1 Introduction

A recent book of Gustafsson and Lin [4] explores the evolution of domains under a Laplacian growth process that starts from a simply connected planar domain with smooth boundary. A key result of theirs, Theorem 5.1, states that this process can be continued indefinitely as a family of simply connected domains on a suitable branched Riemann surface. However, their theorem relies on the validity of a lemma which they believe to be true but are unable to prove. (See also section 8 of [3].) The purpose of this note is to verify their conjecture and so complete the proof of their result.

Let g be a holomorphic function on a connected neighbourhood \(\omega \) of \( \overline{\mathbb {D}}\), where \(\mathbb {D}\) denotes the unit disc, and let \( \lambda \) denote planar Lebesgue measure. (We assume that \(g\not \equiv 0\) and assign g the value 1, say, outside \(\omega \) to make it globally defined.) For each \(t>0\) we define \(\Omega (t)=\{u_{t}>0\}\), where

$$\begin{aligned} u_{t}=\inf \{w\in C({\mathbb {R}}^{2}\backslash \{0\}):w\ge 0,\;\Delta w\le \left| g\right| ^{2}\lambda |_{{\mathbb {R}}^{2}\backslash \mathbb {D} }-t\delta _{0}\} \end{aligned}$$
(1)

in the sense of distributions and \(\delta _{0}\) is the unit measure at 0. The conjecture of Gustafsson and Lin is that the domains \(\Omega (t)\) are simply connected for all sufficiently small \(t>0\). Their difficulty in verifying it arises when the function g has one or more zeros on \(\partial \mathbb {D}\). Indeed, they remark that the same issue was also left unresolved in earlier work of Sakai [7]. We prove their conjecture below.

Theorem 1

There exists \(\delta >0\) such that the domains \(\Omega (t)\) \( (0<t<\delta )\) are all starshaped about 0, and so in particular are simply connected.

Our proof of Theorem 1 remains valid if we replace \(\left| g\right| ^{2}\) in (1) by any \(C^{1}\) function \(f>0\) on a neighbourhood of \(\overline{\mathbb {D}}\). (Indeed, with minor modifications, it also yields the corresponding result in higher dimensions for such functions f.) However, the result may fail if f is allowed to have even one zero, as we now illustrate.

Example 2

There is a \(C^{\infty }\) function \(f:{\mathbb {R}}^{2}\rightarrow [0,\infty )\) with precisely one zero such that, if \(\left| g\right| ^{2}\) is replaced by f in (1), then there are arbitrarily small values of \(t>0\) for which \(\Omega (t)\) is multiply connected.

Thus the geometrical character of \(\Omega (t)\) for small \(t>0\) is highly sensitive to the nature of this function f.

We will establish Theorem 1 and Example 2 in Sects. 3 and 4, respectively, following a brief review of the technique of partial balayage, on which these arguments rely. A survey of related topics, including quadrature domains and free boundary problems, may be found in [6].

2 Partial balayage

If \(\mu \) is a (positive) measure with compact support in \({\mathbb {R}}^{2}\), then we define the logarithmic potential

$$\begin{aligned} U\mu (x)=-\frac{1}{2\pi }\int \log \left| x-y\right| d\mu (y)\quad (x\in {\mathbb {R}}^{2}) \end{aligned}$$

and note that \(-\Delta U\mu =\mu \) (in the sense of distributions). Let \(f: {\mathbb {R}}^{2}\rightarrow [0,\infty )\) be a continuous function such that \(f\ge 1\) outside some compact set. The following construction, known as partial balayage, was developed by Gustafsson and Sakai [5] and also expounded by the authors in [2].

