## 1 Introduction

Let $$N \ge 2$$, $$\lambda \in (0,N)$$ and $$m>0$$ ($$\lambda$$ and m are not necessarily integers). For any measurable set $$\Omega \subset \mathbb {R}^N$$, define

\begin{aligned} \mathcal {E}(\Omega )={{\,\mathrm{Per}\,}}\Omega + D(\Omega ), \quad D(\Omega )=\frac{1}{2}\iint _{\Omega \times \Omega }\frac{\,{\mathrm{d}}x \,{\mathrm{d}}y}{|x-y|^\lambda } \,. \end{aligned}

The perimeter $${{\,\mathrm{Per}\,}}\Omega$$ is taken in the sense of De Giorgi, namely

\begin{aligned} {{\,\mathrm{Per}\,}}\, \Omega =\sup \left\{ \int _\Omega {\mathrm{div}}F(x) \,{\mathrm{d}}x \,|\, F\in C_0^1(\mathbb {R}^3,\mathbb {R}^3), |F|\le 1 \right\} , \end{aligned}

which is simply the surface area of $$\Omega$$ when the boundary is smooth. We consider the minimization problem

\begin{aligned} E(m)=\inf _{|\Omega |=m} \mathcal {E}(\Omega ). \end{aligned}

The most important case is $$\lambda =1$$ in dimension $$N=3$$, which goes back to Gamow’s liquid drop model for atomic nuclei [15]. In this case, a nucleus is thought of consisting of nucleons (protons and neutrons) in a set $$\Omega \subset \mathbb {R}^N$$. The nucleons are assumed to be concentrated with constant density, which implies that the number of nucleons is proportional to $$|\Omega |$$. The perimeter term in the energy functional corresponds to a surface tension, which holds the nuclei together. The second term in the energy functional corresponds to a Coulomb repulsion among the protons. Here for simplicity we have scaled all physical constants to be unity.

In the last decade, this model (for general $$\lambda$$ and N) has gained renewed interest in the mathematics literature. We refer to [6] for a review and, for instance, to [2, 9,10,11,12, 16,17,18,19, 22] and references therein; see also [14, 25]. A variant of the problem with a constant background has also been intensely studied, see, for instance, [1, 3,4,5, 8, 13, 20] and references therein.

In principle, the two terms in $$\mathcal {E}(\Omega )$$ are competing against each other: balls minimize the first term (by the isoperimetric inequality [7], see also [23, Theorem 14.1]) and maximize the second term (by the Riesz rearrangement inequality [26], see also [21, Theorem 3.7]). Thus the question about the existence of a minimizer for E(m) is nontrivial.

Clearly, the existence will depend on the parameter $$m>0$$. By scaling $$\Omega \mapsto m^{1/N}U$$ with $$|U|=1$$, we see that

\begin{aligned} \mathcal {E}(\Omega ) = m^{\frac{N-1}{N}} {{\,\mathrm{Per}\,}}U + m^{\frac{2N-\lambda }{N}} \ D(U) = m^{\frac{N-1}{N}} \Big ( {{\,\mathrm{Per}\,}}U + m^{\frac{N+1-\lambda }{N}}\ D(U) \Big ). \end{aligned}

Note that $$(N+1-\lambda )/N>0$$. This suggests that for small m the short range attraction due to the perimeter term is dominant, whereas for large m the long range repulsion due to the Riesz potential is dominant. Correspondingly, we expect that there is a minimizer for small m and there is no minimizer for large m.

In the case $$\lambda =1$$, $$N=3$$, the physics literature suggests that there is a critical volume $$m_*>0$$ such that balls are unique minimizers for E(m) when $$m\le m_*$$ and there is no minimizer when $$m>m_*$$ . The value $$m_*$$ corresponds to the threshold where the energy of a ball of volume m is equal to that of two balls of mass m/2, spaced infinitely far apart. It can be computed explicitly to be (see [5, 12])

\begin{aligned} m_*=\frac{|B_1|\ {{\,\mathrm{Per}\,}}B_1}{D(B_1)} \cdot \frac{2^{1/3}-1}{1-2^{-2/3}} = 5\frac{2^{1/3}-1}{1-2^{-2/3}}\approx 3.512 \end{aligned}

with $$B_1$$ the unit ball in $$\mathbb {R}^3$$. A mathematical proof of this remains unknown.

