Proof. Notice that $v_N(x,y)$ is not harmonic in $\{v\neq 0\}$ , hence $v_N$ is not a minimizer of $J(-, R_N)$ . However, $\frac {\partial v_N}{\partial x}$ exists a.e. in $(-3N,3N)\times (-1,1)$ (see Lemma 2.1), and, where it exists, $\left |\frac {\partial v_N}{\partial x}\right |\le |f_N'(x)|\le \frac {1+\alpha }{N}\le \frac {2}{N}.$

Since $f(x)\equiv 1-\alpha $ for $|x|<N$ and $f(x)\equiv 2$ for $2N<|x|<3N$ , $\left |\frac {\partial v_N}{\partial x}\right |=0$ on $(-N,N)\times (-1,1)$ , on $(2N,3N)\times (-1,1)$ and on $(-3N,-2N)\times (-1,1)$ . Consequently,

$$\begin{align*}\iint_{\overline{R}_N}\left|\frac{\partial v_N}{\partial x}\right|^2=\int_{-2N}^{-N}\int_{-1}^1\left|\frac{\partial v_N}{\partial x}\right|^2+\int_{N}^{2N}\int_{-1}^1\left|\frac{\partial v_N}{\partial x}\right|^2\le \left(\frac{2}{N}\right)^2(2N)\leq \frac{16}{N}.\\[-36pt] \end{align*}$$

Proof. Since $u_N$ is a minimizer of J,

(2.5) $$ \begin{align} \iint_{\overline{R}_N} |\nabla u_N|^2+|\{u_N\neq 0\}|\le \iint_{\overline{R}_N}|\nabla v_N|^2+|\{v_N\neq 0\}|. \end{align} $$

Moreover, for every $x\in [-3N,3N]$ fixed, $v_N(x,\cdot )$ is a minimizer of $H_x$ , hence

$$\begin{align*}\int_{-1}^1\left|\frac{\partial v_N}{\partial y}\right|^2dy+|\{v_N(x,\cdot)\neq 0\}|\le\int_{-1}^1\left|\frac{\partial u_N}{\partial y}\right|^2dy+|\{u_N(x,\cdot)\neq 0\}|. \end{align*}$$

Integrating the last inequality on $[-3N,3N]$ leads to

(2.6) $$ \begin{align} \iint_{\overline{R}_N}\left|\frac{\partial v_N}{\partial y}\right|^2+|\{v_N \neq 0\}|\le\iint_{\overline{R}_N}\left|\frac{\partial u_N}{\partial y}\right|^2+|\{u_N \neq 0\}|. \end{align} $$

Combining equations (2.5) and (2.6), we conclude

$$\begin{align*}\iint_{\overline{R}_N} \left|\frac{\partial u_N}{\partial x}\right|^2\le\iint_R \left|\frac{\partial v_N}{\partial x}\right|^2.\\[-44pt] \end{align*}$$

Proof. Let $w\in W^{1,2}(\overline {R}_N)$ . Notice that

(2.10) $$ \begin{align} S(w)= \iint_{\overline{R}_N}\left|\frac{\partial w}{\partial y}\right|^2+|\{w\neq 0\}|\le J(w), \quad S(w)+\iint_{\overline{R}_N}\left|\frac{\partial w}{\partial x}\right|^2=J(w). \end{align} $$

Using equation (2.10) and the fact that $H_x(v_N(x,\cdot ))\le H_x(u_N(x,\cdot ))$ a.e., we obtain

$$\begin{align*}S(v_N)\le S(u_N)\le J(u_N). \end{align*}$$

Combining this with the equality in equation (2.10) and with equation (2.3), we conclude that if $u_N$ is a minimizer of J, then

$$\begin{align*}S(v_N)\le S(u_N)\le J(u_N)\le J(v_N)=S(v_N)+\iint_{\overline{R}_N}\left|\frac{\partial v_N}{\partial x}\right|^2\le S(v_N)+\frac{16}{N}.\\[-44pt] \end{align*}$$

Proof. Given a subset $X\subset [-3N,3N]\times [-1,1]$ and defining

$$\begin{align*}S_X(u_N)=\iint_X\left|\frac{\partial u_N}{\partial y}\right|^2dxdy+|\{u_N\neq 0\}\cap X|, \end{align*}$$

we obtain $S(u_N)=S_Q(u_N)+S_{{\overline {R}_N}\setminus Q}(u_N)$ . If we had $S_Q(u_N)>S_Q(v_N)+\frac {16}{N}$ , then

(2.12) $$ \begin{align} S(u_N)>S_Q(v_N)+\frac{16}{N}+ S_{\overline{R}_N\setminus Q}(u_N). \end{align} $$

Since for a.e. $x\in [-3N,3N]$ we have $H_x(v_N(x,\cdot ))\le H_x(u_N(x,\cdot ))$ , integrating this inequality we obtain $S_{{\overline {R}_N}\setminus Q}(v_N)\leq S_{{\overline {R}_N}\setminus Q}(u_N)$ . Together with equation (2.12), this gives

$$\begin{align*}S(u_N)>S_Q(v_N)+\frac{16}{N}+S_{{\overline{R}_N}\setminus Q}(v_N)=S(v_N)+\frac{16}{N}, \end{align*}$$

contradicting equation (2.9).

Proof. From Lemma 2.6, we have that $S_Q(u_N) \leq S_Q(v_N) + \frac {16}{N}$ and from equation (2.4) we have $\iint _Q |\partial _x u_N|^2\, dA \leq \frac {16}{N}$ . Putting this together, we get that

$$ \begin{align*}J_Q(u_N) \leq S_Q(v_N) + \frac{32}{N}.\end{align*} $$

Thus, it suffices to show that $S_Q(v_N)$ grows proportionally to $|b-a|$ with a constant of proportionality independent of $a,b, N$ . Indeed, by Lemma 2.1 $H_x(v_N) \leq \max \{2f_N(x)^2 + 2, 4f_N(x)\} \leq 10$ . Integrating that across $[a,b]$ gives the desired result.

Proof. As $u_N$ is a minimizer, we can apply [Reference David and Toro10, Theorem 8.1] (c.f. the discussion at the bottom of page 504 in [Reference David and Toro10]) which gives a Lipschitz bounds on almost-minimizers depending only on the distance from the boundary and the $L^2$ -norm of the gradient (see also [Reference David, Engelstein and Toro13, Remark 2.2]). Corollary 2.7 gives uniform bounds on the $L^2$ -norm of the gradient of $u_N$ inside of any rectangle and a covering argument finishes the proof.

Proof. The result follows from the following simple observations:

• ○ When $|x|\ge \frac {(1+2\alpha )N}{1+\alpha }$ , $f_N(x)\ge 1$ . In this case $v_N(x,y)=yf_N(x)$ .

• ○ When $|x|\le N$ , $f_N(x)\equiv 1-\alpha $ , and $v_N(x,y)=\text {sgn}(y)(|y|-\alpha )_+$ .

• ○ In the remaining interval, $f_N(x)=\frac {|x|(1+\alpha )}{N}-2\alpha $ and $v_N(x,y)=\text {sgn}(y)(|y|-1+\frac {|x|(1+\alpha )}{N}-2\alpha )_+$ .

Proof. Without loss of generality, assume $u_N(x,-)$ is both equal to $f_N(x)$ and $\delta $ on an interval of length  $\varepsilon $ . The lowest energy way to do this is assuming these values are achieved at the endpoints of the interval and that $u_N$ interpolates between them linearly. Thus, $H_x(u_N(x, -)) \geq \frac {(f_N(x) - \delta )^2}{\varepsilon }$ . The second result follows from the first once we remember that $f_N(x) \geq \frac {3}{4}$ for all x, and $H_x(v_N(x,-)) \leq 10$ for all x.

