# Non-degeneracy and quantitative stability of half-harmonic maps from $\R$ to
$\S$

Bin Deng  School of Mathematical Sciences, University of Science and
Technology of China, Hefei, Anhui Province, P.R. China, 230026
<EMAIL_ADDRESS>,  Liming Sun  Academy of Mathematics and Systems
Science, the Chinese Academy of Sciences, Beijing 100190, China.
<EMAIL_ADDRESS>and  Jun-cheng Wei  Department of Mathematics, University of
British Columbia, Vancouver, BC, V6T 1Z2, CA<EMAIL_ADDRESS>
(Date: (Last Typeset))

###### Abstract.

We consider half-harmonic maps from $\mathbb{R}$ (or $\mathbb{S}$) to
$\mathbb{S}$. We prove that all (finite energy) half-harmonic maps are non-
degenerate. In other words, they are integrable critical points of the energy
functional. A full description of the kernel of the linearized operator around
each half-harmonic map is given. The second part of this paper devotes to
studying the quantitative stability of half-harmonic maps. When its degree is
$\pm 1$, we prove that the deviation of any map
$\boldsymbol{u}:\mathbb{R}\to\mathbb{S}$ from Möbius transformations can be
controlled uniformly by
$\|\boldsymbol{u}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}-\text{deg\,}\boldsymbol{u}$.
This result resembles the quantitative rigidity estimate of degree $\pm 1$
harmonic maps $\mathbb{R}^{2}\to\mathbb{S}^{2}$ which is proved recently.
Furthermore, we address the quantitative stability for half-harmonic maps of
higher degree. We prove that if $\boldsymbol{u}$ is already near to a Blaschke
product, then the deviation of $\boldsymbol{u}$ to Blaschke products can be
controlled by
$\|\boldsymbol{u}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}-\text{deg\,}\boldsymbol{u}$.
Additionally, a striking example is given to show that such quantitative
estimate can not be true uniformly for all $\boldsymbol{u}$ of degree 2. We
conjecture similar things happen for harmonic maps
$\mathbb{R}^{2}\to\mathbb{S}^{2}$.

###### Key words and phrases:

Fractional derivative, half-harmonic map, Blaschke product, stability, non-
degeneracy

###### 2020 Mathematics Subject Classification:

Primary 35B38; 35B06. Secondary 58E20

##  1\. Introduction

###  1.1. Motivation and main results

The analysis of critical points of conformal invariant lagrangians has drawn
much attention since 1950, due to their important applications in physics and
geometry. One of the prominent examples is harmonic maps
$\boldsymbol{u}:\Omega\to\mathbb{S}^{n}$, which are critical points of the
Dirichlet energy

$\displaystyle\mathcal{E}(\boldsymbol{u})=\int_{\Omega}|\nabla\boldsymbol{u}|^{2}dx.$
(1.1)

When the domain $\Omega$ is a subset of $\mathbb{R}^{2}$,
$\mathcal{E}(\boldsymbol{u})$ is conformally invariant and this plays a
crucial role in the regularity theory of such maps (see Hélein [21], Rivière
[30] and references therein). The theory has been generalized to even-
dimensional domains whose critical are called poly-harmonic maps.

In the recent years, many authors are interested in the analog of Dirichlet
energy in odd-dimensional cases, for instance, Da Lio [11, 12], Da Lio and
Rivière [13, 14], Millot and Sire [27], Schikorra [31] and the references
therein. In these works, a special but quite interesting case is the so-called
half-harmonic maps from $\mathbb{R}$ into $\mathbb{S}$ which can be defined as
critical points of the following line energy

$\displaystyle\mathcal{E}(\boldsymbol{u})=\frac{1}{2\pi}\int_{\mathbb{R}}|(-\Delta_{\mathbb{R}})^{\frac{1}{4}}\boldsymbol{u}|^{2}dx.$
(1.2)

The functional $\mathcal{E}$ enjoys invariance under the Möbius group which is
the trace of conformal maps keeping invariant the half-space
$\mathbb{R}^{2}_{+}$. In fact, $\mathcal{E}(\boldsymbol{u})$ coincides with
$\|\boldsymbol{u}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}$, see (2.3) below.
Computing the associated Euler-Lagrange equation for (1.2), it is easy to see
that if $\boldsymbol{u}:\mathbb{R}\to\mathbb{S}$ is a half-harmonic map, then
$\boldsymbol{u}$ satisfies

$\displaystyle(-\Delta)^{\frac{1}{2}}\boldsymbol{u}(x)=\left(\frac{1}{2\pi}P.V.\int_{\mathbb{R}}\frac{|\boldsymbol{u}(x)-\boldsymbol{u}(s)|^{2}}{|x-s|^{2}}ds\right)\boldsymbol{u}(x)\quad\text{
in }\mathbb{R}.$ (1.3)

Fundamental regularity of half-harmonic maps has been obtained in [13, 14]. A
complete classification has been known by [27, 12] (cf. Theorem 2.3 below).
Associating111Throughout this paper, bold font $\boldsymbol{u}$ denotes a
vector-valued map, while $u$ means a complex-valued map.
$\boldsymbol{u}=(u_{1},u_{2})$ to a complex function
$u=u_{1}+\textup{i\,}u_{2}$, all half-harmonic maps consist of the following
products and their complex conjugates

$\displaystyle\psi_{\vartheta,\vec{\alpha}}:=e^{\textup{i\,}\vartheta}\prod_{k=1}^{d}\frac{x-\alpha_{k}}{x-\bar{\alpha}_{k}}$
(1.4)

where $\vartheta\in\mathbb{S}$, $\vec{\alpha}=(\alpha_{1},\cdots,\alpha_{d})$
with $\alpha_{k}=x_{k}+\textup{i\,}\lambda_{k}$ lies in the upper half plane
$\mathbb{H}$, and $d=\text{deg\,}\boldsymbol{u}$ (see (2.19) for its
definition). Modulo a Cayley transformation, the above expressions are
equivalent to Blaschke products or their conjugates.

Apart from the strong analogy to harmonic maps on $\mathbb{R}^{2}$, half-
harmonic maps have intricate connections to minimal surfaces with free
boundary, for instance see [17, 25, 27, 23]. On the other hand, in recent
years, several papers were devoted to the study of the fractional Sobolev
space $H^{1/2}$ with values into the circle, in particular in the framework of
the Ginzburg-Landau model, see the paper Mironescu and Pisante [28] and
reference therein. The simplest of such spaces is
$H^{1/2}\left(\mathbb{R};\mathbb{S}\right)$.

The main purpose of this paper is twofold: First, we prove that each (finite
energy)222According to [26], we do have some maps
$\boldsymbol{u}:\mathbb{R}\to\mathbb{S}$ satisfies (1.3) but has infinite line
energy. We do not study it here. half-harmonic map is non-degenerate by
characterising the kernel of the linearized operator around each half-harmonic
map. Second, we study the quantitative stability estimates near each half-
harmonic map. The non-degeneracy and stability are crucial to the half-
harmonic map heat flow, which is an analogy of harmonic map heat flow. There
are vibrant researches along this direction by Sire et al. [33], Schikorra et
al. [32], Wettstein [36]. An interesting conjecture in [33] states that half-
harmonic map heat flow only blow-ups in infinite time, which is quite
different from what we know about harmonic map heat flow.

Evidently, differentiating (1.4) with $\vartheta$, $\textup{Re\,}\alpha_{k}$,
$\textup{Im\,}\alpha_{k}$, for $k=1,\cdots,d$ respectively, generates kernel
maps for the linearized operators $L_{\psi_{\vartheta,\vec{\alpha}}}$ as

$\displaystyle\begin{split}L_{\psi_{\vartheta,\vec{\alpha}}}(\boldsymbol{v})=(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{v}(x)&-\left(\frac{1}{2\pi}P.V.\int_{\mathbb{R}}\frac{|\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}(x)-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}(s)|^{2}}{|x-s|^{2}}ds\right)\boldsymbol{v}(x)\\\
&-\left(\frac{1}{\pi}\int_{\mathbb{R}}\frac{(\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}(x)-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}(y))\cdot(\boldsymbol{v}(x)-\boldsymbol{v}(y))}{|x-y|^{2}}dy\right)\boldsymbol{u}(x)\end{split}$
(1.5)

where $\boldsymbol{v}\in\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$ with
$\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\cdot\boldsymbol{v}=0$. Conversely,
if the kernels maps of $L_{\psi_{\vartheta,\vec{\alpha}}}$ are all generated
by differentiating nearby half-harmonic maps, then we call
$\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}$ is non-degenerate. Such property
is also known as integrability in the context of minimal surfaces [1] and
harmonic maps [20]. The non-degeneracy of harmonic maps from $\mathbb{R}^{2}$
to $\mathbb{S}^{2}$ has been established in [20, 8].

The study to non-degeneracy of half-harmonic maps is initiated by Sire et al.
[34] and Lenzmann and Schikorra [26]. The authors in [34] confirm the case
when $\text{deg\,}\boldsymbol{u}=\pm 1$. The authors in [26] can deal with
very special case of higher degree, more precisely, when
$u=(x-\textup{i\,})^{m}/(x+\textup{i\,})^{m}$. After a stereographic
projection (or Cayley transformation) such $u$ is equivalent to $z^{m}$,
$z\in\mathbb{S}$. Their approach depend on the symmetry of $z^{m}$ crucially.
In the present paper, we completely solve the non-degeneracy for all half-
harmonic maps.

######  Theorem 1.1.

Suppose $u$ is a half-harmonic map $\mathbb{R}\to\mathbb{S}$, then all the
$\dot{H}^{1/2}(\mathbb{R})$ maps in the kernel of $L_{u}$ are generated by
differentiating half-harmonic maps close to $u$. More precisely,
$\text{dim}_{\mathbb{R}}\ker L_{\boldsymbol{u}}=2|\text{deg\,}u|+1$. In
particular, if
$u=e^{\textup{i\,}\vartheta}\prod_{j=1}^{k}(\frac{x-\alpha_{j}}{x-\bar{\alpha}_{j}})^{d_{j}}$
with $\\{\alpha_{j}\\}_{j=1}^{k}$ are distinct and $d_{j}\geq 1$, then

$\displaystyle\ker
L_{u}=\textup{span}_{\mathbb{R}}\left\\{1,\textup{Re\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}},\textup{Im\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}}:s=1,\cdots,d_{j},j=1,\cdots,k.\right\\}\textup{i\,}u.$

The second part of this paper deals with the quantitative stability of half-
harmonic maps. To describe it, we note that half-harmonic maps achieve the
minimum of $\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$ norm in its homotopy class.
Namely

######  Theorem 1.2.

Suppose that $u\in\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$. Then
$\|u\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\geq|\text{deg\,}u|$. If the equality
holds, then $u$, or its complex conjugate, must be the form of (1.4).

A natural question is that whether the discrepancy
$\|u\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}-|\text{deg\,}u|$ can control
quantitatively the difference of $u$ from the half-harmonic maps. Naively, one
expects a linear control as

$\displaystyle\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{d}}\|u-\psi_{\vartheta,\vec{\alpha}}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\leq
C\left(\|u\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}-|\text{deg\,}u|\right)$ (1.6)

where $C$ is independent of $u$.

Such type of question has been raised for many other topics. For instance,
Brezis and Lieb [5] ask a similar question of the classical Sobolev inequality
on $\mathbb{R}^{n}$. Later Bianchi and Egnell [4] obtain a quantitative
stability estimate in the spirit of (1.6). Fusco et al. [18] prove a sharp
quantitative stability about isoperimetric inequality. To authors’ knowledge,
other types of quantitative stability estimates include (but not limited to)
[16, 24, 2]. Reader can see their papers and reference therein.

Recently Bernand-Mantel et al. [3] prove a quantitative stability for degree
$\pm 1$ harmonic maps from $\mathbb{R}^{2}\to\mathbb{S}^{2}$ similar to (1.6),
whose proof is simplified by [22, 35]. Due to the strong analogy between
harmonic maps and half-harmonic ones, their works inspire us to prove the
following theorem. Denote

$\displaystyle D(u)=\|u\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}-|\text{deg\,}u|.$
(1.7)

######  Theorem 1.3.

For $u\in\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$ with $\text{deg\,}u=1$, there
exists an $\alpha\in\mathbb{H}$ and $\vartheta\in\mathbb{S}$ such that

$\displaystyle\|u-\psi_{\vartheta,\alpha}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\leq
36D(u).$ (1.8)

If $\text{deg\,}u=-1$, then the above statement holds with $\bar{u}$.

The above theorem leaves us an intriguing question for half-harmonic maps with
higher degree. The answer to this could shed some light to harmonic maps with
higher degree. We find that there are some fundamental differences between the
case degree $\pm 1$ and higher degree. For instance, in degree 1,
$\psi_{\vartheta,\alpha}$ is equivalent to $\psi_{0,\textup{i\,}}$ after some
harmless rotation and Möbius transformation of $\mathbb{H}$, while both sides
of (1.8) is invariant under these operations. Essentially, the stability
estimates near $\psi_{\vartheta,\alpha}$ is equivalent to that of
$\psi_{0,\textup{i\,}}$. Thus we have a uniform constant in (1.8). However, in
higher degree we do not have such equivalence. For instance, consider
$\psi_{\vartheta,(\alpha_{1},\alpha_{2})}$ in degree 2. One can use Möbius
transformation to bring $\alpha_{1}$ to i , but there is no control of
$\alpha_{2}$, which might be very near to the boundary of $\mathbb{H}$.
Indeed, we prove the following dichotomy for $\text{deg\,}=2$.

######  Theorem 1.4.

For any $M>0$, there exists $u\in\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$ with
$\text{deg\,}u=2$ such that

$\displaystyle\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{2}}\|u-\psi_{\vartheta,\vec{\alpha}}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\geq
M\left(\|u\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}-2\right).$ (1.9)

The function $u$ we choose is a perturbation of $\psi_{0,\vec{\alpha}}$ with
$\alpha_{1}=j^{2}+\textup{i\,}$ and $\alpha_{2}=-j^{2}+\textup{i\,}$ with
$j\to\infty$. This exactly captures the dilemma in higher degree mentioned
above. This example shows that one should not hope the stability for higher
degree as simple as (1.6). Nevertheless, we can prove a local version of
quantitative stability. Namely, if $u$ is already sufficiently close to some
half-harmonic map, then (1.6) still holds.

######  Theorem 1.5.

For any compact set $\Omega\Subset\mathbb{H}$, there exists
$\delta_{d,\Omega,\varepsilon}$ with the following significance. Suppose
$u\in\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$ satisfies $\text{deg\,}u=d>0$ and

$\displaystyle\|u-\psi_{\vartheta,\vec{\beta}}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\leq\delta_{d,\Omega,\varepsilon}$
(1.10)

for some $\vartheta\in\mathbb{S}$ and $\vec{\beta}\in\Omega^{d}$. Then there
exists a constant $C_{d,\Omega,\varepsilon}>0$ (independent of $u$),
$\vartheta^{\prime}\in\mathbb{S}$ and
$\vec{\alpha}\in\Omega_{\varepsilon}^{d}\Subset\mathbb{H}^{d}$ such that

$\displaystyle\|u-\psi_{\vartheta^{\prime},\vec{\alpha}}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\leq
C_{d,\Omega,\varepsilon}D(u).$ (1.11)

Here $\Omega_{\varepsilon}=\cup_{p\in\Omega}B_{\varepsilon}(p)$ with
$B_{\varepsilon}(p)$ are open balls in $\mathbb{D}$.

For harmonic maps, one can expect similar things happening here. We conjecture
that there is local version of quantitative stability, while no uniform one as
simple as (1.6) holds for higher degree.

###  1.2. Comments on the proofs

We briefly sketch the main idea behind the proofs of Theorem 1.1-1.5. To show
the non-degeneracy, we start with the harmonic extension (denote as
$\boldsymbol{U}$) of any half-harmonic map $\boldsymbol{u}$ to
$\mathbb{R}^{2}_{+}$. It has been proven by [27, 26] that the Hopf
differential of $\boldsymbol{U}$

$\displaystyle\Phi=(|\partial_{x}\boldsymbol{U}|^{2}-|\partial_{y}\boldsymbol{U}|^{2})-2\textup{i\,}\partial_{x}\boldsymbol{U}\cdot\partial_{y}\boldsymbol{U}\equiv
0.$ (1.12)

This actually implies $U$ must be a holomorphic or anti-holomorphic function
on $\mathbb{H}$. If $v\in\ker L_{u}$, then we anticipate its harmonic
extension $V$ is also holomorphic or anti-holomorphic on $\mathbb{H}$. This is
done by considering

$\displaystyle\tilde{\Phi}=2U_{z}\bar{V}_{z}=\partial_{x}\boldsymbol{U}\cdot\partial_{x}\boldsymbol{V}-\partial_{y}\boldsymbol{U}\cdot\partial_{y}\boldsymbol{V}-\textup{i\,}(\partial_{x}\boldsymbol{U}\cdot\partial_{y}\boldsymbol{V}+\partial_{x}\boldsymbol{V}\cdot\partial_{y}\boldsymbol{U}).$
(1.13)

Note that $\tilde{\Phi}$ can be thought of the derivative of $\Phi$ in (1.12).
One can show that $\tilde{\Phi}\equiv 0$ which implies $V$ is
(anti-)holomorphic on $\mathbb{H}$ when $U$ is (anti-)holomorphic. A crucial
step is defining $W=V/U$. Since $\boldsymbol{u}\cdot\boldsymbol{v}=0$ on
$\mathbb{R}$, then $W$ is purely imaginary on $\partial\mathbb{H}$. By Scharwz
reflection, we can extend $W$ to a meromorphic function on $\mathbb{C}$. One
can show that $W$ has no essential pole at infinity and thus $W$ is a rational
functions. By counting the dimension of admissible rational functions, we get
$\dim_{\mathbb{R}}\ker L_{u}=2|\text{deg\,}u|+1$. This exactly matches the
dimension of parameters generating nearby half-harmonic maps.

The proof of Theorem 1.3 follows closely the approach in [22]. They use
harmonic polynomials to do the computation there, while here we use Fourier
series of functions $\mathbb{S}\to\mathbb{S}$ instead. Thanks to the fact that
the $\dot{H}^{1/2}(\mathbb{S})$ norm and degree of any map
$u\in\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$ can be written explicitly using the
coefficients of Fourier series, the proof here is much shorter than that in
[22]. The proof of Theorem 1.5 is a carefully refinement of the case degree
$\pm 1$ and induction. Non-degeneracy can be used to prove quantitative
stability estimates in some cases. The reason behind this is that the
linearized operator has a spectral gap on the orthogonal space of its kernel.
For instance, one can see [4]. For similar approaches on fractional Sobolev
inequality, one can see [9]. There might be possible to prove Theorem 1.5
using the non-degeneracy result we have shown. Since using Fourier series is
more direct, we did not pursue this direction.