We define, for \(t>0\),

$$\begin{aligned} V_{t,f}=\sup \left\{ v\in C({\mathbb {R}}^{2}\backslash \{0\}):-\Delta v\le f\lambda |_{{\mathbb {R}}^{2}\backslash \mathbb {D}},v\le tU\delta _{0}\right\} \end{aligned}$$

and \(u_{t,f}=tU\delta _{0}-V_{t,f}\), whence \(u_{t,f}\ge 0\). Then

$$\begin{aligned} -\Delta V_{t,f}=f\lambda |_{\Omega _{f}(t){\setminus } {\mathbb {D}}},\quad {\text {where}}\quad \Omega _{f}(t)=\{u_{t,f}>0\}\supset \overline{\mathbb {D}}, \end{aligned}$$
(2)

and so \(V_{t,f}=U(f\lambda |_{\Omega _{f}(t){\setminus } \mathbb {D}})\). It follows easily, using the assumption that \(f\ge 1\) outside a compact set, that \(\Omega _{f}(t)\) is bounded. Also,

$$\begin{aligned} \int _{\Omega _{f}(t)\backslash \mathbb {D}}f(y)d\lambda (y)=t, \end{aligned}$$
(3)

since \(tU\delta _{0}=V_{t,f}\) outside \(\Omega _{f}(t)\).

Here are some more basic properties that we will need.

Proposition 3

Let \(t>0\) and \(f,f_{n}:{\mathbb {R}}^{2}\rightarrow [0,\infty )\) \((n\ge 1)\) be continuous functions that exceed 1 outside some compact set.

  1. (a)

    If \(f_{1}\le f_{2}\), then \(V_{t,f_{1}}\le V_{t,f_{2}}\), \( u_{t,f_{1}}\ge u_{t,f_{2}}\) and \(\Omega _{f_{2}}(t)\subset \Omega _{f_{1}}(t)\).

  2. (b)

    If \((f_{n})\) decreases to f, then \(V_{t,f_{n}}\rightarrow V_{t,f}\), \(u_{t,f_{n}}\rightarrow u_{t,f}\) and

    $$\begin{aligned} \cup _{n=1}^{\infty }\Omega _{f_{n}}(t)=\Omega _{f}(t). \end{aligned}$$
  3. (c)

    If \((f_{n})\) increases to f, then \(V_{t,f_{n}}\rightarrow V_{t,f}\), \(u_{t,f_{n}}\rightarrow u_{t,f}\),

    $$\begin{aligned} \Omega _{f}(t)\subset \cap _{n=1}^{\infty }\Omega _{f_{n}}(t)\quad {\text {and}}\quad \int _{\cap _{n=1}^{\infty }\Omega _{f_{n}}(t){\setminus } \Omega _{f}(t)}fd\lambda =0. \end{aligned}$$

Proof

(a) This follows immediately from the definition of \(V_{t,f}\).

(b) By part (a) the sequence \((u_{t,f_{n}})\), which equals \(\left( tU\delta _{0}-U\left( f_{n}\lambda |_{\Omega _{f_{n}}(t)\backslash \mathbb {D}}\right) \right) \), increases to the limit

$$\begin{aligned} v=tU\delta _{0}-U(f\lambda |_{\left( \cup _{n}\Omega _{f_{n}}(t)\right) {\setminus } \mathbb {D}}), \end{aligned}$$

where

$$\begin{aligned} 0\le v\le u_{t,f}=tU\delta _{0}-U\left( f\lambda |_{\Omega _{f}(t)\backslash \mathbb {D}}\right) . \end{aligned}$$

Since \(v=u_{t,f}\) outside \(\Omega _{f}(t)\), this equality must hold everywhere. The other assertions follow immediately.

(c) The argument is similar to part (b), except that \((\Omega _{f_{n}}(t))\) is now decreasing. \(\square \)