In the present paper, we consider the general case $$N\ge 2$$ and $$\lambda \in (0,N)$$. We define the critical volume $$m_*$$ to be the unique value such that

\begin{aligned} \mathcal {E}\Big ( \Big ( \frac{m_*}{|B_1|}\Big )^{ \frac{1}{N}} B_1 \Big )= 2\, \mathcal {E}\Big ( \Big ( \frac{m_*}{2|B_1|}\Big )^{1/N} B_1\Big ) \,, \end{aligned}

namely,

\begin{aligned} m_* = \Big ( \frac{2^{1/N}-1}{1-2^{(\lambda -N)/N}} \cdot \frac{{{\,\mathrm{Per}\,}}B_1 }{D(B_1)}\Big )^{N/(N+1-\lambda )} |B_1|. \end{aligned}
(1)

Here $$B_1$$ is the unit ball in $$\mathbb {R}^N$$ (hence, $$(m/|B_1|)^{1/N} B_1$$ is a ball of measure m). Thus, just like in the special case $$\lambda =1$$, $$N=3$$, this is the critical value where the energy of a ball of volume $$m_*$$ is equal to that of two balls of mass $$m_*/2$$, spaced infinitely far apart, and it is natural to conjecture that $$m_*$$ divides the regime where minimizers are balls from the regime where there are no minimizers.

The following results were proved by Knüpfer and Muratov [18, 19]:

1. (a)

For every $$N\ge 2$$ and $$\lambda \in (0,N)$$, there exists a constant $$m_{c_1}>0$$ such that E(m) has a minimizer for every $$m \le m_{c_1}$$.

2. (b)

For every $$N\ge 2$$ and $$\lambda \in (0,2)$$, there exists a constant $$m_{c_2}>0$$ such that E(m) has no minimizer for every $$m > m_{c_2}$$.

3. (c)

If $$N=2$$ and $$\lambda >0$$ is sufficiently small, then $$m_{c_1}=m_{c_2}=m_*$$ and balls are unique minimizers for E(m) with $$m\le m_*$$.

4. (d)

if $$N=2$$ and $$\lambda <2$$, or if $$3\le N \le 7$$ and $$\lambda <N-1$$, then there exists a constant $$0<m_{c_1}' \le m_{c_1}$$ such that balls are unique minimizers for E(m) with $$m < m_{c_1}'$$.

In the most important case $$\lambda =1$$, $$N=3$$, see also [11, 22] for alternative proofs of the non-existence result (b) and [16] for a short proof of the uniqueness result (d). In [24], Muratov and Zaleski proved (c) for the explicit range $$0<\lambda \le 0.034$$ and $$N=2$$. In [2], Bonacini and Cristoferi extended (c) and (d) to all $$N\ge 2$$. In [9], Figalli, Fusco, Maggi, Millot and Morini extended (d) to all $$N\ge 2$$ and $$\lambda \in (0,N)$$.

Our first new result concerns the existence in (a). Except when $$\lambda >0$$ is small, the existence of minimizers for E(m) is known only for small m. In this paper, we extend the existence to what is conjectured to be the optimal range of parameters.

### Theorem 1

Let $$N\ge 2$$ and $$\lambda \in (0,N)$$. Then the variational problem E(m) has a minimizer for every $$0<m\le m_*$$, where $$m_*$$ is defined in (1).

We will prove Theorem 1 by establishing the strict binding inequality [12]

\begin{aligned} E(m)< E(m_1) + E(m-m_1), \quad \forall 0<m_1<m \end{aligned}
(2)

for all $$m<m_*$$. As a by product of our proof, we obtain the following conditional uniqueness of minimizers.

### Theorem 2

Let $$N\ge 2$$ and $$\lambda \in (0,N)$$. If E(m) has no minimizer when $$m>m_*$$, then balls are minimizers for E(m) when $$m\le m_*$$ and they are unique minimizers when $$m<m_*$$.

So far, the non-existence result in the sharp range $$m>m_*$$ is only available for $$\lambda >0$$ small [2, 18, 24]. For larger $$\lambda$$ and a nonexplicit range of m, we have

### Theorem 3

Let $$N\ge 2$$ and $$\lambda \in (0,N)$$ and $$\lambda \le 2$$. Then there exists a constant $$m_{c_2} \ge m_*$$ such that E(m) does not have a minimizer for all $$m>m_{c_2}$$.

This result is due to [18, 19, 22] for $$\lambda <2$$ and seems to be unpublished for $$\lambda =2$$. We will combine the methods in [11] and [18, 19]. It is an open problem whether the nonexistence result also holds for $$2<\lambda <N$$ when $$N\ge 3$$.

## 2 Existence

In this section we prove Theorem 1. We will deduce Theorem 1 from the following strict binding inequality.

### Theorem 4

Let $$N\ge 2$$ and $$\lambda \in (0,N)$$. Then for every $$0<m<m_*$$ with $$m_*$$ in (1), we have

\begin{aligned} E(m)< E(m_1) + E(m-m_1), \quad \forall 0<m_1<m. \end{aligned}
(3)

Thanks to [12, Theorem 3.1], the strict binding inequality (3) is a sufficient condition for the existence of minimizers of E(m). Moreover, by [12, Theorem 3.4], the set $$\{m>0: E(m) \text { has a minimizer}\}$$ is closed in $$(0,\infty )$$. Hence, Theorem 4 implies the existence of minimizers of E(m) for all $$0<m\le m_*$$. Note that the proofs of Theorems 3.1 and 3.4 in [12] extend, without modifications, to the case $$\lambda \ne 1$$; see Remark 3.7 in that paper.

