Proof. We only establish the first point, the second being similar. Let $u\in SBV^2_{\text {loc}}(\Omega ,\mathbb {C})$ such that $(u^+)^2=(u^-)^2$ on its jump set $J_u$ . By the chain rule for $BV$ -functions (see [Reference Ambrosio, Fusco and PallaraAFP00, Theorem 3.96]), for every $f\in C^1(\mathbb {C})$ with $\|Df(z)\|_{\infty }\le C(1+|z|)$ for some $C\ge 0$ , $f\circ u\in SBV_{\text {loc}}(\Omega )$ and

Setting $v:=u^2$ and applying this formula with $f(z)=z^2$ , we get

In particular, $v\in W^{1,1}_{\text {loc}}(\Omega )$ . We first consider the domain, $\Omega ':=\{x\in \Omega :u(x)\ne 0\}$ . With the abuse of notation $(\partial _1 u, \partial _2u):=\nabla _a u$ , we have,

(2.2) $$ \begin{align} \operatorname{\mathrm{curl}}_a u = \partial_1 u_2-\partial_2 u_1. \end{align} $$

Now, we express the components of v in terms of the components of u and use the chain rule to compute their partial derivatives. Almost everywhere in $\Omega '$ , there hold

$$\begin{align*}v_1=u_1^2-u_2^2,\qquad v_2=2u_1u_2,\qquad |v|=u_1^2 + u_2^2, \end{align*}$$

which yield

$$ \begin{align*} \partial_1 v_2 - \partial_2 v_1&=2\left[u_1\partial_1 u_2 + u_2\partial_1 u_1 - u_1\partial_2 u_1 + u_2\partial_2 u_2 \right],\\ \partial_1 v_1 + \partial_2 v_2&=2\left[u_1\partial_1 u_1 - u_2\partial_1 u_2 + u_1\partial_2 u_2 + u_2\partial_2 u_1 \right]. \end{align*} $$

Using these identities in the formula of Definition 1.4 and simplifying, we get, almost everywhere in $\Omega '$ and with $(\rho _1,\rho _2):=\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v$ ,

$$ \begin{align*} |v|\rho_1&=u_1^2\left[u_1\partial_1 u_2 + u_2\partial_1 u_1 - u_1\partial_2 u_1 + u_2\partial_2 u_2 \right]\\&\qquad\qquad -u_1u_2\left[u_1\partial_1 u_1 - u_2\partial_1 u_2 + u_1\partial_2 u_2 + u_2\partial_2 u_1 \right]\\&=|v|u_1(\partial_1u_2 -\partial_2u_1)\, \stackrel{(2.2)}=\, |v|u_1\operatorname{\mathrm{curl}}_a u,\\|v|\rho_2&=u_1u_2\left[u_1\partial_1 u_2 + u_2\partial_1 u_1 - u_1\partial_2 u_1 + u_2\partial_2 u_2 \right]\\&\qquad\qquad -u_2^2\left[u_1\partial_1 u_1 - u_2\partial_1 u_2 + u_1\partial_2 u_2 + u_2\partial_2 u_1 \right]\\&=|v|u_2(\partial_1u_2 -\partial_2u_1)\, \stackrel{(2.2)}=\, |v|u_2\operatorname{\mathrm{curl}}_a u. \end{align*} $$

We obtain the desired identity $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v = (\operatorname {\mathrm {curl}}_a u)\boldsymbol {u}$ almost everywhere in $\Omega '$ .

In the remaining region $\Omega ":=\{x\in \Omega :u(x)=0\}$ , we have $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ by definition and $D_au=0$ , hence

(2.3) $$ \begin{align}\operatorname{\mathrm{curl}}_a u=0\text{ almost everywhere in }\Omega" \end{align} $$

(see below). Therefore, the identity $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v = (\operatorname {\mathrm {curl}}_a \boldsymbol {u})\boldsymbol {u}$ holds true almost everywhere in $\Omega $ and since both sides of the identity define locally integrable functions, the identity holds in $L^1_{\text {loc}}(\Omega )$ . This ends the proof of the lemma.

Justification of equation (2.3): Let $T\in C^1(\mathbb {C})$ such that $T(z)=z$ for $|z|\le 1$ , $T(z)=z/|z|$ for $|z|\ge 2$ . For $\delta>0$ , let us set $u_{\delta }:=\delta T(\delta ^{-1}u)$ . On one hand, we have $u_{\delta } \to 0$ in $L^1_{\text {loc}}(\Omega )$ as $\delta \downarrow 0$ so $D u_{\delta }\to 0$ in the sense of distributions. On the other hand, by the chain rule

(2.4)

As $\delta \downarrow 0$ , the second term in the right-hand side goes to $0$ in $\mathcal {D}'(\Omega )$ and the first term converges to $ \boldsymbol {1}_{\Omega "}D_a u$ in $L^1_{\text {loc}}(\Omega )$ by the dominated convergence theorem. Identifying with the limit $D u_{\delta }\to 0$ in $\mathcal {D}'(\Omega )$ , we get $D_au=0$ in $\Omega "$ .

Proof. (i). Let $v:=u^2$ with u as in (i). The condition $E_{\varepsilon }'(\boldsymbol {u}\otimes \boldsymbol {u})<\infty $ implies $u\in SBV^2(\Omega )$ and from the assumption on $\Omega $ also $u\in L^2(\Omega )$ . As a consequence $v\in L^1(\Omega )$ . By Lemma 2.1(i) (replacing $SBV_{\text {loc}}^2$ by $SBV^2$ in the hypothesis), we have $v\in W^{1,1}(\Omega ,\mathbb {C})$ with $\nabla v=2u\nabla _au$ and $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ . Moreover, using the convention $|\nabla v|^2/|v|=0$ wherever $v=0$ , we have

$$\begin{align*}\int_{\Omega} \dfrac{|\nabla v|^2}{4|v|}\, = \int_{\Omega} |\nabla_au|^2. \end{align*}$$

Eventually, $(1-|v|)^2=(1-|u|^2)^2$ from which (2.5) follows.

(ii). Let v be as in (ii).

Step 1. Selection of a $BV$ square root u of v. We would like to define a square root of v of the form $u_{\varphi }:=e^{i\varphi /2} \sigma (e^{-i\varphi } v)$ for some $\varphi \in \mathbb {R}$ . We follow a strategy similar to the one of [Reference Dávila and IgnatDI03]. To deal with the discontinuity of $\sigma $ through $(-\infty ,0)$ and the fact that $z\mapsto |\sigma (z)|=\sqrt {|z|}$ is not smooth at 0, we need to smooth out $\sigma $ . First, we introduce $\chi \in C^{\infty }(\mathbb {R}_+)$ such that $\chi (s)=s$ for $0\le s\le 1$ , $0\le \chi '\le 1$ and $\chi (s)=1$ for $s\ge 2$ . Next, we consider for $0<\delta <\pi $ a $2\pi $ -periodic odd function $f_{\delta } \in C^{\infty }(\mathbb {R},(-\pi /2,\pi /2))$ such that

$$\begin{align*}\begin{array}{rlrl} &f_{\delta}(\theta)=\theta/2&\text{ for }&\theta\in[0, \pi-\delta],\\ 1/2&\ge f_{\delta}'(\theta)\ge 0 &\text{ for }&\theta\in[\pi -\delta, \pi-\delta/2],\\ 0&\ge f_{\delta}'(\theta)\ge -2\pi/\delta & \text{ for }&\theta\in[ \pi-\delta/2, \pi]. \end{array} \end{align*}$$

For $z=r e^{i\theta }\in \mathbb {C}$ and $0<\delta <\pi $ , we define $\sigma ^{\delta }(z):=\sqrt {r}f_{\delta }(\theta )$ and for $0<\delta <\pi $ , $\lambda \ge 1$ ,

$$\begin{align*}\sigma^{\delta,\lambda}(z):=\chi(\lambda|z|) \sigma^{\delta}(z). \end{align*}$$

Eventually, we set

$$\begin{align*}u_{\varphi}^{\delta,\lambda} := e^{i\varphi/2}\sigma^{\delta,\lambda}(e^{-i\varphi}v). \end{align*}$$

By construction, $u_{\varphi }^{\delta ,\lambda }\in W^{1,1}(\Omega ,\mathbb {C})$ and $(u_{\varphi }^{\delta ,\lambda })^2\to v$ pointwise, uniformly in $\varphi $ as $\delta \downarrow 0$ and $\lambda \uparrow \infty $ . We also easily see that

$$\begin{align*}\int_0^{2\pi}|\nabla u_{\varphi}^{\delta,\lambda}|\, d\varphi \le C \dfrac{|\nabla v|}{\sqrt{|v|}}\qquad\text{almost everywhere in }\Omega, \end{align*}$$

for some universal constant $C>0$ . Integrating on $\Omega $ and using Fubini, we have

$$\begin{align*}\int_0^{2\pi} \left(\int_{\Omega} |\nabla u_{\varphi}^{\delta,\lambda}|\,\right)\, d\varphi \le C \int_{\Omega} \dfrac{|\nabla v|}{\sqrt{|v|}}\le |\Omega|^{1/2}\left(\int_{\Omega} \dfrac{|\nabla v|^2}{|v|}\,\right)^{1/2}<\infty. \end{align*}$$

Let $(\delta _n)\downarrow 0$ and $(\lambda _n)\uparrow \infty $ ; we deduce that there exists a sequence $(\varphi _n)\subset [0,2\pi )$ such that $(u_{\varphi _n}^{\delta _n,\lambda _n})$ is bounded in $W^{1,1}(\Omega ,\mathbb {C})$ . Therefore, there exists $u\in BV(\Omega ,\mathbb {C})$ such that, up to extraction, $u_{\varphi _n}^{\delta _n,\lambda _n}\to u$ almost everywhere and weakly star in $BV(\Omega ,\mathbb {C})$ . Passing to the limit in the relation $({u_{\varphi _n}^{\delta _n,\lambda _n}})^2\to v$ as $n\uparrow \infty $ , we get $u^2=v$ , with $u\in BV(\Omega ,\mathbb {C})$ .

