Strict positivity of logarithmic energy for a zero-mass signed charge

Fix $R$ with $R>\mathrm{diam}(\mathrm{supp}\mu \cup \mathrm{supp}\nu )$, write $r=\mathrm{\mid }z-w\mathrm{\mid }$, and let $0<\epsilon <1$. For $0<a<b$ the identity ${\int }_0^{\mathrm{\infty }}(e^{-at}-e^{-bt})dt\mathrm{/}t=\log (b\mathrm{/}a)$ follows from Tonelli applied to the nonnegative integrand ${\int }_a^b{e}^{-st}ds$ and ${\int }_0^{\mathrm{\infty }}{e}^{-st}dt=1\mathrm{/}s$; taking $a=r^2$, $b=R^2$ gives $\log (R\mathrm{/}r)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}(e^{-t{r}^2}-e^{-t{R}^2})dt\mathrm{/}t$ when $r>0$, while at $r=0$ the integral is $+\mathrm{\infty }$; the truncation $K_{\epsilon }(z,w):=\frac{1}{2}{\int }_{\epsilon }^{1\mathrm{/}\epsilon }(e^{-t\mathrm{\mid }z-w{\mathrm{\mid }}^2}-e^{-t{R}^2})dt\mathrm{/}t$ is continuous on $\mathbb{C}\times \mathbb{C}$, satisfies $0\le K_{\epsilon }\le k_R$ pointwise on $\{\mathrm{\mid }z-w\mathrm{\mid }\le R\}$, and increases to $k_R$ as $\epsilon \downarrow 0$.

Let $t>0$ and let $\rho ,{\rho }^{\mathrm{\prime }}$ be finite signed Borel measures of compact support on $\mathbb{C}$. Put $c_t:={\int }_{\mathbb{C}}{e}^{-4t\mathrm{\mid }u{\mathrm{\mid }}^2}dA(u)$, which satisfies $0<c_t<\mathrm{\infty }$ since $\pi e^{-4t}\le c_t$ and $c_t\le \pi +2\pi {\int }_1^{\mathrm{\infty }}{e}^{-4tr}rdr<\mathrm{\infty }$; $T_t\rho (z):=\int e^{-2t\mathrm{\mid }x-z{\mathrm{\mid }}^2}d\rho (x)$ is a bounded continuous function of $z$ by [F9], and Fubini–Tonelli applied to the triple integral of the nonnegative integrand $\mathrm{\mid }{e}^{-2t\mathrm{\mid }x-z{\mathrm{\mid }}^2}{e}^{-2t\mathrm{\mid }y-z{\mathrm{\mid }}^2}\mathrm{\mid }$, together with the substitution $z=\frac{x+y}{2}+u$ and $\mathrm{\mid }x-z{\mathrm{\mid }}^2+\mathrm{\mid }y-z{\mathrm{\mid }}^2=\frac{1}{2}\mathrm{\mid }x-y{\mathrm{\mid }}^2+2\mathrm{\mid }u{\mathrm{\mid }}^2$, gives $\iint e^{-t\mathrm{\mid }x-y{\mathrm{\mid }}^2}d\rho (x)d{\rho }^{\mathrm{\prime }}(y)=c_t^{-1}{\int }_{\mathbb{C}}{T}_t\rho (z)\overline{T_t{\rho }^{\mathrm{\prime }}(z)}dA(z)$.

In the situation of step 1.1, expanding $\sigma \otimes \sigma =\mu \otimes \mu -\mu \otimes \nu -\nu \otimes \mu +\nu \otimes \nu$ and applying Fubini to each finite positive measure with the bounded integrand $K_{\epsilon }$ gives $\iint K_{\epsilon }d\sigma d\sigma =\frac{1}{2}{\int }_{\epsilon }^{1\mathrm{/}\epsilon }\iint (e^{-t\mathrm{\mid }z-w{\mathrm{\mid }}^2}-e^{-t{R}^2})d\sigma (z)d\sigma (w)dt\mathrm{/}t$; since $\sigma (\mathbb{C})=0$ the second exponential contributes $e^{-t{R}^2}\sigma (\mathbb{C}{)}^2=0$, so by step 1.2 the inner integral is $c_t^{-1}\int \mathrm{\mid }{T}_t\sigma {\mathrm{\mid }}^{2}dA\ge 0$ for every $t>0$, and hence $E_{\epsilon }:=\iint K_{\epsilon }d\sigma d\sigma \ge 0$ for every $\epsilon \in (0,1)$.