We will prove the strict binding inequality using a scaling argument, based on the following key observation which uses only the isoperimetric inequality.

### Lemma 5

If $$0<m_1<m$$, then we have, with $$s= m_1/m \in (0,1)$$ and $$B_1$$ the unit ball in $$\mathbb {R}^N$$,

\begin{aligned} E(m_1) \ge s^{(2N-\lambda )/N} E(m) + (1-s^{(N+1-\lambda )/N}) s^{(N-1)/N} \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1. \end{aligned}

### Proof

Take $$\Omega \subset \mathbb {R}^N$$ such that $$|\Omega |=m_1$$. Then $$|s^{-1/N} \Omega |=m$$, and hence

\begin{aligned} E(m) \le {\mathcal {E}} (s^{-1/N} \Omega )&= s^{-(N-1)/N} {{\,\mathrm{Per}\,}}\Omega + s^{-(2N-\lambda )/N} D (\Omega ) \\&= s^{-(2N-\lambda )/N} {\mathcal {E}}(\Omega ) -\Big ( s^{-(2N-\lambda )/N} - s^{-(N-1)/N} \Big ) {{\,\mathrm{Per}\,}}\ \Omega . \end{aligned}

By the isoperimetric inequality

\begin{aligned} {{\,\mathrm{Per}\,}}\ \Omega \ge \Big ( \frac{m_1}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 = s^{(N-1)/N} \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1. \end{aligned}

Thus

\begin{aligned} E(m) \le s^{-(2N-\lambda )/N} {\mathcal {E}}(\Omega ) - \Big ( s^{-(2N-\lambda )/N} - s^{-(N-1)/N} \Big ) s^{(N-1)/N} \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1. \end{aligned}

Optimizing over all $$\Omega$$ satisfying $$|\Omega |=m_1$$ we get

\begin{aligned} E(m) \le s^{-(2N-\lambda )/N} E(m_1) - \Big ( s^{-(2N-\lambda )/N} - s^{-(N-1)/N} \Big ) s^{(N-1)/N} \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 \end{aligned}

which is equivalent to the desired inequality. $$\square$$

### Proof of Theorem 4

Take $$0<m_1<m<m_*$$. Denote $$s=m_1/m\in (0,1)$$. By Lemma 5 we have

\begin{aligned} E(m_1)&\ge s^{(2N-\lambda )/N} E(m) + (1-s^{(N+1-\lambda )/N}) s^{(N-1)/N} \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1,\\ E(m - m_1)&\ge (1-s)^{(2N-\lambda )/N} E(m) \\&\quad + (1- (1-s)^{(N+1-\lambda )/N})(1-s)^{(N-1)/N} \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1. \end{aligned}

Therefore,

\begin{aligned}&E(m_1)+ E(m-m_1)- E(m) \ge {\left( s^{(2N-\lambda )/N} + (1-s)^{(2N-\lambda )/N} - 1 \right) } E(m) \nonumber \\&\quad +{ \left( (1-s^{(N+1-\lambda )/N}) s^{(N-1)/N} + (1- (1-s)^{(N+1-\lambda )/N})(1-s)^{(N-1)/N} \right) } \nonumber \\&\quad \times {\left( \frac{m}{|B_1|}\right) ^{(N-1)/N}} {{\,\mathrm{Per}\,}}\ B_1. \end{aligned}
(4)

Moreover, by the variational principle,

\begin{aligned} E(m)\le {\mathcal {E}}\Big ( \Big (\frac{m}{|B_1|}\Big )^{1/N} B_1 \Big ) = \Big (\frac{m}{|B|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 + \Big (\frac{m}{|B_1|}\Big )^{(2N-\lambda )/N} D(B_1). \end{aligned}
(5)

Inserting (5) in (4) and using

\begin{aligned} s^{(2N-\lambda )/N} + (1-s)^{(2N-\lambda )/N} - 1<0, \quad \forall s\in (0,1), \end{aligned}
(6)

we find that

\begin{aligned}&E(m_1)+ E(m-m_1)- E(m) \\&\quad \ge \Big ( s^{(2N-\lambda )/N} + (1-s)^{(2N-\lambda )/N} - 1 \Big ) \Big ( \Big (\frac{m}{|B|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 + \Big (\frac{m}{|B_1|}\Big )^{(2N-\lambda )/N} D(B_1) \Big ) \\&\qquad + \Big ( (1-s^{(N+1-\lambda )/N}) s^{(N-1)/N} + (1- (1-s)^{(N+1-\lambda )/N})(1-s)^{(N-1)/N} \Big ) \\&\qquad \times \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}\ B_1\\&\quad = \Big ( s^{(N-1)/N} + (1-s)^{(N-1)/N} -1 \Big ) \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}\ B_1 \\&\qquad + \Big ( s^{(2N-\lambda )/N} + (1-s)^{(2N-\lambda )/N} - 1 \Big ) \Big (\frac{m}{|B_1|}\Big )^{(2N-\lambda )/N} D(B_1) \\&\quad = \Big ( s^{(2N-\lambda )/N} + (1-s)^{(2N-\lambda )/N} - 1 \Big ) \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}\ B_1 \\&\qquad \times \Big ( \frac{D(B_1)}{{{\,\mathrm{Per}\,}}\ B_1} \Big (\frac{m}{|B_1|}\Big )^{(N+1-\lambda )/N} - f(s) \Big ) \end{aligned}

with

\begin{aligned} f(s):= \frac{s^{(N-1)/N} + (1-s)^{(N-1)/N} -1 }{1-s^{(2N-\lambda )/N} - (1-s)^{(2N-\lambda )/N} }. \end{aligned}
(7)

