Proof. By assumption, we have the existence of minimizers, that any minimizer is compactly supported, and that any minimizer \(\rho\) with \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho({\boldsymbol{x}})}=0\) satisfies \(\textrm{supp\,}\rho\subset \overline{B(0;R_*)}\). Suppose there are two distinct minimizers \(\rho_0,\rho_1\) which are not translations of each other, then we may assume \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\rho_0({\boldsymbol{x}})\,\mathrm{d}{{\boldsymbol{x}}}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\rho_1({\boldsymbol{x}})\,\mathrm{d}{{\boldsymbol{x}}}=0\) without loss of generality, by translation. It follows that \(\textrm{supp\,}\rho_i\subset \overline{B(0;R_*)},\,i=0,1\). Then the assumption \(R_{\textrm{LIC}}[W]> R_*\) implies that \(t\mapsto E [(1-t)\rho_0+t\rho_1]\) is strictly convex on \(t\in [0,1]\). Since \(\rho_0\) and \(\rho_1\) are minimizer, we have \(E[\rho_0]=E[\rho_1]\), and thus the probability measure \(\frac{1}{2}\rho_0+\frac{1}{2}\rho_1\) has a smaller energy, contradicting the minimizing property of \(\rho_0,\rho_1\). ◻

Proof. By Lemma 1, the minimizer of \(E\) is unique up to translation. Since \(R_*\) is an upper bound of minimizer size, we see that there is a unique minimizer \(\rho_\infty\) of \(E\) satisfying \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho_\infty({\boldsymbol{x}})}=0\), and we have \(\textrm{supp\,}\rho_\infty\subset \overline{B(0;R_*)}\).

If we assume the contrary that \(\rho\) is not a minimizer, then \(E[\rho]>E[\rho_\infty]\). Since \(R_{\textrm{LIC}}[W]> R_*\), we see that \(E\) is strictly convex along the linear interpolation between \(\rho\) and \(\rho_\infty\), and the same is true if \(\rho_\infty\) is slightly mollified. Then one can proceed similarly as the proof of [5] to get a contradiction (where one notices that ?? is effective in the support of a slightly mollified version of \(\rho_\infty\)). ◻

Proof. First, using ?? and ?? , one can proceed as in the proof of Lemma 2 to prove that \(\rho\) is the unique minimizer of \(E\) within the class of probability measures \(\{\rho'\in \mathcal{M}(\overline{B(0;R_0+\frac{2}{3}\epsilon)}):\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho'({\boldsymbol{x}})}=0\}\).

Then let \(\rho_1\ne \rho\) satisfy \(d_\infty(\rho,\rho_1)<\frac{1}{3}\epsilon\). It implies that \(\textrm{supp\,}\rho_1\subset \overline{B(0;R_0+\frac{1}{3}\epsilon)}\) and \({\boldsymbol{x}}_c:=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho_1({\boldsymbol{x}})}\) satisfies \(|{\boldsymbol{x}}_c|\le \frac{1}{3}\epsilon\). Therefore \(\tilde{\rho}_1({\boldsymbol{x}}) := \rho_1({\boldsymbol{x}}+{\boldsymbol{x}}_c)\) satisfies \(E[\tilde{\rho}_1]=E[\rho_1]\), \(\textrm{supp\,}\tilde{\rho}_1\subset \overline{B(0;R_0+\frac{2}{3}\epsilon)}\) and \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\tilde{\rho}_1({\boldsymbol{x}})}=0\). Then it follows from the previous paragraph \(E[\rho_1]=E[\tilde{\rho}_1]\ge E[\rho]\). ◻

Proof. \(W_{a,b}\) with \(-d<b<a\) satisfies the assumptions in [1]. Applying this theorem shows that there exists a minimizer of \(E_{a,b}\), and any minimizer of \(E_{a,b}\) is compactly supported. Then we proceed to estimate the size of a minimizer \(\rho\) of \(E_{a,b}\). We will only treat the case \(b\ne 0\), because the case of \(b=0\) (logarithmic repulsion) can be handled in a similarly way.

First notice that \[\begin{align} E_{a,b}\Big[\frac{\chi_{B(0;1/2)}}{|B(0;1/2)|}\Big] = & \frac{1}{2a|B(0;1/2)|^2}\int_{B(0;1/2)} \int_{B(0;1/2)} |{\boldsymbol{x}}-{\boldsymbol{y}}|^a \,\mathrm{d}{{\boldsymbol{x}}}\,\mathrm{d}{{\boldsymbol{y}}} \\ & - \frac{1}{2b|B(0;1/2)|^2}\int_{B(0;1/2)} \int_{B(0;1/2)} |{\boldsymbol{x}}-{\boldsymbol{y}}|^b \,\mathrm{d}{{\boldsymbol{x}}}\,\mathrm{d}{{\boldsymbol{y}}} \\ \le & {\frac{1}{2a} + C_1 \le \frac{1}{2a_0} + C_1 =: C_2\,,} \end{align}\] where \(C_1\) denotes the integral term involving \(b\), which does not depends on \(a\). As a consequence, \(C_2\) depends on \(b\) and \(a_0\), independent of \(a\), and \[E_{a,b}[\rho] \le E_{a,b}\Big[\frac{\chi_{B(0;1/2)}}{|B(0;1/2)|}\Big] \le C_2\,,\] since \(\rho\) is a minimizer of \(E_{a,b}\).

Then we claim that there exists \(r_*\ge 2\), depending on \(b\) and \(a_0\), such that \[\label{claim95t} \min_{t\ge r} (\frac{t^a}{a}-\frac{t^b}{b}) \ge \frac{r^a}{2a},\quad \forall a\ge a_0,\,r\ge r_*\,.\tag{4}\] In fact, if \(b<0\) then 4 is trivial. If \(b>0\), then for any \(a\ge a_0\) and \(t\ge r_*\) we estimate \[\frac{t^b}{b}\cdot \frac{a}{t^a} = \frac{a}{b}\cdot t^{b-a} \le \frac{a}{b}\cdot r_*^{b-a} \le \frac{1}{b} \cdot 2^{(b-a)/2}a \cdot r_*^{(b-a_0)/2}\,.\] Notice that \(\lim_{a\rightarrow\infty}2^{(b-a)/2}a = 0\), and thus \[C_3:=\sup_{a\ge a_0}2^{(b-a)/2}a<\infty\,,\] is independent of \(a\). Then we choose \[r_* = \Big(\frac{b}{2C_3}\Big)^{2/(b-a_0)}\,,\] so that \(\frac{t^b}{b}\cdot \frac{a}{t^a}\le \frac{1}{2}\) for any \(a\ge a_0\) and \(t\ge r_*\), and 4 follows.

Then, similar to [1], we take \[\label{R1} R_1 = \max\left\{r_*,\Big(\frac{2}{C_4}\big(2C_2-\min\big\{-\frac{1}{b},0\big\}+1\big)\Big)^{2/a_0}\right\}\,,\tag{5}\] where \[\label{C5} C_4 = \min_{a>0}\frac{2^{a/2}}{2a} > 0\,.\tag{6}\] Then \(R_1\) is independent of \(a\) and satisfies \(R_1\ge r_* \ge 2\). We claim that \[\label{claim95r} \rho(B({\boldsymbol{x}}_0;R_1)) \ge \frac{1}{2},\quad \forall {\boldsymbol{x}}_0\in\textrm{supp\,}\rho\,.\tag{7}\] Assume the contrary, i.e., \(\rho(B({\boldsymbol{x}}_0;R_1)^c) > \frac{1}{2}\) for some \({\boldsymbol{x}}_0\in\textrm{supp\,}\rho\), and estimate \[\begin{align} (W_{a,b}*\rho)({\boldsymbol{x}}_0) = & \int_{B({\boldsymbol{x}}_0;R_1)} W_{a,b}({\boldsymbol{x}}_0-{\boldsymbol{y}})\,\mathrm{d}{\rho({\boldsymbol{y}})} + \int_{B({\boldsymbol{x}}_0;R_1)^c} W_{a,b}({\boldsymbol{x}}_0-{\boldsymbol{y}})\,\mathrm{d}{\rho({\boldsymbol{y}})} \\ \ge & \min W_{a,b}\int_{B({\boldsymbol{x}}_0;R_1)} \,\mathrm{d}{\rho({\boldsymbol{y}})} + \min_{t\ge R_1} \big(\frac{t^a}{a}-\frac{t^b}{b}\big)\int_{B({\boldsymbol{x}}_0;R_1)^c} \,\mathrm{d}{\rho({\boldsymbol{y}})} \\ \ge & \min\big\{\frac{1}{a}-\frac{1}{b},0\big\} + \frac{R_1^a}{4a} \\ \ge & \min\big\{-\frac{1}{b},0\big\} + \frac{2^{a/2}}{4a}\cdot R_1^{a_0/2} \\ \ge & \min\big\{-\frac{1}{b},0\big\} + \frac{1}{2}C_4\cdot R_1^{a_0/2}\,, \\ \end{align}\] by using 4 . Then, by 5 , we see that \[(W_{a,b}*\rho)({\boldsymbol{x}}_0) > 2C_2 \ge 2E_{a,b}[\rho]\,,\] which violates the Euler-Lagrange condition ?? . Therefore 7 is proved.

Finally we will derive the conclusion. First, we notice the estimate \[W_{a,b}({\boldsymbol{x}}) \ge \frac{R_1^a}{2a} \ge \frac{2^{a/2}}{2a} \cdot R_1^{a/2} \ge C_4 R_1^{a_0/2} \ge 2\Big(2C_2-\min\big\{-\frac{1}{b},0\big\}+1\Big) ,\quad \forall |{\boldsymbol{x}}|\ge R_1\,,\] due to 4 . Take a point \({\boldsymbol{x}}_0\in\textrm{supp\,}\rho\). Then for any \({\boldsymbol{x}}\) with \(|{\boldsymbol{x}}-{\boldsymbol{x}}_0|>2R_1\), since the ball \(B({\boldsymbol{x}}_0;R_1)\) is contained in \(B({\boldsymbol{x}};R_1)^c\), we have \[\begin{align} (W_{a,b}*\rho)({\boldsymbol{x}}) = & \int_{B({\boldsymbol{x}};R_1)} W_{a,b}({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{\rho({\boldsymbol{y}})} + \int_{B({\boldsymbol{x}};R_1)^c} W_{a,b}({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{\rho({\boldsymbol{y}})} \\ \ge & \min\big\{-\frac{1}{b},0\big\} + \frac{1}{2}\cdot2\Big(2C_2-\min\big\{-\frac{1}{b},0\big\}+1\Big) > 2C_2 \ge 2E[\rho]\,, \end{align}\] due to 7 . This implies \(\textrm{supp\,}\rho\subset \overline{B({\boldsymbol{x}}_0;2R_1)}\) because otherwise the Euler-Lagrange condition ?? would be violated. This implies that the center of mass \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho({\boldsymbol{x}})}\) is also in \(\overline{B({\boldsymbol{x}}_0;2R_1)}\), and thus obtain an upper bound of minimizer size \[R_*(b,a_0)= 4R_1\,.\] ◻

Proof. Since the proof is analogous to that of Theorem 8, we only give an outline of the most important steps. Let \(\rho\) be a minimizer of \(E_{b,\ln}\) (the energy associated to \(W_{b,\ln}\)). First notice that \[\begin{align} E_{b,\ln}\Big[\frac{\chi_{B(0;1/2)}}{|B(0;1/2)|}\Big] \le 0\,, \end{align}\] since \(W_{b,\ln}({\boldsymbol{x}}) = \frac{|{\boldsymbol{x}}|^b}{b}\ln|{\boldsymbol{x}}| \le 0\) for any \(|{\boldsymbol{x}}|\le 1\) if \(b>0\). It follows that \(E_{b,\ln}[\rho]\le 0\).

Then notice that \[\label{claim95tlog} \min_{t\ge r} \frac{t^b}{b}\ln t \ge \frac{r^b}{b},\quad \forall b>0,\,t\ge e\,.\tag{8}\] Then we take \[R_1 = \max\left\{e,\Big(\frac{1}{C_4}\big(\frac{e^{-1}}{b_0^2}+1\big)\Big)^{2/b_0}\right\}\,,\] where \(C_4\) is defined in 6 , and proceed to prove 7 . We again assume the contrary and estimate \[\begin{align} (W_{b,\ln}*\rho)({\boldsymbol{x}}_0) = & \int_{B({\boldsymbol{x}}_0;R_1)} W_{b,\ln}({\boldsymbol{x}}_0-{\boldsymbol{y}})\,\mathrm{d}{\rho({\boldsymbol{y}})} + \int_{B({\boldsymbol{x}}_0;R_1)^c} W_{b,\ln}({\boldsymbol{x}}_0-{\boldsymbol{y}})\,\mathrm{d}{\rho({\boldsymbol{y}})} \\ \ge & \min W_{b,\ln}\int_{B({\boldsymbol{x}}_0;R_1)} \,\mathrm{d}{\rho({\boldsymbol{y}})} + \min_{t\ge R_1} \frac{t^b}{b}\ln t\int_{B({\boldsymbol{x}}_0;R_1)^c} \,\mathrm{d}{\rho({\boldsymbol{y}})} \\ \ge & -\frac{e^{-1}}{b^2} + \frac{R_1^b}{2b} \ge -\frac{e^{-1}}{b_0^2} + \frac{2^{b/2}}{2b}\cdot R_1^{b_0/2} \ge -\frac{e^{-1}}{b_0^2} +C_4\cdot R_1^{b_0/2} \\ > & 0 \ge 2E_{b,\ln}[\rho]\,. \\ \end{align}\] Therefore we get a contradiction against the Euler-Lagrange condition, which proves 7 .

