{Z}}a_{k}e^{ik\theta},$

where

$a_{k}=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{S}^{1}}u(e^{i\theta})e^{-ik\theta}d\theta.$
(60)

Defining the fractional Laplacian (in this case a half-derivative)

$(-\Delta)^{\frac{1}{4}}u(e^{i\theta}):=\frac{1}{\sqrt{2\pi}}\sum_{k\in\mathbb{Z}}\sqrt{|k|}a_{k}e^{ik\theta},$
(61)

we can set in analogy with (54)

$\begin{split}H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})&=\left\\{u\in
L^{2}(\mathbb{S}^{1},\mathbb{C}):(-\Delta)^{\frac{1}{4}}u\in
L^{2}(\mathbb{S}^{1},\mathbb{C})\right\\}\\\ &=\left\\{u\in
L^{2}(\mathbb{S}^{1},\mathbb{C}):\sum_{k\in\mathbb{Z}}|k||a_{k}|^{2}<\infty\right\\},\end{split}$
(62)

and

$[u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1})}^{2}=\sum_{k\in\mathbb{Z}}|k||a_{k}|^{2}.$

Considering in polar coordinates $(r,\theta)$ the harmonic extension
$\tilde{u}$ of $u$ to $D^{2}$

$\tilde{u}(re^{i\theta})=\frac{1}{\sqrt{2\pi}}\sum_{k\in\mathbb{Z}}a_{k}r^{|k|}e^{ik\theta},$
(63)

we can easily compute

$\|\nabla\tilde{u}\|_{L^{2}(D^{2})}=[u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1})}.$
(64)

On the other hand, we can also compute (see e.g. [7, Proposition A.2])

$(-\Delta)^{\frac{1}{2}}u(e^{i\theta})=\frac{1}{\pi}P.V.\int_{0}^{2\pi}\frac{u(e^{i\theta})-u(e^{it})}{|e^{i\theta}-e^{it}|^{2}}dt,$
(65)

and, similar to (59) obtain

$[u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1})}^{2}=\int_{\mathbb{S}^{1}}\bar{u}(-\Delta)^{\frac{1}{2}}u\,d\theta=\frac{1}{\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{|u(e^{i\theta})-u(e^{it})|^{2}}{|e^{i\theta}-e^{it}|^{2}}dtd\theta.$
(66)

We also recall that the harmonic extension of $u$ can be expressed using the
Poisson kernel:

$\tilde{u}(z)=\frac{1}{2\pi}\int_{\mathbb{S}^{1}}\frac{1-|z|^{2}}{|z-e^{i\theta}|}u(e^{i\theta})d\theta,\quad
z\in D^{2}.$ (67)

Finally, we mention a characterisation of fractional Sobolev spaces in terms
of traces of Sobolev maps on arbitrary bounded Lipschitz domains. For such a
domain $\Omega\subset\mathbb{R}^{2}\simeq\mathbb{C}$ with
$\Gamma:=\partial\Omega$, we define

$[u]_{H^{\frac{1}{2},2}(\Gamma)}=\frac{1}{\pi}\int_{\Gamma}\int_{\Gamma}\frac{|u(z)-u(\zeta)|^{2}}{|z-\zeta|^{2}}d\sigma(z)d\sigma(\zeta),$

where $d\sigma$ is the arc-length on $\Gamma$, and set

$H^{\frac{1}{2},2}(\Gamma):=\left\\{u\in
L^{2}(\Gamma):[u]_{H^{\frac{1}{2},2}(\Gamma)}<\infty\right\\}.$

Then we have the following trace theorem, see e.g. [12, 24]:

###### Proposition A.1

For any bounded Lipschitz domain $\Omega$ with $\Gamma:=\partial\Omega$, the
trace maps $u\mapsto u|_{\Gamma}$ is surjective from $H^{1,2}(\Omega)$ to
$H^{\frac{1}{2},2}(\Gamma)$.