Step 2. Properties of u. From the chain rule for $BV$ functions, we have

By identification, we obtain $\nabla v=2u\nabla _au$ , $\nabla _c u=0$ and $(u^+)^2=(u^-)^2 \ \mathcal {H}^1$ -almost everywhere on $J_u$ . Using $|u|^2=|v|$ , we have

$$\begin{align*}\int_{\Omega} |\nabla_a u|^2\, = \int_{\Omega} \dfrac{|\nabla v|^2}{4|v|}\,<\infty, \end{align*}$$

hence $u\in SBV^2(\Omega ,\mathbb {C})$ . Eventually, by Lemma 2.1 (i), we obtain $\operatorname {\mathrm {curl}}_au=0$ , and from the above identity, we get equation (2.5).

Proof. The first point is the counterpart of [Reference Ignat and MerletIM12, Proposition 4] with the changes $\boldsymbol {u}\leftrightarrow \boldsymbol {m}:=\boldsymbol {u}^{\perp }$ , $\operatorname {\mathrm {curl}} u\leftrightarrow \nabla \cdot \boldsymbol {m}$ .

For the second point, if $\lambda \in C_a^{\infty }(\mathbb {T},\mathbb {C})$ , we easily check from the formula that $\Upsilon [\lambda ]$ is even, hence from the first point, $\Upsilon [\lambda ]\in \mathrm {ENT}_{\mathrm {ev}}$ .

Conversely, if $\Phi \in \mathrm {ENT}_{\mathrm {ev}}$ , by equation (i), there exists $\lambda \in C^{\infty }(\mathbb {T},\mathbb {C})$ such that $\Phi =\Upsilon [\lambda ]$ . Writing $\Phi (e^{i(\theta +\pi )})=\Phi (-e^{i\theta })=\Phi (e^{i\theta })$ and taking the dot product with $(-\sin \theta ,\cos \theta )$ , we obtain $-\lambda (\theta +\pi )=\lambda (\theta )$ so $\lambda \in C_a^{\infty }(\mathbb {T},\mathbb {C})$ , that is, $\Phi \in \Upsilon [C^{\infty }_a(\mathbb {T},\mathbb {C})]$ .

Proof. The first point is just a transposition of [Reference DeSimone, Kohn, Müller and OttoDKMO01, Lemmas 1 and 2] with the changes $\boldsymbol {u}\leftrightarrow \boldsymbol {u}^{\perp }=:\boldsymbol {m}$ and $\operatorname {\mathrm {curl}} \boldsymbol {u} \leftrightarrow \nabla \cdot \boldsymbol {m}$ . The statement about the supports of $\alpha $ and $\Psi $ is obvious from the explicit definitions of $\Psi $ and then $\alpha $ given there. We now assume that $\Phi $ is even, that is, $\Phi \in \mathrm {ENT}_{\mathrm {ev}}$ and establish (ii).

Step 1. Symmetrization. Differentiating the identity $\Phi (-z)=\Phi (z)$ and using the first part of the lemma, we obtain

$$\begin{align*}-D\Phi(-z)= -2\Psi(-z)\otimes \mathbf{z} -\alpha(-z)J = -2\Psi(z)\otimes \mathbf{z} + \alpha(z)J. \end{align*}$$

In view of these identities, we can substitute the mean value $\frac 12\left (\Psi (z)+\Psi (-z)\right )$ for $\Psi (z)$ and the quantity $\frac 12\left (\alpha (z)-\alpha (-z)\right )$ for $\alpha (z)$ . The new functions still comply to the conclusions of point (i), and we now have

(3.2) $$ \begin{align} \left\{\begin{array}{rcl}\Psi(-z)&=&\Psi(z),\\ \alpha(-z)&=&-\alpha(z),\end{array} \right. \ \text{for }z\in \mathbb{C}. \end{align} $$

From now on, we assume that $\Psi $ and $\alpha $ satisfy equation (3.2) and the properties stated in (i).

Step 2. Smoothing. Let $v\in W^{1,1}(\Omega ,\mathbb {C})$ . Let $(v_k)\subset C^{\infty }(\Omega ,\mathbb {C})$ be a sequence of approximations of v such that, as $k\uparrow \infty $ , $(\nabla v_k,v_k)\to (\nabla v,v)$ in $L^1(\Omega )$ and pointwise at every Lebesgue point of the mapping

$$\begin{align*}x\in\Omega\mapsto (\nabla v(x),v(x))\in\mathbb{C}^3.\end{align*}$$

By definitions (1.7) & (1.12) of $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v$ and $\widehat \mu _{\Phi }$ , the assumptions on the sequence $(v_k)$ yield

(3.3) $$ \begin{align} \begin{cases}\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v_k\to\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v,\\ \widehat\mu_{\Phi}[v_k]\to \widehat\mu_{\Phi}[v],\end{cases} \ \text{ at every Lebesgue point of }(\nabla v,v). \end{align} $$

Let us fix x, Lebesgue point of $(\nabla v,v)$ with $v(x)\ne 0$ , let us fix k large enough so that $v_k(x)\ne 0$ and let $\xi :=v_k(x)/|v_k(x)|$ . By continuity, there exists $0<r<\operatorname {\mathrm {dist}}(x,\mathbb {R}^2\backslash \Omega )$ (depending on k) such that $\boldsymbol {v}_k\cdot \boldsymbol {\xi }>0$ in $B_{r}(x)$ . The function $v_k/\xi $ is smooth and takes values in $\{z\in \mathbb {C}:\Re z>0\}$ in $B_r(x)$ so $u_k:=\sigma (\xi )\sigma (v_k/\xi )$ is smooth in $B_{r}(x)$ and $(u_k)^2=v_k$ . Using Step 1 and recalling that $\Phi $ is even, we write in $B_r(x)$

$$\begin{align*}\widehat\mu_{\Phi}[v_k]= \mu_{\Phi}[u_k] = \Psi(u_k)\cdot\nabla(1-|u_k|^2) + \alpha(u_k)\operatorname{\mathrm{curl}} u_k. \end{align*}$$

Using Lemma 2.1 (i) and the identities, $u_k^2=v_k$ , $|v_k|=|u_k|^2$ , we get

(3.4) $$ \begin{align} \widehat\mu_{\Phi}[v_k]&=\Psi(u_k)\cdot\nabla(1-|v_k|)+ \dfrac{\alpha(u_k)}{|v_k|} \boldsymbol{u}_k\cdot\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v_k. \end{align} $$

Now, taking into account the symmetries (3.2), we define for $z\in \mathbb {C}$ ,

$$\begin{align*}{\widehat\Psi}(z):=\Psi(\sigma(z))=\Psi(-\sigma(z)),\qquad {\widehat\alpha}(z):=\dfrac{\alpha(\sigma(z))}{|z|}{\boldsymbol \sigma}(z)=\dfrac{\alpha(-\sigma(z))}{|z|}(-\boldsymbol \sigma(z)). \end{align*}$$

These mappings are smooth, supported in $B_{\sqrt {2}}\backslash B_{1/\sqrt {2}}$ and by construction, ${\widehat \alpha }(z)$ is collinear to $\boldsymbol \sigma (z)$ . Moreover, by equation (3.4), the identity (3.1) holds true in $B_r(x)$ with $v_k$ in place of v.

Step 3. Sending k to $+\infty $ and concluding.

Writing equation (3.1) with $v=v_k$ at point x, sending k to $+\infty $ and recalling

$$\begin{align*}(\nabla v_k(x),v_k(x))\ \stackrel{k\uparrow\infty}\longrightarrow\ (\nabla v(x),v(x)), \end{align*}$$

and equation (3.3), we obtain equation (3.1) at point x. This holds at every Lebesgue point x of $(\nabla v,v)$ such that $v(x)\ne 0$ . At Lebesgue points with $v(x)=0$ , both sides of the identity vanish (because $\chi _0,{\widehat \Psi },{\widehat \alpha }= 0$ in the neighborhood of 0) so equation (3.1) holds almost everywhere in $\Omega $ . Eventually, since both sides define locally integrable functions, the identity holds in $L^1_{\text {loc}}(\Omega )$ .

Proof. Let $v\in W^{1,1}_{\text {loc}}(\Omega ,\mathbb {S}^1)$ and $\lambda _0\in C_a^{\infty }(\mathbb {T},\mathbb {C})$ as in the statement of the lemma. The implication (iii) $\Rightarrow $ (i) is obvious; we prove below the implications (i) $\Rightarrow $ (ii) and then (ii) $\Rightarrow $ (iii). In both cases, we use a $BV_{\text {loc}}$ lifting of v.

Step 1. Choose a good lifting of v and expressions of $\widehat \mu _{\Phi }[v]$ and $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v$ .