Finally we notice that \[W_{b,\ln}({\boldsymbol{x}}) \ge \frac{R_1^b}{b} \ge 2C_4 R_1^{b_0/2} \ge 2\Big(\frac{e^{-1}}{b_0^2}+1\Big) ,\quad \forall |{\boldsymbol{x}}|\ge R_1\,,\] by 8 . This leads to \(\textrm{supp\,}\rho\subset \overline{B({\boldsymbol{x}}_0;2R_1)}\) and thus an upper bound of minimizer size \(R_{*,\ln}(b_0)= 4R_1\) similar as the proof of Theorem 8. ◻

Proof. We rewrite as \[2E [\mu*\phi_\epsilon] = \int_{\mathbb{R}^d} (W*(\mu*\phi_\epsilon))(\mu*\phi_\epsilon)\,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^d} (W*\psi_\epsilon*\mu))\mu\,\mathrm{d}{{\boldsymbol{x}}}\] where \(\psi_\epsilon:=\phi_\epsilon*\phi_\epsilon\) is another mollifier with \(\psi=\phi*\phi\) supported on \(\overline{B(0;2)}\). Since \(\mu\) is compactly supported, we may use (W1) with \(R=2\textrm{diam\,}(\textrm{supp\,}\mu)\) to see that the last integrand is bounded by \(((C_1+C_2 W)*|\mu|)|\mu|\) whose integral is finite by the assumption \(E [|\mu|]<\infty\). Therefore, applying the dominated convergence theorem, we get the conclusion since (W1) implies that \(\lim_{\epsilon\rightarrow 0}(W*\psi_\epsilon)({\boldsymbol{x}}) = W({\boldsymbol{x}})\in \mathbb{R}\cup\{\infty\}\) for any \({\boldsymbol{x}}\in\mathbb{R}^d\). ◻

Proof. By adding constant and modifying \(W\) at long distance, we may assume \(\inf W=0\) without loss of generality.

There must hold \(W(0)=\infty\) because otherwise \(W\) would be continuous on \(\mathbb{R}^d\) due to (W1), and \(E [\rho]=\infty\) cannot hold. If \[\iint_{{\boldsymbol{x}}={\boldsymbol{y}}} W({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{(\rho({\boldsymbol{x}})\otimes\rho({\boldsymbol{y}}))} = \infty\,,\] then \(\rho\) contains a Dirac mass, and the conclusion follows easily from \(\lim_{{\boldsymbol{y}}\rightarrow0}W({\boldsymbol{y}})=W(0)=\infty\). Otherwise one must have \[\iint_{{\boldsymbol{x}}={\boldsymbol{y}}} W({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{(\rho({\boldsymbol{x}})\otimes\rho({\boldsymbol{y}}))} = 0\,,\] and then \[\lim_{\alpha\rightarrow 0^+}\iint_{|{\boldsymbol{x}}-{\boldsymbol{y}}|>\alpha} W({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{(\rho({\boldsymbol{x}})\otimes\rho({\boldsymbol{y}}))} = 2E [\rho] = \infty\,.\] Let \(M>0\) be given. Then, we first take \(\alpha>0\) such that \(\iint_{|{\boldsymbol{x}}-{\boldsymbol{y}}|>\alpha} W({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{(\rho({\boldsymbol{x}})\otimes\rho({\boldsymbol{y}}))} \ge M+1\). Then, using the continuity of \(W\) on \(\mathbb{R}^d\backslash \{0\}\), we have \[(W*\psi_\epsilon)({\boldsymbol{x}}) \ge W({\boldsymbol{x}})-1,\quad \forall \alpha \le |{\boldsymbol{x}}| \le \textrm{diam\,}(\textrm{supp\,}\rho)\,,\] for sufficiently small \(\epsilon>0\), where \(\psi_\epsilon\) is the same as in the proof of Lemma 4. This implies \[2E [\rho*\phi_\epsilon] = \int_{\mathbb{R}^d} (W*\psi_\epsilon*\rho))\rho\,\mathrm{d}{{\boldsymbol{x}}} \ge \iint_{|{\boldsymbol{x}}-{\boldsymbol{y}}|>\alpha} (W({\boldsymbol{x}}-{\boldsymbol{y}})-1)\,\mathrm{d}{(\rho({\boldsymbol{x}})\otimes\rho({\boldsymbol{y}}))} \ge M\,.\] This leads to the conclusion. ◻

Proof. We first prove ?? . For \(|{\boldsymbol{x}}|>3\), we write \[(\psi_{1,R}*\psi_2)({\boldsymbol{x}}) = \int_{\mathbb{R}^d} R^d\psi_1(R{\boldsymbol{y}})\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} = \int_{|{\boldsymbol{y}}|< |{\boldsymbol{x}}|/2} + \int_{|{\boldsymbol{x}}|/2 \le |{\boldsymbol{y}}| < 2|{\boldsymbol{x}}|} + \int_{|{\boldsymbol{y}}|\ge 2|{\boldsymbol{x}}|} \,,\\None\] and then estimate the three parts separately.

When \(|{\boldsymbol{y}}|< |{\boldsymbol{x}}|/2\), we have \(|{\boldsymbol{x}}|/2 \le |{\boldsymbol{x}}-{\boldsymbol{y}}| \le 2|{\boldsymbol{x}}|\) where \(|{\boldsymbol{x}}|/2 > 1\), and thus \(|\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})| \lesssim |{\boldsymbol{x}}|^M\). Therefore \[\begin{align} \left|\int_{|{\boldsymbol{y}}|< |{\boldsymbol{x}}|/2}R^d\psi_1(R{\boldsymbol{y}})\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \right| \lesssim & |{\boldsymbol{x}}|^M \int_{|{\boldsymbol{y}}|< |{\boldsymbol{x}}|/2}R^d|\psi_1(R{\boldsymbol{y}})|\,\mathrm{d}{{\boldsymbol{y}}} \\ = & |{\boldsymbol{x}}|^M \int_{|{\boldsymbol{y}}|< R|{\boldsymbol{x}}|/2}|\psi_1({\boldsymbol{y}})|\,\mathrm{d}{{\boldsymbol{y}}} \lesssim |{\boldsymbol{x}}|^M\,, \end{align}\] since \(\psi_1 \in L^1(\mathbb{R}^d)\).

When \(|{\boldsymbol{x}}|/2 \le |{\boldsymbol{y}}| < 2|{\boldsymbol{x}}|\), we have \(|\psi_1(R{\boldsymbol{y}})| \le C_N (1+R|{\boldsymbol{x}}|)^{-N}\), and thus obtain \[\begin{align} \left|\int_{|{\boldsymbol{x}}|/2 \le |{\boldsymbol{y}}| < 2|{\boldsymbol{x}}|} R^d\psi_1(R{\boldsymbol{y}})\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \right| \lesssim & C_N R^d(1+R|{\boldsymbol{x}}|)^{-N}\int_{|{\boldsymbol{x}}|/2 \le |{\boldsymbol{y}}| < 2|{\boldsymbol{x}}|} |\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})|\,\mathrm{d}{{\boldsymbol{y}}}\,. \\ \end{align}\] First notice that \(R^d \le (1+R|{\boldsymbol{x}}|)^d\) since \(|{\boldsymbol{x}}| > 3\). Then we estimate \[\begin{align} \int_{|{\boldsymbol{x}}|/2 \le |{\boldsymbol{y}}| < 2|{\boldsymbol{x}}|} & |\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})|\,\mathrm{d}{{\boldsymbol{y}}} \le \int_{B(0;3|{\boldsymbol{x}}|)} |\psi_2({\boldsymbol{z}})|\,\mathrm{d}{{\boldsymbol{z}}} \\ \le & \left( \int_{B(0;1)} |\psi_2({\boldsymbol{z}})|\,\mathrm{d}{{\boldsymbol{z}}} + \int_{1\le |{\boldsymbol{z}}| \le 3|{\boldsymbol{x}}|}C|{\boldsymbol{z}}|^M\,\mathrm{d}{{\boldsymbol{z}}} \right) \lesssim 1+|{\boldsymbol{x}}|^{M+d} \,, \end{align}\] (the last quantity becomes \(1+\ln|{\boldsymbol{x}}|\), if \(M=-d\), same for the inequality below). By taking \(N=d+\max\{d+1,-M\}\) and using \(R\ge 1\), we conclude that \[\begin{align} \left|\int_{|{\boldsymbol{x}}|/2 \le |{\boldsymbol{y}}| < 2|{\boldsymbol{x}}|}\psi_1({\boldsymbol{y}})\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \right| \lesssim & (1+R|{\boldsymbol{x}}|)^{\min\{-d-1,M\}} + (1+R|{\boldsymbol{x}}|)^{\min\{-d-1,M\}} |{\boldsymbol{x}}|^{M+d} \\ \le & (1+|{\boldsymbol{x}}|)^{M} + (1+|{\boldsymbol{x}}|)^{-d-1} |{\boldsymbol{x}}|^{M+d} \\ \lesssim & |{\boldsymbol{x}}|^M + |{\boldsymbol{x}}|^{M-1}\,. \end{align}\] Here the extra \(-1\) on the last exponent can help absorb the extra \(\ln|{\boldsymbol{x}}|\) factor in case it appears.

When \(|{\boldsymbol{y}}|\ge 2|{\boldsymbol{x}}|\), we have \(|{\boldsymbol{y}}|/2 \le |{\boldsymbol{x}}-{\boldsymbol{y}}| \le 2|{\boldsymbol{y}}|\) where \(|{\boldsymbol{y}}|/2>1\), and thus \(|\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})| \lesssim |{\boldsymbol{y}}|^M\). Combining with \(|\psi_1(R{\boldsymbol{y}})| \le C_N (1+R|{\boldsymbol{y}}|)^{-N}\), \(R^d \le (1+R|{\boldsymbol{y}}|)^d\) and choosing \(N=d + \max\{M+d+1,d\}\), we obtain \[\begin{align} \left|\int_{|{\boldsymbol{y}}|\ge 2|{\boldsymbol{x}}|} R^d\psi_1(R{\boldsymbol{y}})\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \right| \lesssim & \int_{|{\boldsymbol{y}}|\ge 2|{\boldsymbol{x}}|} R^d(1+R|{\boldsymbol{y}}|)^{-d - \max\{M+d+1,d\}}|{\boldsymbol{y}}|^M \,\mathrm{d}{{\boldsymbol{y}}} \\ \le & \int_{|{\boldsymbol{y}}|\ge 2|{\boldsymbol{x}}|} (1+R|{\boldsymbol{y}}|)^{\min\{-M-d-1,-d\}}|{\boldsymbol{y}}|^M \,\mathrm{d}{{\boldsymbol{y}}} \\ \le & \int_{|{\boldsymbol{y}}|\ge 2|{\boldsymbol{x}}|} |{\boldsymbol{y}}|^{\min\{-d-1,M-d\}} \,\mathrm{d}{{\boldsymbol{y}}} \lesssim |{\boldsymbol{x}}|^{M}\,. \\ \end{align}\]

Combining the above estimates, ?? follows.

The proof of ?? for continuous \(\psi_2\) is a standard ‘approximation of identity’ argument. Indeed, for a fixed \({\boldsymbol{x}}\), write \[\begin{align} (\psi_{1,R} & *\psi_2)({\boldsymbol{x}}) - \psi_2({\boldsymbol{x}})\int_{\mathbb{R}^d}\psi_1({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \\ = & \int_{\mathbb{R}^d} R^d\psi_1(R{\boldsymbol{y}})\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} - \psi_2({\boldsymbol{x}})\int_{\mathbb{R}^d}R^d\psi_1(R{\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \\ = & \int_{|{\boldsymbol{y}}|< 1/\sqrt{R}}R^d\psi_1(R{\boldsymbol{y}})(\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})-\psi_2({\boldsymbol{x}})) \,\mathrm{d}{{\boldsymbol{y}}} \\ & + \int_{|{\boldsymbol{y}}|\ge 1/\sqrt{R}}R^d\psi_1(R{\boldsymbol{y}})(\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})-\psi_2({\boldsymbol{x}})) \,\mathrm{d}{{\boldsymbol{y}}}\,. \\ \end{align}\] We first estimate \[\begin{align} \left| \int_{|{\boldsymbol{y}}|< 1/\sqrt{R}}R^d\psi_1(R{\boldsymbol{y}})(\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})-\psi_2({\boldsymbol{x}})) \,\mathrm{d}{{\boldsymbol{y}}} \right| \le \|\psi_1\|_{L^1}\sup_{{\boldsymbol{z}}\in B({\boldsymbol{x}};1/\sqrt{R})} |\psi_2({\boldsymbol{z}})-\psi_2({\boldsymbol{x}})|\,, \end{align}\] which goes to zero as \(R\rightarrow\infty\), by the continuity of \(\psi_2\). Then notice that ?? and the continuity of \(\psi_2\) imply that \(\psi_2({\boldsymbol{z}}) \lesssim (1+|{\boldsymbol{z}}|)^M\) for any \({\boldsymbol{z}}\in\mathbb{R}^d\). Recalling that \({\boldsymbol{x}}\) is fixed, this allows us to estimate \[\begin{align} & \left| \int_{|{\boldsymbol{y}}|\ge 1/\sqrt{R}}R^d \psi_1(R{\boldsymbol{y}})(\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})-\psi_2({\boldsymbol{x}})) \,\mathrm{d}{{\boldsymbol{y}}} \right| \\ \lesssim & \int_{|{\boldsymbol{y}}|\ge 1/\sqrt{R}}R^d|\psi_1(R{\boldsymbol{y}})| (1+|{\boldsymbol{y}}|)^M \,\mathrm{d}{{\boldsymbol{y}}} \\ \lesssim & C_N\int_{|{\boldsymbol{y}}|\ge 1/\sqrt{R}}R^d(1+R|{\boldsymbol{y}}|)^{-N} (1+|{\boldsymbol{y}}|)^M \,\mathrm{d}{{\boldsymbol{y}}} \\ = & C_N\int_{|{\boldsymbol{y}}|\ge \sqrt{R}}(1+|{\boldsymbol{y}}|)^{-N} (1+|\frac{{\boldsymbol{y}}}{R}|)^M \,\mathrm{d}{{\boldsymbol{y}}} \,.\\ \end{align}\] Notice that \(|\frac{{\boldsymbol{y}}}{R}| \le |{\boldsymbol{y}}|\) for any \(R\ge 1\), and thus \((1+|\frac{{\boldsymbol{y}}}{R}|)^M \le (1+|{\boldsymbol{y}}|)^{\max\{M,0\}}\). Therefore, by choosing \(N = d+1+\max\{M,0\}\), we obtain \[\begin{align} & \left| \int_{|{\boldsymbol{y}}|\ge 1/\sqrt{R}}R^d \psi_1(R{\boldsymbol{y}})(\psi_2({\boldsymbol{x}}-{\boldsymbol{y}})-\psi_2({\boldsymbol{x}})) \,\mathrm{d}{{\boldsymbol{y}}} \right| \lesssim \int_{|{\boldsymbol{y}}|\ge \sqrt{R}}(1+|{\boldsymbol{y}}|)^{-d-1} \,\mathrm{d}{{\boldsymbol{y}}} \lesssim R^{-1/2}\,, \\ \end{align}\] which also goes to zero as \(R\rightarrow\infty\). Therefore ?? follows. ◻