###  A.2 Density results

###### Proposition A.2

The space of functions $u\in C^{\infty}(\mathbb{R},\mathbb{S}^{1})$ which are
constant outside a compact set is dense in
$\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$. Equivalently, the space
of functions $u\in C^{\infty}(\mathbb{S}^{1},\mathbb{S}^{1})$ which are
constant in a neighborhood (depending on $u$) of any fixed point
$z_{0}\in\mathbb{S}^{1}$ are dense in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

In the proof of Proposition A.2 we will borrow ideas from [6]. We first need
two lemmas.

###### Lemma A.3

Let $(u_{k})\subset H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$ be such that
$u_{k}\to u$ a.e. in $\mathbb{S}^{1}$ for some $u\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. If $u_{k}\to u$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$ and $|u_{k}|\geq\delta$ a.e. in
$\mathbb{S}^{1}$ for some $\delta>0$ independent of $k$, then
$\frac{u_{k}}{|u_{k}|}\to u$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

###### Proof.

Setting $\varphi(z)=\frac{z}{|z|}-z$ we get that $\varphi$ is smooth in
$\mathbb{C}^{2}\setminus\\{0\\}$, and $|\nabla\varphi|\leq C(\delta)$ on
$\mathbb{C}\setminus B_{\delta}$. The lemma follows if we show that
$\varphi(u_{k})\to\varphi(u)\equiv 0$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$.

We set

$g_{k}(z,\zeta):=\frac{|\varphi(u_{k}(z))-\varphi(u_{k}(\zeta))|^{2}}{|z-\zeta|^{2}},\quad
z,\zeta\in\mathbb{S}^{1}.$

Then

$\displaystyle|g_{k}(z,\zeta)|$ $\displaystyle\leq
C\frac{|u_{k}(z)-u_{k}(\zeta)|^{2}}{|z-\zeta|^{2}}\leq
C\frac{|(u_{k}(z)-u(z))-(u_{k}(\zeta)-u(\zeta))|^{2}}{|z-\zeta|^{2}}+C\frac{|u(z)-u(\zeta)|^{2}}{|z-\zeta|^{2}}.$

This shows that the family $(g_{k})$ is uniformly integrable in
$\mathbb{S}^{1}\times\mathbb{S}^{1}$. In particular, as $g_{k}\to 0$ a.e. in
$\mathbb{S}^{1}\times\mathbb{S}^{1}$, by Vitali’s convergence theorem we get
that $g_{k}\to 0$ in $L^{1}(\mathbb{S}^{1}\times\mathbb{S}^{1})$, which gives
$\varphi(u_{k})\to 0$ in $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$. ∎

###### Lemma A.4

Given $u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ and
$\varepsilon>0$ we set

$u_{\operatorname{av},\varepsilon}(e^{i\theta}):=\fint_{\theta-\varepsilon}^{\theta+\varepsilon}u(e^{it})\,dt.$

Then $|u_{\operatorname{av},\varepsilon}|=1+o_{\varepsilon}(1)$,
$u_{\operatorname{av},\varepsilon}\in C^{0,1}(\mathbb{S}^{1},\mathbb{C})$,
$u_{\operatorname{av},\varepsilon}\to u$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$ and
$\frac{u_{\operatorname{av},\varepsilon}}{|u_{\operatorname{av},\varepsilon}|}\to
u$ in $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

###### Proof.

By the Jensen’s inequality

$\begin{split}(1-|u_{\operatorname{av},\varepsilon}(e^{i\theta})|)^{2}&\leq\fint_{\theta-\varepsilon}^{\theta+\varepsilon}|u(e^{it})-u_{\operatorname{av},\varepsilon}(e^{i\theta})|^{2}dt\\\
&=\fint_{\theta-\varepsilon}^{\theta+\varepsilon}\left|\fint_{\theta-\varepsilon}^{\theta+\varepsilon}u(e^{it})-u(e^{is})\,ds\right|^{2}dt\\\
&\leq\fint_{\theta-\varepsilon}^{\theta+\varepsilon}\fint_{\theta-\varepsilon}^{\theta+\varepsilon}|u(e^{it})-u(e^{is})|^{2}\,dsdt\\\
&\leq\int_{\theta-\varepsilon}^{\theta+\varepsilon}\int_{\theta-\varepsilon}^{\theta+\varepsilon}\frac{|u(e^{it})-u(e^{is})|^{2}}{|e^{it}-e^{is}|^{2}}\,dsdt.\\\
\end{split}$