To get an example violating the uniform quantitative stability, we choose to
perturb $\psi_{\vec{\alpha}}$ with $\alpha_{1}=j^{2}+\textup{i\,}$ and
$\alpha_{2}=-j^{2}+\textup{i\,}$. For any $u$ near to $\psi_{\vec{\alpha}}$,
we formally decompose it to the sum of one part in the kernel of linearized
operator at $\psi_{\vec{\alpha}}$ and the other part $u_{\perp}$ in the
orthogonal space. If $u_{\perp}$ is almost orthogonal to the kernel of the
linearized operator at $\psi_{\alpha_{1}}$ or $\psi_{\alpha_{2}}$, then one
still gets (1.6). As $j\to\infty$, $\psi_{\alpha_{1}}$ and $\psi_{\alpha_{2}}$
has very weak interaction. The four elements of $\ker L_{\psi_{\vec{\alpha}}}$
split to two elements of $\ker L_{\psi_{\alpha_{1}}}$ and two elements of
$\ker L_{\psi_{\alpha_{2}}}$. However, the element $1\in\ker
L_{\psi_{\vec{\alpha}}}$, which corresponds to the rotation, can not split.
This gives us some hope to find $u_{\perp}$ which is almost orthogonal to
$\ker L_{\psi_{\vec{\alpha}}}$ but lies in $\ker L_{\psi_{\alpha_{1}}}$ and
$\ker L_{\psi_{\alpha_{2}}}$ approximately. One typical example is that $u$ is
1 near $j^{2}$ and is $-1$ near $-j^{2}$, which are centers of
$\psi_{\alpha_{1}}$ and $\psi_{\alpha_{2}}$ respectively, as constructed in
(5.21). Once realizing this, the only job left is making sure the infimum in
(1.9) can be achieved by $\psi_{\vec{\alpha}}$. This is done by using the
implicit function theorem near $\psi_{\vec{\alpha}}$. We are inspired by the
modulation analysis near Talenti bubbles, for instance see [10].

###  1.3. Structure of the paper

In Section 2, we give a detailed preliminary on
$(-\Delta_{\mathbb{R}})^{\frac{1}{2}}$, $\dot{H}^{1/2}(\mathbb{R})$ and half-
harmonic maps. The equivalence of half-harmonic maps defined on $\mathbb{R}$
and $\mathbb{S}$ is used implicitly in the following sections. In Section 3,
we prove the non-degeneracy of the linearized operator at each half-harmonic
map. We divert to consider the stability from Section 4. There we prove the
quantitative stability for degree $\pm 1$ and a local result for higher
degree. Section 5 devotes to constructing an example losing uniform stability.
Finally, we put some tedious computations in the Appendix which are needed in
Section 5.

##  2\. Preliminary

In this section, we lay some foundations for half-harmonic maps. There are
various perspective to define them.

###  2.1. Formulations on the real line

######  Definition 2.1.

For any $f:\mathbb{R}\to\mathbb{R}$, we define

$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}f(x)=\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{f(x)-f(y)}{|x-y|^{2}}dy.$
(2.1)

We call $f\in\dot{H}^{1/2}(\mathbb{R})$ if

$\displaystyle\|f\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}:=\frac{1}{4\pi^{2}}\iint_{\mathbb{R}\times\mathbb{R}}\frac{|f(x)-f(y)|^{2}}{|x-y|^{2}}dxdy<\infty.$
(2.2)

Suppose $\boldsymbol{u}=(u_{1},u_{2})$ is a map from $\mathbb{R}$ into
$\mathbb{S}$. Here and the following we always assume $\mathbb{S}$ is embedded
in $\mathbb{R}^{2}=\mathbb{C}$. Define the energy of
$\boldsymbol{u}:\mathbb{R}\to\mathbb{S}$ by

$\displaystyle\mathcal{E}(\boldsymbol{u}):=\frac{1}{2\pi}\int_{\mathbb{R}}\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u}dx.$
(2.3)

Using $|\boldsymbol{u}|=1$, it is easy to verify that

$\displaystyle\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u}(x)=\frac{1}{2\pi}P.V.\int_{\mathbb{R}}\frac{|\boldsymbol{u}(x)-\boldsymbol{u}(y)|^{2}}{|x-y|^{2}}dy.$
(2.4)

Consequently

$\displaystyle\mathcal{E}(\boldsymbol{u})=\|\boldsymbol{u}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=\|u_{1}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}+\|u_{2}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}.$
(2.5)

The functional $\mathcal{E}$ is invariant under the trace of conformal maps
keeping invariant the half-space $\mathbb{R}^{2}_{+}$: the Möbius group.

The critical points of $\mathcal{E}$ are called half-harmonic maps.

######  Definition 2.2.

A map $\boldsymbol{u}\in\dot{H}^{1/2}(\mathbb{R},\mathbb{S})$ is called a weak
half-harmonic map if for any
$\boldsymbol{\phi}\in\dot{H}^{1/2}\left(\mathbb{R},\mathbb{R}^{2}\right)\cap$
$L^{\infty}\left(\mathbb{R},\mathbb{R}^{2}\right)$ there holds

$\frac{d}{dt}\bigg{|}_{t=0}\mathcal{E}\left(\frac{\boldsymbol{u}+t\boldsymbol{\phi}}{|\boldsymbol{u}+t\boldsymbol{\phi}|}\right)=0.$

Computing the associated Euler-Lagrange equation of (2.3), we obtain that if
$\boldsymbol{u}:\mathbb{R}\rightarrow\mathbb{S}$ is a half-harmonic map, then
$\boldsymbol{u}$ satisfies the following equation:

$\displaystyle\left(-\Delta_{\mathbb{R}}\right)^{\frac{1}{2}}\boldsymbol{u}(x)=\left(\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u}\right)\boldsymbol{u}(x)\quad\text{
in }\mathbb{R}.$ (2.6)

All the half-harmonic maps (with finite energy) have been classified by Millot
and Sire [27]. In the sequel, we identify $\mathbb{R}^{2}$ with the complex
plane $\mathbb{C}$ writing $z=x_{1}+\textup{i\,}x_{2}$. We shall write
$u=u_{1}+\textup{i\,}u_{2}$ for any map $\boldsymbol{u}=(u_{1},u_{2})$.

######  Theorem 2.3 (Millot and Sire [27]).

Let $u\in\dot{H}^{1/2}\left(\mathbb{R};\mathbb{S}\right)$ be a non-constant
half-harmonic map into $\mathbb{S}$. Let $U$ be the harmonic extension of $u$
to $\mathbb{R}_{+}^{2}$. There exist some
$d\in\mathbb{N},\vartheta\in\mathbb{R}$ and
$\left\\{\alpha_{k}\right\\}_{k=1}^{d}\subseteq\mathbb{H}=\\{z\in\mathbb{C}:\textup{Im\,}z>0\\}$
such that $U(z)$ or its complex conjugate equals

$\displaystyle\psi_{\vartheta,\vec{\alpha}}:=e^{i\vartheta}\prod_{k=1}^{d}\frac{z-\alpha_{k}}{z-\bar{\alpha}_{k}}.$
(2.7)

Furthermore,

$\displaystyle\|u\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=\frac{1}{2\pi}\int_{\mathbb{R}_{+}^{2}}\left|\nabla
U\right|^{2}dz=d.$ (2.8)

The above theorem says that the Stereographic projection

$\displaystyle\mathcal{S}(x)=\left(\frac{2x}{x^{2}+1},\frac{x^{2}-1}{x^{2}+1}\right):\mathbb{R}\rightarrow\mathbb{S}$
(2.9)

is a half-harmonic map. Actually one can verify it directly. It follows from
(2.1) and some computations that

$\displaystyle(-\Delta)^{\frac{1}{2}}\mathcal{S}(x)=\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{\mathcal{S}(x)-\mathcal{S}(y)}{|x-y|^{2}}dy$
$\displaystyle=\left(\frac{4x}{\left(1+x^{2}\right)^{2}},\frac{2\left(x^{2}-1\right)}{\left(1+x^{2}\right)^{2}}\right),$
(2.10)
$\displaystyle\frac{1}{2\pi}\int_{\mathbb{R}}\frac{|\mathcal{S}(x)-\mathcal{S}(y)|^{2}}{|x-y|^{2}}dy$
$\displaystyle=\frac{2}{1+x^{2}}.$ (2.11)

It is easy to verify that $\mathcal{S}$ satisfies (2.6). Therefore it is a
half-harmonic map from $\mathbb{R}\to\mathbb{S}$.

###  2.2. Formulations on the unit circle

Using the Stereographic projection in (2.9), one can reformulate the problem
by the maps from $\mathbb{S}$ to $\mathbb{S}$. To that end, we parametrize
$\mathbb{S}=\\{e^{\textup{i\,}\theta}:\theta\in[0,2\pi)\\}$. Let
$\mathcal{S}(x)=e^{\textup{i\,}\theta}$,
$\mathcal{S}(y)=e^{\textup{i\,}\vartheta}$, then

$|e^{\textup{i\,}\theta}-e^{\textup{i\,}\vartheta}|^{2}=|\mathcal{S}(x)-\mathcal{S}(y)|^{2}=\frac{4(x-y)^{2}}{(x^{2}+1)(y^{2}+1)}.$

Using $x=\mathcal{S}^{-1}(\theta):=\mathcal{S}^{-1}(e^{\textup{i\,}\theta})$,
we can write $f(x)$ defined on $\mathbb{R}$ to $\tilde{f}(\theta)$ defined on
$\mathbb{S}$. By the above identity and (2.1), we obtain

$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}f(x)=\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{f(x)-f(y)}{|x-y|^{2}}dy=\frac{2}{1+x^{2}}\frac{1}{\pi}P.V.\int_{\mathbb{S}}\frac{\tilde{f}(\theta)-\tilde{f}(\vartheta)}{|e^{\textup{i\,}\theta}-e^{\textup{i\,}\vartheta}|^{2}}d\vartheta.$
(2.12)

This leads to the definition of $(-\Delta_{\mathbb{S}})^{\frac{1}{2}}$.

######  Definition 2.4.

For any $\tilde{f}:\mathbb{S}\to\mathbb{R}$, we define the

$\displaystyle(-\Delta_{\mathbb{S}})^{\frac{1}{2}}\tilde{f}(\theta):=\frac{1+x^{2}}{2}(-\Delta_{\mathbb{R}})^{\frac{1}{2}}f(x)=\frac{1}{\pi}P.V.\int_{\mathbb{S}}\frac{\tilde{f}(\theta)-\tilde{f}(\vartheta)}{|e^{\textup{i\,}\theta}-e^{\textup{i\,}\vartheta}|^{2}}d\vartheta.$
(2.13)

We call $\tilde{f}\in\dot{H}^{1/2}(\mathbb{S})$ if and only if

$\displaystyle\|\tilde{f}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}:=\frac{1}{4\pi^{2}}\iint_{\mathbb{S}\times\mathbb{S}}\frac{|\tilde{f}(\theta)-\tilde{f}(\vartheta)|^{2}}{|e^{\textup{i\,}\theta}-e^{\textup{i\,}\vartheta}|^{2}}d\vartheta
d\theta<\infty.$

Using (2.12), it is easy to see that for any $f:\mathbb{R}\to\mathbb{S}$,

$\displaystyle\|f\|_{\dot{H}^{1/2}(\mathbb{R})}=\|f\circ\mathcal{S}^{-1}\|_{\dot{H}^{1/2}(\mathbb{S})}.$
(2.14)

Therefore $\mathcal{S}$ is an isometric isomorphism of
$\dot{H}^{1/2}(\mathbb{R})$ and $\dot{H}^{1/2}(\mathbb{S})$.

Now for any map $\boldsymbol{u}=(u_{1},u_{2}):\mathbb{S}\to\mathbb{S}$, we
still adopt the notation $u=u_{1}+\textup{i\,}u_{2}$ is a complex-valued
function, we define
$(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u=(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u_{1}+\textup{i\,}(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u_{2}$
and

$\displaystyle\|u\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}:=\|u_{1}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+\|u_{2}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\textup{Re\,}\int_{\mathbb{S}}\bar{u}(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u.$
(2.15)

Here we have used the multiplication of two complex numbers and Re denotes the
real part. One readily derive that
$\|\text{id}_{\mathbb{S}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=1$.

One also define the energy as
$\mathcal{E}(u)=\|u\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}$ and the the critical
points of $\mathcal{E}$ are called half-harmonic maps
$\mathbb{S}\to\mathbb{S}$, which satisfy

$\displaystyle(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u(\theta)=\left(\frac{1}{2\pi}P.V.\int_{\mathbb{S}}\frac{|u(\theta)-u(\vartheta)|^{2}}{|e^{\textup{i\,}\theta}-e^{\textup{i\,}\vartheta}|^{2}}d\vartheta\right)u(\theta)\quad\text{on
}\mathbb{S}.$ (2.16)

The isometry in (2.14) infers the one-to-one correspondence of half-harmonic
maps between $\mathbb{R}\to\mathbb{S}$ and $\mathbb{S}\to\mathbb{S}$ through
the stereographic projection.

One of the great advantages of working on $\mathbb{S}$ is that we have the
Fourier expansion. Namely for any $u:\mathbb{S}\to\mathbb{C}$, we formally
have

$u=\sum_{-\infty}^{\infty}c_{k}e^{\textup{i\,}k\theta}=\sum_{k\in\mathbb{Z}}c_{k}z^{k},\quad\text{where
}c_{k}=\fint_{\mathbb{S}}uz^{-k}d\theta.$

Using the fact that $(-\Delta_{\mathbb{S}})^{\frac{1}{2}}z^{k}=|k|z^{k}$ for
any $k\in\mathbb{Z}$ (for instance, see [34]), we have the following
interpretation of $(-\Delta_{\mathbb{S}})^{\frac{1}{2}}$,

$(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u=\sum_{k\in\mathbb{Z}}|k|c_{k}z^{k},$

consequently (2.15) implies

$\displaystyle\|u\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\sum_{k\in\mathbb{Z}}|k||c_{k}|^{2}.$
(2.17)

Furthermore, if $u$ maps $\mathbb{S}$ to $\mathbb{S}$, then $|u|=1$ is
equivalent to

$\displaystyle\sum_{j\in\mathbb{Z}}|c_{j}|^{2}=1,\quad\sum_{j\in\mathbb{Z}}\bar{c}_{j}c_{j+k}=0,\quad\forall\,k\in\mathbb{Z}.$
(2.18)

Recall that the degree (or winding number) is well defined for
$\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$ by de Monvel-Berthier et al. [15] (see
also Brezis and Nirenberg [6]). More precisely,

$\displaystyle\text{deg\,}u=\frac{1}{2\pi\textup{i\,}}\int_{\mathbb{S}}\bar{u}du=\frac{1}{2\pi}\int_{\mathbb{S}}u\wedge
u_{\theta}d\theta=\sum_{-\infty}^{\infty}k|c_{k}|^{2}.$ (2.19)

Throughout this paper, we use the notation
$(z_{1}+\textup{i\,}z_{2})\wedge(w_{1}+\textup{i\,}w_{2})=z_{1}w_{2}-z_{2}w_{1}$.
From the above equation and (2.17), one can see that the degree is continuous
in $\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$ topology. For
$u\in\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$, we shall define
$\text{deg\,}u=\text{deg\,}u\circ\mathcal{S}^{-1}$. We have the following
simple fact.

######  Lemma 2.5.

If $u,v\in\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$, then
$\text{deg\,}(uv)=\text{deg\,}u+\text{deg\,}v$.

###### Proof.

Since $u,v$ take values in $\mathbb{S}$, then using (2.19) to obtain

$\displaystyle\text{deg\,}(uv)=\frac{1}{2\pi\textup{i\,}}\int_{\mathbb{S}}\bar{u}\bar{v}d(uv)=\frac{1}{2\pi\textup{i\,}}\int_{\mathbb{S}}\bar{u}du+\bar{v}dv=\text{deg\,}u+\text{deg\,}v.$

∎

Define $\mathbb{D}=\\{\xi=x+\textup{i\,}y:|\xi|<1\\}$ and consider
$\mathbb{S}=\partial\mathbb{D}$. Recall that $f\in\dot{H}^{1/2}(\mathbb{S})$
if and only if it is the trace of some function in $\dot{H}^{1}(\mathbb{D})$.
The energy for $u\in\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$ has a tractable
representation

$\displaystyle\|u\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\inf\left\\{\frac{1}{2\pi}\int_{\mathbb{D}}|\nabla
U|^{2}:U\in\dot{H}^{1}(\mathbb{D};\mathbb{D})\text{ with
}\text{Tr\,}U=u\right\\}.$ (2.20)

This infimum is achieved by the harmonic extension of $u$. Since $\partial
U/\partial\nu=(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u$ on $\mathbb{S}$, where
$\nu$ is the outward unit normal of $\partial\mathbb{D}$, then (2.16) is
equivalent to

$\displaystyle\begin{cases}\Delta U=0&\text{in }\mathbb{D},\\\ \frac{\partial
U}{\partial\nu}\wedge U=0&\text{on }\mathbb{S}.\end{cases}$ (2.21)

All half-harmonic maps, as classified in Theorem 2.3, are Blaschke products of
$d$ Möbius transformations of $\mathbb{D}$ or their complex conjugate. This
can seen from Cayley transformation. Recall that Cayley transform is a bi-
holomorphic mapping $\mathfrak{C}:{\mathbb{H}}\rightarrow{\mathbb{D}}$ which
is defined as

$\displaystyle\xi=\mathfrak{C}(z):=\textup{i\,}\frac{z-\textup{i\,}}{z+\textup{i\,}}.$
(2.22)

Here and the following, we use $\xi$ to denote the complex coordinates of
$\mathbb{D}$ and $z$ to denote that of $\mathbb{H}$. Note that the boundary
mapping of $\mathfrak{C}(z)$ is just the stereographic projection (2.9).