Looking at the proof of Proposition 2.2 (or directly using [Reference Dávila and IgnatDI03, Reference MerletMer06]), we see that there exists $\theta \in BV_{\text {loc}}(\Omega )$ such that $v=e^{2i\theta }$ . By the chain rule for $BV$ -functions, we have with obvious notation

Identifying, we have $D_c\theta =0$ , $\theta ^+-\theta ^-\in \pi \mathbb {Z}\ \mathcal {H}^1$ -almost everywhere on $J_{\theta }$ and, denoting $(\partial _1\theta ,\partial _2\theta ):=\nabla _a \theta $ ,

(3.5) $$ \begin{align} \partial_j v = 2(-\sin(2\theta) + i \cos(2\theta))\partial_j\theta,\qquad \text{for }j=1,2. \end{align} $$

Let $\Phi \in \mathrm {ENT}_{\mathrm {ev}}$ , and $\lambda \in C^{\infty }_a(\mathbb {T},\mathbb {C})$ be such that $\Phi =\Upsilon [\lambda ]$ . Using the chain rule and denoting $R[\theta ]:=\cos (\theta )\partial _1\theta + \sin (\theta )\partial _2\theta $ , we compute

(3.6) $$ \begin{align}\widehat\mu_{\Phi}[v]=\nabla\cdot\left[\Upsilon[\lambda](e^{i\theta})\right] =-\left[(\lambda"+\lambda)(\theta)\right] \,R[\theta]. \end{align} $$

Similarly, using again the notation $(\rho _1,\rho _2):=\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v$ and (3.5), there holds

$$ \begin{align*} 2\rho_1&=\left(\cos(2\theta) + 1\right)(\cos(2\theta)\partial_1\theta + \sin(2\theta)\partial_2\theta) \\ &~\qquad\qquad \qquad\qquad\qquad\qquad-\sin(2\theta) (-\sin(2\theta)\partial_1\theta + \cos(2\theta)\partial_2\theta)\\ & =\left[\left(\cos(2\theta) + 1\right)\cos(2\theta) + \sin(2\theta) \sin(2\theta) \right]\partial_1\theta \\ &~\qquad\qquad\qquad\qquad\qquad\qquad+\left[\left(\cos(2\theta) + 1\right)\sin(2\theta) - \sin(2\theta) \cos(2\theta)\right]\partial_2\theta\\ &=\left(\cos(2\theta) + 1\right)\partial_1\theta + \sin(2\theta) \partial_2\theta\\ &=2\cos(\theta)R[\theta], \end{align*} $$

and

$$ \begin{align*} 2\rho_2&=\sin(2\theta)\cos(2\theta)\partial_1\theta + \sin(2\theta)\partial_2\theta\\ &~\qquad\qquad \qquad\qquad\qquad\qquad + \left( \cos(2\theta) -1\right) (-\sin(2\theta)\partial_1\theta + \cos(2\theta)\partial_2\theta )\\ & =\left[\sin(2\theta)\cos(2\theta) - \left(\cos(2\theta) - 1\right)\sin(2\theta)\right]\partial_1\theta \\ &~\qquad\qquad\qquad\qquad\qquad\qquad+\left[\sin^2(2\theta)+\left(\cos(2\theta) - 1\right)\cos(2\theta)\right]\partial_2\theta\\ &=\sin(2\theta)\partial_1\theta +\left(1-\cos(2\theta)\right) \partial_2\theta\\ &=2\sin(\theta)R[\theta]. \end{align*} $$

Hence, we have the identity

(3.7) $$ \begin{align} \operatorname{\mathrm{\mathbf{\widehat{curl}}}} v = R[\theta]\binom{\cos\theta}{\sin\theta}. \end{align} $$

Step 2. (i) $\Rightarrow $ (ii). Let us assume that $\mu _{\Upsilon [\lambda _0]}[v]=0$ . Let

$$\begin{align*}Z_0:=\left\{\varphi\in(-\pi,\pi]:(\lambda_0^{\prime\prime}+\lambda_0)(\varphi)=0\right\}.\end{align*}$$

Using equation (3.6) with $\Phi =\Upsilon [\lambda _0]$ , we obtain that $R[\theta ]=0$ almost everywhere in $\Omega \backslash \theta ^{-1}\left (Z_0\right )$ and by equation (3.7), $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ almost everywhere in $\Omega \backslash \theta ^{-1}\left (Z_0\right )$ .

The complement of $\theta ^{-1}\left (Z_0\right )$ is the union of the sets $v^{-1}(e^{2i\varphi })$ for $\varphi \in Z_0$ . By assumption, this family of sets is at most countable and on each one $\nabla v$ vanishes almost everywhere by a standard property of Sobolev functions (see the justification of equation (2.3) where the case of $SBV$ functions is treated). We conclude that $\nabla v=0$ in $\theta ^{-1}\left (Z_0\right )$ , hence $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ almost everywhere in $\Omega $ which is (ii).

Step 3. (ii) $\Rightarrow $ (iii). We assume that $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ so that equation (3.7) implies $R[\theta ]=0$ . From equation (3.6), we deduce that $\widehat \mu _{\Phi }[v]=0$ for every $\Phi \in \mathrm {ENT}_{\mathrm {ev}}$ , as required. We have established $(iii)\Rightarrow (i)\Rightarrow (ii)\Rightarrow (iii)$ , hence the lemma.

Properties 4.1.

1. (i) For $n\in \mathbb {Z}$ , we have $\Phi ^{-n}=-\overline {\Phi ^n}$ . Indeed, from equation (1.13), the mapping $\lambda \mapsto \Upsilon [\lambda ]$ is $\mathbb {C}$ -linear and $\Upsilon [\lambda ]$ is real valued whenever $\lambda $ is, consequently, $\Upsilon [\overline \lambda ]=\overline {\Upsilon [\lambda ]}$ . Hence,

   $$\begin{align*}\Phi^{-n}=\Upsilon[2ie_{-n}]=\Upsilon[-\overline{2ie_n}]=-\overline{\Upsilon[2ie_n]}=-\overline{\Phi^n}. \end{align*}$$

2. (ii) By point (i) of Lemma 3.2 and density of the trigonometric polynomials in $C^{\infty }(\mathbb {T},\mathbb {C})$ , $\operatorname {\mathrm {span}}\{\Phi ^n\}_{n\in \mathbb {Z}}$ is dense in $\operatorname {\mathrm {ENT}}$ .

3. (iii) We have $e_n\in C^{\infty }_a(\mathbb {T},\mathbb {C})$ if and only if n is odd. In fact, $\operatorname {\mathrm {span}} \{e_{2m+1}\}_{m\in \mathbb {Z}}$ is dense in $C_a^{\infty }(\mathbb {T},\mathbb {C})$ . From Lemma 3.2 (ii), we see that $\operatorname {\mathrm {span}}\{\Phi ^{2m+1}\}_{m\in \mathbb {Z}}$ is dense in $\mathrm {ENT}_{\mathrm {ev}}$ .

Proof. Let us establish the first point. We write $z=e^{i\theta }$ . Using $\frac {d}{d\theta }e^{in\theta }=ine^{in\theta }$ and then the formulas $2i\sin \theta =e^{i\theta }-e^{-i\theta }$ , $2\cos \theta =e^{i\theta }+e^{-i\theta }$ , we get

$$ \begin{align*} \Phi^n(z)&= e^{in\theta} \left[2i \binom{-\sin\theta}{\cos\theta}+2n\binom{\cos\theta}{\sin\theta}\right]=e^{in\theta}\left[\binom{-e^{i\theta}+e^{-i\theta}}{ie^{i\theta}+ie^{-i\theta}}+n\binom{e^{i\theta}+e^{-i\theta}}{-ie^{i\theta}+ie^{-i\theta}}\right]\\ &= e^{in\theta}\binom{(n-1)e^{i\theta}+(n+1)e^{-i\theta}}{-i(n-1)e^{i\theta}+i(n+1)e^{-i\theta}} = e^{in\theta}\left[(n-1)e^{i\theta} \binom1{-i} +(n+1)e^{-i\theta} \binom1i\right]. \end{align*} $$

Substituting back $z=e^{i\theta }$ , $z^n=e^{in\theta }$ , we obtain equation (4.1).

Identity (4.2) is obtained by substituting $n=2m+1$ and $z=\sigma (w)$ in the first one and then using $z^2=w$ .

Properties 4.3.

1. (i) For $n=0$ , we have $\Phi ^0(z)=2i \boldsymbol {z}^{\perp }$ , so $\mu _{\Phi ^0}[u]=2i\operatorname {\mathrm {curl}} u$ for $u\in W^{1,1}(\Omega ,\mathbb {S}^1)$ .

2. (ii) For $n=\pm 1$ , we have

   $$\begin{align*}\Phi^{\pm1}(z)=\pm 2 \binom1{\pm i}, \end{align*}$$

   and the pair $\Phi ^{-1}$ , $\Phi ^{1}$ spans the space of constant vector fields $\mathbb {C}\to \mathbb {C}^2$ .

3. (iii) For $n=2$ and $z\in \mathbb {S}^1$ , we have $\Phi ^{\pm 2}(z)=\pm 6\,\Sigma _2(z)+ i\,6\,\Sigma _1(z)$ where for $z\in \mathbb {S}^1$ ,

   $$\begin{align*}\Sigma_1(z):=(z_2(1- (2/3)z_2^2), z_1(1-(2/3)z_1^2)),\qquad \Sigma_2(z):=(2/3)(z_1^3,-z_2^3) \end{align*}$$

   define the Jin–Kohn entropies.