where in the last inequality we used that $|e^{it}-e^{is}|\leq 2\varepsilon$
for $s,t\in[\theta-\varepsilon,\theta+\varepsilon]$. By (66) and the absolute
continuity of the Lebesgue integral, the last integral goes to $0$ as
$\varepsilon\to 0$ uniformly with respect to $\theta$, hence
$|u_{\operatorname{av},\varepsilon}|\to 1$ uniformly as $\varepsilon\to 0$.

We now prove that $u_{\operatorname{av},\varepsilon}\to u$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. To this end we write

$\displaystyle
u_{\operatorname{av},\varepsilon}(e^{i\theta_{1}})-u_{\operatorname{av},\varepsilon}(e^{i\theta_{2}})=\fint_{-\varepsilon}^{\varepsilon}\left(u(e^{i(\theta_{1}+t)})-u(e^{i(\theta_{2}+t)})\right)dt.$

Then for any $A_{1},A_{2}\subset(0,2\pi)$,

$\displaystyle\int_{A_{1}}\int_{A_{2}}\frac{|u_{\operatorname{av},\varepsilon}(e^{i\theta_{1}})-u_{\operatorname{av},\varepsilon}(e^{i\theta_{2}})|^{2}}{|e^{i\theta_{1}}-e^{i\theta_{2}}|^{2}}d\theta_{2}d\theta_{1}$
$\displaystyle\leq\fint_{-\varepsilon}^{\varepsilon}\int_{A_{1}}\int_{A_{2}}\frac{|u(e^{i(\theta_{1}+t)})-u(e^{(i(\theta_{2}+t)})|^{2}}{|e^{i(\theta_{1}+t)}-e^{i(\theta_{2}+t)}|^{2}}d\theta_{2}d\theta_{1}dt$
$\displaystyle=\fint_{-\varepsilon}^{\varepsilon}\int_{A_{1}+t}\int_{A_{2}+t}\frac{|u(e^{i\theta_{1}})-u(e^{(i\theta_{2}})|^{2}}{|e^{i\theta_{1}}-e^{i\theta_{2}}|^{2}}d\theta_{2}d\theta_{1}dt.$

This shows that the family
$\frac{|u_{\operatorname{av},\varepsilon}(e^{i\theta_{1}})-u_{\operatorname{av},\varepsilon}(e^{i\theta_{2}})|^{2}}{|e^{i\theta_{1}}-e^{i\theta_{2}}|^{2}}$
is uniformly integrable in $(0,2\pi)\times(0,2\pi)$. Therefore, as
$u_{\operatorname{av},\varepsilon}\to u$ a.e., by the Vitali’s convergence
theorem we get that $u_{\operatorname{av},\varepsilon}\to u$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$.

Now the convergence
$\frac{u_{\operatorname{av},\varepsilon}}{|u_{\operatorname{av},\varepsilon}|}\to
u$ in $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ follows from Lemma
A.3. ∎

_Proof of Proposition A.2_ We first prove that the space of functions $u\in
C^{\infty}(\mathbb{S}^{1},\mathbb{S}^{1})$ which are constant in a
neighborhood of a fixed point, which we can assume to be $i$, is dense in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

Let $u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. It follows from
Lemma A.4 that it suffices to consider $u\in
C^{0,1}(\mathbb{S}^{1},\mathbb{S}^{1})$. Let $\tilde{u}\in H^{1,2}(D^{2})$ be
the harmonic extension of $u$ in $D^{2}$. Writing the Poisson kernel as