By the one-to-one correspondence of half-harmonic maps in
$\dot{H}^{1/2}(\mathbb{R};\mathbb{S})$ and
$\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$, one can obtain all half-harmonic maps
in $\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$ from (2.7). More precisely, it
consists of Blaschke products
$\\{\phi_{\vartheta^{\prime},\vec{a}}:\vartheta^{\prime}\in\mathbb{S},\,\vec{a}=(a_{1},\cdots,a_{d})\in\mathbb{D}^{d}\\}$
and their complex conjugates, where

$\displaystyle\phi_{\vartheta^{\prime},\vec{a}}(\xi):=e^{i\vartheta^{\prime}}\prod_{j=1}^{d}\frac{\xi-
a_{j}}{1-\bar{a}_{j}\xi}.$ (2.23)

The parameters in (2.7) and the above equation are related by

$\displaystyle
a_{j}=\mathfrak{C}(\alpha_{j}),\quad\vartheta^{\prime}=\vartheta+\sum_{j=1}^{d}\theta_{j},\quad
e^{\textup{i\,}\theta_{j}}=\frac{-\alpha_{j}-\textup{i\,}}{1+\textup{i\,}\bar{\alpha}_{j}}.$
(2.24)

Using some integration by parts, the degree of $u$ (see (2.19)) can also be
defined by

$\displaystyle\text{deg\,}u=\frac{1}{\pi}\int_{\mathbb{D}}U_{x}\wedge
U_{y}dxdy.$ (2.25)

Notice the following identity when $U=U^{1}+\textup{i\,}U^{2}$

$\displaystyle|\nabla
U|^{2}=\left(U_{x}^{1}-U_{y}^{2}\right)^{2}+\left(U_{y}^{1}+U_{x}^{2}\right)^{2}+2U_{x}\wedge
U_{y}\geqslant 2U_{x}\wedge U_{y}.$ (2.26)

Since $\phi_{\vartheta,\vec{a}}$ is holomorphic in $\mathbb{D}$,
$\phi_{\vartheta,\vec{a}}$ achieves the identity for the above equation.
Combining this with (2.20) and (2.25), we just proved (2.8) since

$\displaystyle\|\phi_{\vartheta,\vec{a}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\text{deg\,}\phi_{\vartheta,\vec{a}}=d.$
(2.27)

We show the proof of Theorem 1.2 to end this section.

###### Proof of Theorem 1.2.

The inequality is trivial by (2.26). The proof of equality is essentially
contained in [28]. For the reader’s convenience, we include it here. Suppose
$U$ is the harmonic extension of $u$. It follows from
$\|u\|_{\dot{H}^{1/2}(\mathbb{S})}=|\text{deg\,}u|$ and (2.26) that

$U_{x}^{1}=U_{y}^{2}\quad\text{ and }\quad U_{y}^{1}=U^{2}_{x}\quad\text{ in
}L^{2}(\mathbb{D}).$

By Weyl’s lemma, this is equivalent to $U$ being holomorphic from $\mathbb{D}$
to $\mathbb{D}$. Moreover, $|U(z)|\to 1$ uniformly as $|z|\to 1$ (see [6]).
Then the result of Burckel [7] implies all such maps must be Blaschke
products. ∎

##  3\. Non-degeneracy of the linearized operator

Consider the half-harmonic map
$u=u_{1}+\textup{i\,}u_{2}:\mathbb{R}\to\mathbb{S}$, equation (2.6) is
equivalent to

$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}u\wedge u=0.$ (3.1)

The linearized operator at $u$ is

$\displaystyle L_{u}[v]=(-\Delta_{\mathbb{R}})^{\frac{1}{2}}u\wedge
v+(-\Delta_{\mathbb{R}})^{\frac{1}{2}}v\wedge u.$ (3.2)

We define

$\ker
L_{u}=\\{v=v_{1}+\textup{i\,}v_{2}\in\dot{H}^{1/2}(\mathbb{R};\mathbb{C}):L_{u}[v]=0,u_{1}v_{1}+u_{2}v_{2}=0\\}.$

Suppose $V=V_{1}+\textup{i\,}V_{2}$ is the harmonic extension of $v$ to the
upper half space $\mathbb{H}$ by Possion kernel, then $v\in\ker L_{u}$ is
equivalent to

$\displaystyle\begin{cases}\Delta V=0&\text{in }\mathbb{H},\\\
\partial_{y}U\wedge v+\partial_{y}V\wedge u=0&\text{on
}\partial\mathbb{H}.\end{cases}$ (3.3)

The stereographic projection induces an isometry between
$\dot{H}^{1/2}(\mathbb{R})$ and $\dot{H}^{1/2}(\mathbb{S})$. For a half-
harmonic map $u:\mathbb{S}\to\mathbb{S}$, one can also define the linearized
operator

$\displaystyle\tilde{L}_{u}[v]=(-\Delta_{\mathbb{S}})^{\frac{1}{2}}u\wedge
v+(-\Delta_{\mathbb{S}})^{\frac{1}{2}}v\wedge u$ (3.4)

and
$\ker\tilde{L}_{u}=\\{v\in\dot{H}^{1/2}(\mathbb{S};\mathbb{C}):\tilde{L}_{u}[v]=0,\textup{Re\,}u\bar{v}=0\\}$.

######  Lemma 3.1.

For any half-harmonic map $u:\mathbb{R}\to\mathbb{S}$, let
$\tilde{u}=u\circ\mathcal{S}^{-1}$, we have

$L_{u}=J_{\mathcal{S}}(\tilde{L}_{\tilde{u}}\circ\mathcal{S}^{-1})$

where $J_{\mathcal{S}}=(1-\sin\theta)$ is the Jacobian of $\mathcal{S}$.
Consequently $\text{dim}_{\mathbb{R}}\ker
L_{u}=\text{dim}_{\mathbb{R}}\ker\tilde{L}_{\tilde{u}}$.

###### Proof.

It is easy to show that the Jacobian of $\mathcal{S}$ is

$J_{\mathcal{S}}=\frac{2}{x^{2}+1}=1-\sin\theta.$

Recall from (2.1) that

$(-\Delta_{\mathbb{R}})^{\frac{1}{2}}u(\mathcal{S}(\theta))=(1-\sin\theta){(-\Delta_{\mathbb{S}})^{\frac{1}{2}}\tilde{u}(\theta)}.$

For any $v\in\dot{H}^{1/2}(\mathbb{R};\mathbb{C})$, denote
$\tilde{v}=v\circ\mathcal{S}^{-1}\in\dot{H}^{1/2}(\mathbb{S};\mathbb{C})$,
then

$L_{u}[v](x)=(-\Delta_{\mathbb{R}})^{\frac{1}{2}}u\wedge
v+(-\Delta_{\mathbb{R}})^{\frac{1}{2}}v\wedge
u={(1-\sin\theta)}\tilde{L}_{\tilde{u}}[\tilde{v}].$

∎

The operator $L_{u}$ (or $\tilde{L}_{u}$) arise naturally in the linearization
of the energy functional $\mathcal{E}$. To see that, let us use
$\boldsymbol{u}^{\perp}$ denote the vector rotating $\boldsymbol{u}$
counterclockwise by $\pi/2$. Any variation $\boldsymbol{\phi}$ can be written
as $\tilde{h}\boldsymbol{u}+h\boldsymbol{u}^{\perp}$ for some real-valued
function $\tilde{h}$ and $h$ in $\dot{H}^{1/2}(\mathbb{R})$.

######  Lemma 3.2.

Suppose that $\boldsymbol{u}$ is a half-harmonic map $\mathbb{R}\to\mathbb{S}$
and $\boldsymbol{\phi}=\tilde{h}\boldsymbol{u}+h\boldsymbol{u}^{\perp}$ is a
variation satisfying $|\boldsymbol{u}+\boldsymbol{\phi}|=1$. Then

$\displaystyle\mathcal{E}(\boldsymbol{u}+\boldsymbol{\phi})=\mathcal{E}(\boldsymbol{u})+\int_{\mathbb{R}}h\mathcal{L}_{\boldsymbol{u}}[h]+O(h^{3})$
(3.5)

where the operator $\mathcal{L}_{\boldsymbol{u}}$ is given by

$\displaystyle\mathcal{L}_{\boldsymbol{u}}=(-\Delta_{\mathbb{R}})^{\frac{1}{2}}-(\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u})+R$
(3.6)

with the integral operator

$\displaystyle(Rf)(x)=\frac{1}{2\pi}\int_{\mathbb{R}}\frac{|\boldsymbol{u}(x)-\boldsymbol{u}(y)|^{2}}{|x-y|^{2}}f(y)dy.$
(3.7)

###### Proof.

The constraint $|\boldsymbol{u}+\boldsymbol{\phi}|^{2}=1$ a.e. implies that
$2\tilde{h}+\tilde{h}^{2}+h^{2}=0$. Then
$\tilde{h}=-\frac{1}{2}h^{2}+O(h^{3})$. We have

$\displaystyle\begin{split}\mathcal{E}(\boldsymbol{u}+\boldsymbol{\phi})=&\
\mathcal{E}(\boldsymbol{u})+2\int_{\mathbb{R}}\boldsymbol{\phi}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u}+\int_{\mathbb{R}}\boldsymbol{\phi}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\phi}\\\
=&\
\mathcal{E}(\boldsymbol{u})-\int_{\mathbb{R}}(\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u})h^{2}+\int_{\mathbb{R}}\boldsymbol{\phi}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\phi}+O(h^{3}).\end{split}$
(3.8)

It follows from (2.1) that

$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(h\boldsymbol{u}^{\perp})(x)=((-\Delta_{\mathbb{R}})^{\frac{1}{2}}h(x))\boldsymbol{u}^{\perp}(x)+\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{\boldsymbol{u}^{\perp}(x)-\boldsymbol{u}^{\perp}(y)}{|x-y|^{2}}h(y)dy.$
(3.9)

Using
$\boldsymbol{u}^{\perp}(x)\cdot(\boldsymbol{u}^{\perp}(x)-\boldsymbol{u}^{\perp}(y))=1-\boldsymbol{u}(x)\cdot\boldsymbol{u}(y)=\frac{1}{2}|\boldsymbol{u}(x)-\boldsymbol{u}(y)|^{2}$,
we conclude that

$\displaystyle\int_{\mathbb{R}}\boldsymbol{\phi}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\phi}=$
$\displaystyle\int_{\mathbb{R}}(\tilde{h}\boldsymbol{u}+h\boldsymbol{u}^{\perp})\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(\tilde{h}\boldsymbol{u}+h\boldsymbol{u}^{\perp})$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}h\boldsymbol{u}^{\perp}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(h\boldsymbol{u}^{\perp})+O(h^{3})$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}h(-\Delta_{\mathbb{R}})^{\frac{1}{2}}h+\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}\frac{|\boldsymbol{u}(x)-\boldsymbol{u}(y)|^{2}}{|x-y|^{2}}h(x)h(y)dxdy+O(h^{3}).$

Plugging it into (3.8), we get (3.5). ∎

For any $v=v_{1}+\textup{i\,}v_{2}\in\dot{H}^{1/2}(\mathbb{R};\mathbb{C})$
satisfying $u_{1}v_{1}+u_{2}v_{2}=0$, we can write $v=h\textup{i\,}u$ for some
real-valued function $h$ on $\mathbb{R}$. Then one can verify that

$\displaystyle\begin{split}&L_{u}[v]=(-\Delta_{\mathbb{R}})^{\frac{1}{2}}u\wedge(h\textup{i\,}u)+(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(h\textup{i\,}u)\wedge
u\\\
=&[(-\Delta_{\mathbb{R}})^{\frac{1}{2}}u\wedge\textup{i\,}u]h-(-\Delta_{\mathbb{R}})^{\frac{1}{2}}h(\textup{i\,}u\wedge
u)+\frac{1}{\pi}P.V.\int_{\mathbb{R}}\frac{(\textup{i\,}u(x)-\textup{i\,}u(y))\wedge
u(x)}{|x-y|^{2}}h(y)dy\\\
=&(\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{u})h-(-\Delta_{\mathbb{R}})^{\frac{1}{2}}h-Rh\\\
=&-\mathcal{L}_{\boldsymbol{u}}[h].\end{split}$

######  Lemma 3.3.

Suppose $u$ is a half-harmonic map from $\mathbb{R}$ to $\mathbb{S}$, then
$\text{dim}_{\mathbb{R}}\ker L_{u}\geq 2|\text{deg\,}u|+1$. More precisely,
for instance, if
$u=e^{\textup{i\,}\vartheta}\prod_{j=1}^{k}(\frac{x-\alpha_{j}}{x-\bar{\alpha}_{j}})^{d_{j}}$
with $\\{\alpha_{j}\\}_{j=1}^{k}$ are distinct and $d_{j}\geq 1$, then

$\displaystyle\ker
L_{u}\supset\textup{span}_{\mathbb{R}}\left\\{1,\textup{Re\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}},\textup{Im\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}}:s=1,\cdots,d_{j},j=1,\cdots,k.\right\\}\textup{i\,}u.$

###### Proof.

Assume $d=\text{deg\,}u>0$. It is known that $u$ takes form (1.4) if and only
if there exists $c_{0},\cdots,c_{d-1}\in\mathbb{C}$ and
$\vartheta\in\mathbb{S}$ such that

$\displaystyle
u=e^{\textup{i\,}\vartheta}\frac{x^{d}+c_{d-1}x^{d-1}+\cdots+c_{1}x+c_{0}}{x^{d}+\bar{c}_{d-1}x^{d-1}+\cdots+\bar{c}_{1}x+\bar{c}_{0}}$
(3.10)

with $x^{d}+c_{d-1}x^{d-1}+\cdots+c_{1}x+c_{0}$ has zeros all in $\mathbb{H}$.
This fact comes from the theorem 3.3.2 in [19] (page 43) and conformal
equivalence between $\mathbb{D}$ and $\mathbb{H}$.

It is clear that changing the parameters $\vartheta,c_{d-1},\cdots,c_{0}$
(complex numbers) of $u$ continuously yields a family of half-harmonic maps.
Therefore, it generates kernel maps for the linearized operators $L_{u}$.
Taking derivatives of $u$ on $\vartheta$, we get
$\\{\textup{i\,}u\\}\subset\ker L_{u}$. It is easy to see that for
$l=0,\cdots,d-1$,

$\displaystyle\partial_{c_{j}}u=e^{\textup{i\,}\vartheta}\frac{x^{l}}{x^{d}+\bar{c}_{d-1}x^{d-1}+\cdots+\bar{c}_{1}x+\bar{c}_{0}}=\frac{x^{l}u}{\prod_{j=1}^{k}(x-\alpha_{j})^{d_{j}}},$
(3.11)
$\displaystyle\partial_{\bar{c}_{j}}u=\frac{-x^{l}u}{x^{d}+\bar{c}_{d-1}x^{d-1}+\cdots+\bar{c}_{1}x+\bar{c}_{0}}=\frac{-x^{l}u}{\prod_{j=1}^{k}(x-\bar{\alpha}_{j})^{d_{j}}}.$
(3.12)

Note that $x\in\mathbb{R}$. Taking derivative of $u$ on the real part of
$c_{j}$, we get

$\displaystyle\begin{split}\partial_{c_{j}}u+\partial_{\bar{c}_{j}}u=2\textup{i\,}u\
\text{Im\,}\frac{x^{l}}{\prod_{j=1}^{k}(x-\bar{\alpha}_{j})^{d_{j}}}.\end{split}$
(3.13)

Taking derivative of $u$ on the imaginary part of $c_{j}$, we get

$\displaystyle\textup{i\,}(\partial_{c_{j}}u-\partial_{\bar{c}_{j}}u)=2\textup{i\,}u\
\textup{Re\,}\frac{x^{l}}{\prod_{j=1}^{k}(x-\bar{\alpha}_{j})^{d_{j}}}.$
(3.14)

Therefore, the $\mathbb{R}$-linear combination of them must belong to $\ker
L_{u}$.

$\displaystyle\begin{split}\ker L_{u}\supset&\
\text{span}_{\mathbb{R}}\left\\{1,\textup{Re\,}\frac{x^{l}}{\prod_{j=1}^{k}(x-\bar{\alpha}_{j})^{d_{j}}},\text{Im\,}\frac{x^{l}}{\prod_{j=1}^{k}(x-\bar{\alpha}_{j})^{d_{j}}}:l=0,\cdots,d-1\right\\}\textup{i\,}u\\\
=&\
\textup{span}_{\mathbb{R}}\left\\{1,\textup{Re\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}},\textup{Im\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}}:s=1,\cdots,d_{j},j=1,\cdots,k.\right\\}\textup{i\,}u.\end{split}$

Obviously they are all $\mathbb{R}$-linearly independent, therefore
$\text{dim}_{\mathbb{R}}\ker L_{u}\geq 2|d|+1$.

If $d=0$, then obviously $\\{\textup{i\,}u\\}\subset\ker L_{u}$ and thus
$\dim_{\mathbb{R}}\ker L_{u}\geq 1$. If $d<0$, one can prove similarly as
above by working on the conjugate of $u$. ∎

Suppose that $\boldsymbol{V}=(V_{1},V_{2})$. We abuse the notation
$V=V_{1}+\textup{i\,}V_{2}$ and denote it a complex-valued function defined on
upper half plane $\mathbb{H}$. We also adopt the notation
$\partial_{z}=\frac{1}{2}(\partial_{x}-\textup{i\,}\partial_{y})$,
$\partial_{\bar{z}}=\frac{1}{2}(\partial_{x}+\textup{i\,}\partial_{y})$.

######  Lemma 3.4.

Suppose $u$ is half-harmonic map from $\mathbb{R}$ to $\mathbb{S}$. If
$v\in\ker L_{u}$ with $\text{deg\,}u>0$ ($\text{deg\,}u<0$), then $V$ is
holomorphic (anti-holomorphic) in $\mathbb{H}$. Moreover, $V$ can be extended
to a meromorphic (anti-meromorphic) function with possible poles at poles of
$U$. In addition, $V\circ\mathfrak{C}^{-1}$ is smooth on
$\overline{\mathbb{D}}$.