Proof. Fix $\delta> 0$ . Recall from Theorem 2.8 that there exists $L> 0$ (independent of N but dependent on $\delta> 0$ ) such that if

$$\begin{align*}\Omega = \left[-3N + \delta, 3N- \delta\right] \times \left[-1+\frac{1}{100}, 1-\frac{1}{100}\right], \end{align*}$$

then $\|u_N\|_{\mathrm {Lip}(\Omega )} \leq L$ .

We first check that $|u_N(x_0,y_0)| \geq \frac 18$ on $R_\pm $ . Note that, if $|u_N(x_0,y_0)| < \frac 18$ for some $(x_0,y_0) \in [-3N + \delta , 3N-\delta ]\times ([-1 +\frac {1}{88}, -1+\frac {1}{44}]\cup [1-\frac {1}{44}, 1-\frac {1}{88}])$ , then by Lipschitz continuity there exists an interval $I \subset [-3N + \delta , 3N-\delta ]$ of length $\frac {1}{8L}$ such that $|u(x,y_0)| < \frac {1}{4}$ for all $x\in I$ .

We apply Lemma 3.3 to conclude that for almost every $x\in I$ we have $H_x(u_N(x,-)) \geq H_x(v_N(x,-)) + 1$ . If $Q = I \times [-1,1]$ , then this implies that $S_Q(u_N) \geq S_Q(v_N) + |I| = S_Q(v_N) + \frac {1}{8L}.$ Of course, this contradicts equation (2.11) as long as $\frac {16}{N} < \frac {1}{8L}\Leftrightarrow 128L < N$ .

Now, we check that u has the right sign on $R_\pm $ . Suppose for instance that $u(x,y) \leq -1/8$ somewhere on $R_+$ . Since u is continuous, $u(x,y) \leq -1/8$ everywhere on $R_+$ . Then for all $x \in \left [-3N + \delta , 3N-\delta \right ] \setminus X_0$ , $u(x,-)$ is a Sobolev function that goes from $-1/8$ to at least $1/4$ in an interval of length at most $1/88$ , a direct computation shows that $H_x(u_N(x,-)) \geq \frac {3}{8} \times 88 \geq 33$ , and we reach a contradiction as above.

Proof. Assuming by contradiction this were not the case, without loss of generality there would exist a point in $(x_0, y_0)$ with $|x_0| < N-1, y_0 \leq \alpha /8$ and $u_N(x_0, y_0)> 0$ .

By continuity of $u_N$ , the connected component of $\{u_N> 0\}$ containing $(x_0, y_0)$ must be separated from $\{y = -1\}$ by $\{u_N = 0\}$ . This implies that there is a connected subset of $\{(x,y)\mid |x| \leq N, u_N(x,y) = 0\}$ which touches the sets $\{x = N\}$ and $\{x= -N\}$ and which separates $(x_0, y_0)$ from $\{y =-1\}$ . By Lemma 3.4, this connected component cannot intersect the set $\{(x,y) \mid |x| < N, -1+1/44> y > -1+1/88\}$ . If $y_0 < -1+1/44$ , then this connected component lies below $y = -1+1/88$ and the length of its projection onto the x-axis is at least $2N$ . Such a configuration has too much energy by Lemma 3.3, and thus we can assume $y_0> -1+1/44$ .

We can also assume that $x_0 \notin X_0$ (since $\{u_N> 0\}$ is open), so by continuity of $u_N$ on the slice $\{x = x_0\}$ and Lemma 3.4 there exists a point $(x_0, \tilde {y})$ with $-1+1/44<\tilde {y} < y_0$ and $u_N(x_0, \tilde {y}) = 0$ and $(x_0, \tilde {y}) \in \partial \{u_N> 0\}$ .

We note that $\{u_N> 0\}\cap ([-N, N]\times [-1+1/88, 1-1/88])$ is a locally non-tangentially accessible (NTA) domain, uniformly in N (i.e., for any $K \subset \subset ([-N, N]\times [-1+1/88, 1-1/88])$ , $\{u_N>0\}$ satisfies the corkscrew conditions at $Q\in \partial \{u_N> 0\}\cap K$ with constants and at scales that depend only on K not N c.f. [Reference David, Engelstein and Toro13, Theorem 2.3])). In particular, there exists a point $(x_1, y_1) \in \{u_N> 0\}$ such that $\|(x_1, y_1) - (x_0, \tilde {y})\| \leq r_0 = r_0(K) \leq \alpha /8$ and $\mathrm {dist}((x_1, y_1), \{u_N \leq 0\}) \geq r_0/M$ for some $M> 1$ (where both $r_0, M$ are independent of N large).

By the nondegeneracy of $u_N$ (c.f. [Reference Alt and Caffarelli1, Lemma 3.4]) and the Lipschitz continuity of $u_N$ , Theorem 2.8, there exists a constant $C> 1$ (again uniform for large N) such that $u_N \geq r_0/C$ in the ball $B((x_1, y_1), r_0/(3CL))$ (where L is the Lipschitz constant).

Therefore, there exists an interval $I \subset [-N, N]$ of length $2r_0/(3CL)$ such that for each $x\in I$ there exists a $y < \alpha /2$ with $u_N(x, y)> r_0/C$ . Invoking the computations of Section 6.2 (c.f. Claim 6.1), we get that $H_x(u_N(x,-)) \geq H_x(v_N(x, -)) + \eta $ for every $x\in I \backslash X_0$ , where $\eta = \eta (\alpha , C, r_0)> 0$ is independent of N.

Let $Q = I \times [-1,1]$ ; we get (from equation (2.11))

$$ \begin{align*}S_Q(v_N) + \frac{16}{N} \geq S_Q(u_N) \geq S_Q(v_N) + \mathcal H^1(I)\eta.\end{align*} $$

This gives a contradiction if $N> 0$ is large enough (since $|I| = 2r_0/(3CL), \eta> 0$ are independent of N).

Proof. Recall that for x as in the statement of Lemma 3.6, we have $v_N(x,y) = yf_N(x)$ . Our plan is to first modify w, reducing energy, and show the desired inequality for w, then estimate the $L^\infty $ distance between the original w and our modified functional. We know that $w\in W^{1,2}([-1,1])$ , so by Sobolev embedding we have that w is Hölder continuous and thus

$$ \begin{align*}-1 < \alpha := \inf\{t\mid w(t) =0\} \leq \sup\{t\mid w(t) = 0\} =: \beta < 1.\end{align*} $$

We construct

$$ \begin{align*}\hat{w}(t) = \begin{cases} 0& t\in [a,b]\\ \frac{f_N(x)(t-a)}{1+a}& t\in [-1, a]\\ \frac{f_N(x)(t-b)}{1-b}& t\in [b, 1].\end{cases}\end{align*} $$

We observe that $H_x(w) \geq H_x(\hat {w})$ (as we have enlarged the zero set and minimized Dirichlet energy where w is positive).

We can compute that

$$ \begin{align*}H_x(\hat{w}) = f_N^2(x)\left(\frac{1}{1-b} + \frac{1}{a+1}\right) + 2-(b-a).\end{align*} $$

Recall that $H_x(v_N(x, -)) = 2f_N^2(x) + 2$ and rewrite

$$\begin{align*}\begin{aligned} H_x(\hat{w}) - H_x(v_N(x,-)) =\,& f_N^2(x) \left(\frac{1}{1-b} + \frac{1}{a+1} -2\right) -(b-a)\\ \geq\,& \left(\frac{1}{1-b} + \frac{1}{a+1} -2\right) -(b-a) =: F(a,b),\end{aligned}\end{align*}$$

where the last inequality follows because $f_N(x) \geq 1$ in the salient range and $\frac {1}{1-b} + \frac {1}{a+1} - 2 \geq 0$ as long as $1> b \geq a >-1$ .