Proof of Theorem 11. Since \(W\) is a tempered distribution, \(\hat{W}\) is well-defined as a tempered distribution. Let \(\mu\) be a compactly supported signed measure with \(E[|\mu|]<\infty\). We take a mollifier \(\phi_\epsilon\) and a truncation function \(\Phi_R\) (see the last paragraph of Section 1). Define \[\label{fepsR1} f_{\epsilon,R} = \int_{\mathbb{R}^d} ((W\Phi_R)*(\mu*\phi_\epsilon))(\mu*\phi_\epsilon)\,\mathrm{d}{{\boldsymbol{x}}}\,.\tag{10}\] Notice that both \((W\Phi_R)*(\mu*\phi_\epsilon)\) and \(\mu*\phi_\epsilon\) are \(L^2\) functions. Therefore we may apply the Plancherel identity to get \[\label{fepsR2}\begin{align} f_{\epsilon,R} = & \int_{\mathbb{R}^d} \mathcal{F}((W\Phi_R)*(\mu*\phi_\epsilon))\cdot \overline{\mathcal{F}(\mu*\phi_\epsilon)}\,\mathrm{d}{\xi} \\ = & \int_{\mathbb{R}^d}(\hat{W}*\hat{\Phi}_R)|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d}\hat{W}\big((|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2) *\hat{\Phi}_R\big)\,\mathrm{d}{\xi}\,. \end{align}\tag{11}\] We first fix \(\epsilon\), and notice that \(f_{\epsilon,R}\) is constant in \(R\) for sufficiently large \(R\), and equal to \[\label{feps1} f_{\epsilon} = \int_{\mathbb{R}^d} (W*(\mu*\phi_\epsilon))(\mu*\phi_\epsilon)\,\mathrm{d}{{\boldsymbol{x}}}\,,\tag{12}\] since \(\mu*\phi_\epsilon\) is compactly supported. Then, in 11 we send \(R\rightarrow\infty\). By assumption \(\hat{W}\) is a locally integrable function which is also a tempered distribution, and thus

\[\label{hatWb} \int_{B(0;1)^c}|\hat{W}(\xi)|\cdot|\xi|^{-k}\,\mathrm{d}{{\boldsymbol{x}}}<\infty\,,\tag{13}\] for sufficiently large \(k\).

Now we treat the convolution on the RHS of 11 . Notice that \(|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\) is a Schwartz function, and \(\hat{\Phi}_R(\xi) = R^d \hat{\Phi}(R\xi)\), with \(\hat{\Phi}\) a Schwartz function. Applying ?? in Lemma 6 with \(\psi_1 = \hat{\Phi}\) and \(\psi_2 = |\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\) with \(M=-k\) (\(k\) as given in 13 ), we see that \[\big((|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2) *\hat{\Phi}_R\big)(\xi) \lesssim (1+|\xi|)^{-k},\quad \forall |\xi|>3,\quad \forall R\ge 1\,.\] This inequality also holds for \(|\xi|\le 3\) by Young’s inequality, since \(|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\in L^\infty\) and \(\|\hat{\Phi}_R\|_{L^1}=\|\hat{\Phi}\|_{L^1}<\infty\). Therefore we see that the last integrand \(\hat{W}\big((|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2) *\hat{\Phi}_R\big)\) of 11 is dominated by an integrable function \(C|\hat{W}(\xi)|(1+|\xi|)^{-k}\) for any \(R\ge 1\).

Applying ?? in Lemma 6, combining with \(\int_{\mathbb{R}^d} \hat{\Phi}\,\mathrm{d}{\xi}= \Phi(0)=1\), gives \(\lim_{R\rightarrow\infty}\big((|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2) *\hat{\Phi}_R\big)(\xi) = |\hat{\mu}(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\) for every \(\xi\). Therefore, applying the dominated convergence theorem to 11 , we see that \[\label{feps2} f_{\epsilon} = \int_{\mathbb{R}^d}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi}\,.\tag{14}\]

Then we send \(\epsilon\rightarrow 0\). In 12 , since \(W\) satisfies (W1) and \(E[|\mu|]<\infty\), we apply Lemma 4 to get \[\lim_{\epsilon\rightarrow 0}f_{\epsilon} = \int_{\mathbb{R}^d} (W*\mu)\mu\,\mathrm{d}{{\boldsymbol{x}}} \,.\]

In 14 , notice that \(|\hat{\mu}|^2\in L^\infty\), and \(\hat{\phi}_\epsilon=\hat{\phi}(\epsilon\cdot)\) is uniformly bounded and converges uniformly to 1 on compact sets. Therefore, if \(\hat{W}\in L^1(\mathbb{R}^d)\), then the integrand in 14 is dominated by an integrable function \(C|\hat{W}(\xi)|\) for any \(\epsilon>0\). Therefore, by the dominated convergence theorem, we get \[\label{phiepsconv} \lim_{\epsilon\rightarrow 0}\int_{\mathbb{R}^d}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d}\hat{W}|\hat{\mu}|^2\,\mathrm{d}{\xi} \,,\tag{15}\] and thus 9 is proved.

If we assume that \(\hat{W}\) is one-signed on \(B(0;R_0)^c\) for some \(R_0>0\), then we may still apply the dominated convergence theorem on \(B(0;R_0)\) to get \[\lim_{\epsilon\rightarrow 0}\int_{B(0;R_0)}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi} = \int_{B(0;R_0)}\hat{W}|\hat{\mu}|^2\,\mathrm{d}{\xi} \,.\] Next, using the assumption that \(\hat{W}\) is one-signed on \(B(0;R_0)^c\), we may obtain that \[\label{R0converge} \lim_{\epsilon\rightarrow 0}\int_{B(0;R_0)^c}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi} = \int_{B(0;R_0)^c}\hat{W}|\hat{\mu}|^2\,\mathrm{d}{\xi} \,.\tag{16}\] To see this, we first consider the case \(\hat{W}\ge 0\) on \(B(0;R_0)^c\). If \(\int_{B(0;R_0)^c}\hat{W}|\hat{\mu}|^2\,\mathrm{d}{\xi}=\infty\), then one clearly has \(\lim_{\epsilon\rightarrow 0}\int_{B(0;R_0)^c}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi}=\infty\) since \(\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\ge 0\) and \(\hat{\phi}_\epsilon\) converges uniformly to 1 on compact sets. If \(\int_{B(0;R_0)^c}\hat{W}|\hat{\mu}|^2\,\mathrm{d}{\xi}<\infty\) then 16 follows from the dominated convergence theorem. The case that \(\hat{W}\le 0\) on \(B(0;R_0)^c\) can be treated similarly. Therefore we again obtain 15 and thus 9 is proved. ◻

Proof. Let \(R\) be sufficiently large so that \(\textrm{supp\,}\mu\subset B(0;R)\), and ?? is valid for \(|{\boldsymbol{x}}|\ge R\). For any \({\boldsymbol{x}}\) with \(|{\boldsymbol{x}}|\ge 2R\), we write \[(f*\mu)({\boldsymbol{x}}) = \int_{\textrm{supp\,}\mu} f({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{\mu({\boldsymbol{y}})} = \int_{\textrm{supp\,}\mu} (f({\boldsymbol{x}}-{\boldsymbol{y}})-f({\boldsymbol{x}}))\,\mathrm{d}{\mu({\boldsymbol{y}})}\,.\] For each \({\boldsymbol{y}}\in\textrm{supp\,}\mu\), applying the mean value theorem to \(t\mapsto f({\boldsymbol{x}}-t{\boldsymbol{y}})\) on \(t\in [0,1]\) gives \[f({\boldsymbol{x}}-{\boldsymbol{y}})-f({\boldsymbol{x}}) = -\nabla f({\boldsymbol{x}}-t_{\boldsymbol{y}}{\boldsymbol{y}}) \cdot {\boldsymbol{y}}\,,\] for some \(t_{\boldsymbol{y}}\in (0,1)\) (depending on \({\boldsymbol{y}}\)). Since \(|{\boldsymbol{x}}|\ge 2R\) and \(|{\boldsymbol{y}}|\le R\), we have \(\frac{|{\boldsymbol{x}}|}{2}\le |{\boldsymbol{x}}-t_{\boldsymbol{y}}{\boldsymbol{y}}|\le 2|{\boldsymbol{x}}|\), which implies \(|\nabla f({\boldsymbol{x}}-t_{\boldsymbol{y}}{\boldsymbol{y}})| \lesssim |{\boldsymbol{x}}|^p\). Therefore we conclude that \[|(f*\mu)({\boldsymbol{x}})| \lesssim |{\boldsymbol{x}}|^p \int_{\textrm{supp\,}\mu} |{\boldsymbol{y}}| \,\mathrm{d}{|\mu|({\boldsymbol{y}})} \lesssim |{\boldsymbol{x}}|^p\,.\] ◻

Proof. This lemma can be proved similarly as the previous one, by using \[(f*\mu)({\boldsymbol{x}}) = \int_{\textrm{supp\,}\mu} (f({\boldsymbol{x}}-{\boldsymbol{y}})-f({\boldsymbol{x}}) + \nabla f({\boldsymbol{x}})\cdot {\boldsymbol{y}})\,\mathrm{d}{\mu({\boldsymbol{y}})}\,,\] and applying Taylor’s formula to \(t\mapsto f({\boldsymbol{x}}-t{\boldsymbol{y}})\) to get \[f({\boldsymbol{x}}-{\boldsymbol{y}}) = f({\boldsymbol{x}}) - \nabla f({\boldsymbol{x}})\cdot {\boldsymbol{y}}+ \frac{1}{2}{\boldsymbol{y}}^\top \nabla^2 f({\boldsymbol{x}}-t_{\boldsymbol{y}}{\boldsymbol{y}}){\boldsymbol{y}}\,.\] ◻

Proof of Theorem 12. Let \(\mu\) be a compactly supported signed measure with \(E[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\). We first assume \(\mu\) is a smooth function. Denote \[g = -\Delta^{-1}\mu = -\frac{1}{|S^{d-1}|} \cdot \frac{|{\boldsymbol{x}}|^{2-d}}{2-d}*\mu\,.\] Then \(g\) is a locally integrable smooth function satisfying \[|g({\boldsymbol{x}})| \lesssim |{\boldsymbol{x}}|^{1-d},\quad |\nabla g({\boldsymbol{x}})| \lesssim |{\boldsymbol{x}}|^{-d}\,,\] for large \(|{\boldsymbol{x}}|\), due to \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\). Indeed, these two estimates follow by applying Lemma 7 with \(f\) being \(|{\boldsymbol{x}}|^{2-d}\) and \(\nabla|{\boldsymbol{x}}|^{2-d}=C{\boldsymbol{x}}|{\boldsymbol{x}}|^{-d}\), respectively.

We also have \[|(W*\mu)({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-1}\,,\] due to \(|\nabla W({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-1}\) in ?? and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), in view of Lemma 7.