###### Proof.

We just prove the case $\text{deg\,}u>0$. Define the Hopf differential

$\displaystyle\Phi=2U_{z}\bar{V}_{z}=\partial_{x}\boldsymbol{U}\cdot\partial_{x}\boldsymbol{V}-\partial_{y}\boldsymbol{U}\cdot\partial_{y}\boldsymbol{V}-\textup{i\,}(\partial_{x}\boldsymbol{U}\cdot\partial_{y}\boldsymbol{V}+\partial_{x}\boldsymbol{V}\cdot\partial_{y}\boldsymbol{U}).$
(3.15)

Here in the middle we are using complex multiplication and
$\bar{V}_{z}=\partial_{z}\bar{V}$. Since $U$ is holomorphic, then

$\Phi_{\bar{z}}=2U_{z\bar{z}}\bar{V}_{z}+2U_{z}\bar{V}_{z\bar{z}}=2U_{z}\overline{V_{z\bar{z}}}=\frac{1}{2}U_{z}\overline{\Delta
V}=0.$

Therefore $\Phi$ is holomorphic in $\mathbb{H}$. We claim that
$\text{Im}\,\Phi=0$ on $\partial\mathbb{H}$. To see that, the boundary
condition in (3.3) means

$\displaystyle 0=\partial_{y}V\wedge
u+(\boldsymbol{u}\cdot\partial_{y}\boldsymbol{U})u\wedge
v=(\partial_{y}V-(\boldsymbol{u}\cdot\partial_{y}\boldsymbol{U})v)\wedge
u\quad\text{on }\partial\mathbb{H},$ (3.16)

which means

$\displaystyle\partial_{y}V=(\boldsymbol{u}\cdot\partial_{y}\boldsymbol{U})v+(\boldsymbol{u}\cdot\partial_{y}\boldsymbol{V})u,\quad\text{
on }\partial\mathbb{H}.$ (3.17)

Since $\partial_{x}\boldsymbol{u}\cdot\boldsymbol{u}=0$ and
$\partial_{x}\boldsymbol{u}\cdot\boldsymbol{v}+\boldsymbol{u}\cdot\partial_{x}\boldsymbol{v}=0$
on $\partial\mathbb{H}$, then

$\displaystyle\partial_{x}\boldsymbol{u}\cdot\partial_{y}\boldsymbol{V}+\partial_{x}\boldsymbol{v}\cdot\partial_{y}\boldsymbol{U}=(\boldsymbol{u}\cdot\partial_{y}\boldsymbol{U})\partial_{x}\boldsymbol{u}\cdot\boldsymbol{v}+(\boldsymbol{u}\cdot\partial_{y}\boldsymbol{U})\partial_{x}\boldsymbol{v}\cdot\boldsymbol{u}=0.$
(3.18)

Thus we have shown that $g(z):=\text{Im}\,\Phi$ vanishes on
$\partial\mathbb{H}$. By odd reflection across $\partial\mathbb{H}$, we can
extend the harmonic function $g$ to all of $\mathbb{C}$. However, since $g$ is
harmonic and $g\in L^{1}(\mathbb{R}^{2}_{+})$ because
$U,V\in\dot{H}^{1}(\mathbb{R}^{2}_{+})$, we conclude that $g\equiv 0$ on
$\mathbb{C}$. Thus $\Phi$ is real-valued and holomorphic, which implies that
$\Phi$ is constant. Since $\Phi\in L^{1}(\mathbb{R}^{2}_{+})$, we deduce that
$\Phi(z)\equiv 0$.

Now we have $\Phi=2U_{z}\bar{V}_{z}=0$. Since $U_{z}$ only have isolated zeros
in $\mathbb{H}$, then $\bar{V}_{z}=0$. Consequently, $V$ is holomorphic.

We shall rewrite $U$ as

$\displaystyle
U(z)=e^{\textup{i\,}\vartheta}\prod_{j=1}^{k}\left(\frac{z-\alpha_{j}}{z-\bar{\alpha}_{j}}\right)^{d_{j}}$
(3.19)

with $\alpha_{j}\in\mathbb{H}$ which are different from each other. Then
$d_{1}+\cdots+d_{k}=d$. Now define $W=V/U$. Then the previous argument implies
that $W$ is a meromorphic function on $\mathbb{H}$ with possible poles at
$\\{\alpha_{1},\cdots,\alpha_{k}\\}$. The orthogonality condition says that
$u_{1}v_{1}+u_{2}v_{2}=0$ on $\partial\mathbb{H}$. Therefore the real part of
$W$ vanishes on $\partial\mathbb{H}$ because

$\displaystyle
W=\frac{v_{1}+\textup{i\,}v_{2}}{u_{1}+\textup{i\,}u_{2}}=\frac{u_{1}v_{1}+u_{2}v_{2}}{u_{1}^{2}+u_{2}^{2}}+\textup{i\,}\frac{u_{1}v_{2}-u_{2}v_{1}}{u_{1}^{2}+u_{2}^{2}}.$
(3.20)

By Schwarz reflection principle, we can extend $W$ to a meromorphic function
on $\mathbb{C}$ (we still denote it as $W$) by $W(z)=-\overline{W(\bar{z})}$
for the lower half plane. Now $W$ has possible poles at
$\\{\alpha_{1},\cdots,\alpha_{k},\bar{\alpha}_{1},\cdots,\bar{\alpha}_{k}\\}$.
The order of $W$ at $\alpha_{j}$ (or $\bar{\alpha}_{j}$) is at most $d_{j}$.
Since $U$ is a meromorphic function on $\mathbb{C}$, then so does $V=WU$. The
poles of $V$ are contained in
$\\{\bar{\alpha}_{1},\cdots,\bar{\alpha}_{k}\\}$.

Note that $\bar{\alpha}_{j}$, $j=1,\cdots,k$, are all away from the
$\partial\mathbb{H}$. Therefore $V$ is holomorphic at any point on
$\partial\mathbb{H}$. Since $\mathfrak{C}$ is a holomorphic on
$\overline{\mathbb{H}}$, then $V\circ\mathfrak{C}^{-1}$ is holomorphic in
$\overline{\mathbb{D}}\setminus\\{\text{i}\\}$. Because
$\dot{H}^{1/2}(\mathbb{R};\mathbb{C})$ is isometric to
$\dot{H}^{1/2}(\mathbb{S};\mathbb{C})$ through the stereographic projection,
then $v\circ\mathfrak{C}^{-1}\in\dot{H}^{1/2}(\mathbb{S};\mathbb{C})$. Thus
$v\circ\mathfrak{C}^{-1}\in\ker\tilde{L}_{u\circ\mathfrak{C}^{-1}}$. However,
we can repeat the above whole process by using another bi-holomorphic mapping
between $\mathbb{H}$ and $\mathbb{D}$, say
$(z-\textup{i\,})/(z+\textup{i\,})$, to show that $V\circ\mathfrak{C}^{-1}$ is
holomorphic in $\overline{\mathbb{D}}\setminus\\{1\\}$. Combining with the
previous statement, we know $V\circ\mathfrak{C}^{-1}$ is holomorphic in
$\overline{\mathbb{D}}$. ∎

Now we can prove the Theorem 1.1.

###### Proof of Theorem 1.1.

We shall assume $\text{deg\,}u>0$ and $U$ takes the form (3.19). Suppose
$v\in\ker L_{u}$ and $V$ is the harmonic extension to $\mathbb{R}^{2}_{+}$.
Define $W=V/U$. Lemma 3.4 implies that $W$ is meromorphic on $\mathbb{C}$ with
possible poles at
$\\{\alpha_{1},\cdots,\alpha_{k},\bar{\alpha}_{1},\cdots,\bar{\alpha}_{k}\\}$.
The order of $W$ at $\alpha_{j}$ (or $\bar{\alpha}_{j}$) is at most $d_{j}$.

We claim that $W$ is bounded at infinity. Indeed, this just follows from the
fact that $V\circ\mathfrak{C}^{-1}$ is holomorphic in $\overline{\mathbb{D}}$
and consequently $V$ is bounded on $\overline{\mathbb{H}}$.

Thus $W$ is a rational function. There exists some polynomial $P(x)$ such that

$\displaystyle W(z)=\frac{P(z)}{Q(z)},\quad
Q(x)=\prod_{j=1}^{k}(z-a_{j})^{d_{j}}(z-\bar{a}_{j})^{d_{j}}.$ (3.21)

The boundedness of $W$ implies $\text{deg\,}P\leq 2d$. Since
$W(z)=-\overline{W(\bar{z})}$ and $Q(z)=\overline{Q(\bar{z})}$, it holds that
$P(z)=-\overline{P(\bar{z})}$. If $P(z)=c_{0}+c_{1}z+\cdots+c_{2d}z^{2d}$,
then one must have $c_{j}=-\bar{c}_{j}$. The dimension of all such polynomials
is $2d+1$. Thus $\dim_{\mathbb{R}}\ker L_{u}\leq 2|\text{deg\,}u|+1$. Applying
Lemma 3.1 and Lemma 3.3, we conclude $\dim_{\mathbb{R}}\ker
L_{u}=2|\text{deg\,}u|+1$. Furthermore, if
$u=e^{\textup{i\,}\vartheta}\prod_{j=1}^{k}(\frac{x-\alpha_{j}}{x-\bar{\alpha}_{j}})^{d_{j}}$
with $\\{\alpha_{j}\\}_{j=1}^{k}$ are distinct and $d_{j}\geq 1$, then

$\displaystyle\ker
L_{u}=\textup{span}_{\mathbb{R}}\left\\{1,\textup{Re\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}},\textup{Im\,}\frac{1}{(x-\bar{\alpha}_{j})^{s}}:s=1,\cdots,d_{j},j=1,\cdots,k.\right\\}\textup{i\,}u.$

∎

######  Remark 3.5.

Translating the above results to half-harmonic map
$\tilde{u}:\mathbb{S}\to\mathbb{S}$ (see (2.23)), we can get

$\displaystyle\ker\tilde{L}_{\tilde{u}}=\textup{span}_{\mathbb{R}}\left\\{1,\textup{
Im\,}\frac{\xi^{j}}{\prod_{i=1}^{d}(\xi-
a_{i})},\textup{Re\,}\frac{\xi^{j}}{\prod_{i=1}^{d}(\xi-
a_{i})}:j=0,\cdots,d-1\right\\}\textup{i\,}\tilde{u}.$

In particular, if $\tilde{u}=\xi^{m}$, then

$\displaystyle\ker\tilde{L}_{\xi^{m}}=\textup{i\,}\xi^{m}\
\textup{span}_{\mathbb{R}}\\{1,\textup{Im\,}\xi^{j-m},\textup{Re\,}\xi^{j-m}:j=0,\cdots,m-1\\}.$
(3.22)

Using $\xi=\cos\theta+\textup{i\,}\sin\theta$ on $\mathbb{S}$, we can rewrite
it as

$\displaystyle\ker\tilde{L}_{\xi^{m}}=\textup{i\,}\xi^{m}\
\textup{span}_{\mathbb{R}}\\{1,\sin j\theta,\cos j\theta:j=1,\cdots,m\\}.$
(3.23)

This recovers the result of Lenzmann and Schikorra [26, Proposition 6.2].

##  4\. Stability of half-harmonic map

We only need to show the Theorem 1.3 for $d=1$. The negative case follows from
taking complex conjugate in the proof. The case $d=\pm 1$ has a flavor of the
main theorem proved by Hirsch and Zemas [22] about the stability of harmonic
maps $\mathbb{S}^{2}\to\mathbb{S}^{2}$ with degree $\pm 1$. We will use
induction to prove a local quantitative stability for higher degree.

Denote the set of all maps in $\dot{H}^{1/2}(\mathbb{S};\mathbb{S})$ with
degree $d$ by

$\displaystyle\mathcal{A}_{\mathbb{S}}^{d}=\\{u\in\dot{H}^{1/2}(\mathbb{S};\mathbb{S}):\text{deg\,}\,u=d\\}.$
(4.1)

######  Lemma 4.1.

For any $u\in\mathcal{A}_{\mathbb{S}}^{d}$ with $|d|\geq 1$, there exists
$a_{0}\in\mathbb{D}$ such that $\fint_{\mathbb{S}}u\circ\phi_{0,a_{0}}=0$.

###### Proof.

Assume first $u\in\dot{H}^{1/2}(\mathbb{S};\mathbb{S})\cap C^{\infty}$. Since
$|\text{deg\,}u|>1$, then $u$ is surjective. We will write
$\phi_{0,a}=\phi_{a}$ for short. Define

$\displaystyle
F(a)=\fint_{\mathbb{S}}u\circ\phi_{-a}:=\frac{1}{2\pi}\int_{\mathbb{S}}u\circ\phi_{-a}.$
(4.2)

One readily knows $F$ is continuous and maps $\mathbb{D}$ to $\mathbb{D}$ and
$F(0)=\fint_{\mathbb{S}}u$. For any $\zeta\in\partial\mathbb{D}=\mathbb{S}$,
one has

$\lim_{r\to 1^{-}}\phi_{-r\zeta}(\theta)=\zeta,\quad\forall\,\theta\neq-\zeta$

and this convergence is uniform on
$\\{a\in\overline{\mathbb{D}}:|a+\zeta|\geq\epsilon\\}$ for any $\epsilon>0$.
This implies that $\lim_{r\to 1^{-}}F(r\zeta)=u(\zeta)\in\mathbb{S}$ uniformly
in $\zeta$. Hence $F$ is a continuous function on $\overline{\mathbb{D}}$.
Moreover, the Lerary-Schauder degree of $F$ with respect to 0 is the same as
the winding number of $u$ (see [29]). By Leray-Schauder degree theory, there
exists $a_{0}\in\mathbb{D}$ such that $F(a_{0})=0$.

In the general case of a map $u\in\mathcal{A}^{1}_{\mathbb{S}}$, one can
approximate $u$ by a sequence of $u_{j}\in C^{\infty}(\mathbb{S};\mathbb{S})$
with the property that

$u_{j}\to u\text{ strongly with
}u\in\dot{H}^{1/2}(\mathbb{S};\mathbb{S})\quad\text{and
}\quad\text{deg\,}u_{j}=\text{deg\,}u.$

Going to a subsequence if necessary, we can assume $u_{j}\to u$ a.e.
$\mathbb{S}$. By the previous paragraph, there exists $a_{j}\in\mathbb{D}$
such that

$\fint_{\mathbb{D}}u_{j}\circ\phi_{a_{j}}=0.$

We must have $|a_{j}|<1-\epsilon_{0}$. Suppose not, then there is a
subsequence, which we still label it as $a_{j}$, converging to some
$-\zeta\in\partial\mathbb{D}$. Then $\phi_{a_{j}}\to\zeta$ a.e on $\mathbb{S}$
and $u_{j}\circ\phi_{a_{j}}\to u(\zeta)$ a.e. on $\mathbb{S}$. So we can use
Dominated Convergence Theorem to infer that

$u(\zeta)=\fint_{\mathbb{S}}u(\zeta)=\lim_{j\to\infty}\fint_{\mathbb{S}}u_{j}\circ\phi_{a_{j}}=0.$

However, since $|u_{j}\circ\phi_{a_{j}}|\equiv 1$,

$|u(\zeta)|=\fint_{\mathbb{S}}|u(\zeta)|=\lim_{j\to\infty}\fint_{\mathbb{S}}|u_{j}\circ\phi_{a_{j}}|=1.$

This is a contradiction. Now we can assume $a_{j}\to a_{0}$ where
$a\in\mathbb{D}$ and consequently

$\fint_{\mathbb{S}}u\circ\phi_{a_{0}}=0.$

∎

###### Proof of Theorem 1.3.

By the Lemma 4.1, it suffices to prove the stability (1.8) for
$u\in\mathcal{A}_{\mathbb{S}}^{1}$ with the additional assumption
$\fint_{\mathbb{S}}u=0$. This is due to the invariance of energy and the
degree under Möbius transformation of $\mathbb{D}$, namely

$\displaystyle\mathcal{E}(u)=\mathcal{E}(u\circ\phi_{a}),\quad\text{deg\,}u\circ\phi_{a}=\text{deg\,}u=1$
(4.3)

holds for any $a\in\mathbb{D}$ and the group structure of Möbius
transformations

$\displaystyle\phi_{b}\circ\phi_{a}=\phi_{\frac{a+b}{1+\bar{a}b}}.$ (4.4)

If $D(u)\geq 1$, then

$\displaystyle D(u)\geq
1=\text{deg\,}u=\|u\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}-D(u),$ (4.5)

which means $\|u\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq 2D(u)$. Consequently
(1.8) holds. Thus for the following, we can assume $D(u)<1$.

Suppose that $u=\sum_{k\in\mathbb{Z}}c_{k}z^{k}$. By the previous step, we can
assume $c_{0}=0$. Since $u\equiv 1$ a.e. $\mathbb{S}$, then

$\displaystyle D(u)=$
$\displaystyle\|u\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}-\fint_{\mathbb{S}}u^{2}=\sum_{k\geq
1}[k-1](|c_{k}|^{2}+|c_{-k}|^{2})$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\sum_{k\geq
2}k(|c_{k}|^{2}+|c_{-k}|^{2})=\frac{1}{2}\|u-c_{1}e^{\textup{i\,}\theta}-c_{-1}e^{-\textup{i\,}\theta}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}.$

That is

$\displaystyle\|u-c_{1}e^{\textup{i\,}\theta}-c_{-1}e^{-\textup{i\,}\theta}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
2D(u).$ (4.6)

Again $|u|\equiv 1$ a.e. $\mathbb{S}$ and (2.18) implies

$\sum_{k\geq 1}|c_{k}|^{2}+|c_{-k}|^{2}=1.$

Combining with the consequence of (2.19), that is

$1=\text{deg\,}u=\sum_{k\geq 1}k(|c_{k}|^{2}-|c_{-k}|^{2}),$

one obtains

$\displaystyle 1-|c_{1}|^{2}\leq\sum_{k\geq 2}(|c_{k}|^{2}+|c_{-k}|^{2})\leq
D(u),$ (4.7) $\displaystyle|c_{-1}|^{2}\leq\frac{1}{2}\sum_{k\geq
2}k(|c_{k}|^{2}-|c_{-k}|^{2})\leq D(u).$ (4.8)

Therefore plugging in (4.7) and (4.8) to (4.6), we obtain

$\displaystyle\|u-e^{\textup{i\,}(\theta+\arg
c_{1})}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}$ $\displaystyle\leq
9\left[2D(u)+\|c_{-1}e^{-\textup{i\,}\theta}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+\|(e^{\textup{i\,}\arg
c_{1}}-c_{1})e^{\textup{i\,}\theta}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}\right]$
$\displaystyle\leq 9[3D(u)+(1-|c_{1}|)^{2}]\leq 9[3D(u)+D(u)^{2}].$

Since we assume $D(u)<1$, then the above inequality implies

$\|u-e^{\textup{i\,}\vartheta}\text{id}_{\mathbb{S}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
36D(u),$

for some $\vartheta$. Therefore (1.8) is established.

∎

Now we plan to prove Theorem 1.5. Before that, let us make some preparations.
First, we can expand $\phi_{\vartheta,\vec{b}}$ to be

$\displaystyle\phi_{\vartheta,\vec{b}}=e^{\textup{i\,}\vartheta}\sum_{k\geq
0}A_{k}(\vec{b})z^{k}$ (4.9)

where $A_{k}(\vec{b})$ is some function for $\vec{b}$ for each $k$. For
instance, for any $a\in\mathbb{D}$,

$\displaystyle\begin{split}\phi_{a}=\frac{z-a}{1-\bar{a}z}=(z-a)\sum_{k\geq
0}\bar{a}^{k}z^{k}=-a+\sum_{k\geq 1}(1-|a|^{2})\bar{a}^{k-1}z^{k}.\end{split}$
(4.10)

Since
$\phi_{\vartheta,\vec{b}}=e^{\textup{i\,}\vartheta}\phi_{b_{1}}\cdots\phi_{b_{d}}$,
one knows $A_{0}(\vec{b})=(-1)^{d}\prod_{j=1}^{d}b_{j}$. For any vector
$\vec{b}\in\mathbb{D}^{d}$ and a set $\Omega\in\mathbb{D}$, we abuse the
notation that $\vec{b}\in\Omega^{d}$ if $b=(b_{1},\cdots,b_{d})$ with
$b_{i}\in\Omega$ for $i=1,\cdots,d$.