We compute that $F(0,0) = 0, \nabla F(0,0) = (0,0)$ and that

$$ \begin{align*}\nabla^2 F(a_0,b_0) = \begin{pmatrix} \frac{2}{(1+a_0)^3} & 0\\ 0 & \frac{2}{(1-b)^3} \end{pmatrix},\end{align*} $$

is a diagonal matrix with entries between $10^7$ and $\frac {1}{10^7}$ as long as $[a_0,b_0] \in \left [-\frac {99}{100}, \frac {99}{100}\right ]$ . If either of $a_0, b_0$ is outside that range, then Lemma 3.3 gives a contradiction to the assumption on energy. Thus, by the Taylor remainder theorem (and the fact that F is $C^2$ as long as a stays away $-1$ and b stays away from $1$ ) we have that

$$ \begin{align*}|F(a,b) - F(0,0)| = (a,b)^T\nabla^2 F(a_0, b_0)(a,b) \leq C(a^2 + b^2)\end{align*} $$

for some $(a_0, b_0)$ on the segment connecting $(0,0)$ and $(a,b)$ .

On the other hand, since $v_N(x,-)$ is linear in y and $\hat {w}$ is piecewise linear in y we can see that

$$ \begin{align*}\|v_N - \hat{w}\|_{L^\infty} = \max_{t = a,b} |v_N(x, t) - \hat{w}(t)| =f_N(x)\max_{t = a,b} |t|.\end{align*} $$

Chaining everything together, we get that

$$ \begin{align*}\|v_N - \hat{w}\|_{L^\infty} \leq C\sqrt{H_x(\hat{w}) - H_x(v_N(x,-))}.\end{align*} $$

We now estimate $|\hat {w}-w|(t_0)$ for $t_0 \in [-1,1]$ . We have two cases; in the first, assume that $t_0 \in [a,b]$ . Then $2|\hat {w}(t_0) - w(t_0)| = 2|w(t_0)| \leq \int _a^{t_0}|w_y| + \int _{t_0}^b |w_y|$ . Using Jensen’s inequality, we get that

$$ \begin{align*}|w(t_0)| \leq \left(\frac{b-a}{4}\int_a^b |w_y|^2\, dy\right)^{1/2} \leq C\sqrt{H_x(w) - H_x(\hat{w})}.\end{align*} $$

Putting this together, we have that

$$ \begin{align*}|v_N(x, t_0) - w(t_0)| \leq C\sqrt{H_x(w) - H_x(\hat{w})} + C\sqrt{H_x(\hat{w}) - H_x(v_N(x,-))} \leq C\varepsilon^{1/2}.\end{align*} $$

In the second case, we assume that $t_0 \in [-1, a]$ (the case that $t_0 \in [b, 1]$ works the same way). Since $w(-1) = \hat {w}(-1)$ and $w(a) = \hat {w}(a)$ we get that $2|w(t_0) - \hat {w}(t_0)| \leq \int _{-1}^a |\partial _y(w- \hat {w})|\, dy$ . Again applying Jensen’s inequality, we get that

$$ \begin{align*}|w(t_0) - \hat{w}(t_0)| \leq C(a+1) \left(\int_{-1}^a \left|\partial_y w - \frac{f_N}{1+a}\right|^2\, dy\right)^{1/2}.\end{align*} $$

Expanding out the integrand and using the fact that at $w(a) - w(-1) = f_N(x)$ we get

$$ \begin{align*}|w(t_0) - \hat{w}(t_0)| \leq C\left(\int_{-1}^a|\partial_y w|^2\, dy -2\frac{f_N(x)^2}{1+a} + \frac{f_N(x)^2}{1+a}\right)^{1/2} \leq C\left(H_x(w) - H_x(\hat{w})\right)^{1/2}.\end{align*} $$

We can chain the inequalities together as above to get the desired result.

Proof. Fix $\delta> 0$ , and let $N> 0$ be big enough so that, invoking Theorem 2.8, we can say $u_N$ is L-Lipschitz in $\left [-3N + \delta , 3N - \delta \right ]\times \left [-1+\frac {1}{44}, 1-\frac {1}{44}\right ]$ .

Assume there exists a point $(x_0, y_0)$ such that $f_N(x_0) \geq 1$ , $\theta \leq |y_0| \leq 1-\frac {1}{88}$ and $u_N(x_0, y_0) = 0$ . We know that $|y_0| < 1-\frac {1}{44}$ , by Lemma 3.4. Then there exists an interval I of length at least $\frac {\theta }{2L}$ such that $|u_N(x, y_0) - y_0f_N(x)| \geq \frac {\theta }{2}$ and $f_N(x) \geq 1$ on all $x\in I$ .

By Lemma 3.6, this implies that $H_x(u_N(x,-)) \geq H_x(v_N(x,-)) + \frac {\theta ^4}{4C}$ for almost every $x\in I$ . Integrating and letting $Q = I\times [-1,1]$ , we get that $S_Q(u_N) \geq S_Q(v_N) + C^{-1}\theta ^5$ , where $C= C(\delta )> 0$ is independent of $N, \theta $ . This contradicts equation (2.11) as long as N is large enough (depending on $C, \theta $ and thus on $\delta , \theta $ ).

Proof of Theorem 4.1 assuming Lemma 4.2

By Lemma 3.5, $u_N$ vanishes in the region, where $|x| \leq N-1$ and $|y| \leq 1/8$ . Denote by ${\mathcal O}_0$ the interior of $\{ u_N = 0 \}$ , and let $\mathcal O$ the connected component of $B(0, 1/8)$ in ${\mathcal O}_0$ .

Let us first check that $\partial \mathcal O$ contains one-phase points of both types. Consider the line segment $\ell $ from the origin to $z_+ = (0,1/44)$ ; we know that $u(z_+) \geq 1/8$ (by equation (3.2)), so $\ell $ meets $\partial \mathcal O$ . At the first point of intersection $z=(0,y)$ (going up from $0$ ), Lemma 3.5 says that $y> \alpha /8$ , and then Lemma 3.5 says that $u(w) \geq 0$ for w near z; hence, z is a (positive) one-phase point. Similarly, the first point of $\partial \mathcal O$ on the interval from $0$ to $z_- = (0,-1/44)$ is a negative one-phase point.

Next, we want to show that ${\mathcal O} \cap R_{ext} = \emptyset $ . Suppose not, and let $\gamma $ be a path in ${\mathcal O}$ that goes from $0$ to some point of $\mathcal O_0\cap R_{ext}$ . Certainly, $\gamma $ does not get close to the top and bottom boundaries, that is, where $|y| = 1-\frac {1}{88}$ , by equation (3.2). So there is a point $(x_0,y_0)\in \gamma $ such that $|x_0|> 2N+1$ and $|y_0| \leq 1-1/44$ . Then, as above with the origin, we can find a point $P = (x_0,y_0')$ above $(x_0,y_0)$ , which lies in $\partial \mathcal O$ . This vertical segment lies inside of $\mathcal O$ (and is nonempty by the openness of $\mathcal O_0$ ). By Lemma 4.2, this point is a two-phase point. But in fact the proof of Lemma 4.2 will say more: near P, the free boundary is a Lipschitz graph, with a small constant, and then the nondegeneracy of u shows that on the vertical line that goes through P, u is (strictly) positive on one side of P, and negative on the other side; this contradicts the fact that the open segment between $(x_0,y_0)$ and P lies in $\mathcal O$ . Hence, ${\mathcal O} \cap R_{ext} = \emptyset $ . Note this, with Lemma 3.4, implies that

$$ \begin{align*}\mathcal O \subset R_{in} := \{(x,y) \in R_N \mid |x| \leq 2N+1, |y| \leq 1-\frac{1}{44}\}.\end{align*} $$

We still need to show that ${\mathcal O}$ contains a branch point. Suppose not, and let $z\in \partial \mathcal O$ be given. Obviously, u takes nonzero values near z, so z lies in the free boundary, and by assumption z is a one-phase point (since only one-phase points or branch points can be on the boundary of an open subset of $\{u =0\}$ ). Suppose $z \in \Gamma ^+(u)$ . In the present situation (and even in ambient dimension $3$ ), the free boundary in a neighborhood of z is a smooth hypersurface $\Gamma $ , with $u> 0$ on one side of $\Gamma $ , and $u=0$ on the other side. Thus, $B_r(z) \cap \partial \mathcal O \subset \Gamma ^+(u)$ for some $r> 0$ small enough (depending on z).