Put $A_{\epsilon }:=\iint K_{\epsilon }d\mu d\mu$, $B_{\epsilon }:=\iint K_{\epsilon }d\mu d\nu$, $C_{\epsilon }:=\iint K_{\epsilon }d\nu d\nu$, so that $E_{\epsilon }=A_{\epsilon }-2B_{\epsilon }+C_{\epsilon }$ by step 2.1. Since $0\le K_{\epsilon }\uparrow k_R$ pointwise by step 1.1, [F4] gives the monotone limits $A_{\epsilon }\uparrow A:=\iint k_{R}d\mu d\mu =I(\mu )+M^2\log R<\mathrm{\infty }$, $C_{\epsilon }\uparrow C:=I(\nu )+M^2\log R<\mathrm{\infty }$ and $B_{\epsilon }\uparrow B:=\iint k_{R}d\mu d\nu \in [0,\mathrm{\infty }]$, using [F1]. From $0\le E_{\epsilon }=A_{\epsilon }-2B_{\epsilon }+C_{\epsilon }\le A+C-2B_{\epsilon }$ we get $2B_{\epsilon }\le A+C$ for every $\epsilon$, so $B\le (A+C)\mathrm{/}2<\mathrm{\infty }$: the mixed energy $I(\mu ,\nu )=B-M^2\log R$ is finite, and $I(\sigma ):=I(\mu )-2I(\mu ,\nu )+I(\nu )=A-2B+C={\lim }_{\epsilon \downarrow 0}{E}_{\epsilon }$ is a well-defined real number with $I(\sigma )\ge 0$.

Combining the integral representation of $E_{\epsilon }$ in step 2.1 with the convergence $E_{\epsilon }\to I(\sigma )$ of step 3.1 shows $\frac{1}{2}{\int }_{\epsilon }^{1\mathrm{/}\epsilon }{Q}_t(\sigma )dt\mathrm{/}t\uparrow I(\sigma )$ as $\epsilon \downarrow 0$, where $Q_t(\sigma )=\iint e^{-t\mathrm{\mid }z-w{\mathrm{\mid }}^2}d\sigma d\sigma =c_t^{-1}\int \mathrm{\mid }{T}_t\sigma {\mathrm{\mid }}^{2}dA\ge 0$ for every $t>0$ by step 1.2; since $t\mapsto Q_t(\sigma )$ is measurable and nonnegative, [F4] gives $\frac{1}{2}{\int }_0^{\mathrm{\infty }}{Q}_t(\sigma )dt\mathrm{/}t=I(\sigma )$.