4. (iv) For $n\in \mathbb {Z}$ and $\theta \in \mathbb {R}$ ,

   (4.3) $$ \begin{align} \left\|\Phi^n(e^{i\theta})\right\|_{\ell^2(\mathbb{C}^2)}=2\sqrt{n^2+1}\ \quad\text{and}\ \quad \left\|\frac d{d\theta}\left[\Phi^{n}(e^{i\theta})\right]\right\|_{\ell^2(\mathbb{C}^2)}\!=2|n^2-1|. \end{align} $$

Proof. The first equality is a direct application of equation (4.1), $\Phi ^{-n}=-\overline {\Phi ^n}$ and of the identities

(4.7) $$ \begin{align} \binom1i\wedge\binom1i= \binom1{-i}\wedge\binom1{-i}=0,\quad\binom1{-i}\wedge\binom1i= -\binom1i\wedge\binom1{-i}=2i. \end{align} $$

We get the second identity by taking $w=z$ in equation (4.4). For the last one, we use the bilinearity of the wedge product and the two firsts to obtain that the left-hand side of equation (4.6) is equal to:

$$ \begin{align*} 2i(n+1)^2 &\left[2-(\overline z w)^{n-1}-(z\overline w)^{n-1}\right] - 2i (n-1)^2 \left[2-(\overline z w)^{n+1}-(z\overline w)^{n+1}\right] \\ &=2i(n+1)^2 \left|z^{n-1}- w^{n-1}\right|{}^2- 2i(n-1)^2\left|z^{n+1}- w^{n+1}\right|{}^2. \notag \end{align*} $$

We used $z,w\in \mathbb {S}^1$ for the last equality. This proves equation (4.6) and the lemma.

Proof. Case of assumption (ii). Let m be a positive integer and set $n:= 2m+1$ . Integrating identity (4.5) of Lemma 4.4 with respect to $\nu $ we obtain for the left-hand side of equation (4.8),

(4.9) $$ \begin{align} \dfrac1{8i}\int_{\mathbb{S}^1}\left[\Phi^{n}\circ\sigma\wedge\Phi^{-n}\circ\sigma\right] d\nu= \dfrac14\left[(n+1)^2-(n-1)^2\right]=(m+1)^2-m^2. \end{align} $$

Next, using equation (4.2) and integrating with respect to $\nu $ , we get

$$\begin{align*}\dfrac{1}{2}\int_{\mathbb{S}^1} \Phi^{n}\circ\sigma\, d\nu = m \left[\int_{\mathbb{S}^1} w^{m+1} \, d\nu(w)\right] \binom1{-i} + (m+1) \left[\int_{\mathbb{S}^1} w^{m} \, d\nu(w)\right] \binom1i. \end{align*}$$

Denoting by

$$\begin{align*}c_k:=c_k(\nu)=\int_0^{2\pi} e^{-ik\theta}\,d\nu(e^{i\theta})=\int_{\mathbb{S}^1} w^{-k }\, d\nu(w), \end{align*}$$

the $k^{\text {th}}$ Fourier coefficient of the probability measure $\nu $ , the last identity writes as

$$\begin{align*}\dfrac{1}{2}\int_{\mathbb{S}^1} \Phi^{n}\circ\sigma\, d\nu =m\,c_{-(m+1)} \binom1{-i} + (m+1) c_{-m} \binom1i. \end{align*}$$

Using $\Phi ^{-n}=-\overline {\Phi ^{n}}$ , the identities (4.7) and the relations $c_{-k}=\overline {c_k}$ for $k\in \mathbb {Z}$ (because $\nu $ is real valued), we get, for the right-hand side of equation (4.8),

$$ \begin{align*} &\dfrac1{8i}\left[\int \Phi^{n}\circ\sigma\, d\nu\right]\wedge\left[\int\Phi^{-n}\circ\sigma\, d\nu\right]= \dfrac{-1}{2i} \left[\dfrac12\int \Phi^{n}\circ\sigma\, d\nu\right]\wedge\overline{\left[\dfrac12\int \Phi^{n}\circ\sigma\, d\nu\right]} \\ =& \dfrac{-1}{2i}\left[m\,c_{-(m+1)} \binom1{-i} + (m+1) c_{-m} \binom1i\right]\wedge\left[m\,\overline{c_{-(m+1)}} \binom1i + (m+1) \overline{c_{-m}} \binom1{-i}\right]\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad =(m+1)^2|c_m|^2 -m^2|c_{m+1}|^2. \end{align*} $$

By assumption this quantity is equal to equation (4.9). We deduce the relations

$$\begin{align*}m^2(1- |c_{m+1}|^2) = (m+1)^2(1-|c_m|^2)\quad\text{ for }m\ge1. \end{align*}$$

This leads by induction to $1-|c_m|^2=m^2(1-|c_1|^2)$ for $m\ge 1$ . Since $\nu $ is a (finite) measure, the sequence $(c_k)$ is bounded and we must have $|c_1|=|c_1(\nu )|=1$ . We conclude that the probability measure $\nu $ is a Dirac mass (we can for instance compute the variance $\operatorname {\mathrm {Var}}(\nu )=\int |w|^2\, d\nu (w) - \left |\int w\, d\nu (w)\right |{}^2=1-|c_1(\nu )|^2=0$ ).

Case of assumption (i). Performing the same computations with $n=2m$ in place of $n=2m+1$ and $\Phi ^{2m}$ in place of $\Phi ^{2m+1}\circ \sigma $ leads to the identities

$$\begin{align*}1-|c_{2m-1}(\nu)|^2=(2m-1)^2(1-|c_1(\nu)|^2)\quad\text{ for }m\ge 1.\end{align*}$$

We conclude again that $|c_1(\nu )|=1$ and then that $\nu $ is a Dirac mass.

Proof. It is enough to prove the claim for either $(\lambda _0,\lambda _1)=(1,0)$ or $(\lambda _0,\lambda _1)=(0,1)$ . Let ${\widehat \Psi }$ , ${\widehat \alpha }$ denote the functions given by Lemma 3.3 (ii) with the entropy $\Phi $ . Applying equation (3.1) to $v\in W^{1,1}(\Omega ,\mathbb {C})$ , we write

$$ \begin{align*} \nabla\cdot\left[\Phi(\sigma(v))\right] & = \nabla\cdot\left[\Phi(\sigma(v)) - (1-|v|){\widehat\Psi}(v)\right] + \nabla\cdot\left[(1-|v|){\widehat\Psi}(v)\right] \\ &=-(1-|v|) D{\widehat\Psi}(v) D v + {\widehat\alpha}(v)\cdot\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v+ \nabla\cdot\left[ (1-|v|){\widehat\Psi}(v)\right]\\ & =:\, f_1 + f_2 + f_3. \end{align*} $$

Let $\zeta \in C^1_c(\Omega ,\mathbb {C})$ . We estimate successively the terms $\left |\int f_j\zeta \,\right |$ for $j=1,2,3$ . For $f_2$ , we consider two different bounds.

1. (1) Since $D{\widehat \Psi }$ is bounded and supported in $B_{\sqrt {2}}\backslash B_{1/\sqrt {2}}$

   $$\begin{align*}\left|\int_{\Omega} f_1\zeta\,\right|\le C\left\|\chi(|v|)(1-|v|)\nabla v\right\|_{1}\|\zeta\|_{\infty}. \end{align*}$$

2. (2) Next, since ${\widehat \alpha }$ is bounded, we have on the one hand the bound

   $$\begin{align*}\left|\int_{\Omega} f_2\zeta\,\right| \leq C \|\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v\|_{1} \|\zeta\|_{\infty}. \end{align*}$$
   On the other hand, by definition of the $\|\cdot \|_{H^{-1}}$ norm, we also have
   $$ \begin{align*} \left|\int_{\Omega} f_2\zeta\,\right|&\le \|\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v\|_{H^{-1}(\Omega)} \left(\int_{\Omega}|\nabla[{\widehat\alpha}(v)\zeta]|^2\,\right)^{1/2}\\ &\le \|\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v\|_{H^{-1}(\Omega)} \left[\left(\int_{\Omega} |D{\widehat\alpha}(v)Dv|^2\,\right)^{1/2} \|\zeta\|_{\infty} + \|{\widehat\alpha}(v)\|_{\infty} \|\nabla \zeta\|_{2}\right]. \end{align*} $$
   As for ${\widehat \Psi }$ , the function ${\widehat \alpha }$ is bounded and supported in $B_{\sqrt {2}}\backslash B_{1/\sqrt {2}}$ and thus
   $$\begin{align*}\left|\int_{\Omega} f_2\zeta\,\right|\le C \|\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v\|_{H^{-1}(\Omega)} \Big[\left\|\chi(|v|)\nabla v\right\|_{2} \|\zeta\|_{\infty} + \|\nabla \zeta\|_{2}\Big]. \end{align*}$$

3. (3) In the last term, we integrate by parts and use the Cauchy–Schwarz inequality to get

   $$\begin{align*}\left|\int_{\Omega} f_3\zeta\,\right|=\left|\int_{\Omega} (1-|v|){\widehat\Psi}(v)\cdot\nabla\zeta\right|\le C \left\| \chi(|v|)(1-|v|)\right\|_{2}\|\nabla \zeta\|_{2}. \end{align*}$$
   Summing the estimates for $j=1,2,3$ , we conclude the proof of equation (5.2).

Proof. Step 1. Convergence of $(v_k)$ as a sequence of Young measures.

Up to extraction, there exists a positive Radon measure $\gamma $ over $\Omega \times \mathbb {C}$ such that, for every $\varphi \in C_c(\Omega \times \mathbb {C},\mathbb {R})$ ,

(5.3) $$ \begin{align} \int_{\Omega} \varphi(x,v_k(x))\,dx\ \stackrel{k\uparrow\infty}\longrightarrow\ \int_{\Omega\times\mathbb{C}} \varphi\,d\gamma. \end{align} $$

Moreover, $\gamma $ disintegrates as $\gamma =\nu _x\otimes \mathcal {L}^2$ where, for almost every $x\in \Omega $ , $\nu _x$ is a positive Radon measure on $\mathbb {C}$ . By assumption (i) and the fundamental theorem on Young measures (see, e.g., [Reference MüllerMül99, Theorem 3.1, (iii)&(iv)]), $\nu _x$ is a probability measure supported in $\mathbb {S}^1$ . Moreover, in order to prove the strong $L^1$ convergence of $(v_k)$ , it is enough to establish

(5.4) $$ \begin{align} \nu_x\text{ is a Dirac mass for almost every }x\in\Omega. \end{align} $$

This follows from assumption (i) again, [Reference MüllerMül99, Corollary 3.2] and Vitali convergence theorem. Let us now establish equation (5.4).