More globally, the curve $\Gamma \subset \partial \mathcal O$ that contains z is a Jordan curve (it is disjoint from $R_{ext}$ , does not touch the boundary of R and is locally smooth), so $\mathcal O$ , which is connected, is contained in one of the two components of U = ${\mathbb {R}}^2 \setminus \Gamma $ . If $\mathcal O$ is contained in the unbounded component of U, then we can replace u with $0$ on the bounded component of U and keep u a valid competitor (because $\Gamma \subset \partial \mathcal O \subset R_{in} \subset \subset R_N$ ). However, this will strictly decrease energy which is a contradiction. Thus, $\mathcal O$ is contained in the bounded component of U. Arguing as before, if $\mathcal O$ is not the entirety of this bounded component, then we could replace u by $0$ on the rest of this bounded component and decrease energy. So it must be that $\mathcal O$ is one of the connected components of $\mathbb R^2 \setminus \Gamma $ .

This contradicts the fact that $\partial {\mathcal O}$ meets $\Gamma ^-$ too. So $\partial {\mathcal O}$ contains a branch point, and the theorem follows from the proof of the lemma.

In ambient dimension $3$ , the same proof would work, using the fact that a connected smooth orientable hypersurface in ${\mathbb {R}}^n$ always separates ${\mathbb {R}}^n$ in exactly two connected component, as in the Jordan curve theorem.

Proof. Recall that $f_N(x) = 2$ for $2N \leq |x| \leq 3N$ , so $v_N(x,y) = 2 y$ and $\partial _y v_N(x,-) = 2$ . Also, $y\mapsto v_N(x,y)$ is harmonic on $[-1,1]$ and (when $x \notin X_0$ ) $y \mapsto u_N(x,y)$ is a $W^{1,2}$ function with the same trace as $v_N$ . Fix such an x, and set $I(w) = \int _{-1}^1 |w_y|^2\, dy$ for $w\in W^{1,2}([-1,1])$ . Amongst $w\in W^{1,2}([-1,1])$ with the same boundary values $\pm 2$ as $v_N(x,-)$ , $v_N(x,-)$ minimizes I. Thus, the Euler–Lagrange equation shows that $\int _{-1}^1 \partial _y v_N (\partial _y u_N-\partial _y v_N) dy=0$ . Hence, by Pythagoras

$$\begin{align*}\int_{-1}^1|\partial_y u_N|^2 = \int_{-1}^1 |\partial_y v_N|^2dy + \int_{-1}^1|\partial_y u_N-\partial_y v_N|^2dy. \end{align*}$$

Now, we also care about the functional $H_x$ , so we need to add $|\{y \mid w(y) \neq 0\}|$ to $I(w)$ . For $v_N(x,-)$ , we get $2$ , since $v_N(x,y)=2y$ only vanishes at $0$ . For $u_N(x,-)$ , Corollary 3.7 says that, for any given $\theta> 0$ , and if N is large enough $u_N(-)$ can only vanish for $|y| < \theta $ . Then $2- |\{y \mid u_N(x,y) \neq 0\}| \leq 2\theta $ and

$$ \begin{align*} \int_{-1}^1|\partial_y u_N-2|^2dy &= \int_{-1}^1|\partial_y u_N-\partial_y v_N|^2dy = \int_{-1}^1|\partial_y u_N|^2\, dy - \int_{-1}^1 |\partial_y v_N|^2\, dy \\ &= H_x(u_N(x,-)) - H_x(v_N(x,-)) + 2 -|\{y \mid u_N(x,y) \neq 0\}| \\ &\leq H_x(u_N(x,-)) - H_x(v_N(x,-)) + 2 \theta. \end{align*} $$

We now integrate over x such that $2N < |x| < 3N-\frac {1}{4}$ , use equation (2.9) and get the desired result for $N\ge N_0(\varepsilon )$ .

Proof. Pick some $r_0> 0$ small, but independent of $N, \delta $ . For points $x_0$ such that $B(x_0, r_0) \cap \Gamma _{\mathrm {TP}}(u_N) = \emptyset $ , the result follows from uniform Reifenberg flatness (Corollary 3.7) and standard ‘flat-implies-smooth’ regularity for the one-phase problem (c.f. [Reference Alt and Caffarelli1]).

If $B(x_0, r_0) \cap \Gamma _{\mathrm {TP}}(u_N) \neq \emptyset $ , then we can consider $y\in B(x_0, r_0) \cap \Gamma _{\mathrm {TP}}(u_N)$ , and look at $B(y, 2r_0)$ . Then the regularity follows from the $\theta $ -Reifenberg flatness at scale 1 (Corollary 3.7), the uniform Lipschitz continuity (Theorem 2.8) and [Reference De Philippis, Spolaor and Velichkov20, Theorem 3.1].

In both instances, the dependence on $r_0, \theta $ is such that the $C^{1,1/4}$ -norm of the graph(s) goes to zero as $r_0> 0$ stays constant but the Reifenberg flatness parameter, $\theta $ , goes to zero. Corollary 3.7 says we can take $\theta $ arbitrarily small, at the expense of making N larger. The regularity of the gradient is a consequence of standard elliptic regularity once we know the regularity of the free boundary.

Proof. Let $\varepsilon> 0$ be small, to be chosen later and assume by contradiction that there exists a one-phase point $(x_0, y_0) \in \partial \{u_N> 0\}\cap \{2N\leq |x| \leq 3N-1\}$ . By Lemma 4.3, we may choose N large enough (depending only on $\varepsilon $ ) such that for any square, Q, of side length $\ell (Q) \leq 3/4$ centered at the point $(x_0,y_0)$ we have

(4.1) $$ \begin{align} \int_Q \left|\frac{\partial u_N}{\partial y}- 2\right|^2\, dx dy \leq \varepsilon. \end{align} $$

Using Fubini, we also get that

$$ \begin{align*}\int_0^{3/4}\int_{\partial Q}\left|\frac{\partial u_N}{\partial y} - 2\right|^2 \, d\mathcal H^{1} d\ell(Q) \leq \varepsilon,\end{align*} $$

where the integration is occurring over squares all centered at $(x_0, y_0)$ with increasing side lengths. Thus, we can pick a cube $Q_0$ (which will be fixed going forward) with $1/2 \leq \ell (Q_0) \leq 3/4$ such that

(4.2) $$ \begin{align}\int_{Q_0} \left|\frac{\partial u_N}{\partial y}- 2\right|^2\, dx dy, \int_{\partial Q_0} \left|\frac{\partial u_N}{\partial y}- 2\right|^2\, d\mathcal H^1 \leq 4\varepsilon.\end{align} $$

Note that $u_N|_{Q_0}$ is L-Lipschitz (by Theorem 2.8), where L is independent of $N, Q_0$ .