Let

$$\begin{aligned} D_{r}(w)=\left\{ z\in \mathbb {C}:\left| z-w\right| <r\right\} \quad (w\in \mathbb {C},r>0) \end{aligned}$$

and \(D_{r}=D_{r}(0)\), so that \(\mathbb {D}=D_{1}\). We identify \(\mathbb {C}\) with \({\mathbb {R}}^{2}\) in the usual way. The function g in Sect. 1 is holomorphic on a neighbourhood \(\omega \) of \(\overline{\mathbb {D}}\). We choose \(R>1\) such that \(\overline{D}_{R}\subset \omega \) and g has no zeros in \(\overline{D}_{R}\backslash \overline{\mathbb {D}}\). In the next section we choose f such that \(f=\left| g\right| ^{2}\) on \( \overline{D}_{R}\) and \(f=1\) outside \(D_{R+1}\), and will drop the symbol f from the subscripts in the notation \(V_{t,f}\), \(u_{t,f}\), \(\Omega _{f}(t)\) where no confusion can arise. We claim that there exists \(\varepsilon >0\) such that

$$\begin{aligned} \Omega (t)\subset D_{R}\quad (0<t<\varepsilon ). \end{aligned}$$

To see this we note that, if \(1<r_{1}<r_{2}<R\), then there exists \(c\in (0,1] \) such that \(f\ge c\) on the set \(A=\left( D_{r_{2}}\backslash D_{r_{1}}\right) \cup \left( {\mathbb {R}}^{2}\backslash D_{R+1}\right) \). Hence \(\Omega _{f}(t)\subset \Omega _{c\chi _{A}}(t)\). The latter set is of the form \(D_{\rho (t)}\) for some \(\rho (t)>1\), and \(\rho (t)\rightarrow r_{1} \) as \(t\rightarrow 0+\), in view of (3). Indeed, there exists \(r(t)>1\) such that \(r(t)\rightarrow 1\) as \(t\rightarrow 0+\) and \( \Omega _{f}(t)\subset D_{r(t)}\).

3 Proof of Theorem 1

Let g, f and R be as described above.

Lemma 4

Let \(x_{1},x_{2},\ldots ,x_{k}\) denote the zeros (if any) of g on \( \partial \mathbb {D}\). Then, for each \(i\in \{1,2,\ldots ,k\}\), there exist \( r_{i}\in (0,R-1)\) and a positive constant \(C_{i}\) such that

$$\begin{aligned} \nabla f(x)\cdot x\ge -C_{i}f(x)\quad (x\in D_{r_{i}}(x_{i})\backslash \overline{\mathbb {D}}). \end{aligned}$$

Proof

Suppose that g has a zero of order m at \(x_{i}\). Then \(f(x)=\left| x-x_{i}\right| ^{2m}h(x)\) on \(\omega \), where \(h\ge 0\) is smooth and \( h(x_{i})>0\). It follows that

$$\begin{aligned} \nabla f(x)\cdot x= & {} 2m|x-x_{i}|^{2m-2}h(x)(x-x_{i})\cdot x+\left| x-x_{i}\right| ^{2m}\nabla h(x)\cdot x \\= & {} h(x)\left| x-x_{i}\right| ^{2m}\left( 2m\frac{(x-x_{i})\cdot x}{ |x-x_{i}|^{2}}+\frac{\nabla h(x)\cdot x}{h(x)}\right) \\\ge & {} f(x)\frac{\nabla h(x)\cdot x}{h(x)}\quad (x\in D_{R}\backslash \overline{\mathbb {D}}), \end{aligned}$$

since

$$\begin{aligned} (x-x_{i})\cdot x=\left| x\right| ^{2}-x_{i}\cdot x>0\quad (\left| x\right| >\left| x_{i}\right| =1). \end{aligned}$$

The result follows on noting that \(h>0\) on a neighbourhood of \(x_{i}\). \(\square \)

Lemma 5

There exists \(C_{0}>0\) such that

$$\begin{aligned} \nabla f(x)\cdot x+(C_{0}+2)f(x)\ge 0\quad (x\in D_{R}\backslash \overline{\mathbb {D}}). \end{aligned}$$

Proof

Let \(x_{i},r_{i},C_{i}\) \((i=1,\ldots ,k)\) be as in Lemma 4 and define

$$\begin{aligned} A=D_{R}\backslash \left( \overline{\mathbb {D}}\cup D_{r_{1}}(x_{1})\cup \cdots \cup D_{r_{k}}(x_{k})\right) . \end{aligned}$$