Step 2. $H^{-1}_{\text {loc}}$ compactness of the sequences of entropy productions. We observe that from assumption (ii) of the proposition and a simple variant of a lemma by Murat [Reference MuratMur81] (see also [Reference TartarTar79, Lemma 28] and [Reference DeSimone, Kohn, Müller and OttoDKMO01, Lemma 6]) applied to the (uniformly bounded) sequence of mappings $\Phi (\sigma (v_k))$ , for any $\Phi \in \mathrm {ENT}_{\mathrm {ev}}$ , the sequence $(\widehat \mu _{\Phi }[v_k])$ is relatively compact in $H^{-1}_{\text {loc}}(\Omega )$ .

Step 3. End of the proof.

Let $n=2m+1$ and $\Phi ^n$ be the corresponding trigonometric entropy. By Step 2, the sequences with terms $\widehat \mu _{\Phi ^{\pm n}}[v_k]=\nabla \cdot \left [\Phi ^{\pm n}(\sigma (v_k))\right ]$ are relatively compact in $H^{-1}_{\text {loc}}(\Omega )$ and we can apply the div-curl lemma of Murat and Tartar [Reference MuratMur78, Reference TartarTar79] to the pair of sequences

$$\begin{align*}\left(\Phi^{-n}(\sigma(v_k))\right),\qquad \left(\left[\Phi^n(\sigma(v_k))\right]^{\perp}\right), \end{align*}$$

to get

$$\begin{align*}\lim_{k\uparrow\infty}\Phi^{-n}(\sigma(v_k)) \wedge \Phi^n(\sigma(v_k))=\left[\lim_{k\uparrow\infty} \Phi^{-n}(\sigma(v_k))\right]\wedge \left[\lim_{k\uparrow\infty} \Phi^n(\sigma(v_k))\right], \end{align*}$$

where the above limits are weak limits in $L^2_{\text {loc}}(\Omega )$ . Rewriting this identity with the limit Young measure, we obtain, for almost every $x\in \Omega $ ,

$$\begin{align*}\int_{\mathbb{S}^1} \left[\Phi^{-n}\circ\sigma\wedge\Phi^n\circ\sigma\right]d\nu_x = \left[ \int_{\mathbb{S}^1} \Phi^{-n}\circ\sigma\,d\nu_x\right]\wedge\left[\int_{\mathbb{S}^1} \Phi^n\circ\sigma\,d\nu_x \right]. \end{align*}$$

This identity holds true almost everywhere for any fixed n. Since we consider a countable number of entropies, it holds true for any odd n on a set of full measure. For every x in this set, we can apply Lemma 4.8 (ii) to the probability $\nu _x$ and deduce that it is a Dirac mass, that is equation (5.4). This concludes the proof.

Proof. We first notice that by definition of $\widehat {E}_{\varepsilon }$ , and

(5.6) $$ \begin{align} \int_{\Omega} W(v)+\lambda_{\varepsilon}^1\|\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v\|^2_{H^{-1}(\Omega)}\le \varepsilon \widehat{E}_{\varepsilon}(v). \end{align} $$

Using equation (1.10) for W, this yields in particular

$$\begin{align*}\int_{\Omega} \min ((1-|v|)^2,|1-|v||) \le \varepsilon \widehat{E}_{\varepsilon}(v)/\kappa. \end{align*}$$

Since $\Omega $ is assumed to have finite area, we get first that if $ \widehat {E}_{\varepsilon _k}(v_k)$ is bounded, then $(|v_k|)$ converges to $1$ in $L^1(\Omega )$ . Moreover, recalling the definition (5.1) of $R(v)$ , equation (5.6) also gives

(5.7) $$ \begin{align} R(v)\le \left(\varepsilon \widehat{E}_{\varepsilon}(v)/\kappa\right)^{1/2}. \end{align} $$

Next, using Young’s inequality we find

$$ \begin{align*} \int_{\Omega} g^{1/2}(v)W^{1/2}(v) |\nabla v|+\lambda_{\varepsilon}^0\int_{\Omega} |\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v| + &\left(\lambda_{\varepsilon}^1 \|\operatorname{\mathrm{\mathbf{\widehat{curl}}}} v\|^2_{H^{-1}(\Omega)}\right)^{1/2}\left(\int_{\Omega} g(v) |\nabla v|^2\right)^{1/2}\\ &\qquad\qquad\qquad \le C \widehat{E}_{\varepsilon}(v). \end{align*} $$

Using again equation (1.10) and recalling the definition (5.1) of Q, we get

(5.8) $$ \begin{align} Q(v)\le C \widehat{E}_{\varepsilon}(v). \end{align} $$

From equations (5.7) and (5.8), we conclude that if $\widehat {E}_{\varepsilon _k}(v_k)$ is bounded, then $Q(v_k)$ is also bounded and $R(v_k)$ goes to zero. We may therefore apply Corollary 5.3 to conclude.

Proof of Proposition 6.1

Since the result is local, it is enough to prove it for $\omega $ a ball (we will only use the fact that $\omega $ is simply connected). Set $r:= \min \left (1/2,\operatorname {\mathrm {dist}}(\omega ,\mathbb {R}^2\backslash \Omega )/2\right )$ and $\omega ':= \omega +B_r\subset \subset \Omega $ (which is also simply connected). We fix in the proof a cutoff function $\zeta \in C^1_c(\Omega ,\mathbb {R}_+)$ supported in $\omega '$ and such that $\zeta \ge 1$ in $\omega $ .

We first introduce some notation.

1. (a) For every $x\in \omega $ , we denote by $\delta _h(x)$ the unique element of $(-1,1]$ such that $v(x+h)=e^{i\pi \delta _h(x)}v(x)$ .

2. (b) For $k\ge 0$ , we define

   $$\begin{align*}\omega_k:=\left\{x\in\omega : \dfrac12<2^k|\delta_h(x)|\le1\right\}. \end{align*}$$

3. (c) For positive odd integers n and $x\in \omega $ , we define (recall Definition 1.7 of the trigonometric entropies $\Phi ^n$ )

   $$\begin{align*}q_n(x):=\dfrac1{2i}\dfrac1{(n-1)^2(n+1)^2} \left[D_h\left[\Phi^{n}(\sigma(v))\right](x)\right]\wedge\left[D_h\left[\Phi^{-n}(\sigma(v))\right](x)\right]. \end{align*}$$

4. (d) For $k\ge 0$ and $x\in \omega $ , we set

   $$\begin{align*}Q_k(x):=\sum_{m=2^k}^{2^{k+1}-1}q_{2m+1}(x). \end{align*}$$

5. (e) Eventually, for $k\ge 0$ , we define the quantities

   $$\begin{align*}\mathcal{Q}_k:=\int_{\Omega} Q_k\zeta^2. \end{align*}$$
   (In this formula, the quantity $Q_k(x)$ is arbitrary for $x\in \Omega \backslash \omega $ (say $Q_k(x)=0$ ). This makes no difference since $\operatorname {\mathrm {supp}}\zeta \subset \omega $ .)

6. (f) In this proof, we use the notation $A\lesssim B$ to indicate that there is a universal constant $C\ge 0$ such that $A\le C B$ .

We split the proof into four steps. In the first two steps, we obtain bounds on the quantities $\mathcal {Q}_k$ from below and then from above. In Step 3, we multiply these bounds by $2^{k/2}$ and sum over $k\ge 0$ . Using Hölder and sending $|h|$ to $0$ leads to $v\in W^{1,3/2}(\omega )$ . This establishes point (i) of the lemma. By Lemma 3.4, we then get $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ , which is equation (ii). In the last step, we sum the bounds of Steps 1 and 2 over $k\in \{0,\dots ,k_0\}$ where $k_0:=\lceil \log _2(1/|h|)\rceil $ . Using $v\in W^{1,1}_{\text {loc}}(\Omega )$ , we obtain equation (6.1).