######  Lemma 4.2.

Suppose $\Omega\Subset\mathbb{D}$ is compact. There exists a
$\epsilon_{\Omega}>0$ and $N_{\Omega}$ such that, for any Blaschke product
$\phi_{\vartheta,\vec{b}}$ with $\vec{b}\in\Omega^{d}$, we have

$\displaystyle\max_{1\leq i\leq
N_{\Omega}}\\{|A_{i}(\vec{b})|\\}\geq\epsilon_{\Omega}.$ (4.11)

###### Proof.

Let $\epsilon(N)=\sum_{1\leq k\leq N}|A_{k}(\vec{b})|^{2}$. By the assumption,
there exists $\epsilon_{1}=\epsilon_{1}(\Omega)$ such that

$\displaystyle|A_{0}(\vec{b})|^{2}\leq
1-\epsilon_{1},\quad\forall\,\vec{b}\in\Omega^{d}.$

By the property (2.18), we get $\sum_{k\geq 0}|A_{k}(\vec{b})|^{2}=1$. Then

$\displaystyle\sum_{k\geq
N+1}|A_{k}(\vec{b})|^{2}\geq\epsilon_{1}-\epsilon(N).$

Since $\text{deg\,}\phi_{\vartheta,\vec{b}}=d$ and (2.19), we get $\sum_{k\geq
1}k|A_{k}(\vec{b})|^{2}=d$. Then

$\displaystyle\sum_{k\geq N+1}k|A_{k}(\vec{b})|^{2}\leq d-\epsilon(N).$

Compare the above two inequalities, we get

$\displaystyle d-\epsilon(N)>(N+1)(\epsilon_{1}-\epsilon(N)).$

Now we choose $N_{\Omega}$ such that $(N_{\Omega}+1)\epsilon_{1}>d$. One
readily see that

$\displaystyle\epsilon(N_{\Omega})>\frac{1}{N_{\Omega}}[(N_{\Omega}+1)\epsilon_{1}-d].$

This implies (4.11) holds for
$\epsilon_{\Omega}=\left(\epsilon(N_{\Omega})/N_{\Omega}\right)^{1/2}$. ∎

######  Proposition 4.3.

Suppose $\Omega\Subset\mathbb{D}$ is compact. There exists a constant
$C_{\Omega}$ such that, for any $u\in\mathcal{A}_{\mathbb{S}}^{d}$ and
$\vec{b}\in\Omega^{d}$, we have

$\displaystyle\|u-\phi_{\vartheta,\vec{b}}\|_{L^{2}(\mathbb{S})}^{2}\leq
C_{\Omega}\left(\|u-\phi_{\vartheta,\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+D(u)+D(u)^{2}\right).$
(4.12)

###### Proof.

Without loss of generality, we will assume $\vartheta=0$ in the assumption,
and write $\phi_{\vartheta,\vec{b}}=\phi_{\vec{b}}$ for short. Suppose that
$u\in\mathcal{A}_{\mathbb{S}}^{d}$ with the Fourier expansion
$u=\sum_{k\in\mathbb{Z}}c_{k}z^{k}$. By (2.17), (2.19) and (1.7),

$\displaystyle
D(u)=\sum_{k\in\mathbb{Z}}|k||c_{k}|^{2}-\sum_{k\in\mathbb{Z}}k|c_{k}|^{2}=2\sum_{k\leq-1}|k||c_{k}|^{2}.$
(4.13)

Now since $u-\phi_{\vec{b}}=\sum_{k\geq
0}(c_{k}-A_{k})z^{k}+\sum_{k\leq-1}c_{k}z^{k}$, then

$\displaystyle\|u-\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\sum_{k\geq
1}k|c_{k}-A_{k}(\vec{b})|^{2}+\sum_{k\leq-1}|k||c_{k}|^{2}$ (4.14)

and

$\displaystyle\|u-\phi_{\vec{b}}\|_{L^{2}(\mathbb{S})}^{2}=2\pi\sum_{k\geq
0}|c_{k}-A_{k}(\vec{b})|^{2}+2\pi\sum_{k\leq-1}|c_{k}|^{2}.$ (4.15)

Combining (4.13), (4.14), we get

$\displaystyle\|u-\phi_{\vec{b}}\|_{L^{2}(\mathbb{S})}^{2}\leq
2\pi\left(|c_{0}-A_{0}(\vec{b})|^{2}+\|u-\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+D(u)\right).$
(4.16)

It suffices to establish the following claim.

######  Claim 1.

There exists a constant $C_{\Omega}$ independent of $u$ and $\vec{b}$ such
that

$\displaystyle|A_{0}(\vec{b})-c_{0}|\leq
C_{\Omega}\left(\|u-\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}+D(u)+\sqrt{D(u)}\right).$
(4.17)

In fact, Lemma 4.2 implies that there exists $1\leq k\leq N_{\Omega}$ such
that $|A_{k}(\vec{b})|\geq\epsilon_{\Omega}$. Without loss of generality, we
assume $k=1$ and $|A_{1}(\vec{b})|\geq\epsilon_{\Omega}$. The following proof
still works for other $k$ with minor modification.

Since $|u|=1$ and $|\phi_{\vec{b}}|=1$ a.e. on $\mathbb{S}$, (2.18) implies

$\displaystyle\bar{c}_{0}c_{1}=-\sum_{j\in\mathbb{Z}\setminus\\{0\\}}\bar{c}_{j}c_{j+1},\quad\bar{A}_{0}A_{1}=-\sum_{j\geq
1}\bar{A}_{j}A_{j+1}.$ (4.18)

Here for the time being we write $A_{k}(\vec{b})=A_{k}$ for short. Subtract
two equations and make interpolation

$\displaystyle A_{1}(\bar{c}_{0}-\bar{A}_{0})=$
$\displaystyle-\bar{c}_{0}(c_{1}-A_{1})-\sum_{j\geq
1}[\bar{c}_{j}c_{j+1}-\bar{A}_{j}A_{j+1}]-\sum_{j\leq-1}\bar{c}_{j}c_{j+1}$
$\displaystyle=$ $\displaystyle-\bar{c}_{0}(c_{1}-A_{1})-\sum_{j\geq
1}[(\bar{c}_{j}-\bar{A}_{j})c_{j+1}+\bar{A}_{j}(c_{j+1}-A_{j+1})]-\sum_{j\leq-1}\bar{c}_{j}c_{j+1}.$

Applying Hölder’s inequality to get

$\displaystyle\begin{split}|A_{1}||c_{0}-A_{0}|\leq&\ \left(\sum_{j\geq
1}|c_{j}-A_{j}|^{2}\right)^{\frac{1}{2}}\left(\sum_{j\geq
0}|c_{j}|^{2}+|A_{j}|^{2}\right)^{\frac{1}{2}}+|D(u)|+|c_{-1}c_{0}|\\\ \leq&\
2\|u-\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}+|D(u)|+\sqrt{D(u)}.\end{split}$
(4.19)

By our assumption $|A_{1}|\geq\epsilon_{\Omega}$, the inequality (4.17) holds.

∎

######  Lemma 4.4.

Suppose $\Omega\Subset\mathbb{D}$ is compact. For any $\varepsilon>0$, there
exists $\delta_{\Omega,\varepsilon}>0$ such that if
$u\in\mathcal{A}_{\mathbb{S}}^{d}$ and

$\|u-\phi_{\vartheta,\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq\delta_{\Omega,\epsilon}$

for some $\vartheta\in\mathbb{S}$ and $\vec{b}\in\Omega^{d}$, then the
harmonic extension of $u$ has at least one zero at
$\Omega_{\varepsilon}=\cup_{p\in\Omega}B_{\varepsilon}(p)$. Here
$B_{\varepsilon}(p)$ denotes an open ball in $\mathbb{D}$.

###### Proof.

Suppose $u=\sum_{k\in\mathbb{Z}}c_{k}z^{k}$. The harmonic extension $U$ of $u$
are simply $U(z)=\sum_{k\geq 0}c_{k}z^{k}+\sum_{k<0}c_{k}\bar{z}^{k}$ where
$z\in\mathbb{D}$. Therefore

$\displaystyle\|U-\phi_{\vartheta,\vec{b}}\|_{L^{2}(\mathbb{D})}^{2}=\pi^{2}\sum_{k\geq
0}|c_{k}-A_{k}(\vec{b})|^{2}+\pi^{2}\sum_{k\leq-1}|c_{k}|^{2}=\frac{\pi}{2}\|u-\phi_{\vartheta,\vec{b}}\|_{L^{2}(\mathbb{S})}^{2}.$
(4.20)

Suppose $\varepsilon>0$ is small enough such that
$\Omega_{\varepsilon}\Subset\mathbb{D}$. Otherwise, Leray-Schauder degree
theory tells us the conclusion holds obviously.

Since $U-\phi_{\vartheta,\vec{b}}$ is a harmonic function in $\mathbb{D}$,
then by the interior estimates

$\displaystyle\|U-\phi_{\vartheta,\vec{b}}\|_{C^{0}(\overline{\Omega_{\varepsilon}})}\leq
C_{\Omega,\varepsilon}\|U-\phi_{\vartheta,\vec{b}}\|_{L^{2}(\mathbb{D})}.$
(4.21)

By the property of Leray-Schauder degree, there exists
$\eta=\eta(\Omega,\varepsilon)$ such that if
$|U-\phi_{\vartheta,\vec{b}}|_{C^{0}(\partial\Omega_{\varepsilon})}<\eta$,
then

$\text{deg\,}(U,\Omega_{\varepsilon},0)=\text{deg\,}(\phi_{\vartheta,\vec{b}},\Omega_{\varepsilon},0).$

Thus, we choose $\delta_{\Omega,\varepsilon}=2\eta^{2}/(\pi
C_{\Omega,\varepsilon}^{2})$. The above equation will be true for any $u$
satisfying
$\|u-\phi_{\vartheta,\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}<\delta_{\Omega,\varepsilon}$.
However, the zeros of $\phi_{\vartheta,\vec{b}}$ are
$b_{1},b_{2},\cdots,b_{d}$, which all contained in $\Omega$. Therefore
$\text{deg\,}(\phi_{\vartheta,\vec{b}},\Omega_{\varepsilon},0)=d$. The above
identity means $u$ has at least one zero in $\Omega_{\varepsilon}$. ∎

Finally, we give the poof of the local version of quantitative stability.

###### Proof of Theorem 1.5.

Here we work on the norm $\dot{H}^{1/2}(\mathbb{S})$, but one can easily
translate results to norm $\dot{H}^{1/2}(\mathbb{R})$ by (2.14). Without loss
of generality, we will assume $\vartheta=0$ in the assumption, and write
$\phi_{\vartheta,\vec{b}}=\phi_{\vec{b}}$ for short. Choose any
$\varepsilon<\text{dist}(\Omega,\partial\mathbb{D})/2$ and fix it for the rest
of proof. Obviously $\Omega_{\varepsilon}\Subset\mathbb{D}$. Suppose
$u\in\mathcal{A}_{\mathbb{S}}^{d}$ and

$\displaystyle\|u-\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq\delta_{d,\Omega,\epsilon}$
(4.22)

holds for some small $\delta_{d,\Omega,\epsilon}$, which is to be chosen
later. If $D(u)\geq\delta_{d,\Omega,\epsilon}$, then we can take
$\vec{a}=\vec{b}$ and $\vartheta^{\prime}=0$. Then

$\displaystyle\|u-\phi_{\vartheta^{\prime},\vec{a}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
C_{d,\Omega,\epsilon}D(u)$ (4.23)

holds for $C_{d,\Omega,\epsilon}=1$. Therefore, we will assume
$D(u)<\delta_{d,\Omega,\epsilon}$ for the rest of proof.

######  Claim 2.

It suffices to prove the theorem for $u\in\mathcal{A}_{\mathbb{S}}^{d}$ with
the additional assumption $\fint_{\mathbb{S}}u=0$.

Indeed, it follows from Proposition 4.3 and Lemma 4.4 that there exists
$\tilde{\delta}_{d,\Omega,\varepsilon}$ such that if
$\delta_{d,\Omega,\epsilon}<\tilde{\delta}_{d,\Omega,\varepsilon}$ then the
harmonic extension $U$ of $u$ has a zero, say $a$, within
$\Omega_{\varepsilon}$. Now consider $u\circ\phi_{a}$. It satisfies
$\int_{\mathbb{S}}u\circ\phi_{a}=0$, because
$u\circ\phi_{a}=\sum{c_{k}}\phi_{a}^{k}$, and

$\displaystyle\int_{\mathbb{S}}u\circ\phi_{a}=\sum_{k\geq
0}c_{k}(-a)^{k}+\sum_{k\leq-1}c_{k}(-\bar{a})^{k}=U(-a)=0.$ (4.24)

Since composing Möbius transformation does not alter the
$\dot{H}^{1/2}(\mathbb{S})$ norm,

$\|u\circ\phi_{a}-\phi_{\vec{b}}\circ\phi_{a}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq\delta_{d,\Omega,\epsilon}.$

Moreover $\phi_{\vec{b}}\circ\phi_{a}=\phi_{\vec{e}}$ with

$e_{i}=\frac{b_{i}+a}{1+\bar{a}b_{i}}=\phi_{-a}(b_{i}),\quad i=1,\cdots,d.$

Note that $a$ belongs to a compact set $\Omega_{\varepsilon}$, which just
depend on $\Omega$. Then $\Omega\circ\phi_{a}:=\phi_{-a}(\Omega)$ is uniformly
away from $\partial\mathbb{D}$, because for any $z\in\Omega$

$1-|\phi_{-a}(z)|^{2}=1-\left|\frac{z+a}{1+\bar{a}z}\right|^{2}=\frac{(1-|a|^{2})(1-|z|^{2})}{|1+\bar{a}z|^{2}}\geq\frac{1}{6}\text{dist}(\Omega,\partial\mathbb{D})^{2}.$

Suppose for $\Omega\circ\phi_{a}$ and $u\circ\phi_{a}$, we have find
$\delta_{d,\Omega\circ\phi_{a},\tilde{\varepsilon}}$ such that there exists
$\vec{a}^{\prime}\in\Omega\circ\phi_{a}+B_{\tilde{\varepsilon}}(0)$ and
$\vartheta^{\prime}\in\mathbb{S}$

$\|u\circ\phi_{a}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
C_{d,\Omega\circ\phi_{a},\tilde{\varepsilon}}D(u\circ\phi_{a}).$

Here $\tilde{\varepsilon}$ is chosen to satisfy

$\displaystyle\phi_{a}(\Omega\circ\phi_{a}+B_{\tilde{\varepsilon}}(0))\subset\Omega_{\varepsilon},\quad\forall
a\in\Omega_{\varepsilon}.$ (4.25)

Take
$\delta_{d,\Omega,\varepsilon}=\min\\{\tilde{\delta}_{d,\Omega,\epsilon},\
\inf\\{\delta_{d,\Omega\circ\phi_{a},\tilde{\varepsilon}}:a\in\Omega_{\varepsilon}\\}\\}$
and
$C_{d,\Omega,\varepsilon}=\sup\\{C_{d,\Omega\circ\phi_{a},\tilde{\varepsilon}}:a\in\Omega_{\varepsilon}\\}$.
Again since $\Omega\circ\phi_{a}$ is uniformly away from $\partial\mathbb{D}$,
one must have $\delta_{d,\Omega,\varepsilon}>0$ and
$C_{d,\Omega,\varepsilon}<\infty$. Then for $\delta_{d,\Omega,\varepsilon}$,
we can find $\vec{a}$ such that (4.23) holds with $C_{d,\Omega,\varepsilon}$.
Moreover, (4.25) means that $\vec{a}$ can be chosen in $\Omega_{\varepsilon}$.
Therefore Claim 2 is proved.

With the above claim in hand, we will prove the statement by induction.

Suppose $d=1$, we write $\vec{b}=b_{1}$ in (4.22). Theorem 1.3 guarantee the
existence of $\vec{a}=a_{1}$ such that (4.23) holds with
$C_{1,\Omega,\epsilon}=36$. It suffices to show $|a_{1}-b_{1}|<\varepsilon$ if
$\delta_{1,\Omega,\varepsilon}$ is chosen small enough. If fact, if (4.22) and
(4.23) holds, then

$\displaystyle\|\phi_{\vartheta,b_{1}}-\phi_{\vartheta^{\prime},a_{1}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
2\|u-\phi_{\vartheta,b_{1}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+2\|u-\phi_{\vartheta^{\prime},a_{1}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
100\delta_{1,\Omega,\epsilon}.$ (4.26)

Direct computation shows that

$\displaystyle\|\phi_{\vartheta,b_{1}}-$
$\displaystyle\phi_{\vartheta^{\prime},a_{1}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\|\phi_{\vartheta-\vartheta^{\prime},\frac{b_{1}-a_{1}}{1-\bar{a}_{1}b_{1}}}-\text{id}_{\mathbb{S}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}$
$\displaystyle=$
$\displaystyle\left|e^{\textup{i\,}(\vartheta-\vartheta^{\prime})}\left(1-\left|\frac{b_{1}-a_{1}}{1-\bar{a}_{1}b_{1}}\right|\right)-1\right|^{2}+\left|\frac{b_{1}-a_{1}}{1-\bar{a}_{1}b_{1}}\right|^{2}\left(2-\left|\frac{b_{1}-a_{1}}{1-\bar{a}_{1}b_{1}}\right|^{2}\right).$

Now it is easy to deduce that

$|b_{1}-a_{1}|\leq 20\sqrt{\delta_{1,\Omega,\epsilon}}.$

Choose $\delta_{1,\Omega,\varepsilon}$ small so that
$|a_{1}-b_{1}|<\varepsilon$. This proves the case $d=1$. Now by induction, we
can assume the theorem holds for the case $d-1$ with
$\delta_{d-1,\Omega,\varepsilon}$ and $C_{d-1,\Omega,\varepsilon}$.