Using again (6), we find that the strict binding inequality

\begin{aligned} E(m_1)+ E(m-m_1)- E(m) >0 \end{aligned}

holds true if

\begin{aligned} f(s)> \frac{D(B_1)}{{{\,\mathrm{Per}\,}}\ B_1} \Big (\frac{m}{|B_1|}\Big )^{(N+1-\lambda )/N} ,\quad \forall s\in (0,1). \end{aligned}
(8)

On the other hand, we can show that (see Lemma 6 below)

\begin{aligned} \min _{s\in (0,1)} f(s)= f(1/2)= \frac{2^{1/N}-1}{1-2^{(\lambda -N)/N}}. \end{aligned}
(9)

Therefore, (8) holds true when

\begin{aligned} \frac{2^{1/N}-1}{1-2^{(\lambda -N)/N}} > \frac{D(B_1)}{{{\,\mathrm{Per}\,}}\ B_1} \Big (\frac{m}{|B_1|}\Big )^{(N+1-\lambda )/N} \end{aligned}

which is equivalent to $$m<m_*$$. $$\square$$

It remains to prove (9). We have

### Lemma 6

For all $$0<a<1<b<2$$ and $$0<s<1$$ we have

\begin{aligned} \frac{s^a + (1-s)^a -1 }{2^{1-a}-1} \ge - \frac{s \log s + (1-s)\log (1-s)}{\log 2} \ge \frac{s^b + (1-s)^b -1 }{2^{1-b}-1}. \end{aligned}

The inequality (9) follows from Lemma 6 with $$a=(N-1)/N$$ and $$b=(2N-\lambda )/N$$.

### Proof

We will prove that for all $$\alpha \in (0,2)$$ and all $$s \in (0,1)$$,

\begin{aligned} g(s):= s^\alpha + (1-s)^\alpha -1 + \frac{2^{1-\alpha }-1}{\log 2} \Big ( s \log s + (1-s)\log (1-s) \Big ) \ge 0. \end{aligned}
(10)

Then the desired conclusion follows from (10) and the fact that $$2^{1-\alpha }-1>0$$ if $$\alpha \in (0,1)$$ while $$2^{1-\alpha }-1<0$$ if $$\alpha \in (1,2)$$ (Fig. 1).

By the symmetry $$s \leftrightarrow 1-s$$, it suffices to prove (10) for $$s\in (0,1/2]$$. Also, (10) is trivial when $$\alpha =1$$, so we will distinguish two cases $$\alpha \in (0,1)$$ and $$\alpha \in (1,2)$$.

Case 1: $$\alpha \in (0,1)$$. We have

\begin{aligned} g'(s)&= \alpha \Big (s^{\alpha -1} - (1-s)^{\alpha -1}\Big ) + \frac{2^{1-\alpha }-1}{\log 2} \Big ( \log s -\log (1-s) \Big ),\\ g''(s)&= \alpha (\alpha -1) \Big (s^{\alpha -2} + (1-s)^{\alpha -2} \Big ) + \frac{2^{1-\alpha }-1}{\log 2} \Big ( s^{-1}+(1-s)^{-1}\Big ). \end{aligned}

Define $$h:(0,1/2]\rightarrow \mathbb {R}$$ by

\begin{aligned} h(s)&:= s(1-s) g''(s) \\&= \alpha (\alpha -1) s(1-s) \Big (s^{\alpha -2} + (1-s)^{\alpha -2} \Big ) + \frac{2^{1-\alpha }-1}{\log 2} \\&= \alpha (\alpha -1) \Big (s^{\alpha -1} + (1-s)^{\alpha -1} - s^\alpha - (1-s)^\alpha \Big ) + \frac{2^{1-\alpha }-1}{\log 2}. \end{aligned}

Note that for all $$s\in (0,1/2)$$ we have

\begin{aligned} h'(s)= \alpha (\alpha -1)^2 \Big ( s^{\alpha -2} - (1-s)^{\alpha -2} \Big ) + \alpha ^2 (1-\alpha ) \Big ( s^{\alpha -1} - (1-s)^{\alpha -1}\Big )> 0 \end{aligned}

since

\begin{aligned} s^{\alpha -2} - (1-s)^{\alpha -2}> 0, \quad (1-\alpha ) ( s^{\alpha -1} - (1-s)^{\alpha -1}) >0. \end{aligned}