Step 1. Lower bound. We notice that since $\Phi ^{\pm (2m+1)}\in \mathrm {ENT}_{\mathrm {ev}}$ , we have

$$\begin{align*}\Phi^{\pm(2m+1)}(\sigma(\xi e^{2i\varphi}))=\Phi^{\pm(2m+1)}(\sigma(\xi)e^{i\varphi})\qquad\text{for any }\xi\in\mathbb{S}^1,\ \varphi\in\mathbb{R}. \end{align*}$$

In particular, for $x\in \omega $ and $m\ge 1$

$$\begin{align*}\Phi^{\pm(2m+1)}\left(\sigma(v(x+h))\right)= \Phi^{\pm(2m+1)}\left(e^{i(\pi/2)\delta_h(x)}\sigma(v(x))\right). \end{align*}$$

With this in mind, we apply equation (4.6) of Lemma 4.4 with $n=2m+1$ , $m\ge 1$ and $z=\sigma (v(x))$ , $w=\sigma (v(x+h))=e^{i(\pi /2)\delta _h(x)}\sigma (v(x))$ . We get in $\omega $

$$ \begin{align*} q_{2m+1}&=\dfrac{\left|e^{i\pi m\delta_h}-1\right|{}^2}{4m^2} - \dfrac{\left|e^{i\pi (m+1)\delta_h}-1\right|{}^2}{4(m+1)^2}\\ &=\dfrac{\sin^2\left(m(\pi/2)\delta_h\right)}{m^2} - \dfrac{\sin^2\left((m+1)(\pi/2)\delta_h\right)}{(m+1)^2}. \end{align*} $$

Summing these identities for m ranging over $\{2^k,\dots ,2^{k+1}-1\}$ , we get for $k\ge 0$

$$\begin{align*}Q_k =\dfrac{\sin^2\left((\pi/2)2^k\delta_h\right)}{2^{2k}}-\dfrac{\sin^2\left((\pi/2)2^{k+1}\delta_h\right)}{2^{2(k+1)}} =\dfrac1{2^{2k}}\left(\sin^2\theta-\dfrac14\sin^2(2\theta)\right), \end{align*}$$

with $\theta :=(\pi /2)2^k\delta _h$ . Using $\sin (2\theta )=2\sin \theta \cos \theta $ and $1-\cos ^2\theta =\sin ^2\theta $ , we obtain

$$\begin{align*}Q_k=\dfrac1{2^{2k}}(\sin^2\theta-\sin^2\theta\,\cos^2\theta)=\dfrac{\sin^4\theta}{2^{2k}} =\dfrac{\sin^4\left((\pi/2)2^k\delta_h\right)}{2^{2k}}. \end{align*}$$

For $x\in \omega _k$ , we have $1\ge 2^k|\delta _h(x)|>1/2$ ; hence, since $|\sin \theta |\ge (2/\pi )|\theta |$ for $|\theta |\le \pi /2$ ,

$$\begin{align*}Q_k\ge \dfrac1{2^{2k}}\left|2^k\delta_h\right|{}^4 =2^{2k}|\delta_h|^4\gtrsim |\delta_h|^2. \end{align*}$$

Multiplying by $\zeta ^2$ and integrating over $\Omega $ , we obtain

(6.2) $$ \begin{align} \mathcal{Q}_k \,\gtrsim\, \int_{\omega_k}|\delta_h|^2 \zeta^2\,. \end{align} $$

Step 2. Upper bound. Let $m\ge 1$ , since by assumption $\nabla \cdot [\Phi ^{-(2m+1)}(\sigma (v))]=0$ in the simply connected domain $\omega '\subset \subset \Omega $ , there exists $F^m\in \operatorname {\mathrm {Lip}}(\omega ',\mathbb {C})$ such that $\nabla ^{\perp } F^m=\Phi ^{-(2m+1)}(\sigma (v))$ . Using an integration by parts and $\nabla \cdot [\Phi ^{2m+1}(\sigma (v))]=0$ , we compute for $m\ge 1$

(6.3) $$ \begin{align} \int_\Omega q_{2m+1}\zeta^2 &= \dfrac1{4m^2(m+1)^2} \int_\Omega \left[D_h\nabla^\perp F^m\right]\wedge \left[D_h\Phi^{2m+1}(\sigma(v))\right]\zeta^2 \nonumber\\ &= -\dfrac1{4m^2(m+1)^2} \int_\Omega \left[\nabla D_hF^m\right] \cdot \left[D_h\Phi^{2m+1}(\sigma(v))\right]\zeta^2 \nonumber\\ &=\dfrac1{2m^2(m+1)^2} \int_\Omega D_hF^m \left[D_h\Phi^{2m+1}(\sigma(v))\right]\cdot [\zeta\ \nabla \zeta]. \end{align} $$

(To make the argument rigorous, we first regularize $\Phi ^{-2m-1}(\sigma (v))$ , perform the integration by parts and pass to the limit.)

Let us recall the identities (4.3). We have, for $\theta \in \mathbb {R}$ ,

$$ \begin{align*} \left\|\Phi^{\pm(2m+1)}(e^{i\theta})\right\|_{\ell^2(\mathbb{C}^2)}&=2 \sqrt{(2m+1)^2+1}\le 4(m+1),\\ \left\|\frac d{d\theta}\left[\Phi^{\pm(2m+1)}(e^{i\theta})\right]\right\|_{\ell^2(\mathbb{C}^2)}&=8m(m+1). \end{align*} $$

In particular, for $x\in \omega '$ ,

$$\begin{align*}\left\|\nabla F^m(x)\right\|_{\ell^2(\mathbb{C}^2)}=\left\|\Phi^{2m+1}(\sigma(v(x)))\right\|_{\ell^2(\mathbb{C}^2)}\lesssim m. \end{align*}$$

We use these bounds in the form

$$\begin{align*}\left\|D_hF^m(x)\right\|_{\ell^2(\mathbb{C}^2)}\lesssim m |h|,\qquad\quad \left\|D_h\Phi^{2m+1}(\sigma(v))\right\|_{\ell^2(\mathbb{C}^2)}\lesssim m \min\left(1,m|\delta_h|\right). \end{align*}$$

Using these inequalities to estimate the right-hand side of equation (6.3), we get

$$\begin{align*}\int_{\Omega} q_{2m+1}\zeta^2\, \lesssim\dfrac{|h|}{m^2} \int_{\Omega} \min\left(1,m|\delta_h|\right)\zeta |\nabla \zeta|\,. \end{align*}$$

Summing over $m\in \{2^k,\dots ,2^{k+1}-1\}$ , we obtain

(6.4) $$ \begin{align} \mathcal{Q}_k=\int_{\Omega} \,\sum_{m=2^k}^{2^{k+1}-1} q_{2m+1}\zeta^2 \lesssim \dfrac{|h|}{2^k} \int_{\Omega} \min\left(1,2^k|\delta_h|\right)\zeta |\nabla \zeta|\,. \end{align} $$

Step 3. Proof of equations (i) and (ii).

Multiplying equations (6.2) and (6.4) by $2^{k/2}$ and summing over $k\ge 0$ , we obtain

$$\begin{align*}\sum_{k\ge0} \int_{\omega_k} 2^{k/2}|\delta_h|^2 \zeta^2\,\lesssim\, \sum_{k\ge0} 2^{k/2}\mathcal{Q}_k\,\lesssim\, |h| \int_{\Omega} \left[\sum_{k\ge0}\dfrac1{2^{k/2}} \min\left(1,2^k|\delta_h|\right)\right]\zeta |\nabla \zeta|\,. \end{align*}$$

We use again $1/(2|\delta _h|)<2^k$ in $\omega _k$ to estimate the left-hand side from below. We get

(6.5) $$ \begin{align} \int_{\Omega} |\delta_h|^{3/2} \zeta^2\,\lesssim\, |h| \int_{\Omega} \left[\sum_{k\ge0}\dfrac1{2^{k/2}} \min\left(1,2^k|\delta_h|\right)\right]\zeta |\nabla \zeta|\,. \end{align} $$

We now estimate the right-hand side. For $K\ge 0$ , we have in $\omega _K$

$$\begin{align*}\sum_{k\ge0}\dfrac1{2^{k/2}} \min\left(1,2^k|\delta_h|\right)\lesssim \pi\sum_{k=0}^K 2^{k/2}|\delta_h| +\sum_{k>K} 2^{-k/2} \lesssim |\delta_h| 2^{K/2} +2^{-K/2} \lesssim |\delta_h|^{1/2} \end{align*}$$

where in the last inequality we used that $2^{-(K+1)}\le |\delta _h|\le 2^{-K}$ in $\omega _K$ .

Plugging this estimate in (6.5) and dividing by $|h|^{3/2}$ , we obtain

$$\begin{align*}\int_{\Omega} \left(\dfrac{|\delta_h|}{|h|}\right)^{3/2}\zeta^2\,\lesssim \int_{\Omega}\left(\dfrac{|\delta_h|}{|h|}\right)^{1/2}\zeta|\nabla\zeta|\,. \end{align*}$$

Applying Hölder inequality with parameters $p=3$ , $q=3/2$ to the functions $f=\left (|\delta _h|/|h|\right )^{1/2}\zeta ^{2/3}$ and $g=\zeta ^{1/3}|\nabla \zeta |$ and simplifying, we get

$$\begin{align*}\left(\int_{\Omega} \left(\dfrac{|\delta_h|}{|h|}\right)^{3/2}\zeta^2\,\right)^{2/3}\,\lesssim \left(\int_{\Omega}\zeta^{1/2}|\nabla\zeta|^{3/2}\,\right)^{2/3}. \end{align*}$$

Using that $|D_hv|\le \pi |\delta _h|$ , and a standard characterization of Sobolev spaces, we deduce that $v\in W^{3/2}_{\text {loc}}(\Omega ,\mathbb {S}^1)$ with the estimate

$$\begin{align*}\left(\int_{\Omega} |\nabla v|^{3/2}\zeta^2\,\right)^{2/3}\,\lesssim \left(\int_{\Omega}\zeta^{1/2}|\nabla\zeta|^{3/2}\,\right)^{2/3}. \end{align*}$$

Point (i) of the Lemma is established. As already explained at the beginning of this proof, since v is a zero-state point (ii) then follows from Lemma 3.4.

Step 4. Proof of equation (iii).

