## 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 $$f\in C^{\infty }({\mathbb {R}}^{2})$$.