Thus h is strictly increasing on (0, 1/2]. Moreover,

\begin{aligned} \lim _{s\rightarrow 0^+} h(s)=-\infty \end{aligned}

and

\begin{aligned} h(1/2)&= \alpha (\alpha -1) 2^{1-\alpha } + \frac{2^{1-\alpha }-1}{\log 2}= 2^{1-\alpha } \Big ( \frac{1}{\log 2} - \alpha (1-\alpha ) \Big ) - \frac{1}{\log 2}\\&\ge \Big (1 + (1-\alpha ) \log 2\Big ) \Big (\frac{1}{\log 2} - \alpha (1-\alpha ) \Big ) -\frac{1}{\log 2}\\&= (1-\alpha )^2 \Big [ 1 - \alpha \log 2 \Big ]>0, \end{aligned}

since

\begin{aligned} 2^{1-\alpha }=e^{(1-\alpha )\log 2} \ge 1 + (1-\alpha )\log 2, \quad \alpha (1-\alpha ) \le \frac{1}{4} < \frac{1}{\log 2}. \end{aligned}

Thus there exists a unique value $$s_1\in (0,1/2)$$ (depending on $$\alpha$$) such that

\begin{aligned} h(s) <0 \text{ on } s\in (0,s_1), \quad h(s)>0 \text { on } s\in (s_1,1/2). \end{aligned}

Putting back the definition $$h(s)= s(1-s) g''(s)$$, we find that

\begin{aligned} g''(s) <0 \text{ on } s\in (0,s_1), \quad g''(s)>0 \text { on } s\in (s_1,1/2). \end{aligned}

Thus $$g'(s)$$ is strictly decreasing on $$s\in (0,s_1)$$ and strictly increasing on $$s\in (s_1,1/2)$$. Combining with

\begin{aligned} \lim _{s\rightarrow 0^+} g'(s)= \infty , \quad g'(1/2)= 0, \end{aligned}

we find that there exists a unique value $$s_2\in (0,1/2)$$ (depending on $$\alpha$$) such that

\begin{aligned} g'(s) >0 \text{ on } s\in (0,s_2), \quad g'(s)<0 \text { on } s\in (s_2,1/2). \end{aligned}

Thus g(s) is strictly increasing on $$s\in (0,s_2)$$ and strictly decreasing on $$s\in (s_2,1/2)$$. Therefore,

\begin{aligned} \inf _{s\in (0,1/2]}g(s) = \min \{ \lim _{s\rightarrow 0^+} g(s), g(1/2) \}=0. \end{aligned}

Case 2: $$\alpha \in (1,2)$$. We can proceed similarly. To be precise, the function $$h(s)=s(1-s)g''(s)$$ also satisfies

\begin{aligned} h'(s)= \alpha (\alpha -1)^2 \Big ( s^{\alpha -2} - (1-s)^{\alpha -2} \Big ) + \alpha ^2 (1-\alpha ) \Big ( s^{\alpha -1} - (1-s)^{\alpha -1}\Big )> 0 \end{aligned}

for all $$s\in (0,1/2)$$. Thus h is also strictly increasing on (0, 1/2]. Moreover,

\begin{aligned} \lim _{s\rightarrow 0^+} h(s)=\frac{2^{1-\alpha }-1}{\log 2}<0 \end{aligned}

and

\begin{aligned} h(1/2)&= \alpha (\alpha -1) 2^{1-\alpha } + \frac{2^{1-\alpha }-1}{\log 2}= \alpha (\alpha -1) 2^{1-\alpha } + \frac{ e^{(1-\alpha ) \log 2} -1 }{\log 2}\\&\ge \alpha (\alpha -1) 2^{1-\alpha } + 1-\alpha = (\alpha -1) (\alpha 2^{1-\alpha }-1)>0. \end{aligned}

Here we have used $$\alpha 2^{1-\alpha }>1$$ for all $$\alpha \in (1,2)$$ (the function $$q(\alpha )=\alpha 2^{1-\alpha }$$ is concave on (1, 2) as $$q''(\alpha )=2^{1-\alpha }(\log 2) ( \alpha \log 2 - 2)<0$$ and $$q(1)=q(2)=1$$).

Thus there exists a unique value $$s_1\in (0,1/2)$$ (depending on $$\alpha$$) such that

\begin{aligned} h(s) <0 \text{ on } s\in (0,s_1), \quad h(s)>0 \text { on } s\in (s_1,1/2). \end{aligned}

The rest is exactly the same as in Case 1. This completes the proof of Lemma 6. $$\square$$