Then we claim that \[\int_{\mathbb{R}^d} (W*\mu)\mu\,\mathrm{d}{{\boldsymbol{x}}} = -\int_{\mathbb{R}^d} (\Delta W*\nabla g)\cdot\nabla g\,\mathrm{d}{{\boldsymbol{x}}}\,,\] where the LHS is well-defined since \(E[|\mu|]<\infty\). To see this, we start by \[\begin{align} (\Delta W*\partial_j g )({\boldsymbol{x}})= & \int_{\mathbb{R}^d} \Delta W({\boldsymbol{x}}-{\boldsymbol{y}})\partial_j g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} = \int_{\mathbb{R}^d} \nabla W({\boldsymbol{x}}-{\boldsymbol{y}})\cdot\nabla \partial_j g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \\ = & \int_{\mathbb{R}^d} \partial_j\nabla W({\boldsymbol{x}}-{\boldsymbol{y}})\cdot\nabla g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} = \int_{\mathbb{R}^d} \partial_jW({\boldsymbol{x}}-{\boldsymbol{y}})\Delta g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \\ =& -\int_{\mathbb{R}^d} \partial_j W({\boldsymbol{x}}-{\boldsymbol{y}})\mu({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}}\,, \\ \end{align}\] where the integration by parts (three times) are justified by using the estimate \(|\nabla W({\boldsymbol{x}}-{\boldsymbol{y}}) \otimes\nabla g({\boldsymbol{y}})| \lesssim |{\boldsymbol{y}}|^{b-1}\cdot |{\boldsymbol{y}}|^{-d},\,b-1-d<-d+1\) for fixed \({\boldsymbol{x}}\) and large \(|{\boldsymbol{y}}|\). Indeed, the first integration by parts can be justified by \[\begin{align} \int_{\mathbb{R}^d} & \Delta W({\boldsymbol{x}}-{\boldsymbol{y}})\partial_j g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} = \lim_{R\rightarrow\infty} \int_{B(0;R)} \Delta W({\boldsymbol{x}}-{\boldsymbol{y}})\partial_j g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} \\ = & \lim_{R\rightarrow\infty} -\int_{\partial B(0;R)} \nabla W({\boldsymbol{x}}-{\boldsymbol{y}}) \partial_j g({\boldsymbol{y}}) \cdot {\boldsymbol{n}}\,\mathrm{d}{S({\boldsymbol{y}})} + \int_{B(0;R)} \nabla W({\boldsymbol{x}}-{\boldsymbol{y}})\cdot \partial_j \nabla g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}}\,. \end{align}\] Here, \(\nabla W({\boldsymbol{x}}-{\boldsymbol{y}}) \partial_j g({\boldsymbol{y}})=O(R^{b-1-d})\), and thus \[\int_{\partial B(0;R)} \nabla W({\boldsymbol{x}}-{\boldsymbol{y}}) \partial_j g({\boldsymbol{y}}) \cdot {\boldsymbol{n}}\,\mathrm{d}{S({\boldsymbol{y}})} = O(R^{b-1-d + (d-1)})\,,\] which vanishes in the limit \(R\rightarrow\infty\). Therefore this integration by parts is justified. The other two integration by parts can be justified similarly.

As a consequence, we obtain \[\Delta W*\nabla g = -\nabla W * \mu\,.\] Finally we integrate by parts to get \[\int_{\mathbb{R}^d} (\nabla W * \mu)\cdot\nabla g\,\mathrm{d}{{\boldsymbol{x}}} = -\int_{\mathbb{R}^d} ( W * \mu)\Delta g\,\mathrm{d}{{\boldsymbol{x}}}= \int_{\mathbb{R}^d} ( W * \mu)\mu\,\mathrm{d}{{\boldsymbol{x}}}\,,\] using the estimate \(|(W*\mu)({\boldsymbol{x}})\nabla g({\boldsymbol{x}})| \lesssim |{\boldsymbol{x}}|^{b-1}\cdot |{\boldsymbol{x}}|^{-d}\). This proves the claim.

Then, applying Theorem 11 to the potential \(-\Delta W\) and the compactly supported signed measure \(\Phi_R\partial_j g\) where \(\Phi_R\) is a truncation function, we get \[\begin{align} -\int_{\mathbb{R}^d} (\Delta W*(\Phi_R\partial_j g))\cdot(\Phi_R\partial_j g)\,\mathrm{d}{{\boldsymbol{x}}} = & \int_{\mathbb{R}^d\backslash\{0\}} \mathcal{F}[-\Delta W](\xi) |\mathcal{F}[\Phi_R\partial_j g]|^2\,\mathrm{d}{\xi} \\ = & \int_{\mathbb{R}^d\backslash\{0\}} \mathcal{F}[-\Delta W](\xi) |2\pi\xi_j\hat{g}*\hat{\Phi}_R|^2\,\mathrm{d}{\xi}\,. \\ \end{align}\] Using \(|\nabla g({\boldsymbol{x}})|\lesssim|{\boldsymbol{x}}|^{-d}\) and \(|\Delta W({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-2}\) in ?? , one can justify that the LHS converges to \(-\int_{\mathbb{R}^d} (\Delta W*\partial_j g)\cdot\partial_j g\,\mathrm{d}{{\boldsymbol{x}}}\) as \(R\rightarrow\infty\). On the RHS, notice that the Fourier transform of the relation \(-\Delta g = \mu\) gives \(4\pi^2|\xi|^2 \hat{g} = \hat{\mu}\), which leads to \[2\pi\xi_j\hat{g} = \frac{1}{2\pi}\cdot\frac{\xi_j}{|\xi|^2}\hat{\mu},\quad \xi\ne 0\,,\] and \(\hat{\mu}\) is a smooth function with rapid decay at infinity, with \(\hat{\mu}(0)=0\). Therefore \(\hat{\mu}(\xi)\approx \nabla \hat{\mu}(0)\cdot \xi\) near 0, and thus \(2\pi\xi_j\hat{g}\) is a bounded function, smooth on \(\mathbb{R}^d\backslash \{0\}\), with rapid decay at infinity. Then, since \(-\Delta W\) satisfies the assumptions of Theorem 11, \(\mathcal{F}[-\Delta W]\) satisfies \(\int_{B(0;1)^c}|\mathcal{F}[-\Delta W](\xi)|\cdot |\xi|^{-k}\,\mathrm{d}{{\boldsymbol{x}}}<\infty\) for sufficiently large \(k\), as discussed in 13 . Therefore, we may proceed similarly as in the proof of Theorem 11 to obtain \[\begin{align} \lim_{R\rightarrow\infty} & \int_{\mathbb{R}^d\backslash\{0\}} \mathcal{F}[-\Delta W](\xi) |2\pi\xi_j\hat{g}*\hat{\Phi}_R|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d\backslash\{0\}} \mathcal{F}[-\Delta W](\xi) |2\pi\xi_j\hat{g}|^2\,\mathrm{d}{\xi} \\ = & \int_{\mathbb{R}^d\backslash\{0\}}4\pi^2|\xi|^2\hat{W}(\xi) \Big|\frac{1}{2\pi}\cdot\frac{\xi_j}{|\xi|^2}\hat{\mu}(\xi)\Big|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d\backslash\{0\}}\frac{\xi_j^2}{|\xi|^2}\hat{W}(\xi) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,. \\ \end{align}\] Then summing in \(j\), the last quantity becomes \(\int_{\mathbb{R}^d\backslash\{0\}}\hat{W}(\xi) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\), which finishes the proof of 9 in case \(\mu\) is a smooth function.

For a general signed measure \(\mu\), one can first mollify it into \(\mu*\phi_\epsilon\), so that the latter satisfies 9 , i.e., \[\int_{\mathbb{R}^d} (W*(\mu*\phi_\epsilon))(\mu*\phi_\epsilon)\,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^d\backslash\{0\}}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi}.\] Then sending \(\epsilon\rightarrow 0\), similar to the proof of Theorem 11, we first use (W1) to derive that \(\lim_{\epsilon\rightarrow0}\int_{\mathbb{R}^d} (W*(\mu*\phi_\epsilon))(\mu*\phi_\epsilon)\,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^d} (W*\mu)\mu\,\mathrm{d}{{\boldsymbol{x}}}\). Then, since \(\hat{\mu}(\xi)\approx \nabla \hat{\mu}(0)\cdot \xi\) near 0 and we assumed that \(-\Delta W\) satisfies the assumptions of Theorem 11, the function \(\hat{W}|\hat{\mu}|^2 = \mathcal{F}[-\Delta W]\frac{1}{4\pi^2|\xi|^2}|\hat{\mu}|^2\) is locally integrable on \(\mathbb{R}^d\). If \(\mathcal{F}[-\Delta W]\in L^1(\mathbb{R}^d)\) or \(\mathcal{F}[-\Delta W]\) is one-signed on \(B(0;R_0)^c\), then the same is true for \(\hat{W}|\hat{\mu}|^2\). Therefore we derive \(\lim_{\epsilon\rightarrow 0}\int_{\mathbb{R}^d}\hat{W}|\hat{\mu}|^2|\hat{\phi}_\epsilon|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d}\hat{W}|\hat{\mu}|^2\,\mathrm{d}{\xi}\) in the same way as in the proof of Theorem 11, and thus obtain 9 for \(\mu\).  ◻

Proof. Let \(\mu\) be a compactly supported signed measure with \(E[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\). We first assume \(\mu\) is a smooth function. Define \(g\) as in the proof of Theorem 12, which satisfies \[|g({\boldsymbol{x}})| \lesssim |{\boldsymbol{x}}|^{-d},\quad |\nabla g({\boldsymbol{x}})| \lesssim |{\boldsymbol{x}}|^{-d-1}\,,\] for large \(|{\boldsymbol{x}}|\), due to \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), in view of Lemma 8. We also have \[|(W*\mu)({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-2},\quad |(\nabla W*\mu)({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-3}\,,\] by \(|\nabla^2 W({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-2}\) and \(|\nabla^3 W({\boldsymbol{x}})|\lesssim |{\boldsymbol{x}}|^{b-3}\), in view of Lemma 8.

Then we claim that \[\int_{\mathbb{R}^d} (W*\mu)\mu\,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^d} (\Delta^2 W* g) g\,\mathrm{d}{{\boldsymbol{x}}}\,.\] To see this, we start by \[\begin{align} (\Delta^2 W* g)({\boldsymbol{x}}) = & \int_{\mathbb{R}^d} \Delta^2 W({\boldsymbol{x}}-{\boldsymbol{y}})g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}} = \int_{\mathbb{R}^d} \nabla\Delta W({\boldsymbol{x}}-{\boldsymbol{y}})\cdot \nabla g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}}\\ = & \int_{\mathbb{R}^d} \Delta W({\boldsymbol{x}}-{\boldsymbol{y}})\Delta g({\boldsymbol{y}})\,\mathrm{d}{{\boldsymbol{y}}}\,, \end{align}\] where the integration by parts (two times) are justified by using the estimates \(|\nabla\Delta W({\boldsymbol{x}}-{\boldsymbol{y}}) g({\boldsymbol{y}})| \lesssim |{\boldsymbol{y}}|^{b-3}\cdot |{\boldsymbol{y}}|^{-d},\,b-3-d<-d+1\) and \(|\Delta W({\boldsymbol{x}}-{\boldsymbol{y}}) \nabla g({\boldsymbol{y}})| \lesssim |{\boldsymbol{y}}|^{b-2}\cdot |{\boldsymbol{y}}|^{-d-1},\,b-2-d-1<-d+1\) for fixed \({\boldsymbol{x}}\) and large \(|{\boldsymbol{y}}|\). Then we integrate by parts to get \[\begin{align} \int_{\mathbb{R}^d} (\Delta^2 W* g) g\,\mathrm{d}{{\boldsymbol{x}}} = & \int_{\mathbb{R}^d} (\Delta W* \Delta g) g\,\mathrm{d}{{\boldsymbol{x}}}= \int_{\mathbb{R}^d} (\Delta W* \mu) g\,\mathrm{d}{{\boldsymbol{x}}} \\ = & -\int_{\mathbb{R}^d} (\nabla W*\mu) \cdot \nabla g \,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^d} (W*\mu) \Delta g\,\mathrm{d}{{\boldsymbol{x}}}= \int_{\mathbb{R}^d} (W*\mu) \mu\,\mathrm{d}{{\boldsymbol{x}}}\,, \end{align}\] where the integration by parts (two times) are justified by using the estimates \(|(\nabla W*\mu)g| \lesssim |{\boldsymbol{x}}|^{b-3}\cdot |{\boldsymbol{x}}|^{-d}\) and \(|(W*\mu)\nabla g| \lesssim |{\boldsymbol{x}}|^{b-2}\cdot |{\boldsymbol{x}}|^{-d-1}\). This proves the claim.

Then, similar to the proof of Theorem 12, we may calculate \[\begin{align} \int_{\mathbb{R}^d} (\Delta^2 W*(\Phi_R g))\cdot(\Phi_R g)\,\mathrm{d}{{\boldsymbol{x}}} = & \int_{\mathbb{R}^d\backslash\{0\}} \mathcal{F}[\Delta^2 W](\xi) |\hat{g}*\hat{\Phi}_R|^2\,\mathrm{d}{\xi}\,. \\ \end{align}\] Notice that \[\hat{g} = \frac{1}{4\pi^2|\xi|^2}\hat{\mu},\quad \xi\ne 0\,,\] and \(\hat{\mu}\) is a smooth function with rapid decay at infinity, with \(\hat{\mu}=\nabla\hat{\mu}=0\). Therefore \(\hat{\mu}(\xi)\approx \frac{1}{2}\xi^\top\nabla^2 \hat{\mu}(0)\cdot \xi\) near 0, and thus \(\hat{g}\) is a bounded function, smooth on \(\mathbb{R}^d\backslash \{0\}\), with rapid decay at infinity. Then one can pass to the limit \(R\rightarrow\infty\) similar to the proof of Theorem 12 to get \[\int_{\mathbb{R}^d} (\Delta^2 W* g) g\,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^d\backslash\{0\}} \mathcal{F}[\Delta^2 W](\xi) |\hat{g}(\xi)|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d\backslash\{0\}} \hat{W}(\xi) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,,\] which finishes the proof for smooth \(\mu\). The case of a general \(\mu\) can be treated by mollification trick similar to the proof of Theorem 12. ◻

Proof. Let \(\mu\ne 0\) be a compactly supported signed measure on \(\mathbb{R}^d\) with \(E[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}{\boldsymbol{x}}^{\otimes j}\,\mathrm{d}{\mu({\boldsymbol{x}})} = 0,\,j=0,1,\dots,k-1\). Then, denote \[\tilde{\mu} = \mu(x_1,\dots,x_d) \otimes \delta(x_{d+1},\dots,x_D)\,,\] which is a compactly supported signed measure on \(\mathbb{R}^D\) with \(\int_{\mathbb{R}^D}{\boldsymbol{x}}^{\otimes j}\,\mathrm{d}{\tilde{\mu}({\boldsymbol{x}})} = 0,\,j=0,1,\dots,k-1\). We have \(\tilde{E}[|\tilde{\mu}|] = E[|\mu|]<\infty\) due to ?? . Then the Fourier representable property at level \(k\) for \(\tilde{W}\) gives \[\int_{\mathbb{R}^D} (\tilde{W}*\tilde{\mu})\tilde{\mu}\,\mathrm{d}{{\boldsymbol{x}}} = \int_{\mathbb{R}^D\backslash \{0\}} \hat{\tilde{W}}(\xi)|\hat{\tilde{\mu}}(\xi)|^2\,\mathrm{d}{\xi}\,.\] Notice that the LHS is exactly \(\int_{\mathbb{R}^d} (W*\mu)\mu\,\mathrm{d}{{\boldsymbol{x}}}\) due to ?? . For the RHS, we have \[\hat{\tilde{\mu}}(\xi_1,\dots,\xi_D) = \hat{\mu}(\xi_1,\dots,\xi_d)\,.\] Therefore, using ?? , we see that it is \(\int_{\mathbb{R}^d\backslash\{0\}} \hat{W}(\xi)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\). This gives the conclusion. ◻

Proof. ?? in the case \(0<\Re(s)<d\) is well-known. For \(\Re(s)\le 0\), it suffices to show that \[\label{xsphi} \int_{\mathbb{R}^d}\frac{|{\boldsymbol{x}}|^{-s}-1}{s}\phi({\boldsymbol{x}})\,\mathrm{d}{{\boldsymbol{x}}} = \pi^{s-\frac{d}{2}}\frac{\Gamma(\frac{d-s}{2})}{\Gamma(\frac{s}{2})s} \int_{\mathbb{R}^d\backslash\{0\}}|\xi|^{-d+s}\bar{\hat{\phi}}(\xi)\,\mathrm{d}{\xi}\,,\tag{17}\] for any complex-valued Schwartz function \(\phi\) such that \(0\notin \textrm{supp\,}\hat{\phi}\). Notice that the extra \(-1\) on the LHS gives no contribution because \(0\notin \textrm{supp\,}\hat{\phi}\) implies that \(\phi\) is mean-zero. Also notice that the case \(s=0\) can be derived using a limit procedure \(s\rightarrow 0\) in the above expression.