Clearly \(\inf _{A}f>0\). The result follows on choosing \(C_{0}\) large enough so that \(C_{0}+2\ge C_{i}\) \((i=1,\ldots ,k)\) and

$$\begin{aligned} \inf _{x\in A}\nabla f(x)\cdot x+(C_{0}+2)\inf _{A}f\ge 0. \end{aligned}$$

\(\square \)

Proof of Theorem 1

Let

$$\begin{aligned} v_{t}(x)=\nabla u_{t}(x)\cdot x+C_{0}u_{t}(x)\quad (t>0), \end{aligned}$$

where \(u_{t}\) is as in Sect. 2 and \(C_{0}\) is as in Lemma 5. We choose \(R>1\) and \(\varepsilon >0\) as in Sect. 2, whence \(\Omega (t)\subset D_{R}\) when \(0<t<\varepsilon \). Since

$$\begin{aligned} \Delta \left( \nabla u_{t}(x)\cdot x\right)= & {} 2\Delta u_{t}(x)+\left( \nabla \Delta u_{t}(x)\right) \cdot x \\= & {} 2f(x)+\nabla f(x)\cdot x\quad (x\in \Omega (t)\backslash \overline{\mathbb {D}}), \end{aligned}$$

the function \(v_{t}\) is subharmonic in \(\Omega (t)\backslash \overline{ \mathbb {D}}\).

We know that \(u_{t}\), and hence \(v_{t}\), vanishes outside \(\Omega (t)\). Next, we will show that \(v_{t}\le 0\) on \(\partial \mathbb {D}\) for all sufficiently small t. Suppose that \(x\ne 0\). Since

$$\begin{aligned} u_{t}(x)=-\frac{t}{2\pi }\log \left| x\right| +\frac{1}{2\pi } \int _{\Omega (t)\backslash \mathbb {D}}\log \left| x-y\right| f(y)d\lambda (y), \end{aligned}$$
(4)

we see that

$$\begin{aligned} \nabla u_{t}(x)\cdot x= & {} -\frac{t}{2\pi }\frac{x}{\left| x\right| ^{2}}\cdot x+\frac{1}{2\pi }\int _{\Omega (t)\backslash \mathbb {D}}\frac{x-y}{ \left| x-y\right| ^{2}}\cdot xf(y)d\lambda (y) \nonumber \\= & {} -\frac{t}{2\pi }+\frac{1}{2\pi }\int _{\Omega (t)\backslash \mathbb {D}} \frac{x-y}{\left| x-y\right| ^{2}}\cdot (x-y)f(y)d\lambda (y) \nonumber \\&+\frac{1}{2\pi }\int _{\Omega (t)\backslash \mathbb {D}}\frac{x-y}{ \left| x-y\right| ^{2}}\cdot yf(y)d\lambda (y) \nonumber \\= & {} \frac{1}{2\pi }\int _{\Omega (t)\backslash \mathbb {D}}\frac{x-y}{ \left| x-y\right| ^{2}}\cdot yf(y)d\lambda (y), \end{aligned}$$
(5)

by (3). This last integrand is negative when \(\left| x\right| =1\), since \((x-y)\cdot y=x\cdot y-\vert y\vert ^2 \) and \(\left| y\right| >1\). Let

$$\begin{aligned} A_{x,t}=\{y\in \Omega (t)\backslash \mathbb {D}:x\cdot y\le 0\}\quad (x\in \partial \mathbb {D},t>0). \end{aligned}$$

Then

$$\begin{aligned} \frac{x-y}{\left| x-y\right| ^{2}}\cdot y\le -\frac{\left| y\right| ^{2}}{\left| x-y\right| ^{2}}\le -\frac{1}{4}\quad (y\in A_{x,t}), \end{aligned}$$

and so

$$\begin{aligned} \int _{\Omega (t)\backslash \mathbb {D}}\frac{x-y}{\left| x-y\right| ^{2}}\cdot yf(y)d\lambda (y)\le -\frac{1}{4}\int _{A_{x,t}}fd\lambda \le - \frac{1}{4}\inf _{z\in \partial \mathbb {D}}\int _{A_{z,t}}fd\lambda . \end{aligned}$$
(6)