## 3 Uniqueness

### Proof of Theorem 2

Step 1 We prove that balls are minimizers for $$E(m_*)$$. Assume by contradiction that balls are not minimizers for $$E(m_*)$$, namely

\begin{aligned} E(m_*)< \mathcal {E}( (m_*/|B_1|)^{1/N} B_1). \end{aligned}

Since $$m\mapsto E(m)$$ and $$m\mapsto \mathcal {E}( (m/|B_1|)^{1/N} B_1)$$ are continuous (for the first function, see [12, Proof of Theorem 3.1]), there exists a constant $$\delta \in (0,1)$$ such that for all $$m \in [m_*, m_*+\delta )$$ we have

\begin{aligned} E(m)\le & {} (1-\delta )\mathcal {E}( (m/|B_1|)^{1/N} B_1) \nonumber \\\le & {} \Big (\frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 + \Big (\frac{m}{|B_1|}\Big )^{(2N-\lambda )/N} (1-\delta ) D(B_1). \end{aligned}
(11)

This is similar to (5), but $$D(B_1)$$ is replaced by $$(1-\delta ) D(B_1)$$. Proceeding similarly as in the proof of Theorem 4 and inserting (11) (instead of (5)) in (4), for all $$m\in [m_*, m_*+\delta )$$ and $$0<m_1<m$$ we have

\begin{aligned}&E(m_1)+ E(m-m_1)- E(m) \\&\quad \ge \Big ( s^{(2N-\lambda )/N} + (1-s)^{(2N-\lambda )/N} - 1 \Big ) \Big ( \frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}\ B_1 \times \\&\qquad \times \Big ( \frac{(1-\delta ) D(B_1)}{{{\,\mathrm{Per}\,}}\ B_1} \Big (\frac{m}{|B_1|}\Big )^{(N+1-\lambda )/N} - f(s) \Big ) \end{aligned}

with $$s=m_1/m\in (0,1)$$ and with the same function f(s) in (7). By (9), we conclude that

\begin{aligned} E(m_1)+ E(m-m_1)- E(m) >0 , \quad \forall 0<m_1<m \end{aligned}

provided that

\begin{aligned} \frac{2^{1/N}-1}{1-2^{(\lambda -N)/N}} \ge \frac{(1-\delta )D(B_1)}{{{\,\mathrm{Per}\,}}\ B_1} \Big (\frac{m}{|B_1|}\Big )^{(N+1-\lambda )/N} \end{aligned}

which is equivalent to

\begin{aligned} m\le m_*(1-\delta )^{- N/(N+1-\lambda )}. \end{aligned}

Thus the variational problem E(m) has a minimizer for all

\begin{aligned} m\le \min \{ m_*+\delta , m_*(1-\delta )^{- N/(N+1-\lambda )} \}. \end{aligned}

This is a contradiction to the assumption that E(m) has no minimizer if $$m>m_*$$. Thus we conclude that balls are minimizers for $$E(m_*)$$.

Step 2 Now we prove that if $$m<m_*$$, then balls are unique minimizers for E(m). This fact follows from [2, Theorem 2.10] which states that the set where balls are minimizers is an interval and that one has uniqueness away from the endpoint (note that this part does not require the assumption $$\lambda <N-1$$ which is imposed in the rest of [2]). For the reader’s convenience, we provide a direct proof below.

Consider an arbitrary measurable set $$\Omega \subset \mathbb {R}^N$$ with $$|\Omega |=m<m_*$$. Then proceeding as in the proof of Lemma 5, we find that

\begin{aligned} \mathcal {E}(\Omega ) \ge s^{(2N-\lambda )/N} E(m_*) + (1-s^{(N+1-\lambda )/N}) s^{(N-1)/N} \Big ( \frac{m_*}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 \end{aligned}
(12)

with $$s=m/m_*\in (0,1)$$ and the equality occurs if and only if $$\Omega$$ is a ball. On the other hand, we know that balls are minimizers for $$E(m_*)$$, namely

\begin{aligned} E(m_*) = \mathcal {E}\Big ( \Big (\frac{m_*}{|B_1|}\Big )^{1/N} B_1 \Big ) = \Big (\frac{m_*}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 + \Big (\frac{m_*}{|B_1|}\Big )^{(2N-\lambda )/N} D(B_1). \end{aligned}

Inserting the latter equality in (12), we obtain

\begin{aligned} \mathcal {E}(\Omega )&\ge s^{(2N-\lambda )/N} \Big ( \Big (\frac{m_*}{|B|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 + \Big (\frac{m_*}{|B_1|}\Big )^{(2N-\lambda )/N} D(B_1) \Big ) \\&\qquad + (1-s^{(N+1-\lambda )/N}) s^{(N-1)/N} \Big ( \frac{m_*}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1\\&= \Big (\frac{m}{|B_1|}\Big )^{(N-1)/N} {{\,\mathrm{Per}\,}}B_1 + \Big (\frac{m}{|B_1|}\Big )^{(2N-\lambda )/N} D(B_1) = \mathcal {E}\Big ( \Big ( \frac{m}{B_1}\Big )^{1/N} B_1 \Big ). \end{aligned}

Thus balls are minimizers for E(m); moreover, if $$\Omega$$ is a minimizer for E(m), then the equality occurs in (12) and $$\Omega$$ is a ball. $$\square$$

## 4 Nonexistence

In this section we prove Theorem 3. First, by extending the analysis for $$\lambda =1$$ in [11] to general $$\lambda$$, we have

### Lemma 7

Let $$N\ge 2$$ and $$\lambda \in (0,N)$$. Let $$m>0$$ be arbitrary. Let $$\Omega \subset \mathbb {R}^N$$ be a minimizer for E(m). Then

\begin{aligned} \iint _{\Omega \times \Omega } \frac{dx\,dy}{|x-y|^{\lambda -1}} \lesssim |\Omega | \,, \end{aligned}

with an implied constant depending only on $$\lambda$$ and N.