Fix \(\phi\). The LHS of 17 is a holomorphic function in \(\Re(s)<d\) (which is \(\int_{\mathbb{R}^d}(-\ln|{\boldsymbol{x}}|)\phi({\boldsymbol{x}})\,\mathrm{d}{{\boldsymbol{x}}}\) when \(s=0\)). Since the gamma function has no zero and has simple poles at \(0,-1,-2,\dots\), we see that the RHS is also a holomorphic function in \(\Re(s)<d\), which is understood as \(0\) when \(s=-2,-4,-6,\dots\) and \(\pi^{-\frac{d}{2}}\frac{\Gamma(\frac{d}{2})}{2} \int_{\mathbb{R}^d\backslash\{0\}}|\xi|^{-d}\bar{\hat{\phi}}(\xi)\,\mathrm{d}{\xi}\) when \(s=0\). Since we already know that 17 is true for any \(0<\Re(s)<d\), it is also true for any \(\Re(s)<d\) by analytic continuation.

?? can be obtained from ?? by differentiating with respect to \(s\). In fact, formally differentiating ?? leads to \[\mathcal{F}\Big[-\frac{|{\boldsymbol{x}}|^{-s}\ln|{\boldsymbol{x}}|}{s}-\frac{|{\boldsymbol{x}}|^{-s}}{s^2}\Big](\xi) = c_{\textrm{PF}}'(s)|\xi|^{-d+s} + c_{\textrm{PF}}(s)|\xi|^{-d+s}\ln|\xi|\,.\] Then doing a linear combination with ?? gives ?? . Here, when \(0<\Re(s)<d\), the differentiation with respect to \(s\) inside \(\mathcal{F}[\cdot]\) can be justified by taking a Schwartz test function \(\phi\). When \(\Re(s)\le 0,\,s\ne 0\), we further impose \(0\notin \textrm{supp\,}\hat{\phi}\), and thus justify ?? on \(\mathbb{R}^d\backslash\{0\}\).  ◻

Proof. We first consider \(W_b({\boldsymbol{x}})=-\frac{|{\boldsymbol{x}}|^b}{b}\). If \(-d<b<0\), then Theorem 11 applies, showing that \(W_b\) is Fourier representable at level 0. Now we first assume \(d\ge 3\). If \(0\le b < 2\), then Theorem 12 applies since \[-\Delta W_b = (b-2+d)|{\boldsymbol{x}}|^{b-2}\,,\] showing that \(W_b\) is Fourier representable at level 1. If \(2\le b < 4\), then Theorem 13 applies since \[\Delta^2 (-W_b) = (b-2+d)(b-2)(b-4+d)|{\boldsymbol{x}}|^{b-4}\,,\] showing that \(-W_b\) is Fourier representable at level 2. The same is true for \(W_b\) because \(E_{-W_b}[|\mu|]<\infty\) for any compactly supported signed measure \(\mu\). For the case \(0\le b < 4\) with \(d=1,2\), the same results can be obtained by using Theorem 14 with \(D=3\) and \(\tilde{W}({\boldsymbol{x}})=-\frac{|{\boldsymbol{x}}|^b}{b}\), for which the conditions ?? and ?? can be verified explicitly via Theorem 15.

For \(W_{a,b}({\boldsymbol{x}})=\frac{|{\boldsymbol{x}}|^a}{a}-\frac{|{\boldsymbol{x}}|^b}{b}\) with \(-d<b<a<4\), we notice that its singularity near 0 is comparable to \(W_b\). Therefore \(E_{a,b}[|\mu|]<\infty\) implies \(E_b[|\mu|]<\infty\) (where \(E_b\) denotes the energy associated to \(W_b\)), which further implies \(E_a[|\mu|]<\infty\). Therefore, applying the Fourier representable property at level 2 for \(W_b\) and \(W_a\) and noticing that \(W_{a,b}= W_b-W_a\), we see that \(W_{a,b}\) is Fourier representable at level 2.

For \(W_{b,\ln} = \frac{|{\boldsymbol{x}}|^b}{b}\ln|{\boldsymbol{x}}|\), one can prove its Fourier representable property similarly as \(W_b\). In fact, the case \(-d<b<0\) follows from Theorem 11. Now we first assume \(d\ge 3\). The case \(0<b<2\) follows from Theorem 11 and \[-\Delta W_{b,\ln} = -(b+d-2)|{\boldsymbol{x}}|^{b-2}\ln|{\boldsymbol{x}}| - \frac{1}{b}(2b+d-2)|{\boldsymbol{x}}|^{b-2}\,.\] The case \(2\le b<4\) follows from Theorem 12 and \[\begin{align} \Delta^2 W_{b,\ln} = & (b-2+d)(b-2)(b-4+d)|{\boldsymbol{x}}|^{b-4}\ln|{\boldsymbol{x}}| \\ & + \Big((b+d-2)(2b+d-6)+\frac{1}{b}(2b+d-2)(b-2)(b+d-4)\Big) |{\boldsymbol{x}}|^{b-4}\,. \end{align}\] For the case \(d=1,2\) with \(0<b<4\), the same results can be obtained by using Theorem 14. ◻

Proof. It is clear that these potentials satisfy (W0), (W1). We claim that they satisfy the property: \[\label{claim95rho01}\begin{align} & \text{For any compactly supported \rho_0,\rho_1\in\mathcal{M}(\mathbb{R}^d) with E[\rho_i]<\infty,\,i=0,1,}\\ & \text{there holds E[\rho_0+\rho_1] < \infty. } \end{align}\tag{18}\]

To prove the claim for \(W_{a,b}\) with \(-d<b<a\), it suffices to treat \(W_b({\boldsymbol{x}})=-\frac{|{\boldsymbol{x}}|^b}{b}\) since its singularity at 0 is comparable to \(W_{a,b}\) . The case \(b>0\) is trivial because \(W_b\) is continuous. For \(-d<b\le 0\), \(W_b\) is Fourier representable at level 1 by Theorem 17. Take a mollifier \(\phi_\epsilon\). Then \((\rho_1-\rho_0)*\phi_\epsilon\) is a compactly supported mean-zero smooth function. Therefore its Fourier representation for \(W_b\) gives \[E_b [(\rho_1-\rho_0)*\phi_\epsilon] = \frac{1}{2}\int_{\mathbb{R}^d\backslash\{0\}} \hat{W}_b(\xi)|\hat{\rho}_1(\xi)-\hat{\rho}_0(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\,\mathrm{d}{\xi} \ge 0\,,\] since \(\hat{W}_b>0\). It follows that \[E[(\rho_0+\rho_1)*\phi_\epsilon] \le 2(E[\rho_0*\phi_\epsilon]+E[\rho_1*\phi_\epsilon])\,,\] by convexity. Then sending \(\epsilon\rightarrow 0\), the RHS converges to \(2(E[\rho_0]+E[\rho_1])\) by Lemma 4, which implies \(E[\rho_0+\rho_1] < \infty\) due to Lemma 5.

Then we prove the claim for \(W_{b,\ln}\) with \(b>-d,\,b\ne 0\). The case \(b>0\) is trivial because \(W_{b,\ln}\) is continuous. For \(-d<b<0\), \(W_{b,\ln}\) is Fourier representable at level 0 by Theorem 17. Therefore \[E[\rho_i] = \frac{1}{2}\int_{\mathbb{R}^d\backslash\{0\}} \hat{W}(\xi)|\hat{\rho}_i(\xi)|^2\,\mathrm{d}{\xi},\quad E[\rho_i*\phi_\epsilon] = \frac{1}{2}\int_{\mathbb{R}^d\backslash\{0\}} \hat{W}(\xi)|\hat{\rho}_i(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\,\mathrm{d}{\xi}\,,\] and \[\begin{align} E[(\rho_0+\rho_1)*\phi_\epsilon] = & \frac{1}{2}\int_{\mathbb{R}^d\backslash\{0\}} \hat{W}(\xi)|\hat{\rho}_0(\xi)+\hat{\rho}_1(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\,\mathrm{d}{\xi} \\ \le & \frac{1}{2}\int_{\hat{W}>0} \hat{W}(\xi)|\hat{\rho}_0(\xi)+\hat{\rho}_1(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\,\mathrm{d}{\xi} \\ \le & \int_{\hat{W}>0} \hat{W}(\xi)|\hat{\rho}_0(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\,\mathrm{d}{\xi} + \int_{\hat{W}>0} \hat{W}(\xi)|\hat{\rho}_1(\xi)|^2|\hat{\phi}_\epsilon(\xi)|^2\,\mathrm{d}{\xi}\,.\\ \end{align}\] Sending \(\epsilon\rightarrow 0\), the RHS converges to \(\int_{\hat{W}>0} \hat{W}(\xi)|\hat{\rho}_0(\xi)|^2\,\mathrm{d}{\xi} + \int_{\hat{W}>0} \hat{W}(\xi)|\hat{\rho}_1(\xi)|^2\,\mathrm{d}{\xi}\) which is finite since \(E[\rho_i]\) are finite. Therefore \(E[\rho_0+\rho_1] < \infty\) due to Lemma 5.

Then we take any distinct \(\rho_0,\rho_1\in\mathcal{M}(\overline{B(0;R)})\) with \(E[\rho_0]<\infty,\,E[\rho_1]<\infty\) and \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho_0({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho_1({\boldsymbol{x}})}=0\). For the previously stated potentials, we have \(E[\rho_0+\rho_1]<\infty\), which implies \(E[|\rho_1-\rho_0|]<\infty\). Therefore one can calculate \[\frac{\,\mathrm{d}^2}{\,\mathrm{d}t^2} E[(1-t)\rho_0+t\rho_1] = 2E[\rho_1-\rho_0]\,.\] Since \(\mu=\rho_1-\rho_0\) is a nontrivial signed measure supported on \(\overline{B(0;R)}\) with \(E[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), we see that \(R_{\textrm{LIC}}[W]\) is at least the RHS of ?? . To see that \(R_{\textrm{LIC}}[W]\) is at most this quantity, we take any \(\mu\) as described above, then one can write \(\mu = C(\rho_1-\rho_0)\) where \(C>0\) and \(\rho_0,\rho_1\in\mathcal{M}(\overline{B(0;R)})\) with \(E[\rho_0]<\infty,\,E[\rho_1]<\infty\) and \(\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho_0({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\rho_1({\boldsymbol{x}})}=0\). In fact, one can take \(\rho_1=\frac{1}{C}(\mu_+ + f)\) and \(\rho_0=\frac{1}{C}(\mu_- + f)\) where \(f\) is a nonnegative continuous function supported on \(\overline{B(0;R)}\) such that \(\rho_0,\rho_1\) satisfy the condition on the center of mass, and \(C>0\) is a normalizing constant. Then, using the definition of \(R_{\textrm{LIC}}[W]\), we see that \(R_{\textrm{LIC}}[W]\) is at most the RHS of ?? .

Finally ?? follows from ?? and the Fourier representable property at level 2 for \(W_{a,b}\) with \(-d<b<a<4\) or \(W_{b,\ln}\) with \(-d<b<4,\,b\ne 0\), which was given by Theorem 17. ◻

Proof. By rescaling, we may assume \(R=1\). In fact, for a general \(\mu\) with \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\) and \(\textrm{supp\,}\mu\subset \overline{B(0;R)}\), take \(\tilde{\mu}({\boldsymbol{x}}) = \mu(R{\boldsymbol{x}})\), then \(\textrm{supp\,}\tilde{\mu}\subset \overline{B(0;1)}\), with \(\hat{\tilde{\mu}}(\xi) = R^{-d}\hat{\mu}(\frac{\xi}{R})\), \[\int_{\mathbb{R}^d}|\xi|^{-\alpha}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} = R^{d+\alpha}\int_{\mathbb{R}^d}|\xi|^{-\alpha}|\hat{\tilde{\mu}}(\xi)|^2\,\mathrm{d}{\xi}\,,\] and similar for the \(\beta\) term. Then the conclusion for \(\mu\) may follow from that for \(\tilde{\mu}\).