$K(z,e^{i\theta}):=\frac{1-|z|^{2}}{2\pi|z-e^{i\theta}|^{2}}=\frac{1-r^{2}}{2\pi|re^{it}-e^{i\theta}|^{2}},\quad
z=re^{it}\in D^{2},$

from (67), and using that $u\in C^{0,1}(\mathbb{S}^{1},\mathbb{S}^{1})$, we
have

$\begin{split}\tilde{u}(re^{it})&=u(e^{it})+\int_{\mathbb{S}^{1}}K(re^{it},e^{i\theta})\left(u(e^{i\theta})-u(e^{it})\right)d\theta\\\
&=u(e^{it})+O\left(\int_{\partial
D^{2}}K(re^{it},e^{i\theta})|e^{it}-e^{i\theta}|d\theta\right)\\\
&=u(e^{it})+o(1)\quad\text{as }r\to 1.\end{split}$

In particular, $|\tilde{u}|\to 1$ in $D^{2}\setminus B_{1-\varepsilon}$
uniformly as $\varepsilon\to 0$, and $|\nabla\tilde{u}|\in L^{\infty}(D^{2})$.
Hence, one can find $U_{k}\in C^{\infty}(\bar{D}^{2})$ such that $U_{k}\to U$
in $H^{1,2}(D^{2})$, $|\nabla U_{k}|\leq C$, and $|U_{k}|\to 1$ as
$k\to\infty$ uniformly on $\partial D^{2}$.

For $\varepsilon>0$ let $\eta_{\varepsilon}\in C^{\infty}(\mathbb{C})$ be such
that $\eta_{\varepsilon}\equiv 0$ in $B_{\varepsilon}(i)$,
$\eta_{\varepsilon}\equiv 1$ in $B^{c}_{2\varepsilon}(i)$,
$0\leq\eta_{\varepsilon}\leq 1$, and
$|\nabla\eta_{\varepsilon}|\leq\frac{4}{\varepsilon}$. We now define

$U_{k,\varepsilon}\in C^{\infty}(\overline{D^{2}},\mathbb{C}),\quad
U_{k,\varepsilon}:=U_{k}\eta_{\varepsilon}+U_{k}(i)(1-\eta_{\varepsilon}).$

Clearly each $U_{k,\varepsilon}$ is constant in $B_{\varepsilon}(i)$. Notice
that

$\nabla(U_{k,\varepsilon}-U)=(U_{k}(i)-U_{k})\nabla\eta_{\varepsilon}+\nabla
U_{k}(1-\eta_{\varepsilon}),$

which shows that $U_{k,\varepsilon}\to U_{k}$ in $H^{1,2}(D^{2})$ as
$\varepsilon\to 0$. Moreover, as $|\nabla U_{k}|\leq C$, one easily obtains
that $|U_{k,\varepsilon}|=1+o_{k,\varepsilon}(1)$ on $\partial D^{2}$. Thus,
for some suitable choice of $\varepsilon=\varepsilon_{k}$, setting
$u_{k}(z):=\frac{U_{k,\varepsilon_{k}}(z)}{|U_{k,\varepsilon_{k}}(z)|}$ on
$\mathbb{S}^{1}=\partial D^{2}$ we see that $u_{k}\to u$ in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$, thanks to Lemma A.3.

The second part of the proposition follows from the equivalence of the
Dirichlet energies

$\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{|f\circ\Pi_{+}(e^{it})-f\circ\Pi_{+}(e^{is})|^{2}}{|e^{it}-e^{is}|}dtds=\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{|f(x)-f(y)|^{2}}{|x-y|^{2}}dxdy,$

for any $f\in\dot{H}^{\frac{1}{2},2}(\mathbb{R},\mathbb{S}^{1})$, which is
turn follows from (57), (64) and the fact that $\Pi_{+}$ can be extended to
the conformal map $\Psi:\overline{D^{2}}\setminus\\{i\\}\to\mathbb{R}^{2}_{+}$
given in (34). $\square$

Similar to the proof of Proposition A.2, one can approzimate functions in
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$ through their extension
(recalling that $H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{C})$ is the space of
traces of functions in $H^{1,2}(D^{2},\mathbb{C})$), as in the following
lemma, which is a special case of [6, Lemma 8 and Lemma A.13].