######  Claim 3.

There exists $\hat{\delta}_{d-1,\Omega,\varepsilon}$ such that if
$\delta_{d,\Omega,\epsilon}<\hat{\delta}_{d-1,\Omega,\varepsilon}$, then one
can find $\vec{b}^{\prime}\in\Omega^{d-1}$ such that

$\displaystyle\|u\bar{z}-\phi_{\vec{b}^{\prime}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}<\delta_{d-1,\Omega,\varepsilon}.$
(4.27)

To prove the claim, we notice that $\phi_{\vec{b}}$ is invariant under the
permutation of $b_{i}$. Without loss of generality, we assume
$|b_{1}|\leq\cdots\leq|b_{d}|$333such ordering might not be unique, but it
does not affect the proof. Recalling $|A_{0}(\vec{b})|=\prod_{1\leq k\leq
d}|b_{k}|$ and (4.17), we get $|b_{1}|\leq
10\delta_{d,\Omega,\epsilon}^{1/4d}$. Let
$\vec{b}^{\prime}=(b_{2},\cdots,b_{d})\in\Omega^{d-1}$. By the Cauchy
inequality

$\displaystyle\|u\bar{z}-\phi_{\vec{b}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
2\|u\bar{z}-\bar{z}\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+2\|\bar{z}\phi_{\vec{b}}-\phi_{\vec{b}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}.$
(4.28)

One can compute that

$\displaystyle\|u\bar{z}-\bar{z}\phi_{\vec{b}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}-\|u-\phi_{\vec{b}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}$
$\displaystyle=\sum|k||c_{k+1}-A_{k+1}(\vec{b})|^{2}-\sum|k||c_{k}-A_{k}(\vec{b})|^{2}$
$\displaystyle=-\sum_{k\geq 1}|c_{k}-A_{k}(\vec{b})|^{2}+\sum_{k\leq
0}|c_{k}|^{2}<D(u).$

Notice that

$\displaystyle\|\bar{z}\phi_{\vec{b}}-\phi_{\vec{b}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}=\|\frac{1-b_{1}\bar{z}}{1-\bar{b}_{1}z}\phi_{b_{2}}\cdots\phi_{b_{d}}-\phi_{b_{2}}\cdots\phi_{b_{d}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\to
0$ (4.29)

as $\delta_{d,\Omega,\epsilon}\to 0$. Therefore, choosing
$\delta_{d,\Omega,\epsilon}$ small enough, we can get (4.27).

Therefore, by induction, we have $\vartheta^{\prime}\in\mathbb{S}$ and
$\vec{a}^{\prime}\in\Omega^{d-1}_{\varepsilon/2}$ such that

$\displaystyle\|u\bar{z}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
C_{d-1,\Omega_{\varepsilon/2},\varepsilon}D(u\bar{z}).$ (4.30)

######  Claim 4.

We claim that there exists $C_{d,\Omega,\varepsilon}$ such that

$\displaystyle\|u-z\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
C_{d,\Omega,\varepsilon}D(u).$ (4.31)

Indeed, since $u\bar{z}=\sum c_{k+1}z^{k}$,

$\displaystyle\begin{split}D(u\bar{z})=&2\sum_{k\leq-1}|k||c_{k+1}|^{2}=2\sum_{k\leq-1}|k+1||c_{k+1}|^{2}+2\sum_{k\leq-1}|c_{k+1}|^{2}\\\
=&D(u)+2\sum_{k\leq 0}|c_{k}|^{2}\leq 2D(u).\end{split}$ (4.32)

Here in the last step, we have used $c_{0}=0$. Then (4.30) and (4.32) imply

$\displaystyle\|u\bar{z}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}\leq
2C_{d-1,\Omega_{\varepsilon/2},\varepsilon}D(u).$ (4.33)

Since
$u-z\phi_{\vartheta^{\prime},\vec{a}^{\prime}}=\sum(c_{k}-e^{\textup{i\,}\vartheta^{\prime}}A_{k-1}(\vec{a}^{\prime}))z^{k}$,
then

$\displaystyle\begin{split}\|u-z\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}=&\sum|k||c_{k}-e^{\textup{i\,}\vartheta^{\prime}}A_{k-1}(\vec{a}^{\prime})|^{2}=\sum|k+1||c_{k+1}-e^{\textup{i\,}\vartheta^{\prime}}A_{k}(\vec{a}^{\prime})|^{2}\\\
=&\|u\bar{z}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}^{2}+\sum_{k\geq
0}|c_{k+1}-e^{\textup{i\,}\vartheta^{\prime}}A_{k}(\vec{a}^{\prime})|^{2}-\sum_{k\leq
0}|c_{k}|^{2}.\end{split}$

Since (2.17), we have $\sum_{k\geq
1}k|c_{k+1}-e^{\textup{i\,}\vartheta^{\prime}}A_{k}(\vec{a}^{\prime})|^{2}\leq\|u\bar{z}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}$

$\displaystyle\begin{split}\|u-z\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}\leq&2\|u\bar{z}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}+|c_{1}-e^{\textup{i\,}\vartheta^{\prime}}A_{0}(\vec{a})|^{2}\\\
\leq&4C_{d-1,\Omega_{\varepsilon/2},\varepsilon}D(u)+|c_{1}-e^{\textup{i\,}\vartheta^{\prime}}A_{0}(\vec{a})|^{2}.\end{split}$
(4.34)

Since $|u|=1$ and $|\phi_{\vartheta^{\prime},\vec{a}^{\prime}}|=1$ a.e. on
$\mathbb{S}$, (2.18) implies

$\displaystyle\bar{c}_{1}c_{2}=-\sum_{j\in\mathbb{Z}\setminus\\{1\\}}\bar{c}_{j}c_{j+1},\quad\bar{A}_{0}A_{1}=-\sum_{j\geq
1}\bar{A}_{j}A_{j+1}.$ (4.35)

Here and the following we write $A_{k}(\vec{a}^{\prime})=A_{k}$ for short. We
subtract two equations and make interpolations.

$\displaystyle
e^{\textup{i\,}\vartheta^{\prime}}A_{1}(\bar{c}_{1}-e^{-\textup{i\,}\vartheta^{\prime}}\bar{A}_{0})=$
$\displaystyle-\bar{c}_{1}(c_{2}-e^{\textup{i\,}\vartheta^{\prime}}A_{1})-\sum_{j\geq
2}[\bar{c}_{j}c_{j+1}-\bar{A}_{j-1}A_{j}]-\sum_{j\leq-2}\bar{c}_{j}c_{j+1}$
$\displaystyle=$
$\displaystyle-\bar{c}_{1}(c_{2}-e^{\textup{i\,}\vartheta^{\prime}}A_{1})-\sum_{j\leq-2}\bar{c}_{j}c_{j+1}$
$\displaystyle-\sum_{j\geq
2}[(\bar{c}_{j}-e^{-\textup{i\,}\vartheta^{\prime}}\bar{A}_{j-1})c_{j+1}+e^{-\textup{i\,}\vartheta^{\prime}}\bar{A}_{j-1}(c_{j+1}-e^{\textup{i\,}\vartheta^{\prime}}A_{j})].$

Applying Hölder’s inequality and (4.33), we have

$\displaystyle\begin{split}|A_{1}||c_{1}-e^{\textup{i\,}\vartheta^{\prime}}A_{0}|\leq&\
\left(\sum_{j\geq
2}|c_{j}-e^{\textup{i\,}\vartheta^{\prime}}A_{j-1}|^{2}\right)^{\frac{1}{2}}\left(\sum_{j\geq
1}|c_{j}|^{2}+|A_{j}|^{2}\right)^{\frac{1}{2}}+D(u)\\\ \leq&\
2\|u\bar{z}-\phi_{\vartheta^{\prime},\vec{a}^{\prime}}\|_{\dot{H}^{1/2}(\mathbb{S})}+D(u)\leq\
3\sqrt{2C_{d-1,\Omega_{\varepsilon/2},\varepsilon}D(u)}.\end{split}$ (4.36)

Here we used $D(u)<\delta<1$. Using Lemma 4.2, we have

$\displaystyle|c_{1}-A_{0}|\leq
3\epsilon_{\Omega_{\varepsilon}}^{-1}\sqrt{2C_{d-1,\Omega_{\varepsilon/2},\varepsilon}D(u)}.$
(4.37)

Plugging this back to (4.34) to get

$\displaystyle\|u-z\phi_{\vartheta^{\prime},\vec{a}}\|^{2}_{\dot{H}^{1/2}(\mathbb{S})}\leq
C_{d,\Omega,\varepsilon}D(u)$ (4.38)

with
$C_{d,\Omega,\varepsilon}=(4+18\epsilon_{\Omega_{\varepsilon}}^{-2})C_{d-1,\Omega_{\varepsilon/2},\varepsilon}$.
The (4.31) is proved.

Having established Claim 4, we can see (4.23) holds with the choice of
$\vec{a}=(0,\vec{a}^{\prime})\in\Omega_{\varepsilon}^{d}$,
$\delta_{d,\Omega,\varepsilon}=\min\\{\tilde{\delta}_{d,\Omega,\varepsilon},\hat{\delta}_{d-1,\Omega,\varepsilon}\\}$
and
$C_{\Omega,d,\varepsilon}=\max\\{(4+18\epsilon_{\Omega_{\varepsilon}}^{-2})C_{d-1,\Omega_{\varepsilon/2},\varepsilon},1\\}$.
The induction is complete. ∎

##  5\. A counter example for higher degree

In this section we shall prove that there is no uniform stability. Recall that
all the half-harmonic maps from $\mathbb{R}$ to $\mathbb{S}$ with positive
degree can be written in (1.4). Within this section, we will assume
$\alpha_{i}=x_{i}+\textup{i\,}\lambda_{i}$ with $x_{i}\in\mathbb{R}$ and
$\lambda_{i}>0$. Then

$\displaystyle\psi_{\vartheta,\vec{\alpha}}=e^{\textup{i\,}\vartheta}\frac{x-x_{1}+\textup{i\,}\lambda_{1}}{x-x_{1}+\textup{i\,}\lambda_{1}}\frac{x-x_{2}-\textup{i\,}\lambda_{2}}{x-x_{2}+\textup{i\,}\lambda_{2}}.$
(5.1)

One can equip $\dot{H}^{1/2}(\mathbb{R},\mathbb{R}^{2})$ with the inner
product,

$\displaystyle\langle\boldsymbol{\phi}_{1},\boldsymbol{\phi}_{2}\rangle=\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}\frac{(\boldsymbol{\phi}_{1}(x)-\boldsymbol{\phi}_{1}(y))\cdot(\boldsymbol{\phi}_{2}(x)-\boldsymbol{\phi}_{2}(y))}{|x-y|^{2}}dxdy.$
(5.2)

If $\boldsymbol{\phi}_{2}$ is smooth enough, then RHS of the above equation
can be written
$\int_{\mathbb{R}}\boldsymbol{\phi}_{1}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\phi}_{2}$.
We shall abuse the notation by denoting
$\langle\boldsymbol{\phi}_{1},\boldsymbol{\phi}_{2}\rangle=\int_{\mathbb{R}}\boldsymbol{\phi}_{1}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\phi}_{2}$
for all
$\boldsymbol{\phi}_{1},\boldsymbol{\phi}_{2}\in\dot{H}^{1/2}(\mathbb{R},\mathbb{R}^{2})$.
It should be interpreted as the RHS of the above equation. Let us use
$\boldsymbol{\psi}^{\perp}$ denote the vector rotating $\boldsymbol{\psi}$
counterclockwise by $\pi/2$. If
$\boldsymbol{\phi}_{1}=h_{1}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp}$
and
$\boldsymbol{\phi}_{2}=h_{2}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp}$
with $h_{1},h_{2}\in H^{1/2}(\mathbb{R};\mathbb{R})$, then

$\displaystyle\begin{split}\langle\boldsymbol{\phi}_{1},\boldsymbol{\phi}_{2}\rangle=&\
\int_{\mathbb{R}}h_{1}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(h_{2}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp})\\\
=&\
\int_{\mathbb{R}}h_{1}(-\Delta_{\mathbb{R}})^{\frac{1}{2}}h_{2}+\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{\vartheta,\vec{\alpha}}(x,y)h_{1}(x)h_{2}(y)dxdy,\end{split}$
(5.3)

where

$\displaystyle
Q_{\vartheta,\vec{\alpha}}(x,y)=\frac{|\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}(x)-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}(y)|^{2}}{|x-y|^{2}}.$
(5.4)

######  Remark 5.1.

For instance, in the very special case
$\psi=\frac{x-\lambda\textup{i\,}}{x+\lambda\textup{i\,}}$, then

$\psi(x)-\psi(y)=\frac{2\lambda^{2}(x^{2}-y^{2})}{(\lambda^{2}+x^{2})(\lambda^{2}+y^{2})}+\textup{i\,}\frac{2\lambda(y-x)(\lambda^{2}-xy)}{(\lambda^{2}+x^{2})(\lambda^{2}+y^{2})}.$

Then

$\displaystyle
Q(x,y)=\frac{|\psi(x)-\psi(y)|^{2}}{|x-y|^{2}}=\frac{4\lambda^{2}}{(\lambda^{2}+x^{2})(\lambda^{2}+y^{2})}.$
(5.5)

It follows from the non-degeneracy result that
$\ker\mathcal{L}=\textup{span}\\{K_{1},K_{2},K_{3}\\}$ where

$K_{1}=1,\quad K_{2}=\frac{x^{2}-\lambda^{2}}{x^{2}+\lambda^{2}},\quad
K_{3}=\frac{2\lambda x}{x^{2}+1}.$

Then it is easy to verify that

$\langle
K_{1}\boldsymbol{\psi}^{\perp},K_{2}\boldsymbol{\psi}^{\perp}\rangle=\langle
K_{1}\boldsymbol{\psi}^{\perp},K_{3}\boldsymbol{\psi}^{\perp}\rangle=\langle
K_{2}\boldsymbol{\psi}^{\perp},K_{3}\boldsymbol{\psi}^{\perp}\rangle=0.$

Getting back to degree two case, we will mainly work on the case
$\vartheta=0$, $\alpha_{1}=j^{2}+\textup{i\,}$ and
$\alpha_{2}=-j^{2}+\textup{i\,}$ in this section. For brevity, such
$\psi_{\vartheta,\vec{\alpha}}$ will be written $\psi_{j}$ for short. Denote

$\displaystyle\rho_{j}(x)=\boldsymbol{\psi}_{j}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\psi}_{j},\quad
Q_{j}(x,y)$
$\displaystyle=\frac{|\boldsymbol{\psi}_{j}(x)-\boldsymbol{\psi}_{j}(y)|^{2}}{|x-y|^{2}}.$
(5.6)

######  Lemma 5.2.

One can compute that

$\displaystyle\rho_{j}(x)=\frac{4(1+j^{4}+x^{2})}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]},$
(5.7)
$\displaystyle\begin{split}Q_{j}(x,y)=\frac{16(1+xy+j^{4})^{2}}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}][1+(y-j^{2})^{2}][1+(y+j^{2})^{2}]}.\end{split}$
(5.8)

We always have

$\displaystyle\int_{\mathbb{R}}\rho_{j}=\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}=4\pi.$
(5.9)

###### Proof.

By a straightforward computation, one can get (5.7) and (5.8). Note that
$\|\psi_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=|\text{deg\,}\psi_{j}|=2$, then
we have

$\displaystyle\int_{\mathbb{R}}\rho_{j}=\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}=2\pi\|\psi_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=4\pi.$
(5.10)

∎

Introduce the notation

$\displaystyle\begin{split}K_{1}(x)=1,\quad
K_{2}(x)=\frac{2}{(x-j^{2})^{2}+1},\quad
K_{3}(x)=\frac{2(x-j^{2})}{(x-j^{2})^{2}+1},\\\
K_{4}(x)=\frac{2}{(x+j^{2})^{2}+1},\quad
K_{5}(x)=\frac{2(x+j^{2})}{(x+j^{2})^{2}+1}.\end{split}$ (5.11)

It is easy to know
$\ker\mathcal{L}_{\boldsymbol{\psi}_{j}}=\text{span}\\{K_{1},K_{2},K_{3},K_{4},K_{5}\\}$.

Define $\mathcal{J}=(J_{kl})_{1\leq k,l\leq 5}$ with

$\displaystyle J_{kl}=\langle
K_{k}\boldsymbol{\psi}_{j}^{\perp},K_{l}\boldsymbol{\psi}_{j}^{\perp}\rangle.$
(5.12)

######  Lemma 5.3.

For any $j>0$, one can show

$\displaystyle J_{11}=4\pi,\quad
J_{12}=\frac{2(2+j^{4})\pi}{j^{4}+1}=J_{14},\quad
J_{13}=\frac{-2j^{2}\pi}{j^{4}+1}=-J_{15},$ $\displaystyle
J_{22}=\left(1+\frac{2j^{8}+5j^{4}+5}{(j^{4}+1)^{2}}\right)\pi,\quad
J_{23}=\frac{-2j^{2}}{(j^{4}+1)^{2}}\pi,\quad
J_{24}=\frac{2j^{4}+6}{(j^{4}+1)^{2}}\pi,$
$J_{25}=\frac{2j^{2}(j^{4}+3)}{(j^{4}+1)^{2}}\pi,\quad
J_{33}=\left(1+\frac{3j^{4}+1}{(j^{4}+1)^{2}}\right)\pi=J_{55},\quad
J_{34}=-\frac{2j^{2}(j^{4}+3)}{(1+j^{4})^{2}}\pi,$
$J_{35}=\frac{2-2j^{4}}{(j^{4}+1)^{2}}\pi,\quad J_{44}=J_{22},\quad
J_{45}=\frac{2j^{2}\pi}{(j^{4}+1)^{2}}.$

The determinant of $\mathcal{J}=(J_{kl})_{1\leq k,l\leq 5}$ is

$\displaystyle\begin{split}\det\mathcal{J}&=\frac{j^{8}(3j^{16}+22j^{12}+51j^{8}+48j^{4}+16)}{(j^{4}+1)^{6}}\pi^{5}.\end{split}$
(5.13)

###### Proof.

We shall use (5.3) to compute all the inner products. First, let us compute
$(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{i}$ for $i=1,\cdots,5$. It is easy to
know $(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{1}=0$. By the extension method,
we can obtain

$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{2}$
$\displaystyle=-\frac{\partial}{\partial
y}\Big{|}_{y=0}\frac{2y+2}{(x-j^{2})^{2}+(y+1)^{2}}=\frac{2(1-(x-j^{2})^{2})}{(1+(x-j^{2})^{2})^{2}},$
$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{3}$
$\displaystyle=-\frac{\partial}{\partial
y}\Big{|}_{y=0}\frac{2(x-j^{2})}{(x-j^{2})^{2}+(y+1)^{2}}=\frac{4(x-j^{2})}{(1+(x-j^{2})^{2})^{2}},$
$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{4}$
$\displaystyle=-\frac{\partial}{\partial
y}\Big{|}_{y=0}\frac{2y+2}{(x+j^{2})^{2}+(y+1)^{2}}=\frac{2(1-(x+j^{2})^{2})}{(1+(x+j^{2})^{2})^{2}},$
$\displaystyle(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{5}$
$\displaystyle=-\frac{\partial}{\partial
y}\Big{|}_{y=0}\frac{2(x+j^{2})}{(x+j^{2})^{2}+(y+1)^{2}}=\frac{4(x+j^{2})}{(1+(x+j^{2})^{2})^{2}}.$