There exists \(c>0\) such that \(\Omega (t)\supset D_{1+ct}\), because f is bounded above. Since f has only finitely many zeros on \(\partial \mathbb {D} \), there exists \(C_{*}>0\) such that

$$\begin{aligned} \inf _{z\in \partial \mathbb {D}}\int _{A_{z,t}}fd\lambda \ge C_{*}t\quad (0<t<\varepsilon ), \end{aligned}$$

so we now see from (5) and (6) that

$$\begin{aligned} \nabla u_{t}(x)\cdot x\le -\frac{C_{*}}{8\pi }t<0\quad (x\in \partial \mathbb {D},0<t<\varepsilon ). \end{aligned}$$
(7)

Also, it follows from (4) and (3) that the family \(\left\{ u_{t}/t:0<t<\varepsilon \right\} \) of subharmonic functions on \({\mathbb {R}} ^{2}\backslash \{0\}\) is locally uniformly bounded above. Since

$$\begin{aligned} \limsup _{t\rightarrow 0+}\frac{u_{t}(x)}{t}=0\quad (x\in {\mathbb {R}}^{2}\backslash \overline{\mathbb {D}}), \end{aligned}$$

this upper limit is bounded above by \(-\left( \log \left| x\right| \right) /2\pi \) on \(\overline{\mathbb {D}}\). It follows from Corollary 5.7.2 of [1] that \(u_{t}(x)/t\rightarrow 0\) uniformly on \(\partial \mathbb {D} \) as \(t\rightarrow 0+\). Hence, by (7), there exists \(\delta \in (0,\varepsilon )\) such that

$$\begin{aligned} \nabla u_{t}(x)\cdot x\le -\frac{C_{*}}{8\pi }\frac{t}{u_{t}(x)} u_{t}(x)\le -C_{0}u_{t}(x)\quad (x\in \partial \mathbb {D},0<t<\delta ), \end{aligned}$$

and so \(v_{t}\le 0\) on \(\partial \mathbb {D}\) when \(0<t<\delta \), as claimed.

We can now apply the maximum principle to the subharmonic function \(v_{t}\) on \(\Omega (t)\backslash \overline{\mathbb {D}}\) to see that \(v_{t}<0\) there. Hence

$$\begin{aligned} \nabla u_{t}(x)\cdot x\le -C_{0}u_{t}(x)<0\quad (x\in \Omega (t)\backslash \overline{\mathbb {D}},0<t<\delta ), \end{aligned}$$

and we also know that \(\nabla u_{t}(x)\cdot x=0\) on \({\mathbb {R}} ^{2}\backslash \Omega (t)\). Since \(\overline{\mathbb {D}}\subset \{u_{t}>0\}=\Omega (t)\), and \(u_{t}\) is decreasing in the radial direction from 0 at each point of \(\Omega (t)\backslash \mathbb {D}\), it follows that \(\Omega (t)\) is starshaped about 0, as required. \(\square \)

4 Details of Example 2

Let

$$\begin{aligned} f_{e}(x)=\left\{ \begin{array}{cc} \exp \left( -\left| x-y_{0}\right| ^{-2}\right) &{} (x\in {\mathbb {R}} ^{2}\backslash \{y_{0}\}) \\ 0 &{} (x=y_{0}) \end{array} \right. , \end{aligned}$$

where \(y_{0}\) is the point (1, 0), and let \(\psi :{\mathbb {R}}^{2}\rightarrow [0,1]\) be a \(C^{\infty }\) function such that \(\psi (x)=0\) when \( \left| x\right| \in [{\frac{1}{2}}, {\frac{3}{4}} ]\) and \(\psi (x)=1\) when \(\left| x\right| \in [0, {\frac{1}{4}} ]\cup [1,\infty )\). For each n in \(\mathbb {N}\) we define

$$\begin{aligned} x_{n}=\left( \cos \frac{\pi }{n},\sin \frac{\pi }{n}\right) \quad {\text {and}}\quad r_{n}=\frac{1}{n(n+1)}, \end{aligned}$$

whence the discs \(\overline{D}_{r_{n}}(x_{n})\) are pairwise disjoint, and the closed annulus

$$\begin{aligned} A_{n}=\overline{D}_{3r_{n}/4}(x_{n})\backslash D_{r_{n}/2}(x_{n}). \end{aligned}$$