Let \(\phi\) be a mollifier (see the last paragraph of Section 1). Then \(\hat{\phi}\) is a radial smooth function with rapid decay, with \(\hat{\phi}(0) = 1\). Using the fact that \((\nabla \hat{\phi})(\xi) = \int_{\overline{B(0;1)}}-2\pi i {\boldsymbol{x}}\phi({\boldsymbol{x}})e^{-2\pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}}\), we see that \(\|\nabla \hat{\phi}\|_{L^\infty} \le 2\pi\), and thus \[\label{Phihatlow} \hat{\phi}(\xi) \ge \frac{1}{2},\quad\forall |\xi| \le \frac{1}{4\pi}\,.\tag{22}\]

Define \[\nu = \phi*\mu\,.\] Then \(\nu\) is a smooth function supported on \(\overline{B(0;2)}\), with \(\int_{\mathbb{R}^d}\nu({\boldsymbol{x}})\,\mathrm{d}{{\boldsymbol{x}}}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\nu({\boldsymbol{x}})\,\mathrm{d}{{\boldsymbol{x}}}=0\). It follows that \[\hat{\nu}(0) = \nabla \hat{\nu}(0) = 0\,,\] and \[\left|\frac{\partial^2}{\partial \xi_j \partial \xi_k}\hat{\nu}(\xi)\right| = \left|\int_{\overline{B(0;2)}}-4\pi^2 x_jx_k \nu({\boldsymbol{x}})e^{-2\pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}}\right| \le 4\pi^2 \cdot 4 \|\nu\|_{L^1} \le 16\pi^2 |B(0;2)|^{1/2}\|\nu\|_{L^2}\,,\] for any \(\xi\in\mathbb{R}^d\). It follows that the maximal possible eigenvalue (in absolute value) of \(\nabla^2 \hat{\nu}(\xi)\) is no more than \(16\pi^2 |B(0;2)|^{1/2}\|\nu\|_{L^2}\cdot d\), and thus \[\label{hatnu} |\hat{\nu}(\xi)| \le 8\pi^2 d |B(0;2)|^{1/2}\|\nu\|_{L^2} \cdot |\xi|^2\,,\tag{23}\] for any \(\xi\in\mathbb{R}^d\).

Take \(\lambda>0\) to be determined, and we consider the quantity \[A = \int_{\mathbb{R}^d}(|\xi|^{-\beta}-\lambda|\xi|^{-\alpha}) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,,\] and aim to prove \(A\ge 0\) for properly chosen \(\lambda\). We first notice that \[\label{rlam} |\xi|^{-\beta}-\lambda|\xi|^{-\alpha} \ge \frac{1}{2}|\xi|^{-\beta},\quad \forall |\xi| \ge r_\lambda\,,\tag{24}\] where \[r_\lambda = (2\lambda)^{\frac{1}{\alpha-\beta}}\,.\] We will take \(\lambda\) satisfying \[\lambda \le \frac{1}{2(4\pi)^{\alpha-\beta}}\,,\] so that \[\label{rlam4pi} r_\lambda \le \frac{1}{4\pi}\,.\tag{25}\]

For small \(\xi\) values, we first estimate the negative quantity \[-\lambda \int_{B(0;r_\lambda)}|\xi|^{-\alpha}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,.\] Here, using 22 25 23 , the last integral can be estimated by \[\begin{align} \int_{B(0;r_\lambda)}|\xi|^{-\alpha}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \le & 4\int_{B(0;r_\lambda)}|\xi|^{-\alpha}|\hat{\mu}(\xi)|^2|\hat{\phi}(\xi)|^2\,\mathrm{d}{\xi} = 4\int_{B(0;r_\lambda)}|\xi|^{-\alpha}|\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi} \\ \le & 4(8\pi^2 d |B(0;2)|^{1/2}\|\nu\|_{L^2})^2\int_{B(0;r_\lambda)}|\xi|^{-\alpha} |\xi|^4\,\mathrm{d}{\xi} \\ = & 4(8\pi^2 d)^2 |B(0;2)| \|\nu\|_{L^2}^2 \cdot |\partial B(0;1)| \frac{1}{4-\alpha+d}r_\lambda^{4-\alpha+d}\,. \end{align}\] Therefore we conclude \[\label{Aest1} -\lambda \int_{B(0;r_\lambda)}|\xi|^{-\alpha}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \ge -C_1 \lambda r_\lambda^{4-\alpha+d} \|\nu\|_{L^2}^2 \,,\tag{26}\] where \[\label{eqC1} C_1 = 4(8\pi^2 d)^2 |B(0;2)|\cdot |\partial B(0;1)| \frac{1}{4-\alpha+d}\,.\tag{27}\]

For large \(\xi\) values, together with positive contribution from small \(\xi\), we estimate as \[\int_{B(0;r_\lambda)} |\xi|^{-\beta}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} + \int_{B(0;r_\lambda)^c}(|\xi|^{-\beta}-\lambda|\xi|^{-\alpha}) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \ge \frac{1}{2}\int_{\mathbb{R}^d}|\xi|^{-\beta} |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,,\] by 24 . Notice that \(|\xi|^{2+\beta}|\hat{\phi}(\xi)|^2\) is a continuous radial function which decays at infinity, and thus it is bounded, and then \[\label{eqC20} |\xi|^{-\beta} \ge \frac{2}{C_2} |\xi|^2|\hat{\phi}(\xi)|^2,\quad \forall \xi\ne 0\,,\tag{28}\] where \[\label{eqC2} C_2 = 2\sup_{\xi\in\mathbb{R}^d} |\xi|^{2+\beta}|\hat{\phi}(\xi)|^2\,.\tag{29}\] It follows that \[\begin{align} \int_{B(0;r_\lambda)} & |\xi|^{-\beta}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} + \int_{B(0;r_\lambda)^c}(|\xi|^{-\beta}-\lambda|\xi|^{-\alpha}) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \\ \ge & \frac{1}{C_2}\int_{\mathbb{R}^d} |\xi|^2|\hat{\phi}(\xi)|^2 |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} = \frac{1}{C_2}\int_{\mathbb{R}^d} |\xi|^2 |\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi}\,. \end{align}\] Combining with 26 we get \[A \ge \frac{1}{C_2}\int_{\mathbb{R}^d} |\xi|^2 |\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi} -C_1 \lambda r_\lambda^{4-\alpha+d} \|\nu\|_{L^2}^2\,.\]

Finally we apply the Heisenberg uncertainty principle (c.f. [24]) to \(\nu\), which gives \[\int_{\mathbb{R}^d} |{\boldsymbol{x}}|^2 |\nu({\boldsymbol{x}})|^2\,\mathrm{d}{{\boldsymbol{x}}} \cdot \int_{\mathbb{R}^d} |\xi|^2 |\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi} \ge \frac{d^2}{16\pi^2}\|\nu\|_{L^2}^4\,.\] Since \(\nu\) is supported on \(\overline{B(0;2)}\), we see that \[\int_{\mathbb{R}^d} |{\boldsymbol{x}}|^2 |\nu({\boldsymbol{x}})|^2\,\mathrm{d}{{\boldsymbol{x}}} \le 4\|\nu\|_{L^2}^2\,,\] which implies \[\label{eqHei} \int_{\mathbb{R}^d} |\xi|^2 |\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi} \ge \frac{d^2}{64\pi^2}\|\nu\|_{L^2}^2\,.\tag{30}\] Therefore we get \[A \ge \frac{d^2}{64\pi^2 C_2}\|\nu\|_{L^2}^2 -C_1 \lambda r_\lambda^{4-\alpha+d} \|\nu\|_{L^2}^2 = \frac{d^2}{64\pi^2 C_2}\|\nu\|_{L^2}^2\Big(1- 64\pi^2 d^{-2} C_1C_2 \cdot 2^{\frac{4-\alpha+d}{\alpha-\beta}}\lambda^{1+\frac{4-\alpha+d}{\alpha-\beta}}\Big)\,,\] where the power on \(\lambda\) is \(1+\frac{4-\alpha+d}{\alpha-\beta} = \frac{4-\beta+d}{\alpha-\beta}>0\). By taking \[\label{lamfinal} \lambda = \min\left\{\frac{1}{2(4\pi)^{\alpha-\beta}} , (64\pi^2 d^{-2} C_1C_2 )^{-\frac{\alpha-\beta}{4-\beta+d}}\cdot 2^{-\frac{4-\alpha+d}{4-\beta+d}}\right\}\,,\tag{31}\] we get \(A\ge 0\), which gives the conclusion with \(C_{\alpha,\beta}=\frac{1}{\lambda}\). ◻

Proof. By rescaling, we may assume \(R=1\). In fact, for a general \(\textrm{supp\,}\mu\subset \overline{B(0;R)}\), take \(\tilde{\mu}({\boldsymbol{x}}) = \mu(R{\boldsymbol{x}})\), then \(\textrm{supp\,}\tilde{\mu}\subset \overline{B(0;1)}\), with \(\hat{\tilde{\mu}}(\xi) = R^{-d}\hat{\mu}(\frac{\xi}{R})\), \[\int_{\mathbb{R}^d}|\xi|^{-\beta}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} = R^{d+\beta}\int_{\mathbb{R}^d}|\xi|^{-\beta}|\hat{\tilde{\mu}}(\xi)|^2\,\mathrm{d}{\xi}\,,\] and \[\int_{\mathbb{R}^d}|\xi|^{-\beta}\ln|\xi|\cdot|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} + \ln R \int_{\mathbb{R}^d}|\xi|^{-\beta}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}= R^{d+\beta}\int_{\mathbb{R}^d}|\xi|^{-\beta}\ln|\xi|\cdot|\hat{\tilde{\mu}}(\xi)|^2\,\mathrm{d}{\xi}\,.\] Then the conclusion for \(\mu\) may follow from that for \(\tilde{\mu}\).

Define \(\phi\), \(\nu\) in the same way as the proof of Theorem 21. Take \(\lambda>0\) to be determined, and we consider the quantity \[A = \int_{\mathbb{R}^d}|\xi|^{-\beta}(\ln|\xi|+\lambda) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,,\] and aim to prove \(A\ge 0\) for properly chosen \(\lambda\ge 2\). We first notice that \[\label{rlam1} \ln|\xi|+\lambda \ge \frac{\lambda}{2},\quad \forall |\xi| \ge r_\lambda\,,\tag{32}\] where \[r_\lambda = e^{-\lambda/2}\,.\] We will take \(\lambda\) satisfying \[\lambda \ge 2\ln(4\pi)\,,\] so that \[\label{rlam4pi1} r_\lambda \le \frac{1}{4\pi}\,.\tag{33}\]

For small \(\xi\) values, we first estimate the negative quantity \[\int_{B(0;r_\lambda)}|\xi|^{-\beta}\ln|\xi|\cdot|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,.\] For any \(|\xi|< r_\lambda=e^{-\lambda/2}\), we have \(\ln|\xi| < -\frac{\lambda}{2} < 0\), and thus the above integrand is non-positive. Using 22 33 23 , the last integral can be estimated by \[\begin{align} &\int_{B(0;r_\lambda)}|\xi|^{-\beta}(-\ln|\xi|)\cdot|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\\ \le & 4\int_{B(0;r_\lambda)}|\xi|^{-\beta}(-\ln|\xi|)\cdot|\hat{\mu}(\xi)|^2|\hat{\phi}(\xi)|^2\,\mathrm{d}{\xi} = 4\int_{B(0;r_\lambda)}|\xi|^{-\beta}(-\ln|\xi|)\cdot|\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi} \\ \le & 4(8\pi^2 d |B(0;2)|^{1/2}\|\nu\|_{L^2})^2\int_{B(0;r_\lambda)}|\xi|^{-\beta}(-\ln|\xi|)\cdot |\xi|^4\,\mathrm{d}{\xi} \\ = & 4(8\pi^2 d)^2 |B(0;2)| \|\nu\|_{L^2}^2 \cdot |\partial B(0;1)| \Big(\frac{1}{4-\beta+d}r_\lambda^{4-\beta+d}(-\ln r_\lambda) + \frac{1}{(4-\beta+d)^2}r_\lambda^{4-\beta+d}\Big) \\ \le & 4(8\pi^2 d)^2 |B(0;2)| \|\nu\|_{L^2}^2 \cdot |\partial B(0;1)| \Big(\frac{1}{4-\beta+d}+\frac{1}{(4-\beta+d)^2\ln(4\pi)}\Big)r_\lambda^{4-\beta+d}(-\ln r_\lambda)\,. \\ \end{align}\] Therefore we conclude \[\label{Aest1log} \int_{B(0;r_\lambda)}|\xi|^{-\beta}\ln|\xi|\cdot|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \ge -C_1 r_\lambda^{4-\beta+d}(-\ln r_\lambda) \|\nu\|_{L^2}^2 \,,\tag{34}\] where \[\label{eqC1log} C_1 = 4(8\pi^2 d)^2 |B(0;2)|\cdot |\partial B(0;1)| \Big(\frac{1}{4-\beta+d}+\frac{1}{(4-\beta+d)^2\ln(4\pi)}\Big)\,.\tag{35}\]