Suppose $I(\sigma )=0$. Then ${\int }_0^{\mathrm{\infty }}{Q}_t(\sigma )dt\mathrm{/}t=0$ by step 4.1, so $Q_t(\sigma )=0$ for almost every $t>0$ and in particular there is $t_0\in [1,2]$ with $Q_{t_0}(\sigma )=0$; by step 1.2 this means $\int \mathrm{\mid }{T}_{t_0}\sigma {\mathrm{\mid }}^{2}dA=0$, so $T_{t_0}\sigma =0$ almost everywhere, and since $z\mapsto T_{t_0}\sigma (z)$ is continuous by [F9] one has $T_{t_0}\sigma \equiv 0$ on $\mathbb{C}$. The functions $z\mapsto \int e^{-2t_0\mathrm{\mid }x-z{\mathrm{\mid }}^2}d\sigma (x)$ and their complex derivatives ${\mathrm{\partial }}_z^j{\mathrm{\partial }}_{\bar{z}}^k$ are continuous on $\{\mathrm{\mid }z\mathrm{\mid }\le 1\}$ by [F8] applied iteratively, with dominating functions bounded on $\mathrm{\mid }z\mathrm{\mid }\le 1$ by a constant times a power of $\mathrm{diam}K+1$, which is integrable against the finite signed measure $\sigma$ carried by the compact set $K$; since $T_{t_0}\sigma \equiv 0$, all these derivatives vanish at $z=0$. Expanding $e^{-2t_0\mathrm{\mid }x-z{\mathrm{\mid }}^2}=e^{-2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}{e}^{2t_0\bar{x}z+2t_{0}x\bar{z}-2t_{0}z\bar{z}}$ shows that the $(j,k)$ derivative at zero is $e^{-2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}$ times $(2t_0{)}^{j+k}{\bar{x}}^j{x}^k$ plus a linear combination of terms ${\bar{x}}^{j-r}{x}^{k-r}$ with $1\le r\le \min (j,k)$. Induction on $j+k$ therefore gives $\int {\bar{x}}^j{x}^k{e}^{-2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}d\sigma (x)=0$ for all $j,k\ge 0$. The monomials ${\bar{x}}^j{x}^k$ span the polynomials in the real variables $\mathrm{Re}x,\mathrm{Im}x$ (equivalently, the polynomials in $x$ and $\bar{x}$), so $\int P(x)e^{-2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}d\sigma (x)=0$ for every such polynomial $P$.

Assume $I(\sigma )=0$ and, seeking a contradiction, $\sigma \mathrm{\ne }0$; by [F10] and $M>0$ the set $K$ is a nonempty compact subset of $\mathbb{C}$, and $\sigma$ is carried by $K$ because $\mu (\mathbb{C}\setminus K)=\nu (\mathbb{C}\setminus K)=0$ by [F10]. For every $g\in C(K,\mathbb{R})$ the function $g(x)e^{2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}$ is continuous on $K$, so by [F7] applied to the polynomials in the real variables $\mathrm{Re}x,\mathrm{Im}x$ — a unital subalgebra of $C(K,\mathbb{R})$ separating points of $K$ — there are such polynomials $P_n$ with $P_n\to g{e}^{2t_0\mathrm{\mid }\cdot {\mathrm{\mid }}^2}$ uniformly on $K$; since $\mathrm{\mid }{e}^{-2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}\mathrm{\mid }\le 1$ on $K$, the integrals of bounded Borel functions against the finite signed measure $\sigma$ are bounded by $\mu (K)+\nu (K)<\mathrm{\infty }$ in absolute value, and $\sigma$ is carried by $K$, step 5.1 gives ${\int }_{K}gd\sigma ={\lim }_n{\int }_K{P}_n(x)e^{-2t_0\mathrm{\mid }x{\mathrm{\mid }}^2}d\sigma (x)=0$. For a proper open $U⊊\mathbb{C}$, the functions ${\psi }_j(x):=\min (1,jd(x,\mathbb{C}\setminus U))$ are continuous with $0\le {\psi }_j\uparrow {\mathbf{1}}_U$; hence $\int {\psi }_{j}d\mu =\int {\psi }_{j}d\nu$, and [F4] gives $\mu (U)=\nu (U)$. Equality also holds for $U=\mathbb{C}$ because both measures have mass $M$. By [F6] the finite Borel measures $\mu ,\nu$ are outer regular, so for every Borel $B$ one has $\mu (B)={\inf }_{U\supseteq B}\mu (U)={\inf }_{U\supseteq B}\nu (U)=\nu (B)$, that is, $\sigma =0$, contradicting $\sigma \mathrm{\ne }0$. Hence $I(\sigma )=0$ forces $\sigma =0$.

Conversely $\sigma =0$ means $\mu =\nu$, hence $I(\mu ,\nu )=I(\mu )$ and $I(\sigma )=I(\mu )-2I(\mu )+I(\mu )=0$ by the definition in step 3.1; combined with step 6.1 this proves the equivalence (3), while the finiteness of $I(\mu ,\nu )$ and the well-definedness of $I(\sigma )$ with $I(\sigma )\ge 0$ were proved in step 3.1 and the representation of $I(\sigma )$ in step 4.1.