Using (5.3) and (5.8), we can compute each $J_{kl}=\langle
K_{k}\boldsymbol{\psi}_{j}^{\perp},K_{l}\boldsymbol{\psi}_{j}^{\perp}\rangle$
respectively. Since the integrals here involve only rational functions, we
take advantage of a symbolic software, Mathematica, to aid our computation. ∎

######  Proposition 5.4.

Fix any $j\geq 100$, there exists $\delta_{j}$ such that for any $u$ with
$\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}<\delta_{j}$,
then there exists a unique $\vartheta=\vartheta(\boldsymbol{u})$,
$\vec{\alpha}=\vec{\alpha}(\boldsymbol{u})$ satisfying

$\displaystyle\int_{\mathbb{R}}\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(g\boldsymbol{\psi}^{\perp}_{\vartheta,\vec{\alpha}})=0,\quad\forall\,g\in\ker\mathcal{L}_{\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}}.$
(5.14)

###### Proof.

Define the following function

$\displaystyle\Phi:\dot{H}^{1/2}(\mathbb{R})\times$
$\displaystyle\mathbb{S}\times\mathbb{H}^{2}\to\mathbb{R}^{5}$
$\displaystyle(\boldsymbol{u},\vartheta,\vec{\alpha})$
$\displaystyle\mapsto(\langle\boldsymbol{u},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{1}\rangle,\langle\boldsymbol{u},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{2}\rangle,\langle\boldsymbol{u},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{3}\rangle,\langle\boldsymbol{u},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{4}\rangle,\langle\boldsymbol{u},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{5}\rangle)$

where $\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{i}$ are obtained from $\ker
L_{\psi_{\vartheta,\vec{\alpha}}}=\text{span}\\{\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{1},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{2},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{3},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{4},\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{5}\\}$.
More precisely

$\displaystyle\begin{split}\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{1}=\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp},\quad\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{2}=\frac{2\lambda_{1}^{2}}{(x-x_{1})^{2}+\lambda_{1}^{2}}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp},\quad\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{3}=\frac{2\lambda_{1}(x-x_{1})}{(x-x_{1})^{2}+\lambda_{1}^{2}}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp},\\\
\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{4}=\frac{2\lambda_{2}^{2}}{(x-x_{2})^{2}+\lambda_{2}^{2}}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp},\quad\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{5}=\frac{2\lambda_{2}(x-x_{2})}{(x-x_{2})^{2}+\lambda_{2}^{2}}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp}.\end{split}$
(5.15)

Such $\Phi$ is well-defined, because $\boldsymbol{u}$ and
$\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{i}$ all belong to
$\dot{H}^{1/2}(\mathbb{R};\mathbb{R}^{2})$. Moreover, since
$\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{i}$ depends on
$\vartheta,\vec{\alpha}$ smoothly, then $\Phi$ also depends on
$\vartheta,\vec{\alpha}$ smoothly. Since the inner product (5.3) depends on
its arguments smoothly, consequently $\Phi$ depends on $\boldsymbol{u}$
smoothly. Moreover, since
$\int_{\mathbb{R}}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{i}=\int_{\mathbb{R}}(-\Delta_{\mathbb{R}})^{\frac{1}{2}}\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\cdot\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{i}=0$
for any $\vartheta,\vec{\alpha}$ and $i=1,\cdots,5$, then

$\displaystyle\Phi(\boldsymbol{u},\vartheta,\vec{\alpha})=\Phi(\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}},\vartheta,\vec{\alpha}):=\Phi(\boldsymbol{v}_{\vartheta,\vec{\alpha}},\vartheta,\vec{\alpha}).$
(5.16)

Here we introduced the notation
$\boldsymbol{v}_{\vartheta,\vec{\alpha}}=\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}$.

We intend to apply implicit function theorem to $\Phi$ at
$(\boldsymbol{\psi}_{j},0,(j^{2}+\textup{i\,},-j^{2}+\textup{i\,}))$. The
Jacobian matrix with respect to parameters $\vartheta,\vec{\alpha}$ at
$(\boldsymbol{\psi}_{j},0,(j^{2}+\textup{i\,},-j^{2}+\textup{i\,}))$ is

$\displaystyle
Jac_{\Phi}(\boldsymbol{\psi}_{j},0,(j^{2}+\textup{i\,},-j^{2}+\textup{i\,}))=\
(\partial_{\vartheta}\Phi,\partial_{x_{1}}\Phi,\partial_{\lambda_{1}}\Phi,\partial_{x_{2}}\Phi,\partial_{\lambda_{2}}\Phi)^{T}.$
(5.17)

If $\vartheta=0$ and $\vec{\alpha}=(j^{2}+\textup{i\,},-j^{2}+\textup{i\,})$,
we will write
$\boldsymbol{v}_{j}=\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}$
and $\boldsymbol{K}^{i}_{j}=\boldsymbol{K}_{\vartheta,\vec{\alpha}}^{i}$ for
short. At
$(\boldsymbol{\psi}_{j},0,(j^{2}+\textup{i\,},-j^{2}+\textup{i\,}))$, one has
$\boldsymbol{v}_{j}=0$ and
$\boldsymbol{K}^{i}_{j}=K_{i}\boldsymbol{\psi}_{j}^{\perp}$, where $K_{i}$ are
defined in (5.11). Therefore using (5.16), we have

$\displaystyle\partial_{\vartheta}\Phi=(\langle\partial_{\vartheta}\boldsymbol{v}_{j},K_{1}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\partial_{\vartheta}\boldsymbol{v}_{j},K_{2}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\partial_{\vartheta}\boldsymbol{v}_{j},K_{3}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\partial_{\vartheta}\boldsymbol{v}_{j},K_{4}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\partial_{\vartheta}\boldsymbol{v}_{j},K_{5}\boldsymbol{\psi}_{j}^{\perp}\rangle).$

Similar equality holds for $\partial_{x_{1}}\Phi$,
$\partial_{\lambda_{1}}\Phi$, $\partial_{x_{2}}\Phi$ and
$\partial_{\lambda_{2}}\Phi$. Furthermore, one can compute that
$\partial_{\vartheta}\boldsymbol{v}_{j}=-K_{1}\boldsymbol{\psi}_{j}^{\perp}$
and

$\displaystyle\begin{split}\partial_{x_{1}}\boldsymbol{v}_{j}\
=\frac{-2}{(x-j^{2})^{2}+1}\boldsymbol{\psi}_{j}^{\perp}=-K_{2}\boldsymbol{\psi}_{j}^{\perp},\quad\partial_{\lambda_{1}}\boldsymbol{v}_{j}\
=\frac{-2(x-j^{2})}{(x-j^{2})^{2}+1}\boldsymbol{\psi}_{j}^{\perp}=-K_{3}\boldsymbol{\psi}_{j}^{\perp},\\\
\partial_{x_{2}}\boldsymbol{v}_{j}\
=\frac{-2}{(x+j^{2})^{2}+1}\boldsymbol{\psi}_{j}^{\perp}=-K_{4}\boldsymbol{\psi}_{j}^{\perp},\quad\partial_{\lambda_{2}}\boldsymbol{v}_{j}\
=\frac{-2(x+j^{2})}{(x+j^{2})^{2}+1}\boldsymbol{\psi}_{j}^{\perp}=-K_{5}\boldsymbol{\psi}_{j}^{\perp}.\end{split}$
(5.18)

Plugging in these computations to the Jacobian matrix and using Lemma 5.3, we
have

$\displaystyle\det
Jac_{\Phi}(\boldsymbol{\psi}_{j},0,(j^{2}+\textup{i\,},-j^{2}+\textup{i\,}))=-\det\mathcal{J}=-3\pi^{5}+O(1/j^{4})\neq
0.$ (5.19)

The implicit function theorem gives that there exists $\delta_{j}>0$ and
unique smooth functions
$\vartheta(\boldsymbol{u}),x_{1}(\boldsymbol{u}),x_{2}(\boldsymbol{u}),\lambda_{1}(\boldsymbol{u}),\lambda_{2}(\boldsymbol{u})$
such that for any $\boldsymbol{u}$ with
$\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}<\delta_{j}$
one has $\Phi(\boldsymbol{u},\vartheta,\vec{\alpha})=0$, where
$\vec{\alpha}=(\alpha_{1},\alpha_{2})$,
$\alpha_{1}=x_{1}+\textup{i\,}\lambda_{1}$ and
$\alpha_{2}=x_{2}+\textup{i\,}\lambda_{2}$. That is

$\displaystyle\int_{\mathbb{R}}\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(g\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}^{\perp})=0,\quad\forall\,g\in\ker\mathcal{L}_{\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}}.$
(5.20)

∎

We recall a function defined in [6]. For any $j\in\mathbb{N}$, define $f_{j}$
on $\mathbb{R}$ as

$\displaystyle f_{j}(x)=\begin{cases}1&\text{if }|x|\leq j,\\\
2-\frac{\log|x|}{\log j}&\text{if }j\leq|x|\leq j^{2},\\\ 0&\text{if }|x|\geq
j^{2}.\end{cases}$ (5.21)

######  Lemma 5.5.

There exists some uniform constant $C$ such that
$\|f_{j}\|_{\dot{H}^{1/2}}^{2}\leq C/|\log j|$ for all $j\in\mathbb{N}$.

###### Proof.

We extend $f_{j}$ to $\mathbb{R}^{2}_{+}$ by replacing $|x|$ with $|(x,y)|$.
Since

$\displaystyle\partial_{x}f_{j}=\begin{cases}\frac{x}{x^{2}+y^{2}}\frac{1}{\log
j}&\text{if }j\leq\sqrt{x^{2}+y^{2}}\leq j^{2},\\\ 0&\text{otherwise.
}\end{cases}$ (5.22)

One has a similar expression for $\partial_{y}f_{j}$. Then

$\displaystyle\int_{\mathbb{R}^{2}_{+}}\left(|\partial_{x}f_{j}|^{2}+|\partial_{y}f_{j}|^{2}\right)=\frac{1}{\log^{2}j}\int_{B_{j^{2}}\setminus
B_{j}}\frac{1}{x^{2}+y^{2}}dxdy\leq\frac{C}{\log j}.$ (5.23)

Therefore by (2.20), we get the conclusion. ∎

For any $j\geq 100$ we define

$\displaystyle h(x)=f_{j}(x-j^{2})-f_{j}(x+j^{2}).$ (5.24)

Then $h$ is an odd function. Moreover, $\|h\|_{\dot{H}^{1/2}(\mathbb{R})}\leq
C/\log j$.

######  Lemma 5.6.

For any $j\geq 100$, we have

$\displaystyle\int_{\mathbb{R}}\rho_{j}h^{2}=4\pi+O(1/j),$ (5.25)

and

$\displaystyle\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)h(x)h(y)dxdy=4\pi+O(1/j).$
(5.26)

Consequently we get

$\displaystyle\int_{\mathbb{R}}h\mathcal{L}_{\boldsymbol{\psi}_{j}}h=O(1/\log
j),\quad\langle
h\boldsymbol{\psi}_{j}^{\perp},h\boldsymbol{\psi}_{j}^{\perp}\rangle=4\pi+O(1/\log
j).$ (5.27)

######  Lemma 5.7.

For any $j\geq 100$, we have $\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{1}\boldsymbol{\psi}_{j}^{\perp}\rangle=0$,

$\displaystyle\begin{split}\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{2}\boldsymbol{\psi}_{j}^{\perp}\rangle=\pi+O(1/j),\quad\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{3}\boldsymbol{\psi}_{j}^{\perp}\rangle=O(1/j^{2}),\\\
\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{4}\boldsymbol{\psi}_{j}^{\perp}\rangle=-\pi+O(1/j),\quad\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{5}\boldsymbol{\psi}_{j}^{\perp}\rangle=O(1/j^{2}).\end{split}$
(5.28)

In order not to interrupt the main thread of this section, We will defer the
proof of these two lemmas to the end of this paper.

######  Lemma 5.8.

There exists $\varepsilon_{j}$ with the following significance. For any
$\varepsilon<\varepsilon_{j}$, there exists $h_{\perp}$ such that
$\boldsymbol{u}=\varepsilon
h_{\perp}\boldsymbol{\psi}_{j}^{\perp}+\sqrt{1-\varepsilon^{2}h_{\perp}^{2}}\boldsymbol{\psi}_{j}$
satisfies

$\displaystyle\int_{\mathbb{R}}\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(K_{i}\boldsymbol{\psi}_{j}^{\perp})=0,\quad
i=1,2,3,4,5.$ (5.29)

Furthermore,

$\displaystyle\int_{\mathbb{R}}h_{\perp}\mathcal{L}_{\boldsymbol{\psi}_{j}}h_{\perp}=O(1/\log
j)+O(\varepsilon),$ (5.30) $\displaystyle\langle
h_{\perp}\boldsymbol{\psi}_{j}^{\perp},h_{\perp}\boldsymbol{\psi}_{j}^{\perp}\rangle=\frac{10}{3}\pi+O(1/\log
j)+O(\varepsilon).$ (5.31)

###### Proof.

We can take

$\displaystyle
h_{\perp}=h-c_{1}K_{1}-c_{2}K_{2}-c_{3}K_{3}-c_{4}K_{4}-c_{5}K_{5}$ (5.32)

with $c_{i}$ to be determined. Define a map

$\displaystyle\begin{split}\Phi:\mathbb{R}_{+}\times\mathbb{R}^{5}\to&\
\mathbb{R}^{5}\\\
(\varepsilon,\vec{c})\mapsto&(\langle\boldsymbol{v},K_{1}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\boldsymbol{v},K_{2}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\boldsymbol{v},K_{3}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\boldsymbol{v},K_{4}\boldsymbol{\psi}_{j}^{\perp}\rangle,\langle\boldsymbol{v},K_{5}\boldsymbol{\psi}_{j}^{\perp}\rangle)\end{split}$
(5.33)

where $\vec{c}=(c_{1},c_{2},c_{3},c_{4},c_{5})$ and

$\displaystyle\boldsymbol{v}=h_{\perp}\boldsymbol{\psi}_{j}^{\perp}-\frac{\varepsilon
h_{\perp}^{2}}{\sqrt{1-\varepsilon^{2}h_{\perp}^{2}}+1}\boldsymbol{\psi}_{j}.$
(5.34)

The map $\Phi$ is well defined because $\boldsymbol{v}$ and
$K_{i}\boldsymbol{\psi}_{j}^{\perp}$ all belong to
$\dot{H}^{1/2}(\mathbb{R})$. At $\varepsilon=0$, $\Phi(0,\vec{c})=0$ if and
only if

$\displaystyle\mathcal{J}(c_{1},c_{2},c_{3},c_{4},c_{5})^{T}=(b_{1},b_{2},b_{3},b_{4},b_{5})^{T}$
(5.35)

where $b_{i}=\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{i}\boldsymbol{\psi}_{j}^{\perp}\rangle$,
$i=1,2,3,4,5.$ Since $\mathcal{J}$ is non-degenerate, by Lemma 5.3 and 5.7, we
get $c_{1}=O(1/j)$, $c_{2}=1/3+O(1/j)$, $c_{3}=O(1/j^{2})$,
$c_{4}=-1/3+O(1/j)$, and $c_{5}=O(1/j^{2})$. We denote the solution of the
above equations as $\vec{c}_{*}$.

Note that the Jacobian of $\Phi$ at $(0,\vec{c}_{*})$ is

$(\partial_{c_{1}}\Phi,\partial_{c_{2}}\Phi,\partial_{c_{3}}\Phi,\partial_{c_{4}}\Phi,\partial_{c_{5}}\Phi)(0,\vec{c})=-\mathcal{J}.$

Again the non-degeneracy of $\mathcal{J}$ implies that we can invoke the
implicit function theorem. There exists $\varepsilon_{j}>0$ such that for any
$0\leq\varepsilon<\varepsilon_{j}$, there exists
$\vec{c}=\vec{c}(\varepsilon)=\vec{c}_{*}+O(\varepsilon)$ satisfies
$\Phi(\varepsilon,\vec{c})=0$. That is,

$\int_{\mathbb{R}}\boldsymbol{v}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(K_{i}\boldsymbol{\psi}_{j}^{\perp})=0,\quad
i=1,2,3,4,5.$

Consequently $\boldsymbol{u}=\varepsilon\boldsymbol{v}+\boldsymbol{\psi}_{j}$
also satisfies the above orthogonality. Using the form of $\boldsymbol{v}$,
one readily check $\boldsymbol{u}=\varepsilon
h_{\perp}\boldsymbol{\psi}_{j}^{\perp}+\sqrt{1-\varepsilon^{2}h_{\perp}^{2}}\boldsymbol{\psi}_{j}$
takes the desired form.

Since $K_{i}\in\ker\mathcal{L}_{\boldsymbol{\psi}_{j}}$, then

$\displaystyle\int_{\mathbb{R}}h_{\perp}\mathcal{L}_{\boldsymbol{\psi}_{j}}h_{\perp}=\int_{\mathbb{R}}h\mathcal{L}_{\boldsymbol{\psi}_{j}}h+O(\varepsilon)=O(1/\log
j)+O(\varepsilon),$ (5.36)

where we used (5.27). To establish (5.31), we just need to use the results
from Lemma 5.3, Lemma 5.6 and Lemma 5.7.

$\displaystyle\langle
h_{\perp}\boldsymbol{\psi}_{j}^{\perp},h_{\perp}\boldsymbol{\psi}_{j}^{\perp}\rangle=$
$\displaystyle\ \langle
h\boldsymbol{\psi}_{j}^{\perp}-\sum_{i=1}^{5}c_{i}K_{i}\boldsymbol{\psi}_{j}^{\perp},h\boldsymbol{\psi}_{j}^{\perp}-\sum_{i=1}^{5}c_{i}K_{i}\boldsymbol{\psi}_{j}^{\perp}\rangle$
$\displaystyle=$ $\displaystyle\ \langle
h\boldsymbol{\psi}_{j}^{\perp},h\boldsymbol{\psi}_{j}^{\perp}\rangle-2\sum_{i=1}^{5}c_{i}\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{i}\boldsymbol{\psi}_{j}^{\perp}\rangle+\sum_{k,l=1}^{5}c_{k}c_{l}J_{kl}$
$\displaystyle=$ $\displaystyle\
4\pi-\frac{2}{3}\pi-\frac{2}{3}\pi+\frac{1}{3}\pi+\frac{1}{3}\pi+O(1/\log
j)+O(\varepsilon)$ $\displaystyle=$ $\displaystyle\ \frac{10}{3}\pi+O(1/\log
j)+O(\varepsilon).$

∎

######  Proposition 5.9.

Fix any $j\geq 100$. Suppose that $h_{\perp}$ and $\boldsymbol{u}$ are
obtained from Lemma 5.8. Then there exists $\varepsilon_{j}$ such that for
$\varepsilon<\varepsilon_{j}$ the following infimum is achieved at
$\boldsymbol{\psi}_{j}$.