Let $\tilde {Q} = Q_0\cap \{u_N> 0\}$ . By Lemma 4.4, the domain $\tilde {Q}$ is piecewise $C^{1,1/4}$ and is an NTA domain, with constants uniform in N (c.f. [Reference Jerison and Kenig25] for definitions and details). Let us say a bit more about this; the NTA constants depend on how the vertical edges of Q touch the smooth graph $\partial \{u_N> 0\}$ . However, this graph over $\{y = 0\}$ has very small norm (uniform in N), so this intersection happens (quantitatively) transversely and thus the NTA constants are also uniform in N. These bounds on norm of the graph which gives $\partial \{u_N> 0\}$ also imply that $4 \geq \mathcal H^1(\tilde {\partial Q}) \geq 1$ and $|\tilde {Q}| \geq \frac {1}{8}$ . From this information, using equation (4.2) and Chebyshev, there exists a $A \in \tilde {Q}$ with $\mathrm {dist}(A, \partial \tilde {Q})> \frac {1}{100}$ and $|\partial _y u(A) - 2|^2 < 8\varepsilon $ (this will work as long as $\varepsilon> 0$ is small enough).

Let $\omega _N^A$ be the harmonic measure of $\tilde {Q}$ with a pole at A (where the notation emphasizes the dependence on N). Since $\partial _y u$ is a harmonic function in $\tilde {Q}$ , we have the following integral representation:

$$ \begin{align*}\partial_y u_N(A) = \int_{\partial \tilde{Q}} \partial_y u_N(P)d\omega_N^A(P).\end{align*} $$

Finally, it will be convenient to let $\partial \tilde {Q} = \Gamma _1 \cup \Gamma _2$ , where $\Gamma _1 = \partial \{u_N> 0\} \cap \overline {Q}_0$ and $\Gamma _2 = \partial Q_0 \cap \overline {\{u_N> 0\}}$ ; see the picture below.

Recall that $(x_0, y_0)$ is a one-phase point for $u_N$ . Thus, we can compute $|\partial _y u_N(x_0, y_0)| \leq |\partial _\nu u_N(x_0, y_0)| = 1$ , where the last equality holds due to the free boundary condition at (regular) one-phase points. Because the derivative of $u_N$ restricted to $\partial \{u_N> 0\}$ has $C^{0, 1/4}$ -seminorm less than 1, we also have that

$$ \begin{align*}\begin{aligned} |\partial_y u_N(x_0,y_0) - \partial_y u_N(P)| <\,& \left(\frac{3}{4}\right)^{1/4} < \frac{94}{100} \quad \text{for all } P \in \Gamma_1 \\ \Rightarrow \partial_y u_N(P)\leq\,& 2-\frac{1}{20}\quad\text{for all } P \in \Gamma_1. \end{aligned} \end{align*} $$

Putting things together, we have that

(4.3)

where in the last line we used that $\omega _N^A(\Gamma _2) = 1-\omega _N^A(\Gamma _1)$ and also Bourgain’s Lemma (c.f. [Reference Jerison and Kenig25, Lemma 4.2]), which implies that there is a constant $\tilde {c}> 0$ (independent of $\varepsilon , N$ ) such that $\frac {\omega _N^A(\Gamma _1)}{1-\omega _N^A(\Gamma _1)} \geq \tilde {c}$ and $\tilde {c}^{-1} \geq \frac {1}{1-\omega _N^A(\Gamma _1)}$ .

Recall that u is $L-$ Lipschitz, and write $\Gamma _2 = \Gamma _{2,+} \cup \Gamma _{2,-}$ , with

$$\begin{align*}\Gamma_{2,+} = \{P\in \Gamma_2 \mid L \geq |\partial_y u_N(P)|> 2 + c/2\}\ \text{and}\ \Gamma_{2,-} = \{P \in \Gamma_2 \mid |\partial_y u_N(P)| \leq 2 + c/2\}. \end{align*}$$

After overestimating over each set, equation (4.3), gives us

(4.4) $$ \begin{align} \frac{c}{2} \leq L \, \frac{\omega_N^A(\Gamma_{2,+})}{\omega_N^A(\Gamma_2)}. \end{align} $$

Once more invoking Bourgain’s theorem, we have that

$$ \begin{align*}\theta: = \frac{c\omega_N^A(\Gamma_2)}{2L}\end{align*} $$

is bounded strictly away from zero, independently of N (large enough) or $\varepsilon> 0$ .

Using the condition on the integral of $|\partial _y u_N - 2|$ on $\partial Q_0$ (i.e., equation (4.2)), we see that

$$\begin{align*}\mathcal H^1(\Gamma_{2,+}) < \frac{16\varepsilon}{c^2}. \end{align*}$$

Recall that $\mathcal H^1(\partial \tilde {Q}) \geq 1$ , and we get that

(4.5) $$ \begin{align} \frac{\mathcal H^1(\Gamma_{2,+})}{\mathcal H^1(\partial \tilde{Q})} < \frac{16\varepsilon}{c^2} \ll \theta \equiv \frac{\omega_N^A(\Gamma_{2,+})}{\omega_N^A(\partial \tilde{Q})}. \end{align} $$

We now recall that in $\tilde {Q}$ we have that the harmonic measure $\omega _N^A \in A_\infty (\mathcal H^1)$ (see, e.g., [Reference David and Jerison9, Theorem 2]). In fact, the $A_\infty $ -constants depend on the NTA constants of $\tilde {Q}$ , the $1$ -Ahlfors regularity of $\partial \tilde {Q}$ , the distance from A to $\partial \tilde {Q}$ and the diameter of $\tilde {Q}$ (for more details and definitions of the relevant terms, see [Reference David and Jerison9]). As discussed above, all of these quantities can be taken uniform for N large enough. Thus, we can take $\varepsilon> 0$ small until equation (4.5) contradicts $\omega _N^A \in A_\infty (\mathcal H^1)$ , and we are done.

Proof of Lemma 5.3, assuming Lemma 5.4

We first check the almost-minimization condition (5.3) for $x_0 \in \Gamma ^+ \cap \Gamma ^-$ :

Case 1. Let $x_0 \in \Gamma ^+\cap \Gamma ^-$ and $r> 0$ small enough, and let $v =u$ on $B(0, R)\backslash B(x_0,r)$ .

Since $u^{\pm } \in C^{1,\alpha }(\overline {\Omega ^{\pm }})$ and $\Gamma ^+\cap \Gamma ^-$ is closed and smooth, there exists an $r_0$ such that if $r < r_0$ , then, for $x\in B(x_0, r)$

(5.4) $$ \begin{align}\begin{aligned} q_+(x_0)((x-x_0)\cdot e)^+ -& q_-(x_0)((x-x_0)\cdot e)^- -r^{1+7\alpha/8} \leq u(x)\\ u(x)\leq q_+(x_0)((x-x_0)\cdot e)^+ -& q_-(x_0)((x-x_0)\cdot e)^- + r^{1+7\alpha/8}.\end{aligned}\end{align} $$

Recall that $q_\pm (x_0) = |\nabla u^\pm |(x_0)$ by definition. To give more detail, equation (5.4) follows from the Taylor series expansion of $u^{\pm }$ at the point $x_0$ , where we have used the fact that $\Gamma ^\pm $ share a unit normal, e, at $x_0$ , which we take to point into the set $\{u> 0\}$ .