We further define

$$\begin{aligned} \psi _{n}(x)=\psi \left( \frac{x-x_{n}}{r_{n}}\right) ,\quad \psi _{n,m}(x)=\frac{\psi _{n}(x)+1/m}{1+1/m}\quad (m\in \mathbb {N}) \end{aligned}$$

and

$$\begin{aligned} f_{0}=f_{e}\prod \limits _{n\ge 1}\psi _{n}. \end{aligned}$$

Since \(\int _{\Omega _{f_{0}}(t){\setminus } D_{1}}f_{0}d\lambda =t\) and

$$\begin{aligned} {\int _{D_{r_{1}/4}(x_{1}){\setminus } D_{1}}f_{0}d\lambda =\int _{D_{r_{1}/4}(x_{1}){\setminus } D_{1}}f_{e}d\lambda >0}, \end{aligned}$$

we can choose \(t_{1}>0\) small enough to ensure that

$$\begin{aligned} D_{r_{1}/4}(x_{1})\backslash \Omega _{f_{0}}(t_{1})\ne \emptyset . \end{aligned}$$

In view of (2) the nonnegative function \(u_{t_{1},f_{0}}\) is nonconstant and harmonic on the domain \(\left( {D_{1}\cup A_{1}^{\circ }} \right) \backslash \{0\}\), and so is strictly positive there. Further, \( u_{t_{1},f_{0}}\) cannot take the value 0 at any point y of \(\partial A_{1}\), since this would imply that \(\nabla \) \(u_{t_{1},f_{0}}(y)=0\), which contradicts the Hopf lemma. Hence

$$\begin{aligned} A_{1}\subset \Omega _{f_{0}}(t_{1}) \end{aligned}$$

and the constant \(c_{1}=\left( \inf _{A_{1}}{u_{t_{1},f_{0}}}\right) /2\) is strictly positive. We define

$$\begin{aligned} f_{1,m}=f_{e}\psi _{1,m}\prod \limits _{n\ge 2}\psi _{n}\quad (m\in \mathbb {N}) \end{aligned}$$

and note that the sequence \((f_{1,m})\) decreases to \(f_{0}\), whence by Proposition 3 the sequences \((\Omega _{f_{1},m}(t_{1}))\) and \( (u_{t_{1},f_{1,m}})\) are increasing,

$$\begin{aligned} \lim _{m\rightarrow \infty }u_{t_{1},f_{1,m}}=u_{t_{1},f_{0}}\quad {\text {and}} \cup _{m}\Omega _{f_{1},m}(t_{1})=\Omega _{f_{0}}(t_{1}). \end{aligned}$$

By compactness we can choose \(m_{1}\in \mathbb {N}\) such that \(A_{1}\subset \Omega _{f_{1},m_{1}}(t_{1})\) and \(\inf _{A_{1}}u_{t,f_{1,m_{1}}}>c_{1}\), and then define

$$\begin{aligned} f_{1}=f_{1,m_{1}}=f_{e}\psi _{1,m_{1}}\prod \limits _{n\ge 2}\psi _{n}. \end{aligned}$$

Since \(f_{1}\ge f_{0}\) we note that

$$\begin{aligned} {D_{r_{1}/4}(x_{1}){\setminus } \Omega _{f_{1}}(t_{1})\supset D_{r_{1}/4}(x_{1}){\setminus } \Omega _{f_{0}}(t_{1})\ne \emptyset .} \end{aligned}$$