$\displaystyle\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{2}}\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}=\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}=\varepsilon^{2}\langle
h_{\perp},h_{\perp}\rangle+O(\varepsilon^{3}).$ (5.37)

###### Proof.

Since degree is continuous in $\dot{H}^{1/2}$-topology, there exists
$\varepsilon^{\prime}_{j}$ such that if
$\varepsilon<\varepsilon^{\prime}_{j}$, then
$\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}\ll 1$
and $\text{deg\,}\boldsymbol{u}=\text{deg\,}\boldsymbol{\psi}_{j}=2$. First,
we claim the infimum is achieved. Indeed, take a minimizing sequence,
$\vartheta_{k}$, $\vec{\alpha}_{k}=(\alpha_{1,k},\alpha_{2,k})$ such that

$\displaystyle\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta_{k},\vec{\alpha}_{k}}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\to\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{2}}\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})},\quad
k\to\infty.$ (5.38)

Since
$\|\boldsymbol{\psi}_{\vartheta_{k},\vec{\alpha}_{k}}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=2$,
going to a subsequence necessary,
$\boldsymbol{\psi}_{\vartheta_{k},\vec{\alpha}_{k}}$ converges weakly to
$\boldsymbol{\psi}_{*}$. Due to the specific form of
$\boldsymbol{\psi}_{\vartheta_{j},\vec{\alpha}_{j}}$, $\boldsymbol{\psi}_{*}$
can takes three possible forms
$\boldsymbol{\psi}_{\vartheta_{*},(\alpha_{1}^{*},\alpha_{2}^{*})}$,
$\boldsymbol{\psi}_{\vartheta_{*},\alpha_{*}}$, and
$e^{\textup{i\,}\vartheta_{*}}$. If
$\boldsymbol{\psi}_{*}=\boldsymbol{\psi}_{\vartheta_{*},\alpha_{*}}$ or
$e^{\textup{i\,}\theta_{*}}$, then
$\|\boldsymbol{\psi}_{*}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}\leq 1$.

$\displaystyle\|\boldsymbol{u}-\boldsymbol{\psi}_{*}\|_{\dot{H}^{1/2}(\mathbb{R})}\leq\liminf_{k\to\infty}\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta_{k},\vec{\alpha}_{k}}\|_{\dot{H}^{1/2}(\mathbb{R})}\leq\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}\ll
1.$ (5.39)

On the other hand, by the Young’s inequality and (2.2), we obtain

$\displaystyle\|\boldsymbol{u}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}\leq\frac{3}{2}\|\boldsymbol{\psi}_{*}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}+\frac{8}{3}\|\boldsymbol{u}-\boldsymbol{\psi}_{*}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}<2.$
(5.40)

However, this contradict to the fact that $\text{deg\,}\boldsymbol{u}=2$ and
Theorem 1.2. Therefore we must have
$\boldsymbol{\psi}_{*}=\boldsymbol{\psi}_{\vartheta_{*},(\alpha_{1}^{*},\alpha_{2}^{*})}$,
then $\|\boldsymbol{\psi}_{*}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=2$,
consequently $\boldsymbol{\psi}_{\vartheta_{k},\vec{\alpha}_{k}}$ converges to
$\boldsymbol{\psi}_{*}$ strongly and $\boldsymbol{\psi}_{*}$ is one minimizer.

Suppose $\delta_{j}$ is defined in Proposition 5.4. Apparently, there exists
$\varepsilon_{j}>0$, such that for $\varepsilon<\varepsilon_{j}$, one has
$\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}<\frac{1}{2}\delta_{j}$.
Then any minimizer $\boldsymbol{\psi}$ of the infimum satisfies

$\displaystyle\begin{split}\|\boldsymbol{\psi}-\boldsymbol{\psi}_{j}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}&\leq
2\|\boldsymbol{\psi}-\boldsymbol{u}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}+2\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}\\\
&\leq
4\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}<\delta_{j}^{2}.\end{split}$
(5.41)

Suppose $u$ can be decomposed to

$\displaystyle\boldsymbol{u}=f\boldsymbol{\psi}^{\perp}+\sqrt{1-f^{2}}\boldsymbol{\psi}.$
(5.42)

Since the infimum achieves at $\boldsymbol{\psi}$, then

$\displaystyle\int_{\mathbb{R}}\boldsymbol{u}\cdot(-\Delta_{\mathbb{R}})^{\frac{1}{2}}(g\boldsymbol{\psi}^{\perp})=0,\quad\forall\,g\in\ker\mathcal{L}_{\boldsymbol{\psi}}.$
(5.43)

It follows from Proposition 5.4 that for any $u$ with
$\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}<\frac{1}{2}\delta_{j}$,
there exist unique $\vartheta,\vec{\alpha}$ such that $\boldsymbol{u}$
satisfies (5.42) with
$\boldsymbol{\psi}=\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}$. This implies
the minimizer is unique. Recall that the choice of $f=\varepsilon h_{\perp}$
with $\vartheta=0$ and $\vec{\alpha}=0$ make (5.42) and (5.43) happen at the
same time. Thus the infimum is achieved at $\boldsymbol{\psi}_{j}$.

Finally, we can compute explicitly

$\displaystyle\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=\|\varepsilon
h_{\perp}\boldsymbol{\psi}_{j}^{\perp}+O(\varepsilon^{2}h_{\perp}^{2})\boldsymbol{\psi}_{j}\|_{\dot{H}^{1/2}(\mathbb{R})}^{2}=\varepsilon^{2}\langle
h_{\perp},h_{\perp}\rangle+O(\varepsilon^{3}).$ (5.44)

∎

Finally, we can prove the main theorem of this section.

###### Proof of Theorem 1.4.

We take $\boldsymbol{u}=\varepsilon
h_{\perp}\boldsymbol{\psi}_{j}^{\perp}+\sqrt{1-\varepsilon^{2}h_{\perp}^{2}}\boldsymbol{\psi}_{j}$
as stated in Lemma 5.8. Proposition 5.9 implies that, if
$\varepsilon<\varepsilon_{j}$, then

$\displaystyle\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{2}}\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}=\|\boldsymbol{u}-\boldsymbol{\psi}_{j}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}=\varepsilon^{2}\langle
h_{\perp},h_{\perp}\rangle+O(\varepsilon^{3}).$ (5.45)

Using (5.31), we obtain

$\displaystyle\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{2}}\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}=\frac{10}{3}\pi\varepsilon^{2}+O(\varepsilon^{2}/\log
j+\varepsilon^{3}).$ (5.46)

On the other hand, Lemma 3.2 infers

$\displaystyle\mathcal{E}(\boldsymbol{u})=\mathcal{E}(\boldsymbol{\psi}_{j})+\varepsilon^{2}\int_{\mathbb{R}}h_{\perp}\mathcal{L}_{\boldsymbol{\psi}_{j}}[h_{\perp}]+O(\varepsilon^{3}).$
(5.47)

Note (2.3) implies
$\mathcal{E}(\boldsymbol{u})=\|\boldsymbol{u}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}$
and
$\mathcal{E}(\boldsymbol{\psi}_{j})=\|\boldsymbol{\psi}_{j}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}=\text{deg\,}\boldsymbol{u}=2$.
Combing with (5.30), it leads to

$\displaystyle\|\boldsymbol{u}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}-2=\varepsilon^{2}O(1/\log
j).$ (5.48)

Now compare (5.46) and (5.48) to get

$\displaystyle\inf_{\vartheta\in\mathbb{S},\vec{\alpha}\in\mathbb{H}^{2}}\|\boldsymbol{u}-\boldsymbol{\psi}_{\vartheta,\vec{\alpha}}\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}\geq
C(\log j)\left(\|u\|^{2}_{\dot{H}^{1/2}(\mathbb{R})}-2\right).$ (5.49)

Choosing $j$ sufficiently large such that $C\log j\geq M$, our theorem is
established. ∎

## Acknowledgement

The research of B. Deng is supported by China Scholar Council and Natural
Science Foundation of China (No. 11721101). The research of L. Sun and J. Wei
is partially supported by NSERC of Canada.

##  Appendix A  Proofs of Lemma 5.6 and Lemma 5.7

In this appendix, we give the proofs of Lemma 5.6 and Lemma 5.7.

###### Proof of Lemma 5.6.

Notice that if $|x-j^{2}|\geq j$ and $x>0$, then

$\displaystyle\rho_{j}(x)\leq\frac{C}{1+(x-j^{2})^{2}}.$ (A.1)

Since $\rho_{j}$ is even, then we have

$\displaystyle\begin{split}\int_{\mathbb{R}}\rho_{j}h^{2}=&\int_{0}^{\infty}\frac{8(1+j^{4}+x^{2})h(x)^{2}}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx\\\
=&\int_{j^{2}-j}^{j^{2}+j}\frac{8(1+j^{4}+x^{2})}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx+O(1/j)\\\
=&\int_{0}^{\infty}\frac{8(1+j^{4}+x^{2})}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx+O(1/j)=\
4\pi+O(1/j).\end{split}$ (A.2)

Then we prove (5.25). To prove (5.26), we divide the support of $h(x)h(y)$,
i.e., $\\{|x|\leq 2j^{2}\\}\cap\\{|y|\leq 2j^{2}\\}$ into $\Omega_{i}$,
$i=1,\cdots,9$, according to their types. Let

$\displaystyle\begin{split}\Omega_{1}=\\{|x-j^{2}|\leq j,j\leq|y-j^{2}|\leq
j^{2}\\}\cup\\{j\leq|x-j^{2}|\leq j^{2},|y-j^{2}|\leq j\\},\\\
\Omega_{2}=\\{|x-j^{2}|\leq j,j\leq|y+j^{2}|\leq
j^{2}\\}\cup\\{j\leq|x-j^{2}|\leq j^{2},|y+j^{2}|\leq j\\},\\\
\Omega_{3}=\\{|x+j^{2}|\leq j,j\leq|y+j^{2}|\leq
j^{2}\\}\cup\\{j\leq|x+j^{2}|\leq j^{2},|y+j^{2}|\leq j\\},\\\
\Omega_{4}=\\{|x+j^{2}|\leq j,j\leq|y-j^{2}|\leq
j^{2}\\}\cup\\{j\leq|x+j^{2}|\leq j^{2},|y-j^{2}|\leq j\\}.\end{split}$

Using the expression (5.8), and there $|1+xy+j^{4}|\leq 8j^{4}$, it is easy to
see that

$\displaystyle Q_{j}(x,y)\leq
C\begin{cases}[1+(x-j^{2})^{2}]^{-1}[1+(y-j^{2})^{2}]^{-1},\quad\Omega_{1},\\\
[1+(x-j^{2})^{2}]^{-1}[1+(y+j^{2})^{2}]^{-1},\quad\Omega_{2},\\\
[1+(x+j^{2})^{2}]^{-1}[1+(y+j^{2})^{2}]^{-1},\quad\Omega_{3},\\\
[1+(x+j^{2})^{2}]^{-1}[1+(y-j^{2})^{2}]^{-1},\quad\Omega_{4}.\end{cases}$

Consequently, for example, on $\Omega_{1}$ we have, by the symmetry of $x$ and
$y$,

$\displaystyle\begin{split}\left|\iint_{\Omega_{1}}Q_{j}(x,y)h(x)h(y)dxdy\right|&\leq
C\iint_{\\{|x-j^{2}|\leq j,j\leq|y-j^{2}|\leq
j^{2}\\}}\frac{dxdy}{[1+(x-j^{2})^{2}][1+(y-j^{2})^{2}]}\\\ &=4C\arctan
j\int_{j}^{j^{2}}\frac{dy}{1+y^{2}}=O(1/j).\end{split}$ (A.3)

Similarly, the integral on $\Omega_{i},i=2,3,4$ is also of order $O(1/j)$.

On both of sets $\Omega_{5}=\\{|x-j^{2}|\leq j\\}\cap\\{|y+j^{2}|\leq j\\}$
and $\Omega_{6}=\\{|x+j^{2}|\leq j\\}\cap\\{|y-j^{2}|\leq j\\}$, it holds that

$\displaystyle|1+xy+j^{4}|\leq 4j^{3}.$ (A.4)

It is easy to see that

$\displaystyle
Q_{j}(x,y)\leq\frac{C}{j^{2}}\begin{dcases}[1+(x-j^{2})^{2}]^{-1}[1+(y+j^{2})^{2}]^{-1},\quad\Omega_{5},\\\
[1+(x+j^{2})^{2}]^{-1}[1+(y-j^{2})^{2}]^{-1},\quad\Omega_{6}.\end{dcases}$

Then we obtain

$\displaystyle\left|\iint_{\Omega_{i}}Q_{j}(x,y)h(x)h(y)dxdy\right|=O(1/j^{2}),\quad
i=5,6.$ (A.5)

Let $\Omega_{7}=\\{|x-j^{2}|\leq j\\}\cap\\{|y-j^{2}|\leq j\\}$ and
$\Omega_{8}=\\{|x+j^{2}|\leq j\\}\cap\\{|y+j^{2}|\leq j\\}$, and
$\Omega_{9}=\\{|x|\leq 2j^{2},|y|\leq
2j^{2}\\}\setminus\cup_{i=1}^{8}\Omega_{i}$. On $\Omega_{9}$, we have

$\displaystyle\begin{split}\left|\iint_{\Omega_{9}}Q_{j}(x,y)h(x)h(y)dxdy\right|&\leq
C\int_{j}^{j^{2}}\int_{j}^{j^{2}}\frac{dxdy}{(1+x^{2})(1+y^{2})}=O(1/j^{2}).\end{split}$
(A.6)

Together with (A.3), (A.6) and (A.5), we have

$\displaystyle\begin{split}\frac{1}{2\pi}&\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)h(x)h(y)dxdy\\\
=&\
\frac{1}{2\pi}\iint_{\Omega_{7}}Q_{j}(x,y)dxdy+\frac{1}{2\pi}\iint_{\Omega_{8}}Q_{j}(x,y)dxdy+O(1/j)\\\
=&\
\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)dxdy+O(1/j)=4\pi+O(1/j).\end{split}$
(A.7)

This is (5.26). Once we obtain (5.25), (5.26) and Lemma 5.5, it follows that

$\displaystyle\begin{split}\int_{\mathbb{R}}h\mathcal{L}_{\boldsymbol{\psi}_{j}}h=&\
\int_{\mathbb{R}}h(-\Delta_{\mathbb{R}})^{\frac{1}{2}}h-\int_{\mathbb{R}}\rho_{j}h^{2}+\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)h(x)h(y)dxdy\\\
=&\ O(1/\log j)\end{split}$ (A.8)

and

$\displaystyle\begin{split}\langle
h\boldsymbol{\psi}_{j}^{\perp},h\boldsymbol{\psi}_{j}^{\perp}\rangle&=\int_{\mathbb{R}}h(-\Delta_{\mathbb{R}})^{\frac{1}{2}}h+\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)h(x)h(y)dxdy\\\
&=4\pi+O(1/\log j).\end{split}$ (A.9)

∎

###### Proof of Lemma 5.7.

Since $h$ is odd and $Q(x,y)=Q(y,x)=Q(-x,-y)$, it is easy to get

$\displaystyle\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{1}\boldsymbol{\psi}_{j}^{\perp}\rangle$
$\displaystyle=\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)h(x)dxdy=0.$
(A.10)

By a direct computation, we have

$\displaystyle\begin{split}\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{2}\boldsymbol{\psi}_{j}^{\perp}\rangle&=\int_{\mathbb{R}}h(-\Delta_{\mathbb{R}})^{\frac{1}{2}}K_{2}+\frac{1}{2\pi}\iint_{\mathbb{R}\times\mathbb{R}}Q_{j}(x,y)h(x)K_{2}(y)dxdy\\\
&=\int_{\mathbb{R}}\frac{2(1-(x-j^{2})^{2})h(x)}{[1+(x-j^{2})^{2}]^{2}}dx+\int_{\mathbb{R}}\frac{2(3+(x+j^{2})^{2})h(x)}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx\\\
&=I_{1}+I_{2}.\end{split}$

Since

$\displaystyle\begin{split}I_{1}=\int_{|x|\leq
j}\frac{2(1-x^{2})h(x)}{(1+x^{2})^{2}}&\leq\left|\int_{|x|\leq
j}\frac{2(1-x^{2})}{(1+x^{2})^{2}}dx\right|+\int_{|x|\geq
j}\left|\frac{2(1-x^{2})}{(1+x^{2})^{2}}\right|dx\\\ &=O(1/j),\end{split}$
(A.11)

and

$\displaystyle\begin{split}I_{2}&=\int_{\mathbb{R}}\frac{2h(x)}{1+(x-j^{2})^{2}}dx+\int_{\mathbb{R}}\frac{4h(x)}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx\\\
&=2\arctan j+O(1/j^{2})=\pi+O(1/j).\end{split}$ (A.12)

Then we get $\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{2}\boldsymbol{\psi}_{j}^{\perp}\rangle=\pi+O(1/j)$.

We also have

$\displaystyle\begin{split}\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{3}\boldsymbol{\psi}_{j}^{\perp}\rangle&=\int_{\mathbb{R}}\frac{4(x-j^{2})h(x)}{[1+(x-j^{2})^{2}]^{2}}dx+\int_{\mathbb{R}}\frac{4(x-j^{2})h(x)}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx.\end{split}$

Since

$\displaystyle\begin{split}\left|\int_{\mathbb{R}}\frac{4(x-j^{2})h(x)}{[1+(x-j^{2})^{2}]^{2}}dx\right|\leq\int_{|x|\geq
j}\frac{4x}{(1+x^{2})^{2}}dx=O(1/j^{2}),\end{split}$ (A.13)

and

$\displaystyle\begin{split}&\left|\int_{\mathbb{R}}\frac{4(x-j^{2})h(x)}{[1+(x-j^{2})^{2}][1+(x+j^{2})^{2}]}dx\right|\\\
&\leq\left|\int_{|x-j^{2}|\leq
j}\frac{4(x-j^{2})}{j^{4}[1+(x-j^{2})^{2}]}dx\right|+\left|\int_{|x-j^{2}|\geq
j}\frac{4}{j[1+(x+j^{2})^{2}]}dx\right|=O(1/j^{2}).\end{split}$ (A.14)

Then we get $\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{3}\boldsymbol{\psi}_{j}^{\perp}\rangle=O(1/j^{2})$.
Similarly, note that $h(x)=-1$ when $|x+j^{2}|\leq j$, we can get

$\displaystyle\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{4}\boldsymbol{\psi}_{j}^{\perp}\rangle=-\pi+O(1/j),\quad\langle
h\boldsymbol{\psi}_{j}^{\perp},K_{5}\boldsymbol{\psi}_{j}^{\perp}\rangle=O(1/j^{2}).$
(A.15)

∎

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