It is unclear to us whether the power $$\lambda -1$$ in Lemma 7 is optimal. If we could replace this power by a smaller one, then we would be able to improve the condition $$\lambda \le 2$$ in Theorem 3.

### Proof

For $$\nu \in \mathbb {S}^{N-1}$$ and $$t\in \mathbb {R}$$ we set

\begin{aligned} \Omega ^{\pm }_{\nu ,t} := \Omega \cap \{ x\in \mathbb {R}^N:\ \pm \nu \cdot x > \pm t \} \,. \end{aligned}

For any $$\rho \ge 0$$, the set

\begin{aligned} \Omega ^+_{\nu ,t} \cup \left( \Omega ^-_{\nu ,t} - \rho \nu \right) \end{aligned}

has measure $$|\Omega ^+_{\nu ,t} \cup \left( \Omega ^-_{\nu ,t} - \rho \nu \right) |=|\Omega |=m$$ and therefore, by minimality of $$\Omega$$,

\begin{aligned} \mathcal {E}\!\left( \Omega ^+_{\nu ,t} \cup \left( \Omega ^-_{\nu ,t} - \rho \nu \right) \right) \ge \mathcal {E}(\Omega ) \,. \end{aligned}
(13)

For any $$\rho >0$$, we have

\begin{aligned} {{\,\mathrm{Per}\,}}\left( \Omega ^+_{\nu ,t} \cup \left( \Omega ^-_{\nu ,t} - \rho \nu \right) \right) = {{\,\mathrm{Per}\,}}\Omega ^+_{\nu ,t} + {{\,\mathrm{Per}\,}}\Omega ^-_{\nu ,t} \le {{\,\mathrm{Per}\,}}\Omega + 2 \sigma (\Omega \cap \{\nu \cdot x = t\}) \,, \end{aligned}

where $$\sigma$$ denotes the induced measure on the hyperplane $$\{\nu \cdot x = t\}$$ and where the inequality holds for almost every $$t\in \mathbb {R}$$.

On the other hand, for any $$\rho \ge 0$$,

\begin{aligned} \iint _{ \left( \Omega ^+_{\nu ,t} \cup \left( \Omega ^-_{\nu ,t} - \rho \nu \right) \right) \times \left( \Omega ^+_{\nu ,t} \cup \left( \Omega ^-_{\nu ,t} - \rho \nu \right) \right) } \frac{dx\,dy}{|x-y|^\lambda }&= \iint _{\Omega ^+_{\nu ,t}\times \Omega ^+_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } + \iint _{\Omega ^-_{\nu ,t}\times \Omega ^-_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } \\&\quad + 2 \iint _{\Omega ^+_{\nu ,t}\times \Omega ^-_{\nu ,t}} \frac{dx\,dy}{|x-y+\rho \nu |^\lambda } \,. \end{aligned}

The last double integral tends to zero as $$\rho \rightarrow \infty$$

Inserting these facts into (13) and letting $$\rho \rightarrow \infty$$, we infer

\begin{aligned}&{{\,\mathrm{Per}\,}}\Omega + 2 \sigma (\Omega \cap \{\nu \cdot x = t\}) + \frac{1}{2} \iint _{\Omega ^+_{\nu ,t}\times \Omega ^+_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } + \frac{1}{2} \iint _{\Omega ^-_{\nu ,t}\times \Omega ^-_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } \\&\quad \ge \mathcal {E}(\Omega ) \\&\quad = {{\,\mathrm{Per}\,}}\Omega + \frac{1}{2} \iint _{\Omega ^+_{\nu ,t}\times \Omega ^+_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } + \frac{1}{2} \iint _{\Omega ^-_{\nu ,t}\times \Omega ^-_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } + \iint _{\Omega ^+_{\nu ,t}\times \Omega ^-_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } \,, \end{aligned}

that is,

\begin{aligned} \sigma (\Omega \cap \{\nu \cdot x = t\}) \ge \frac{1}{2} \iint _{\Omega ^+_{\nu ,t}\times \Omega ^-_{\nu ,t}} \frac{dx\,dy}{|x-y|^\lambda } \,. \end{aligned}

Note that the double integral here can be written as $$\iint _{\Omega \times \Omega } |x-y|^{-\lambda } {\mathbb {1}}_{\{\nu \cdot x>t>\nu \cdot y\}}\,dx\,dy$$. Thus, integrating the inequality with respect to $$t\in \mathbb {R}$$ gives, by Fubini’s theorem,

\begin{aligned} |\Omega | \ge \frac{1}{2} \iint _{\Omega \times \Omega } \frac{(\nu \cdot (x-y))_+}{|x-y|^{\lambda }}\,dx\,dy \,. \end{aligned}