Next, arguing as above, we choose \(t_{2}\in (0,t_{1}/2)\) small enough to ensure that

$$\begin{aligned} D_{r_{2}/4}(x_{2})\backslash \Omega _{f_{1}}(t_{2})\ne \emptyset \end{aligned}$$

and, noting that \(f_{1}=f_{0}\) outside \(D_{r_{1}}(x_{1})\), observe that

$$\begin{aligned} A_{2}\subset \Omega _{f_{1}}(t_{2}). \end{aligned}$$

Let \(c_{2}\) denote the positive constant \(\left( \inf _{A_{2}}{ u_{t_{2},f_{1}}}\right) /2\). We define

$$\begin{aligned} f_{2,m}=f_{e}\psi _{1,m_{1}}\psi _{2,m}\prod \limits _{n\ge 3}\psi _{n}\quad (m\in \mathbb {N}) \end{aligned}$$

and note that \((f_{2,m})\) decreases to \(f_{1}\). As before, we can choose \( m_{2}\in \mathbb {N}\) such that

$$\begin{aligned} A_{j}\subset \Omega _{f_{2},m_{2}}(t_{j})\quad {\text {and}}\quad \inf _{A_{j}}u_{t_{j},f_{2,m_{2}}}>c_{j}\quad (j=1,2). \end{aligned}$$

We define

$$\begin{aligned} f_{2}=f_{2,m_{2}}=f_{e}\psi _{1,m_{1}}\psi _{2,m_{2}}\prod \limits _{n\ge 3}\psi _{n} \end{aligned}$$

and note that \(\Omega _{f_{2}}(t)\subset \Omega _{f_{1}}(t)\) \((t>0)\), whence

$$\begin{aligned} D_{r_{1}/4}(x_{1})\backslash \Omega _{f_{2}}(t_{1})\ne \emptyset \quad {\text {and}}\quad D_{r_{2}/4}(x_{2})\backslash \Omega _{f_{2}}(t_{2})\ne \emptyset . \end{aligned}$$

Proceeding inductively in this way, we obtain a sequence of numbers \((t_{j})\) decreasing to 0, a sequence of positive numbers \((c_{j})\), and an increasing sequence of functions \((f_{k})\) such that

$$\begin{aligned} A_{j}\subset \Omega _{f_{k}}(t_{j}),\quad D_{r_{j}/4}(x_{j})\backslash \Omega _{f_{k}}(t_{j})\ne \emptyset \quad {\text {and}}\quad u_{t_{j},f_{k}}>c_{j}\; {\text { on }}\; A_{j}\quad (1\le j\le k). \end{aligned}$$

We define

$$\begin{aligned} f=\lim _{j\rightarrow \infty }f_{j}=f_{e}\prod \limits _{j\ge 1}\psi _{j,m_{j}}. \end{aligned}$$

Clearly

$$\begin{aligned} D_{r_{j}/4}(x_{j})\backslash \Omega _{f}(t_{j})\ne \emptyset \quad (j\in \mathbb {N}). \end{aligned}$$

By Proposition 3 again we note that \((u_{t,f_{k}})\) decreases to \(u_{t,f}\) as \(k\rightarrow \infty \) for every \(t>0\). Since \( u_{t_{j},f_{k}}\ge c_{j}\) on \(A_{j}\) for all \(j\le k\), we see that \( u_{t_{j},f}\ge c_{j}\) on \(A_{j}\) for all j, and so \(A_{j}\subset \Omega _{f}(t_{j})\) for each j. Thus \(\Omega _{f}(t_{j})\) is multiply connected for each \(j\in \mathbb {N}\). Finally, f vanishes precisely at \( y_{0}\) and, since

$$\begin{aligned} \inf \left\{ \frac{r_{j}}{\left| x-y_{0}\right| ^{2}}:x\in D_{r_{j}}(x_{j}),j\ge 1\right\} >0, \end{aligned}$$

we see that \(f\in C^{\infty }({\mathbb {R}}^{2})\).