For large \(\xi\) values, together with positive contribution from small \(\xi\), we estimate as \[\begin{align} \int_{B(0;r_\lambda)} & \lambda|\xi|^{-\beta}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} + \int_{B(0;r_\lambda)^c}|\xi|^{-\beta}(\ln|\xi|+\lambda) |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \\ \ge & \frac{\lambda}{2}\int_{\mathbb{R}^d}|\xi|^{-\beta} |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \ge \frac{\lambda}{C_2}\int_{\mathbb{R}^d} |\xi|^2|\hat{\phi}(\xi)|^2 |\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} = \frac{\lambda}{C_2}\int_{\mathbb{R}^d} |\xi|^2 |\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi}\,, \end{align}\] by 32 and 28 , where \(C_2\) is given by 29 . Combining with 34 we get \[A \ge \frac{\lambda}{C_2}\int_{\mathbb{R}^d} |\xi|^2 |\hat{\nu}(\xi)|^2\,\mathrm{d}{\xi} -C_1 r_\lambda^{4-\beta+d}(-\ln r_\lambda) \|\nu\|_{L^2}^2\,.\]

Finally we use 30 from the Heisenberg uncertainty principle and get \[\begin{align} A \ge & \frac{\lambda}{64\pi^2d^{-2} C_2}\|\nu\|_{L^2}^2 -C_1 r_\lambda^{4-\beta+d}(-\ln r_\lambda) \|\nu\|_{L^2}^2 \\ = & \frac{\lambda}{64\pi^2d^{-2} C_2}\|\nu\|_{L^2}^2\Big(1- 64\pi^2d^{-2} C_1C_2 \cdot \lambda^{-1}e^{-\lambda(4-\beta+d)/2}\frac{\lambda}{2}\Big) \\ = & \frac{\lambda}{64\pi^2d^{-2} C_2}\|\nu\|_{L^2}^2\Big(1- 32\pi^2d^{-2} C_1C_2 \cdot e^{-\lambda(4-\beta+d)/2}\Big) \,, \end{align}\] where in the last exponential one has \(4-\beta+d>0\). By taking \[\label{lamfinal1} \lambda = \max\left\{2\ln(4\pi) , \frac{2}{4-\beta+d}\ln(32\pi^2d^{-2} C_1C_2) \right\}\,,\tag{36}\] we get \(A\ge 0\), which gives the conclusion with \(C_{\beta}=\lambda\). ◻

Proof of Theorem 2. Take \(a_0 = \max\{1,\frac{b+2}{2}\}\). Then Theorem 8 shows that \(R_*(b,a_0)\) is an upper bound of minimizer size for any \(W_{a,b}\) with \(a\ge a_0\).

We claim that for \(a_-(b)\in [a_0,2)\) sufficiently close to 2, we have \[\label{claim95RLIC} R_{\textrm{LIC}}[W_{a,b}] \ge R_*(b,a_0) +1 =: R,\quad \forall a_-(b) \le a < 2\,,\tag{37}\] which would imply the conclusion due to Lemma 1.

By Theorems 19 and 15, it suffices to prove that for any nontrivial signed measure \(\mu\) supported on \(\overline{B(0;R)}\) with \(E_{a,b}[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), there holds \[\label{EabFou} \int_{\mathbb{R}^d\backslash \{0\}} \hat{W}_{a,b}(\xi)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d} \Big(c_{\textrm{PF}}(-b) |\xi|^{-d-b} -c_{\textrm{PF}}(-a) |\xi|^{-d-a}\Big)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} > 0\,.\tag{38}\]

Notice that \(c_{\textrm{PF}}(s)>0\) for any \(-2<s<d\), with \(c_{\textrm{PF}}(-2)=0\) and \[c_{\textrm{PF}}(s) \approx c_{\textrm{PF}}'(-2)(s+2),\quad c_{\textrm{PF}}'(-2)=\pi^{-2-\frac{d}{2}}\frac{\Gamma(\frac{d+2}{2})}{4}>0\,,\] for \(s\) close to \(-2\), to the leading order (c.f. Remark 16). This implies that \(c_{\textrm{PF}}\) is an increasing function on \([-2,-2+\epsilon]\) for some \(\epsilon>0\). Then we apply Theorem 21 to \(\alpha=d+a\) and \(\beta=d+b\) to get \[\int_{\mathbb{R}^d}|\xi|^{-d-a}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \le C_*\int_{\mathbb{R}^d}|\xi|^{-d-b}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,.\] Here \(C_* = \sup_{a\in [a_0,2)}(C_{d+a,d+b}R^{a-b}) >0\) is a uniform constant since \(\alpha=d+a\) is bounded above by \(d+2\) and thus away from \(d+4\) (c.f. Remark 22). Then we may take \(a_-(b)\in [a_0,2)\) sufficiently close to 2 so that \(2-a_-(b)<\epsilon\) and \[c_{\textrm{PF}}(-a_-(b)) \le \frac{1}{2C_*}c_{\textrm{PF}}(-b)\,.\] It follows that \(c_{\textrm{PF}}(-a) \le c_{\textrm{PF}}(-a_-(b)) \le \frac{1}{2C_*}c_{\textrm{PF}}(-b)\) for any \(a\in [a_-(b),2)\), and we see that \[\int_{\mathbb{R}^d\backslash \{0\}} \hat{W}_{a,b}(\xi)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} > 0\,,\] which finishes the proof. ◻

Proof of Theorem 5. Theorem 10 gives \(R_{*,\ln}(1)\) as an upper bound of minimizer size for any \(W_{b,\ln}\) with \(b\ge 1\).

We claim that for \(b_*\in [1,2)\) sufficiently close to 2, we have \[\label{claim95RLIClog} R_{\textrm{LIC}}[W_{b,\ln}] \ge R_{*,\ln}(1) + 1 =: R,\quad \forall b_* \le b < 2\,,\tag{39}\] which would imply the conclusion due to Lemma 1.

By Theorems 19 and 15, it suffices to prove that for any nontrivial signed measure \(\mu\) supported on \(\overline{B(0;R)}\) with \(E_{a,b}[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), there holds \[\label{EabFoulog} \int_{\mathbb{R}^d\backslash \{0\}} \hat{W}_{b,\ln}(\xi)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} = \int_{\mathbb{R}^d} \Big(\tilde{c}_{\textrm{PF}}(-b) |\xi|^{-d-b} +c_{\textrm{PF}}(-b) |\xi|^{-d-b}\ln|\xi|\Big)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} > 0\,.\tag{40}\] The properties of \(c_{\textrm{PF}}(s)\) near \(s=-2\) were analyzed in the proof of Theorem 2. We recall that \(\tilde{c}_{\textrm{PF}}(-2)=c_{\textrm{PF}}'(-2)>0\) from 19 . Therefore one can take \(\epsilon>0\) such that \(c_{\textrm{PF}}\) is increasing on \([-2,-2+\epsilon]\) (as discussed in the proof of Theorem 2), and \(\tilde{c}_{\textrm{PF}}(s) \ge \frac{1}{2}c_{\textrm{PF}}'(-2) > 0\) for \(s\in [-2,-2+\epsilon]\).

Then we apply Theorem 23 to \(\beta=d+b\) to get \[\int_{\mathbb{R}^d}|\xi|^{-d-b}\ln|\xi|\cdot|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} + C_*\int_{\mathbb{R}^d}|\xi|^{-d-b}|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} \ge 0\,,\] for any \(1\le b < 2\). Here \(C_*=\sup_{1\le b < 2}C_{d+b} + \ln R>0\) is a uniform constant since \(\beta\) is bounded above by \(d+2\) and thus away from \(d+4\) (c.f. Remark 24). Then we may take \(b_*\in [1,2)\) sufficiently close to 2 so that \(2-b_*<\epsilon\) and \[c_{\textrm{PF}}(-b_*) \le \frac{1}{4C_*}c_{\textrm{PF}}'(-2)\,.\] It follows that \[c_{\textrm{PF}}(-b) \le c_{\textrm{PF}}(-b_*) \le \frac{1}{4C_*}c_{\textrm{PF}}'(-2) \le \frac{1}{2C_*}\tilde{c}_{\textrm{PF}}(-b)\,,\] for any \(b\in [b_*,2)\), and we see that \[\int_{\mathbb{R}^d\backslash \{0\}} \hat{W}_{b,\ln}(\xi)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi} > 0\,,\] which finishes the proof. ◻

Proof. The function \((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\) is compactly supported and continuous, and thus its Fourier transform is given by \[\label{Fxa4} \mathcal{F}[(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})](\xi) = \int_{\mathbb{R}^d}(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}}) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\,.\tag{41}\]

We first assume \(d\ge 2\). Let \[k = 1+\lfloor \frac{a}{2} \rfloor\,.\] Multiplying by \((-4\pi^2|\xi|^2)^k\) in 41 and integrating by parts, we get \[\begin{align} (-4\pi^2|\xi|^2)^k\mathcal{F}[(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})](\xi) = & \int_{\mathbb{R}^d}(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}}) \Delta^k e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}} \\ = & \int_{\mathbb{R}^d} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\,. \\ \end{align}\] Here the integration by parts is legal because the integrand is compactly supported and \(\Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\) is locally integrable in \(d\ge 2\), by the choice of \(k\). Since \(\Phi\) is supported on \(\overline{B(0;2)}\), the last integrand is zero for any \(|{\boldsymbol{x}}|>2\). Then we estimate its size by separating it into \[\label{Deltaa4}\begin{align} \int_{\mathbb{R}^d} & \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}} \\ = & \int_{|{\boldsymbol{x}}|\le 1} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot\Phi(2{\boldsymbol{x}}) e^{-2\pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}} \\ & + \int_{\frac{1}{2}<|{\boldsymbol{x}}|\le 2} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot(1-\Phi(2{\boldsymbol{x}})) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\,. \end{align}\tag{42}\]

First, since \(\Phi=1\) on \(B(0;1)\) and \(k\ge 3\), \[\begin{align} \int_{|{\boldsymbol{x}}|\le 1} & \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot\Phi(2{\boldsymbol{x}}) e^{-2\pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}} = \int_{|{\boldsymbol{x}}|\le 1} \Delta^k(|{\boldsymbol{x}}|^a)\cdot\Phi(2{\boldsymbol{x}}) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\,. \end{align}\] If \(a>4\) is an even integer, then \(a=2(k-1)\) and the last integrand is zero. Otherwise, we have \(-2<a-2k<0\), and then \[\begin{align} \int_{|{\boldsymbol{x}}|\le 1} & \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot\Phi(2{\boldsymbol{x}}) e^{-2\pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}} \\ = & \prod_{j=0}^{k-1} (a-2j)(a+d-2-2j) \cdot \int_{|{\boldsymbol{x}}|\le 1} |{\boldsymbol{x}}|^{a-2k}\Phi(2{\boldsymbol{x}}) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}} \\ = & \prod_{j=0}^{k-1} (a-2j)(a+d-2-2j) \cdot \mathcal{F}[|{\boldsymbol{x}}|^{a-2k}\Phi(2{\boldsymbol{x}})](\xi)\,. \\ \end{align}\] The Fourier transform of \(|{\boldsymbol{x}}|^{a-2k}\) is a locally integrable function given by Theorem 15, and then \[\mathcal{F}[|{\boldsymbol{x}}|^{a-2k}\Phi(2{\boldsymbol{x}})] = (2k-a)c_{\textrm{PF}}(2k-a)2^{-d}\cdot |\xi|^{-d+2k-a}*\hat{\Phi}(\frac{\xi}{2})\,.\]

Then we claim that there exists \(C_1>0\), uniform in \(a\) for \(-2<a-2k<0\), such that \[\label{claim952ka} \Big|(2k-a)\big(|\cdot|^{-d+2k-a}*\hat{\Phi}(\frac{\cdot}{2})\big)(\xi)\Big| \le C_1(1+|\xi|)^{-d+2k-a}\,.\tag{43}\] In fact, we first apply Lemma 6 with \(\psi_1 = \hat{\Phi}(\frac{\cdot}{2})\), \(\psi_2=(2k-a)|\cdot|^{-d+2k-a}\), \(M=-d+2k-a\), \(R=1\) to obtain the above inequality for \(|\xi|>3\). By tracing through the proof of Lemma 6, the constant \(C_1\) only depends on \(\psi_2\) through the implied constant in ?? and \(\int_{B(0;1)}|\psi_2({\boldsymbol{x}})|\,\mathrm{d}{{\boldsymbol{x}}}\), which are both uniform in \(a\) for our choice of \(\psi_2\). Therefore \(C_1\) can be chosen uniformly in \(a\).

To see this inequality for \(|\xi|\le 3\), we write its LHS as \[|(\psi_1*\psi_2)(\xi)| \le |(\psi_1*(\psi_2\chi_{B(0;1)})(\xi)| + |(\psi_1*(\psi_2\chi_{B(0;1)^c})(\xi)|\,.\] By Young’s inequality, we have \[|(\psi_1*(\psi_2\chi_{B(0;1)})(\xi)| \le \|\psi_1\|_{L^\infty(\mathbb{R}^d)} \cdot \|\psi_2\|_{L^1(B(0;1))} \le C\,,\] and \[|(\psi_1*(\psi_2\chi_{B(0;1)^c})(\xi)| \le \|\psi_1\|_{L^1(\mathbb{R}^d)} \cdot \|\psi_2\|_{L^\infty(B(0;1)^c)} \le C\,,\] uniform in \(a\). Therefore the claim 43 is proved.