Let $k_0:=\lceil \log _2(1/|h|)\rceil $ , that is $k_0$ is the integer defined by $1/2 < 2^{k_0} |h| \le 1$ . Since $|h|\le 1/2$ , we have $k_0\ge 1$ . Summing equations (6.2) and (6.4) over $k\in \{0,\dots ,k_0-1\}$ , we get

(6.6) $$ \begin{align} \sum_{k=0}^{k_0-1} \int_{\omega_k} |\delta_h|^2 \zeta^2\,\lesssim\, \sum_{k=0}^{k_0-1} \mathcal{Q}_k\,\lesssim\, |h| \int_{\Omega} \left[\sum_{k=0}^{k_0-1}\dfrac1{2^{k}} \min\left(1,2^k|\delta_h|\right)\right]\zeta |\nabla \zeta|\,. \end{align} $$

We proceed as in Step 3 to estimate the right-hand side. Let us fix $K\ge 0$ , we estimate the sum in brackets in $\omega _K$ as follows

$$ \begin{align*} \sum_{k=0}^{k_0-1}\dfrac1{2^{k}} \min\left(1,2^k|\delta_h|\right) &\lesssim \sum_{k=0}^{\min(k_0,K)-1} |\delta_h|+ \sum_{k=\min(k_0,K)}^{k_0} \dfrac1{2^{k}}\\ &\lesssim \, k_0 |\delta_h| + 2^{-K} \lesssim \, ( k_0 + 1) |\delta_h| \lesssim\, |\delta_h| \ln(1/|h|). \end{align*} $$

In the last two inequalities, we used $2^{-K}\le 2|\delta _h|$ in $\omega _K$ and $1\le k_0\le \log _2(1/|h|)$ . Putting this inequality in (6.6), we obtain

$$\begin{align*}\sum_{k=0}^{k_0-1} \int_{\omega_k} |\delta_h|^2 \zeta^2\,\lesssim\, |h| \ln(1/|h|) \int_{\Omega} |\delta_h|\,\zeta |\nabla \zeta|\,. \end{align*}$$

Using $|\delta _h| \le 2 |D_h v|$ and recalling that by Step 3, $v\in W^{1,3/2}_{\text {loc}}(\Omega )\subset W^{1,1}_{\text {loc}}(\Omega )$ , we deduce

(6.7) $$ \begin{align}\sum_{k=0}^{k_0-1} \int_{\omega_k} |\delta_h|^2 \zeta^2\,\le C |h|^2 \ln(1/|h|), \end{align} $$

where $C>0$ only depends on $\zeta $ .

Eventually, for $k\ge k_0$ and $x\in \omega _k$ we have $|\delta _h(x)|\le 2^{-k_0}\le 2 |h|$ so that

$$\begin{align*}\sum_{k\ge k_0} \int_{\omega_k} |\delta_h|^2 \zeta^2\,\lesssim |h|^2 \int_{\Omega} \zeta^2. \end{align*}$$

Together with equation (6.7) (and $|h|\le 1/2$ ) this leads to

$$\begin{align*}\int_{\Omega} |\delta_h|^2 \zeta^2 \,\le C |h|^2 \ln(1/|h|). \end{align*}$$

Since $|D_h v|\le \pi |\delta _h|$ , we conclude the proof of equation (6.1).

Proof. Let us first notice that $v:=u^2\in \widehat {\mathcal {A}}_0(\Omega )$ so that Proposition 6.1 provides $v\in W^{1,3/2}_{\text {loc}}(\Omega )$ and a control on $|D_hv|$ of the form (6.1). In order to conclude, we only have to establish that for any simply connected open set $\omega \subset \subset \Omega $ and any $h\in \mathbb {R}^2$ such that $|h|\le \min \left (1/2,\operatorname {\mathrm {dist}}(\omega ,\mathbb {R}^2\backslash \Omega )/2\right )$ there holds

(6.9) $$ \begin{align} \mathcal{L}^2(A_h\cap\omega) \le C(\omega) |h|^2\qquad\text{ where }\qquad A_h:=\left\{x\in \Omega : \ |D_h u(x)|>\sqrt{2}\right\}. \end{align} $$

Indeed, assuming equation (6.9), we have for every $p>0 \ \zeta \in C^1_c(\Omega ,\mathbb {R}_+)$ and $h\in \mathbb {R}^2$ with $\displaystyle |h|\le \min \left (1/2,\operatorname {\mathrm {dist}}(\operatorname {\mathrm {supp}} \zeta ,\mathbb {R}^2\backslash \Omega )/2\right )$ ,

$$\begin{align*}\int_{\Omega} |D_h u|^p\zeta^2\,= \int_{A_h} |D_h u|^p \zeta^2\,+\int_{\Omega\backslash A_h} |D_h u|^p\zeta^2\,\le C(\zeta) |h|^2 + 2^{-p/2}\int_{\Omega} |D_h v|^p\zeta^2. \end{align*}$$

Here, we used equation (6.9) and $\sqrt 2|D_hu|\le |D_h v|$ in $\Omega \backslash A_h$ . Choosing $p=3/2$ and using $v\in W^{1,3/2}_{\text {loc}}(\Omega )$ , we deduce

$$\begin{align*}\int_{\Omega} |D_h u|^{3/2}\zeta^2 \le C(\zeta) |h|^{3/2}\qquad\text{for }h\in \mathbb{R}^2\text{ with }|h|\text{ small enough}, \end{align*}$$

and we conclude that $u\in W^{1,3/2}_{\text {loc}}(\Omega )$ . Similarly, choosing $p=2$ , equation (6.1) yields equation (6.8).

Let $\omega \subset \subset \Omega $ be an open ball, let $\zeta \in C^1_c(\omega ,\mathbb {R}_+)$ and let $h\in \mathbb {R}^2$ with

$$\begin{align*}|h|\le\operatorname{\mathrm{dist}}(\omega,\mathbb{R}^2\backslash\Omega)/2. \end{align*}$$

To establish equation (6.9), we proceed as in the proof of Proposition 6.1 but using the pair of Jin–Kohn entropies $\Phi ^{\pm 2}$ (the first computations below correspond to the beginning of the proof of [Reference Lamy, Lorent and PengLLP20, Lemma 7] rewritten with our notation). Let us set

$$\begin{align*}q_2(x):=\dfrac1{18\,i} \left[D_h\left[\Phi^{2}(u)\right](x)\right]\wedge\left[D_h\left[\Phi^{-2}(u)\right](x)\right]. \end{align*}$$

On the one hand, from equation (4.6) and elementary calculus,Footnote ¹ we have

(6.10) $$ \begin{align} \int_{\Omega} q_2\zeta^2 =\int_{\Omega}\left(\left|D_h u\right|{}^2 - \dfrac{\left|D_h [u^3]\right|{}^2}{9}\right)\zeta^2\, \ge\dfrac89 \int_{\Omega}\left|D_h u\right|{}^4\zeta^2\, \ge\dfrac{32}9 \int_{A_h} \zeta^2\,. \end{align} $$

On the other hand, using $\mu _{\Phi ^{\pm 2}}[u]=0$ we get, as in the proofs of Proposition 6.1 or of equation (1.18),

$$\begin{align*}\int_{\Omega} q_2\zeta^2 \le C\,|h|\int_{\Omega} |D_hu| \zeta\, =C\,|h|\int_{A_h\cup [\Omega\backslash A_h]} \!|D_hu|\zeta\, \le C'\left(|h|\left(\int_{A_h} \zeta^2\,\right)^{1/2} +|h|^2\right), \end{align*}$$

with $C,C'\ge 0$ depending on $\zeta $ . In the last estimate, we used the Cauchy–Schwarz inequality and $\sqrt 2|D_hu|\le |D_h v|$ in $\Omega \backslash A_h$ with $v\in W^{1,1}_{\text {loc}}(\Omega )$ . With equation (6.10), we obtain $\int _{A_h} \zeta ^2\le C(\zeta )|h|^2$ and then equation (6.9) thanks to a covering argument.

Proof. For the reader’s convenience, we recall some classical computations (see for instance [Reference De Lellis and IgnatDLI15, Reference Lorent and PengLP18]). We first compute for $x\in \omega $

$$ \begin{align*} \nabla v_{\eta}(x)&=\int_{B_{\eta}} v(x-y)\nabla \rho_{\eta}(y)\, dy = \int_{B_{\eta}} [v(x-y)-v_{\eta}(x)]\nabla \rho_{\eta}(y)\, dy\\ &=\int_{B_{\eta}\times B_{\eta}} [v(x-y)-v(x-z)]\nabla \rho_{\eta}(y)\, \rho_{\eta}(z) \,dy\, dz, \end{align*} $$

where we used $\int \nabla \rho _{\eta }=0$ . We deduce the estimate

$$\begin{align*}|\nabla v_{\eta}(x)|\le \frac{C}{\eta^3} \int_{B_{\eta}} \int_{B_{\eta}} |v(x-y)-v(x-z)|\rho_{\eta}(z)\, dz\, dy. \end{align*}$$

Squaring, integrating on $\omega $ , using Jensen inequality and Fubini, we obtain

$$\begin{align*}\int_{\omega} |\nabla v_{\eta}|^2 \le \dfrac {C}{\eta^4} \int_{B_{\eta}} \int_{\omega+B_{\eta}} |v(y)-v(y-h)|^2\, dy\, dh. \end{align*}$$

Using equation (6.1) from Proposition 6.1 (iii), we get

(6.12) $$ \begin{align} \int_{\omega} |\nabla v_{\eta}|^2 \le C \ln(1/\eta). \end{align} $$

Next, since v takes values in $\mathbb {S}^1$ , we have for $x\in \omega $ ,

$$ \begin{align*} 0\le 1-|v_{\eta}(x)|^2 &= \int_{B_{\eta}\times B_{\eta}} \left(1-v(x-y)\overline v(x-z)\right)\rho_{\eta}(y)\rho_{\eta}(z)\, dy\, dz\\ &=\dfrac12 \int_{B_{\eta}\times B_{\eta}}\left |v(x-y)-v(x-z)\right|{}^2\rho_{\eta}(y)\rho_{\eta}(z)\, dy\, dz, \end{align*} $$

where we used the trigonometric formulas $1-\cos \theta =2\sin ^2(\theta /2)=(1/2)|1-e^{i\theta }|^2$ . Integrating over $\omega $ , using Jensen inequality, Fubini and equation (6.1) as above, we get

$$\begin{align*}\int_{\omega} (1-|v_{\eta}|^2)^2\le C \eta^2 \ln(1/\eta). \end{align*}$$

Together withequation (6.12) this leads to equation (6.11).