Finally, we average this inequality with respect to $$\nu \in \mathbb {S}^{N-1}$$ and use the fact that

\begin{aligned} \int _{\mathbb {S}^{N-1}} (\nu \cdot (x-y))_+\, \frac{d\nu }{|\mathbb {S}^{N-1}|} = c_N |x-y| \,, \end{aligned}

to obtain the bound in the lemma. $$\square$$

With Lemma 7 at hand, it is easy to finish the proof of Theorem 3 if $$\lambda \le 1$$. In fact, if $$\lambda =1$$, the lemma gives directly $$|\Omega |^2\lesssim |\Omega |$$, which is the claimed bound. If $$0<\lambda <1$$, by Riesz’s rearrangement inequality we have for all $$t>0$$

\begin{aligned} \iint _{\Omega \times \Omega } t^{-1} ( 1- e^{-t|x-y|^{1-\lambda }} ) \,dx\,dy \ge \iint _{\Omega ^*\times \Omega ^*} t^{-1} ( 1- e^{-t|x-y|^{1-\lambda }} ) \,dx\,dy \,, \end{aligned}

where $$\Omega ^*$$ is the ball centered at 0 with volume $$|\Omega ^*|=|\Omega |$$. Taking $$t\rightarrow 0$$ we obtain

\begin{aligned} \iint _{\Omega \times \Omega } \frac{dx\,dy}{|x-y|^{\lambda -1}} \ge \iint _{\Omega ^*\times \Omega ^*} \frac{dx\,dy}{|x-y|^{\lambda -1}} \gtrsim |\Omega ^*|^{(2N-\lambda +1)/N} = |\Omega |^{(2N-\lambda +1)/N} \,. \end{aligned}

Since $$(2N-\lambda +1)/N>1$$, the lemma implies once again $$|\Omega |\lesssim 1$$.

It remains to deal with the case $$1<\lambda \le 2$$. The key is the following bound, which in the special case $$N=3$$ and $$\lambda =1$$ appears in [22, Eq. (2.12)]. The proof there extends immediately to the general case, since the analogues of [22, Lemma 3 (ii) and Lemma 4] hold according to [19, Lemmas 4.1 and 4.3].

### Lemma 8

Let $$N\ge 2$$ and $$\lambda \in (0,N)$$. Let $$m\ge \omega _N$$ and let $$\Omega \subset \mathbb {R}^N$$ be a minimizer for E(m). Then, for $$1\le R\le {{\,\mathrm{diam}\,}}\Omega$$,

\begin{aligned} |\Omega \cap B_R(x)|\gtrsim R \qquad \text {for a.e.}\ x\in \Omega \,, \end{aligned}

with an implied constant depending only on $$\lambda$$ and N.

Here $${{\,\mathrm{diam}\,}}\Omega$$ in the lemma is understood as the diameter of the set $$\{x\in \mathbb {R}^N:\ |\Omega \cap B_r(x)|>0 \ \text {for all}\ r>0\}$$.

We will use this lemma to deduce Theorem 3 for $$1<\lambda \le 2$$. If $${{\,\mathrm{diam}\,}}\Omega \le 2$$, then $$|\Omega |\lesssim 1$$ and we are done. Thus, assuming $${{\,\mathrm{diam}\,}}\Omega > 2$$, we have, by Lemma 8,

\begin{aligned} \iint _{\Omega \times \Omega } \frac{dx\,dy}{|x-y|^{\lambda -1}}&= (\lambda -1) \int _0^\infty \frac{dR}{R^{\lambda }} \,|\{(x,y)\in \Omega \times \Omega :\ |x-y|<R \}| \\&= (\lambda -1) \int _0^\infty \frac{dR}{R^{\lambda }} \int _\Omega dx\, |\Omega \cap B_R(x)| \\&\ge (\lambda -1) \int _1^{{{\,\mathrm{diam}\,}}\Omega } \frac{dR}{R^{\lambda }} \int _\Omega dx\, |\Omega \cap B_R(x)| \\&\gtrsim \int _1^{{{\,\mathrm{diam}\,}}\Omega } \frac{dR}{R^{\lambda -1}}\, |\Omega | \,. \end{aligned}

The right side is bounded from below by a constant times $$|\Omega | ({{\,\mathrm{diam}\,}}\Omega )^{2-\lambda }$$ if $$\lambda <2$$ and by a constant times $$|\Omega | \log {{\,\mathrm{diam}\,}}\Omega$$ if $$\lambda =2$$. Combining this lower bound on the double integral with the upper bound from Lemma 7, we infer in either case that $${{\,\mathrm{diam}\,}}\Omega \lesssim 1$$, which implies $$|\Omega |\lesssim 1$$ and therefore concludes the proof of Theorem 3.