######  Proposition A.5 ([6])

Given $U\in H^{1,2}(D^{2},\mathbb{C})$ such that $U(\partial
D^{2})\subset\mathbb{S}^{1}$, there exists a sequence $(U_{j})\subset
C^{\infty}(\overline{D^{2}},\mathbb{C})$ such that $U_{j}(\partial
D^{2})\subset\mathbb{S}^{1}$ and $U_{j}\to U$ in
$H^{1,2}(D^{2},\mathbb{S}^{1})$.

Via a conformal map from $D$ to $\mathbb{R}^{2}_{+}$, one obtains

###### Proposition A.6

Given $U\in H^{1,2}(\mathbb{R}^{2}_{+},\mathbb{C})$ such that
$U(\mathbb{R}\times\\{0\\})\subset\mathbb{S}^{1}$, there exists a sequence
$(U_{j})\subset C^{\infty}(\overline{R^{2}_{+}},\mathbb{C})$ such that
$U_{j}(\mathbb{R}\times\\{0\\})\subset\mathbb{S}^{1}$ and $\nabla
U_{j}\to\nabla U$ in $L^{2}(\mathbb{R}^{2}_{+},\mathbb{S}^{1})$.

###  A.3 Two examples

###### Lemma A.7

Let $\varphi\in C^{0}_{c}(\mathbb{R})\cap
C^{\infty}(\mathbb{R}\setminus\\{0\\})$ be such that $\varphi(x)=0$ for
$|x|\geq\frac{1}{2}$ and for $x=0$, and

$\varphi(x)=\frac{1}{\sqrt{\log\frac{1}{|x|}}},\quad\text{for
}0<|x|\leq\frac{1}{4}.$

Then $\varphi\in H^{\frac{1}{2},2}(\mathbb{R})$, but
$\varphi\chi_{[0,\infty)}\not\in H^{\frac{1}{2},2}(\mathbb{R})$.

###### Proof.

To prove that $\varphi\in H^{\frac{1}{2},2}(\mathbb{R})$ we consider an
extension $\Phi\in C^{\infty}_{c}(\bar{\mathbb{R}}^{2}_{+}\setminus\\{0\\})$
with

$\Phi(x,y)=\frac{1}{\sqrt{\log\frac{1}{|(x,y)|}}},\quad\text{for
}0<|(x,y)|\leq\frac{1}{4},\ y\geq 0.$

Then $\nabla\Phi\in L^{2}(B_{\frac{1}{4}}\cap\mathbb{R}^{2}_{+})$, and
$\Phi\in H^{1,2}(\mathbb{R}^{2}_{+})$ follows by smoothness and compact
support. Then $\varphi\in H^{\frac{1}{2},2}(\mathbb{R})$ by (57), and
$\|\nabla\tilde{\varphi}\|_{L^{2}}\leq\|\nabla\Phi\|_{L^{2}}$, where
$\tilde{\varphi}$ is the Poisson harmonic extension of $\varphi$.