Recall the functional proved to be monotone by Weiss [Reference Weiss31]:

$$ \begin{align*}W(u, x_0, r) \equiv \frac{1}{r^n} \int_{B(x_0, r)} |\nabla u|^2 + q^2_+(x) \chi_{\{u> 0\}} + q_-^2(x) \chi_{\{u < 0\}}\, dx - \frac{1}{r^{n+1}} \int_{\partial B(x_0, r)} u^2\, d\sigma.\end{align*} $$

It follows from equation (5.4) and the $C^{0,\alpha }$ -character of $|\nabla u^{\pm }|$ that

(5.5) $$ \begin{align} W(u, x_0, r) \leq \frac{1}{2}\mathrm{Vol}(B(0,1))(q^2_+(x_0) + q^2_-(x_0)) + Cr^{7\alpha/8},\end{align} $$

where $C> 0$ depends on the $C^{0,\alpha }$ -norm of $|\nabla u^\pm |$ restricted to $\overline {\Omega }^{\pm }$ .

We now want to show that for any minimizer v to $J_{B(x_0, r)}$ with $v = u$ in $\mathbb R^n \backslash B(x_0, r)$ we have

(5.6) $$ \begin{align} (\partial \{v> 0\}\cup \partial \{v < 0\})\cap B(x_0, r) \subset \{x\in B(x_0, r)\mid |(x- x_0) \cdot e| < r^{1+\alpha/2}\}. \end{align} $$

Assume that equation (5.6) holds. We would like to compare $W(v, x_0, r)$ to $W(u, x_0, r)$ but naïvely underestimating $W(v, x_0, r)$ by $W(v, x_0, 0)$ is problematic as $x_0$ may not be in the free boundary of v. To combat this, let $x^+$ be the closest point to $x_0$ in $\partial \{v> 0\}$ and $x^-$ the closest point to $x_0$ in $\partial \{v < 0\}$ . We note that equation (5.6) implies that $|x_0 - x^{\pm }| < r^{1+\alpha /2}$ . Let $\rho = r/2 - \max \{ |x_+-x_0|, |x_- -x_0|\}$ . Note that $B(x^{\pm }, 2\rho ) \subset B(x_0, r)$ and $r> 2\rho > r(1-r^{\alpha /2})$ . Also, $B(x_0, r/2) \supset B(x^{\pm }, \rho )$ . Hence,

(5.7) $$ \begin{align} W(v, x_0, r/2)\geq W(v^+, x^+, \rho) + W(v^-, x^-, \rho)- Cr^{\alpha/2}, \end{align} $$

because $|2\rho /r - 1| < r^{\alpha /2}$ and v is Lipschitz in $B(x_0, r/2)$ with a constant controlled by $\frac {1}{r^n}\int _{B(x_0, r)} |\nabla v|^2 \leq \frac {1}{r^n} J_{B(x_0, r)}(u)$ , the latter of which is bounded by the Lipschitz norm of u and the supremums of $q_{\pm }$ .

To estimate each term in the summand on equation (5.7), we think of $v^+, v^-$ is being separate critical points to the one-phase problems associated to weights $\tilde {q}_{\pm }|_{\partial \{v^{\pm }> 0\}} := \partial _{\nu ^{\pm }} v^\pm |_{\partial \{v^{\pm } > 0\}}$ . By the regularity theory of the two-phase problem in [Reference De Philippis, Spolaor and Velichkov20] and equation (5.6), we know that $\partial \{\pm v> 0\}$ are $C^{1,\alpha }$ in $B(x_0, r/2)$ . Thus, $\tilde {q}_{\pm } \in C^{0,\alpha }(\partial \{v^{\pm }> 0\})$ and can be extended Hölder continuously to all of $B(x_0, r/2)$ (with norm uniform in the constants we care about).

The free boundary condition satisfied by v being a minimizer to the two-phase problem tells us that $\tilde {q}_{\pm }:= \partial _{\nu ^{\pm }} v^{\pm } = q_{\pm }$ at one-phase points, but at two-phase points the free boundary condition only implies that $\tilde {q}_{\pm }:= \partial _{\nu ^{\pm }} v^{\pm } \geq q_{\pm }$ . These observations tell us that $\|\tilde {q}_{\pm } - q_{\pm }\|_{L^\infty (B(x_0, r/2))} \leq Cr^{\alpha }$ .

By monotonicity, we can underestimate each

$$ \begin{align*}\begin{aligned}W(v^{\pm}, x^{\pm}, \rho) \geq& \frac{1}{\rho^n}\int_{B(x^\pm, \rho)} |\nabla v^{\pm}|^2 + \tilde{q}^2_\pm(x)\chi_{\{\pm v> 0\}}\, dx - \frac{1}{\rho^{n+1}}\int_{\partial B(x^{\pm}, \rho)} (v^{\pm})^2 \, d\sigma - Cr^\alpha\\ \geq& \frac{1}{2}\mathrm{Vol}(B(0,1))\tilde{q}^2_{\pm}(x^{\pm}) -Cr^{\alpha} \\ \geq& \frac{1}{2}\mathrm{Vol}(B(0,1))q^2_{\pm}(x^{\pm}) - Cr^{\alpha} \\ \geq& \frac{1}{2}\mathrm{Vol}(B(0,1))q_{\pm}^2(x_0) - Cr^{\alpha}.\end{aligned}\end{align*} $$

Putting this back into equation (5.7), using monotonicity and equation (5.5) we get that

$$ \begin{align*} W(v, x_0, r) & \geq W(v, x_0, r/2) - Cr^{\alpha/2} \geq \frac{1}{2}\mathrm{Vol}(B(0,1))(q_+^2(x_0) + q_-^2(x_0)) - Cr^{\alpha/2} \\ & \geq W(u, x_0, r) - Cr^{\alpha/2}. \end{align*} $$

Multiplying through by $r^n$ gives the almost-minimization inequality.

So to finish Case 1, it suffices to prove equation (5.6). Here is where we apply Lemma 5.4. We do this on the interval $[a,b] = [r^{1+\alpha /2}, 3r]$ (recall that we can take r small so that $3r < 1$ ). To simplify notation, let us assume that $x_0 = 0$ and $e = e_n$ . We then create the family of sub/supersolutions to the two-phase problem in $B(x_0,r)$ defined by

$$ \begin{align*}\overline{w}_t = M_+(x_n -t)^+ - m_-(x_n-t)^-, \qquad \underline{w}_t = m_+(x_n + t)^+ - M_-(x_n +t)^-,\end{align*} $$

where $m_\pm = \min _{B(x_0, r)} q_\pm $ and $M_\pm = \max _{B(x_0, r)} q_{\pm }$ . Let v be a minimizer to the two-phase problem in $B(x_0, r)$ with boundary values u.

We first verify condition (5) in Lemma 5.4. We note that for any $x\in B(x_0, r)$

$$ \begin{align*}\overline{w}_{3r}(x) \leq -2m_-r \leq -2q_-(x_0)r +Cr^{1+\alpha} \stackrel{(5.4)}{\leq} \inf_{\partial B(x_0, r)} u = \inf_{\partial B(x_0,r)} v \leq v(x)\end{align*} $$

where the last inequality follows from the minimum principle applied to the superharmonic function $-v^-$ and the middle inequalities follow from the $C^{0,\alpha }$ continuity of $q_-$ and the fact that we can take $r_0 \ll \inf _{\overline {B(0, R)}} q_-$ .