To analyze the behavior of the \(c_{\textrm{PF}}\) part,

• If \(d\ge 3\), then \(2k-a\in (0,2)\) is strictly inside \(\{s\in\mathbb{C}:\Re(s) < d\}\) where \(c_{\textrm{PF}}\) is holomorphic, and thus \(c_{\textrm{PF}}(2k-a)\) is uniformly bounded. Therefore \[\begin{align} \left|\int_{|{\boldsymbol{x}}|\le 1} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot\Phi(2{\boldsymbol{x}}) e^{- 2\pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}} \right| \le \eta_{a,1}(1+|\xi|)^{-d+2k-a}\,, \end{align}\] where \[\label{etaa195d3} \eta_{a,1} = \prod_{j=0}^{k-1} (a-2j)(a+d-2-2j)\cdot C_1 \cdot \sup_{ 0<s<2}c_{\textrm{PF}}(s)\,.\tag{44}\]

• If \(d=2\), then \(2k-a\in (0,2)\) is inside \(\{s\in\mathbb{C}:\Re(s) < d\}\) but can come close to the point 2 where \(c_{\textrm{PF}}\) has a simple pole (coming from the expression \(\Gamma(\frac{d-s}{2})\) in its definition). Therefore, using \(c_{\textrm{PF}}(s)\lesssim (2-s)^{-1}\), we get \[\begin{align} \left|\int_{|{\boldsymbol{x}}|\le 1} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot\Phi(2{\boldsymbol{x}}) e^{- 2 \pi i {\boldsymbol{x}}\cdot\xi}\,\mathrm{d}{{\boldsymbol{x}}} \right| \le \eta_{a,1}(1+|\xi|)^{-d+2k-a}\,, \end{align}\] where \[\label{etaa195d2} \eta_{a,1} = \prod_{j=0}^{k-1} (a-2j)^2\cdot C_1 \cdot \frac{1}{2-2k+a}\sup_{ 0<s <2}(2-s)c_{\textrm{PF}}(s)\,.\tag{45}\]

Notice that in both cases we have \[\eta_{a,1} = O(a-4)\,,\] when \(a\) is close to 4. In fact, we have \(k=3\) when \(a\) is close to 4. For \(d\ge 3\), the product term in 44 contains a factor \((a-4)\). For \(d=2\), the product term in 45 contains a factor \((a-4)^2\) and there is another \(\frac{1}{a-4}\) in 45 .

To estimate the last integral in 42 , we observe that \(\Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot(1-\Phi(2{\boldsymbol{x}}))\) is smooth on \(\mathbb{R}^d\) and supported on \(\{\frac{1}{2}\le |{\boldsymbol{x}}| \le 2\}\). Therefore, the last integral in 42 , being the Fourier transform of this function, is a Schwartz function in \(\xi\). Therefore there exists \(\eta_{a,2}>0\) such that \[\begin{align} \left|\int_{\frac{1}{2}<|{\boldsymbol{x}}|\le 2} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot(1-\Phi(2{\boldsymbol{x}})) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\right| \le \eta_{a,2}(1+|\xi|)^{-d+2k-a}\,. \end{align}\] To analyze the behavior of \(\eta_{a,2}\) for \(a\) close to 4, we first write \[\begin{align} & \left|(1+4\pi^2|\xi|^2)^m \int_{\frac{1}{2}<|{\boldsymbol{x}}|\le 2} \Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot(1-\Phi(2{\boldsymbol{x}})) e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\right|\\ = & \left|\int_{\frac{1}{2}<|{\boldsymbol{x}}|\le 2} (1-\Delta)^m\Big[\Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot(1-\Phi(2{\boldsymbol{x}}))\Big] e^{-2\pi i {\boldsymbol{x}}\cdot\xi} \,\mathrm{d}{{\boldsymbol{x}}}\right|\\ \le & \int_{\frac{1}{2}<|{\boldsymbol{x}}|\le 2} \left|(1-\Delta)^m\Big[\Delta^k\big((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\big)\cdot(1-\Phi(2{\boldsymbol{x}}))\Big] \right| \,\mathrm{d}{{\boldsymbol{x}}}\,,\\ \end{align}\] for any integer \(m\ge 0\). The last integrand is a linear combination of products derivatives of \(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4, \Phi({\boldsymbol{x}}), 1-\Phi(2{\boldsymbol{x}})\). Each derivative of \(\Phi({\boldsymbol{x}})\) or \(1-\Phi(2{\boldsymbol{x}})\) is bounded, and each derivative of \(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4\) is of order \(O(a-4)\) for \(\frac{1}{2}<|{\boldsymbol{x}}|\le 2\). Therefore the last integral is \(O(a-4)\) for any fixed \(m\). By taking \(m\ge \frac{d-2k+a}{2}\), we see that \(\eta_{a,2}=O(a-4)\).

Combining the above two parts, we see that the LHS of 42 is bounded by \(\eta_{a,0} (1+|\xi|)^{-d+2k-a}\) with \(\eta_{a,0}=O(a-4)\) when \(a\) is close to 4. Therefore \(|\mathcal{F}[(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})](\xi)| \lesssim \eta_a |\xi|^{-2k}(1+|\xi|)^{-d+2k-a}\) with \(\eta_a=O(a-4)\). Combined with the straightforward estimate \(|\mathcal{F}[(|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})](\xi)| \lesssim a-4\) for \(|\xi|\le 1\), we get the conclusion when \(d\ge 2\).

Then we treat the case \(d=1\) by a similar technique. Let \[k = 1+\lfloor a \rfloor\,.\] Multiplying by \((2\pi i \xi)^k\) in 41 and integrating by parts, we get \[\begin{align} (2\pi i \xi)^k\mathcal{F}[(|x|^a-|x|^4)\Phi(x)](\xi) = & \int_{\mathbb{R}}(|x|^a-|x|^4)\Phi(x) \big(-\frac{\,\mathrm{d}}{\,\mathrm{d}x}\big)^k e^{-2\pi i x \xi} \,\mathrm{d}{x} \\ = & \int_{\mathbb{R}} \big(\frac{\,\mathrm{d}}{\,\mathrm{d}x}\big)^k\big((|x|^a-|x|^4)\Phi(x)\big) e^{-2\pi i x \xi} \,\mathrm{d}{x}\,. \\ \end{align}\] Then the conclusion can obtained by a decomposition similar to 42 . In fact, if \(a\) is not an integer, then \[\label{Deltaa4d1952} \int_{|x|\le 1} \big(\frac{\,\mathrm{d}}{\,\mathrm{d}x}\big)^k\big((|x|^a-|x|^4)\Phi(x)\big)\Phi(2x) e^{-2\pi i x \xi} \,\mathrm{d}{x} = \prod_{j=0}^{k-1}(a-j) \cdot \mathcal{F}[|x|^{a-k}(\textrm{sgn}x)^k\Phi(2x)](\xi)\,,\tag{46}\] where \(-1<a-k<0\). If \(k\) is even, then the last expression is bounded by \(\eta_{a,1}(1+|\xi|)^{-1+k-a}\) similar as its \(d\ge 2\) analog. If \(k\) is odd (which includes the case when \(a\) is close 4, \(k=5\)), we first use the formula \[\mathcal{F}[|x|^{a-k}(\textrm{sgn}x)] = -2i \pi^{k-a-\frac{1}{2}}\frac{\Gamma(\frac{2-k+a}{2})}{(1+a-k)\Gamma(\frac{k-a-1}{2})}|\xi|^{-1+k-a}\textrm{sgn}\xi\,,\] which can be obtained by ?? via differentiating in \(x\). Then \[\mathcal{F}[|x|^{a-k}(\textrm{sgn}x)^k\Phi(2x)](\xi) = -i \pi^{k-a-\frac{1}{2}}\frac{\Gamma(\frac{2-k+a}{2})}{(1+a-k)\Gamma(\frac{k-a-1}{2})}\big(|\xi|^{-1+k-a}\textrm{sgn}\xi\big)* \hat{\Phi}(\frac{\xi}{2})\,,\] is bounded by \(C_1|\xi|^{-1+k-a}\) uniformly in \(a\) for \(-1<a-k<0\). In fact, one technical issue here is that \(|\xi|^{-1+k-a}\textrm{sgn}\xi\) has a sharp singularity near 0 when \(a-k\) is close to 0. It does not affect the uniform estimate for the convolution because \(|\xi|^{-1+k-a}\textrm{sgn}\xi\) is odd and \(\hat{\Phi}(\frac{\xi}{2})\) is smooth.

Then we see that 46 is again bounded by \(\eta_{a,1}(1+|\xi|)^{-1+k-a}\), with \(\eta_{a,1}=O(a-4)\) when \(a\) is close to 4 since the product in 46 contains a factor \((a-4)\).

If \(a\) is an integer, then \(k=1+a\). If \(a\) is even, then the LHS of 46 is zero. If \(a\) is odd, then \((\frac{\,\mathrm{d}}{\,\mathrm{d}x})^k\big((|x|^a-|x|^4)\Phi(x)\big)\Phi(2x) = 2\cdot a! \cdot \delta(x)\) and thus the LHS of 46 is \(2\cdot a!\), which is again bounded by \(\eta_{a,1}(1+|\xi|)^{-1+k-a}\) with \(\eta_{a,1}=2\cdot a!\).

The term \(\int_{1/2 < |x|\le 2} (\frac{\,\mathrm{d}}{\,\mathrm{d}x})^k\big((|x|^a-|x|^4)\Phi(x)\big)\cdot(1-\Phi(2x)) e^{-2\pi i x \xi} \,\mathrm{d}{x}\) can be handle in the same way as the \(d\ge 2\) case, and the conclusion for \(d= 1\) follows. ◻

Proof of Theorem 3. Theorem 8 shows that \(R_*(b,4)\) is an upper bound of minimizer size for any \(W_{a,b}\) with \(a\ge 4\).

We claim that for \(a_+(b)>4\) sufficiently close to 4, we have \[\label{tibpzeso} R_{\textrm{LIC}}[W_{a,b}] \ge R_*(b,4) +1 =: R,\quad \forall 4<a \le a_+(b)\,,\tag{47}\] which would imply the conclusion due to Lemma 1.

By Theorem 19, it suffices to prove that for any nontrivial signed measure \(\mu\) supported on \(\overline{B(0;R)}\) with \(E_{a,b}[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), there holds \[E_{a,b}[\mu] > 0\,.\] Denote \[\tilde{W}_{a,b} = (2R)^{a-b}\frac{|{\boldsymbol{x}}|^a}{a} - \frac{|{\boldsymbol{x}}|^b}{b}\,,\] and its associated energy \(\tilde{E}_{a,b}\). By rescaling, it suffices to prove that for any \(\mu\) supported on \(\overline{B(0;\frac{1}{2})}\) with \(\tilde{E}_{a,b}[|\mu|]<\infty\) and \(\int_{\mathbb{R}^d}\,\mathrm{d}{\mu({\boldsymbol{x}})}=\int_{\mathbb{R}^d}{\boldsymbol{x}}\,\mathrm{d}{\mu({\boldsymbol{x}})}=0\), there holds \[\tilde{E}_{a,b}[\mu] > 0\,.\] For such \(\mu\), we may rewrite \(\tilde{E}_{a,b}[\mu]\) as \[\begin{align} \tilde{E}_{a,b}[\mu] = & E_b[\mu] + \frac{(2R)^{a-b}}{2a}\int_{|{\boldsymbol{x}}|\le 1/2}\int_{|{\boldsymbol{y}}|\le 1/2}|{\boldsymbol{x}}-{\boldsymbol{y}}|^a\,\mathrm{d}{\mu({\boldsymbol{y}})}\,\mathrm{d}{\mu({\boldsymbol{x}})} \\ = & E_b[\mu] + \frac{(2R)^{a-b}}{2a}\int_{|{\boldsymbol{x}}|\le 1/2}\int_{|{\boldsymbol{y}}|\le 1/2}(|{\boldsymbol{x}}-{\boldsymbol{y}}|^a-|{\boldsymbol{x}}-{\boldsymbol{y}}|^4)\Phi({\boldsymbol{x}}-{\boldsymbol{y}})\,\mathrm{d}{\mu({\boldsymbol{y}})}\,\mathrm{d}{\mu({\boldsymbol{x}})} \\ & + \frac{(2R)^{a-b}}{2a}\int_{|{\boldsymbol{x}}|\le 1/2}\int_{|{\boldsymbol{y}}|\le 1/2}|{\boldsymbol{x}}-{\boldsymbol{y}}|^4\,\mathrm{d}{\mu({\boldsymbol{y}})}\,\mathrm{d}{\mu({\boldsymbol{x}})} \,,\\ \end{align}\] where \(\Phi\) is given by the statement of Lemma 9, and \(E_b[\mu]\) denotes the energy 1 with interaction potential \(W({\boldsymbol{x}})=-\frac{|{\boldsymbol{x}}|^b}{b}\). The last integral is nonnegative by the (nonstrict) LIC property of the interaction potential \(|{\boldsymbol{x}}|^4\) (c.f. [4]). Then notice that the potential \((|{\boldsymbol{x}}|^a-|{\boldsymbol{x}}|^4)\Phi({\boldsymbol{x}})\) is Fourier representable at level 0 due to Theorem 11, whose Fourier transform is controlled by Lemma 9. Therefore we get \[\begin{align} \tilde{E}_{a,b}[\mu] \ge \frac{1}{2}\int_{\mathbb{R}^d\backslash\{0\}} \Big(c_{\textrm{PF}}(-b)|\xi|^{-d-b} - \frac{(2R)^{a-b}}{a}\eta_a (1+|\xi|)^{-d-a}\Big)|\hat{\mu}(\xi)|^2\,\mathrm{d}{\xi}\,, \end{align}\] where \(\eta_a = O(a-4)\) when \(a\) is close to 4. Since \(a>4\) and \(-d<b<2\), we see that \(|\xi|^{-d-b} \ge (1+|\xi|)^{-d-b} \ge (1+|\xi|)^{-d-a}\) for any \(\xi\ne 0\). Also, \(c_{\textrm{PF}}(-b)>0\), and \(\frac{(2R)^{a-b}}{a}\) is uniformly bounded when \(a>4\) is close to 4. Therefore, using \(\eta_a = O(a-4)\), one can take \(a_+(b)\) sufficiently close to 4 to guarantee that \(\frac{(2R)^{a-b}}{a}\eta_a \le \frac{1}{2}c_{\textrm{PF}}(-b)\) for any \(4<a\le a_+(b)\), and thus \(\tilde{E}_{a,b}[\mu]>0\). ◻