Proof of Theorem 6.5 (Theorem 1.2)

The fact that $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ is established in equation (ii) of Proposition 6.1 so we only need to prove (i)–(iii).

Step 1. Identification of the singular set. By equation (6.11) of Lemma 6.4 and [Reference Alicandro and PonsiglioneAP14, Theorem 4.1], the Jacobians $\operatorname {\mathrm {curl}} (v_{\eta } \wedge \nabla v_{\eta })$ locally weakly converge as $\eta \downarrow 0$ (for the flat norm) to a measure $\mu =2\pi \sum _{i} z_i \delta _{x_i}$ with $z_i\in \mathbb {Z}$ . Moreover, for every $\omega \subset \subset \Omega $ , there exists $C,C'\ge 0$ depending on $\omega $ and $\Omega $ such that

$$\begin{align*}|\mu|(\omega)\le C \limsup_{\eta\downarrow0}\frac{\operatorname{\mathrm{GL}}_{\eta}(v_{\eta};\omega)}{\ln(1/\eta)} \stackrel{(6.11)}{\le} C'. \end{align*}$$

Therefore, the sum is locally finite. From equation (i) of Proposition 6.1, $v\in W^{1,3/2}_{\text {loc}}(\Omega )$ ; hence, $\nabla v_{\eta }$ converges strongly in $L^{3/2}_{\text {loc}}(\Omega )$ to $\nabla v$ and $v_{\eta }$ converges strongly in $L^6_{\text {loc}}(\Omega )$ to v so that $v_{\eta } \wedge \nabla v_{\eta }$ converges to $v\wedge \nabla v$ in $L^1_{\text {loc}}(\Omega )$ . Hence, $\mu =\operatorname {\mathrm {curl}} (v\wedge \nabla v)$ . We set

$$\begin{align*}S:=\operatorname{\mathrm{supp}} \mu \backslash\{x\in\Omega : v\text{ is continuous in some neighborhood of }x\}. \end{align*}$$

Step 2. Local Lipschitz regularity of v in $\Omega \backslash S$ .

Let $B=B_r(x)$ be an open ball such that $2B\subset \subset \Omega $ and $v_{|2B}\in C(2B,\mathbb {S}^1)$ or $2B\subset \subset \Omega \backslash \operatorname {\mathrm {supp}}\mu $ . Let us show that $v=e^{i\varphi }$ in $2B$ for some $\varphi \in W^{1,1}(2B)$ . Indeed, on the one hand if $v_{|2B}\in C(2B,\mathbb {S}^1)$ , there exists a lifting $\varphi \in C(2B)$ such that $v= e^{i\varphi }$ in $2B$ and since $v\in W^{1,3/2}(2B)$ and $|\nabla \varphi |=|\nabla v|$ , we have $\varphi \in W^{1,1}(2B)$ . On the other hand, if $2B\subset \subset \Omega \backslash \operatorname {\mathrm {supp}}\mu $ , since $v\in W^{1,3/2}(2B)$ and $\mu = \operatorname {\mathrm {curl}} (v\wedge \nabla v)=0$ on $2B$ , we can apply [Reference DemengelDem90, Reference Brezis, Mironescu and PonceBMP05] to get $\varphi \in W^{1,1}(2B)$ with $v= e^{i\varphi }$ in $2B$ .

We now set $\theta :=\varphi /2$ and $u:=e^{i\theta }$ so that $u^2=v$ with $u\in W^{1,1}(2B,\mathbb {S}^1)$ . From point (ii) of Proposition 6.1, we have $\operatorname {\mathrm {\mathbf {\widehat {curl}}}} v=0$ in $2B$ so that Lemma 2.1 (ii) implies $\operatorname {\mathrm {curl}} u=0$ . By the chain rule, this translates into

(6.13) $$ \begin{align} \boldsymbol{u}\cdot\nabla\theta=0\quad\text{almost everywhere in }2B. \end{align} $$

For $\lambda \in \mathbb {R}$ we define the level sets

By equation (6.13) and since $\theta \in W^{1,1}(2B)$ , for almost every $\lambda $ these sets have finite perimeter in $2B$ and the tangent on $\partial E_{\lambda }^{\pm }$ is collinear to $\boldsymbol e^{i\lambda }$ . Therefore, $E_{\lambda }^{\pm }$ is the intersection with $2B$ of a locally finite number of stripes parallel to $\boldsymbol e^{i\lambda }$ . Observing that for $\lambda _-\le \lambda _+$ the sets $E_{\lambda _-}^-$ and $E_{\lambda _+}^+$ do not intersect in $2B$ and taking into account the orientation of the stripes, we claim that for almost every $\lambda _-,\lambda _+$ with $\lambda _-<\lambda _+\le \lambda _-+\pi /2$

(6.14) $$ \begin{align} \operatorname{\mathrm{dist}}(E_{\lambda_-}^-\cap B, E_{\lambda_+}^+\cap B) \ge 2r\sin\left(\dfrac{\lambda_+-\lambda_-}2\right). \end{align} $$

Indeed, let $x_{\lambda _-}\in E_{\lambda _-}^-\cap B$ and $x_{\lambda _+}\in E_{\lambda _+}^+\cap B$ , where $\lambda _-<\lambda _+\le \lambda _-+\pi /2$ . Since $E_{\lambda _-}^-$ is made of stripes parallel to $\boldsymbol e^{i\lambda _-}$ , the line segment

is contained in $E_{\lambda _-}^-$ (see Figure 3). Similarly,

. Since $\lambda _+\not \equiv \lambda _- \pmod {\pi }$ , the lines spanned by $L_{\lambda _-}$ and $L_{\lambda _+}$ intersect, and since $L_{\lambda _-}\cap L_{\lambda _+}\subset E_{\lambda _-}^-\cap E_{\lambda _+}^+=\varnothing $ they do not intersect in $2B$ . Minimizing $\operatorname {\mathrm {dist}}(L_{\lambda _-}\cap B,L_{\lambda _+}\cap B)$ under these constraints, the minimizer is given up to rotation by

$$ \begin{align*} L_{\lambda_-} &= \left(x+\left[-2r+\mathbb{R} \boldsymbol e^{-\frac{i\beta}{2}}\right]\right)\cap 2B,\\ L_{\lambda_+} &= \left(x+\left[-2r+\mathbb{R} \boldsymbol e^{\frac{i\beta}{2}}\right]\right)\cap 2B, \end{align*} $$

with $\lambda _{\pm } = \pm \frac {\beta }{2}$ (see Figure 4). Is is then easy to see that

(6.15) $$ \begin{align} |x_{\lambda_-}-x_{\lambda_+}|\ge\operatorname{\mathrm{dist}}(L_{\lambda_-}\cap B, L_{\lambda_+}\cap B) \ge 2r\sin\left(\dfrac{\lambda_+-\lambda_-}2\right), \end{align} $$

hence the claim (6.14) by the arbitrariness of $x_{\lambda _-}\in E_{\lambda _-}^-\cap B$ and $x_{\lambda _+}\in E_{\lambda _+}^+\cap B$ .

Figure 3

Example of $E_{\lambda _-}^-$ (dashed region) and $L_{\lambda _-}$ .

Figure 4

As in equation (6.15) $d=\operatorname {\mathrm {dist}}(L_{\lambda _-}\cap B,L_{\lambda _+}\cap B)$ .

From equation (6.14), we deduce that $\theta $ admits a Lipschitz continuous representative in $ B$ with $\|\nabla \theta \|_{L^{\infty }(B)}\le 1/r$ . Therefore, $v=e^{2i\theta }$ admits a Lipschitz continuous representative in $\overline B$ with Lipschitz constant at most $2/r$ . Eventually, returning to equation (6.13), we see that for every $x\in B$ there holds

$$\begin{align*}v=v(x)\quad\text{on }[x+\mathbb{R} \boldsymbol e^{i\theta(x)}]\cap B. \end{align*}$$

Equivalently,

$$\begin{align*}v=v(x)\quad\text{on }[x+\mathbb{R}\boldsymbol \sigma(v(x))]\cap B. \end{align*}$$

We have established points (i) and (ii) of the theorem.

Step 3. Structure of v near the singularities. Let $x^0\in S$ and $r>0$ such that $2B=B_{2r}(x^0)\subset \subset \Omega $ with $2B\cap S=\{x^0\}$ . By translation and scaling, we assume that $B= B_1$ and we set $B':=B\backslash \{0\}$ , $2B':=2B\backslash \{0\}$ . Notice that by Step 2 and definition of S, v is continuous in $2B'$ and discontinuous at $0$ .

Step 3.a. Preliminaries. For $x\in 2B'$ , we denote by $L(x)$ the connected component of $[x+\mathbb {R}\boldsymbol \sigma (v(x))]$ in $2B'$ which contains x. $L(x)$ is an open segment in $\mathbb {R}^2$ with $L(x)\subset 2B'$ . There are two cases:

1. (a) Either $2B'\backslash L(x)$ splits into two connected components. In this case, the two endpoints of $L(x)$ lie on $\partial (2B)$ ; see Figure 5,

   Figure 5

   Case (a).

   

2. (b) or $L(x)$ is a radius of the form $(0,2 x/|x|)$ ; see Figure 6.

   Figure 6

   Case (b).