To check condition (6) in Lemma 5.4, we will show that $\overline {w}_t|_{\partial B(x_0, r)} < u|_{\partial B(x_0, r)}$ for all $r^{1+7\alpha /8}< r^{1+\alpha /2} \leq t \leq 3r$ . When $x_n> t$ we have

$$ \begin{align*}\begin{aligned} \overline{w}_t(x) = M_+(x_n -t) \leq\,& q_+(x_0)(x_n - t) + (\mathrm{osc} q_+)(|x_n| + |t|)\\ \leq\,& q_+(x_0) x_n -q_+(x_0)r^{1+\alpha/2} + Cr^{1+\alpha} < q_+(x_0)x_n - r^{1+7\alpha/8} \stackrel{(5.4)}{\leq} u(x),\end{aligned}\end{align*} $$

again assuming $r_0$ is small enough.

When $0 \leq x_n < t$ , we note that $\overline {w}_t(x) < 0$ . So there is only something to prove if $u(x) < 0\stackrel {(5.4)}{\Rightarrow } x_n < Cr^{1+7\alpha /8}$ . In this case $x_n < t/2$ (for r small enough), and we have

$$ \begin{align*}\overline{w}_t(x) \leq m_-(x_n -t) \leq - \frac{m_-}{2}t \leq -\frac{m_-}{2}r^{1+\alpha/2} < -r^{1+7\alpha/8} \stackrel{(5.4)}{\leq} u(x).\end{align*} $$

When $x_n < 0$ , we have

$$ \begin{align*}\overline{w}_t(x) = m_-(x_n -t) \leq q_-(x_0)(x_n -t) + Cr^{1+\alpha} \leq q_-(x_0) x_n - Cr^{1+\alpha/2} \stackrel{(5.4)}{<} u(x).\end{align*} $$

Condition (1) in Lemma 5.4 follows from the fact that $\overline {w}_t$ is linear where it is not zero. Conditions (2) and (3) follow from the observation that $\{\overline {w}_t = 0\} = \{x_n = t\}$ which moves continuously with t. Finally, the free boundary condition, Condition (4), follows from the definition of $M_{\pm }, m_{\pm }$ .

Thus, by Lemma 5.4 above (which will be checked below) it must be the case that $\overline {w}_{r^{1+\alpha /2}} \leq v$ in all of $B(x_0, r)$ . A similar argument shows that $\underline {w}_{r^{1+\alpha /2}} \geq v$ in all of $B(x_0, r)$ . Thus, $\partial \{v> 0\}, \partial \{v < 0\} \subset \{x\in B(x_0, r)\mid |x_n| \leq r^{1+\alpha /2}\}$ .

Case 2. Now, we assume that $x_0 \in \Gamma ^+ \cup \Gamma ^-$ . Let $M> 1$ to be chosen later. If R is such that $B(x_0, Mr) \cap (\Gamma ^+\cap \Gamma ^-) \neq \emptyset $ then note that $B(x_0, r) \subset B(y, 2Mr)$ , for some $y\in \Gamma ^+\cap \Gamma ^-$ . As such, if $u = v$ in $B(0, R) \backslash B(x_0, r)$ , then $u = v$ in $B(0, R) \backslash B(y, 2Mr)$ . So Case 1 tell us that u satisfies the almost-minimization criterion in these balls with some $\tilde {C} = C(2M)^{n+\alpha /2}$ (where $C> 0$ comes from Case 1).

If $B(x_0, Mr) \cap (\Gamma ^+\cap \Gamma ^-) = \emptyset $ , we assume, without loss of generality, that $x_0 \in \Gamma ^+ \backslash \Gamma ^-$ and we assume that $B(x_0, r)\cap \Gamma ^- \neq \emptyset $ (otherwise $u^-$ is identically zero on $B(x_0,r)$ and the analysis is even easier). Pick $y\in B(x_0, r)\cap \Gamma ^-$ . As long as $M> 3$ then $B(y, 2r) \cap (\Gamma ^+\cap \Gamma ^-) = \emptyset $ and $u^+, u^-$ satisfy one-phase versions of equation (5.4) in $B(x_0, 2r)$ and $B(y, 2r)$ respectively. To show that $u^+$ and $u^-$ each satisfy the correct energy inequalities for the one-phase versions of J inside of $B(x_0, r)$ and $B(y, r)$ we proceed as in Case 1 (but the argument is simpler because we need only care about one-phase functionals). The almost-minimization property for u in $B(x_0, r)$ for the two-phase functional then follows since the two-phase energy is the sum of the one-phase energies. We omit these details.

Case 3. Assume that $x_0 \notin \Gamma ^+ \cup \Gamma ^-$ . Let $M> 1$ be as above (perhaps enlarge a bit, to be determined later), and let $\varepsilon \ll 1$ (to be determined later). If $B(x_0, Mr^{1-\varepsilon }) \cap (\Gamma ^+ \cup \Gamma ^-) \neq \emptyset $ , then the almost minimization criterion in $B(y, 2Mr^{1-\varepsilon })$ Footnote ⁵ for some $y\in \Gamma ^+\cup \Gamma ^-$ shows that

$$ \begin{align*}J_{B(x_0, r)}(u) - J_{B(x_0, r)}(v) \leq C(2Mr^{1-\varepsilon})^{n+\alpha/2} = C_{M} r^{(1-\varepsilon)(n+\alpha/2)} \leq C_M r^{n+\alpha/4},\end{align*} $$

for any $u = v$ outside of $B(x_0, r)$ . Note this last inequality above holds for $r< 1$ as long as $(1-\varepsilon )(n+\alpha /2) \geq n+\alpha /4 \Leftrightarrow \frac {\alpha }{2(2n+\alpha )} \geq \varepsilon $ .

Thus, we may assume that $B(x_0, Mr^{1-\varepsilon })\cap (\Gamma ^+ \cup \Gamma ^-) = \emptyset $ . If $B(x_0, r) \subset \{u =0\}$ the almost-minimization criterion follows from the fact that $u \equiv 0$ in $B(x_0, r)$ . So we may assume that $B(x_0, Mr^{1-\varepsilon }) \subset \{u> 0\}$ (the negative phase follows similarly).

Let v minimize $J_{B(x_0,r)}$ with boundary values $u|_{\partial B(x_0, r)}$ . Let us first assume that v is simply a harmonic function in $B(x_0, r)$ (this will be the case when $v> 0$ ). We compute that

$$ \begin{align*}\int_{B(x_0, r)} |\nabla u|^2 - |\nabla v|^2 = -\int_{B(x_0,r)} |\nabla (u-v)|^2 = \int_{B(x_0,r)} \Delta u (u-v).\end{align*} $$

We note that

$$\begin{align*}u-v < \max_{B(x_0, r)} u - \min_{B(x_0, r)} v= \max_{B(x_0, r)} u - \min_{\partial B(x_0, r)} u < \mathrm{osc}_{B(x_0, r)} u < Cr,\end{align*}$$

where we used the maximum principle and the assumption on $|\nabla u|$ . Similarly, with a lower bound so that $\sup _{B(x_0,r)} |u-v| \leq Cr$ . By our assumptions

$$ \begin{align*}|\Delta u(z)| \leq \|\nabla^2 u(z)\| \leq C \mathrm{dist}(z, \{u = 0\})^{-1} \leq Cr^{\varepsilon-1},\qquad \forall z\in B(x_0,r).\end{align*} $$

Note the last inequality above is because $B(x_0, Mr^{1-\varepsilon }) \subset \{u> 0\}$ . Putting all of this together, we get that

(5.8) $$ \begin{align}\int_{B(x_0, r)} |\nabla u|^2 - |\nabla v|^2 \leq Cr^{n+\varepsilon}.\end{align} $$