To show that $\varphi\chi_{[0,\infty)}\not\in H^{\frac{1}{2},2}(\mathbb{R})$
we compute for any $0<\varepsilon\leq\frac{1}{4}$

$\begin{split}\int_{x=0}^{\varepsilon}\int_{y=-\varepsilon}^{0}\frac{(\varphi(x)-\varphi(y))^{2}}{(x-y)^{2}}dydx&=\int_{0}^{\varepsilon}\left(\log\frac{1}{x}\right)^{-1}\int_{-\varepsilon}^{0}\frac{dy}{(y-x)^{2}}dx\\\
&=\int_{0}^{\varepsilon}\left(\log\frac{1}{x}\right)^{-1}\left(\frac{1}{x}-\frac{1}{x+\varepsilon}\right)dx=\infty,\end{split}$
(68)

so that $[\varphi\chi_{[0,\infty)}]_{H^{\frac{1}{2},2}}=+\infty$. ∎

###### Proposition A.8

Let $\varphi\in H^{\frac{1}{2},2}(\mathbb{R})$ be defined as in Lemma A.7.
Then $u:=e^{i\varphi\circ\Pi_{-}}\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})\cap
C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$, while the function

$w(e^{i\theta}):=\begin{cases}u(e^{i\theta})&\text{for
}\frac{\pi}{2}\leq\theta\leq\frac{3\pi}{2}\\\
1&\text{for}-\frac{\pi}{2}<\theta<\frac{\pi}{2}\end{cases}$

belongs to $C^{0}(\mathbb{S}^{1},\mathbb{S}^{1})$ but not to
$H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

###### Proof.

To prove that $u\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$, it
suffices to verify that

$[u]_{H^{\frac{1}{2},2}(\mathbb{S}^{1})}^{2}=\frac{1}{\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{|u(e^{i\theta})-u(e^{it})|^{2}}{|e^{i\theta}-e^{it}|^{2}}dtd\theta<\infty,$

which in turn, considering that $u$ is smooth away from $i\in\mathbb{S}^{1}$,
it reduces to verify that

$\int_{\frac{\pi}{2}-\varepsilon}^{\frac{\pi}{2}+\varepsilon}\int_{\frac{\pi}{2}-\varepsilon}^{\frac{\pi}{2}+\varepsilon}\frac{|u(e^{i\theta})-u(e^{it})|^{2}}{|e^{i\theta}-e^{it}|^{2}}dtd\theta<+\infty,$

for some $\varepsilon>0$, which easily follows from

$\int_{-1}^{1}\int_{-1}^{1}\frac{(\varphi(x)-\varphi(\xi))^{2}}{(x-\xi)^{2}}dxd\xi<+\infty$

(Lemma A.7) and the fact that $\Pi_{-}$ is bi-Lipschitz from an arc of
$\mathbb{S}^{1}$ around $i$ and an interval in $\mathbb{R}$ around $0$.

Similarly, from (68) it follows that
$[w]_{H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})}=+\infty$, hence
$w\not\in H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$. The continuity of
$u$ and $w$ follow easily by construction. ∎

###### Lemma A.9

Define $\psi\in C^{\infty}(\mathbb{R}\setminus\\{0\\})$ such that

$\psi(x)=\log\log\frac{1}{|x|}\quad\text{for }0<|x|<\frac{1}{4},$

and $\psi(x)=0$ for $|x|\geq 1$. Then $\psi\in H^{\frac{1}{2},2}(\mathbb{R})$.

###### Proof.

Similar to Lemma A.7 it suffices to construct a function $\Psi\in
C^{\infty}(\bar{\mathbb{R}}^{2})$ with $\Psi(x,y)\equiv 0$ for $|(x,y)|\geq 1$
and

$\Psi(x,y)=\log\log\frac{1}{|(x,y)|},\quad\text{for }|(x,y)|\leq\frac{1}{4},\
y\geq 0.$

Then $\nabla\Psi\in L^{2}(\mathbb{R}^{2}_{+})$, hence
$\psi=\Psi|_{\mathbb{R}\times\\{0\\}}\in H^{\frac{1}{2},2}(\mathbb{R})$. ∎

With the same proof as for Proposition A.8 we obtain:

###### Proposition A.10

Let $\psi$ be as in Lemma A.9. Then $v:=e^{i\psi\circ\Pi_{-}}\in
H^{\frac{1}{2},2}(\mathbb{S}^{1},\mathbb{S}^{1})$.

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