If v does vanish, the maximum principle still says that $v \geq 0$ . Let h be the harmonic extension of $u|_{B(x_0, r)}$ into $B:= B(x_0, r)$ and compute, recalling that $h> 0$ ,

$$\begin{align*}J_B(v) - J_B(h) &\geq -(\sup_B q^2_+)|\{v= 0\}\cap B| + \int_B |\nabla v|^2 - |\nabla h|^2 \\ &= -(\sup_B q^2_+)|\{v = 0\}\cap B|+ \int_B |\nabla (v-h)|^2\\ &= -(\sup_B q^2_+)|\{v = 0\}\cap B| + \int_{\partial \{v> 0\}\cap B} \partial_\nu v h\, d\mathcal H^{n-1}\\ &\geq-(\sup_B q^2_+)|\{v = 0\}\cap B| + (\inf_{B} h) (\inf_B q_+) \mathcal H^{n-1}(\partial \{v = 0\}\cap B)\\ &= -(\mathrm{osc}_Bq_+^2)|\{v=0\}\cap B|\\ &\quad+ (\inf_B q_+)\left((\inf_{\partial B} u) \mathcal H^{n-1}(\partial \{v = 0\}\cap B) - \inf_Bq_+ |\{v =0\}\cap B|\right).\end{align*}$$

We apply the isoperimetric inequality to $\{v=0\}\cap B$ (recall that $v> 0$ on $\partial B$ ) and estimate $\inf _{\partial B}u \geq C^{-1}(\mathrm {dist}(x_0, \partial \{u>0\}) - r) \geq \tilde {C}^{-1}r^{1-\varepsilon }$ to get

$$ \begin{align*}J_B(v) - J_B(h) \geq -Cr^{\alpha+n} + C|\{v =0\}\cap B|^{\frac{n-1}{n}}(r^{1-\varepsilon} -c|\{v =0\}\cap B|^{1-\frac{1}{n}}).\end{align*} $$

As long as $\varepsilon < 1/n$ and $r < r_0$ is small enough we have $J_B(v) - J_B(h) \geq -Cr^{\alpha +n}$ , which combined with equation (5.8) (which compares u to its harmonic extension) gives us the inequality we want.

Proof of Lemma 5.4

Let $E =\{\rho \in [a,b]\mid \phi _\rho \leq v\}$ . Since $\rho \mapsto \phi _\rho (x_0)$ is continuous for every $x_0 \in \overline {B}$ it must be the case that E is a closed subset of $[a,b]$ . Furthermore, $b\in E$ by assumption. Thus, we will have shown that $E = [a,b]$ (and we will be done) if we can show that E is (relatively) open in $[a,b]$ .

We first observe that for all $t\in E$ we have $\overline {\{\phi _t> 0\}} \cap \overline {B} \subset \overline {B} \cap \{v > 0\}$ . Indeed, otherwise the zero sets of $\phi _t$ and v would touch in the interior of B. If $\phi _{t}(x_1) = v(x_1) = 0$ , then it must be the case that $x_1 \in \partial \{v> 0\}$ (since $\phi _{t} \leq v$ and $x_1 \in \partial \{\phi _{t}> 0\}$ by the condition (2) on the zero set of $\phi _t$ ). If $x_1$ is a two-phase point for v, then Condition (4) implies that $\partial _{\nu ^+} \phi _{t}(x_1) \geq \partial _{\nu ^+} v(x_1)$ or $\partial _{\nu ^-}\phi _{t}(x_1) \leq \partial _{\nu ^-}v(x_1)$ . Either way we get a contradiction to the Hopf maximum principle. If $x_1$ is a one-phase point for v, then $\partial _{\nu ^+}\phi _{t}(x_1) \geq q_+(x_1) = \partial _{\nu ^+}v(x_1)$ , and we again get a contradiction by the Hopf maximum principle. Note in both cases we use the condition that $\phi _{t} < v$ on the boundary whenever they are both nonzero to conclude that $v-\phi _{t} \neq 0$ , and thus the normal derivatives above must have a strict sign.

Since for all $t_0\in E$ we have $\overline {\{\phi _{t_0}> 0\}} \cap \overline {B} \subset \overline {B} \cap \{v > 0\}$ and since $t\mapsto \partial \{\phi _t> 0\}$ is continuous there exists a $\varepsilon> 0$ such that if $|\tilde {t}-t_0| < \varepsilon $ we have $\overline {\{\phi _{\tilde {t}}> 0\}} \cap \overline {B} \subset \overline {B} \cap \{v > 0\}$ .

Thus, in $\{\phi _{\tilde {t}}> 0\}$ we have that both functions are harmonic and $v \geq \phi _{\tilde {t}}$ on $\partial (\{\phi _{\tilde {t}}> 0\}\cap B) = (\partial B \cap \{\phi _{\tilde {t}} > 0\})\cup (\partial \{\phi _{\tilde {t}} > 0\}\cap \overline {B})$ . Thus $v \geq \phi _{\tilde {t}}$ on this set. Furthermore, $v> 0 = \phi _{\tilde {t}}$ on $\{\phi _{\tilde {t}} = 0\} \cap B$ (by the assumption that the zero set is the free boundary).

Finally, on the set $\{v \leq 0\}$ we have $\phi _{\tilde {t}} < 0$ so the difference $v - \phi _{\tilde {t}}$ is superharmonic on $\{v \leq 0\}$ . On the boundary of this set, we have $v- \phi _{\tilde {t}} \geq 0$ so by the maximum principle we have $v \geq \phi _{\tilde {t}}$ on this set and putting this together with the previous parts on all $\overline {B}$ . This shows E is open, and we are done.

Proof. The estimates in equation (5.10) are straightfoward applications of the Ahlfors regularity condition (c.f. [Reference David, Feneuil and Mayboroda14, Reference David, Engelstein and Mayboroda11]). If $\mathrm {spt}\mu $ has a tangent at $x_0$ and if $\Theta ^d(\mu , x_0)$ exists, the fact that the nontangential limit of $|\nabla D_{\mu , \beta }|$ exists at $x_0$ and is equal to $c\Theta ^d(\mu , Q)^{-1/\beta }$ is contained in the proof of [Reference David, Engelstein and Mayboroda11, Theorem 5.3].

Thus, to complete the proof of the lemma, it suffices to show that $|\nabla D_{\mu , \beta }|$ extends to $\mathrm {spt}(\mu )$ in a Hölder continuous fashion. This follows from the $C^2$ -character of $\mathrm {spt}(\mu )$ and the estimates in [Reference David, Engelstein and Mayboroda11, Section 2]; to be slightly more precise the difference between $\nabla D_{\mu , \beta }$ at a point $x \in \mathbb R^n \setminus \mathrm {spt}(\mu )$ and the closest point $Q \in \mathrm {spt}(\mu )$ is controlled by the $\alpha $ numbers (in the sense of Tolsa [Reference Tolsa30]) of $\mu $ at the point Q and the scale $|x-Q|$ (c.f. [Reference David, Engelstein and Mayboroda11, equation (2.19)]). Since, $\Theta ^d(\mu , Q)$ and $\mathrm {spt}(\mu )$ are regular, these $\alpha $ -numbers decay like $r^\alpha $ , which gives the desired result.

Proof of Theorem 5.1

Let $d\mu ^{\pm }(Q) = q_{\pm }^{-1}(Q) d\mathcal H^{n-1}|_{\Gamma ^{\pm }}(Q)$ , and let $u^{\pm } = D_{\mu ^{\pm }, 1}$ in $\Omega ^{\pm }$ and identically equal to zero elsewhere. Note that $\mu ^{\pm }$ are Ahlfors regular by the boundedness and nondegeneracy of $q_{\pm }$ and the $C^2$ -character of $f^{\pm }$ (we don’t need to worry about issues at infinity because $f^{\pm }$ are constant outside of a large ball).

The result immediately follows from Lemmas 5.5 and 5.3.