Lower semicontinuity of logarithmic potential and energy

Let $K$, ${\mu }_n$, $\mu$ and $\Rightarrow$ be as given, fix $R>\mathrm{diam}K$, and set $k(z,w):=\log \frac{1}{\mathrm{\mid }z-w\mathrm{\mid }}$, $k_R:=k+\log R$, $U^{\nu }(z):=\int k(z,w)d\nu (w)$ and $I(\nu ):=\iint kd\nu \otimes d\nu$ in the conventions of [F1]; then $I(\nu )=\iint k_{R}d\nu \otimes d\nu -\log R$ for every Borel probability $\nu$ on $K$.

The finite sums ${\sum }_i{f}_i(z)h_i(w)$ of continuous functions $f_i,h_i$ on $K$ form a unital subalgebra of $C(K\times K,\mathbb{R})$ separating points, so it is uniformly dense by [F7], since $K\times K$ is a nonempty compact metric space.

The shifted kernel $k_R(z,w)=\log \frac{R}{\mathrm{\mid }z-w\mathrm{\mid }}$ is Borel, nonnegative on $K\times K$, and equal to $+\mathrm{\infty }$ exactly on the diagonal.

If nonnegative measurable functions $f,g$ agree off a null set $E$, then $\int f={\int }_{X\setminus E}f+{\int }_{E}f={\int }_{X\setminus E}g+0=\int g$ by [F4]; hence, for each fixed $z$, Borel functions of $w$ agreeing with $k(z,w)$ $\mu$-almost everywhere produce the same value of $U^{\mu }(z)$, and Borel kernels agreeing with $k$ $(\mu \otimes \mu )$-almost everywhere produce the same energy.

Fix $z\in \mathbb{C}$ and choose $R_z>{\sup }_{w\in K}\mathrm{\mid }z-w\mathrm{\mid }$; for $M>0$ the truncation ${\phi }_M(w):=\min \{M,\log \frac{R_z}{\mathrm{\mid }z-w\mathrm{\mid }}\}$ is continuous on $K$ and bounded by $M$, because it equals $M$ near $w=z$ and is a minimum of continuous functions elsewhere, and $U^{\nu }(z)\ge \int {\phi }_{M}d\nu -\log R_z$ for every Borel probability $\nu$ on $K$.

For such a finite sum $h={\sum }_i{f}_i{h}_i$, [F9] and [F10] give $\iint hd{\mu }_n\otimes d{\mu }_n={\sum }_i(\int f_{i}d{\mu }_n)(\int h_{i}d{\mu }_n)\to {\sum }_i(\int f_{i}d\mu )(\int h_{i}d\mu )=\iint hd\mu \otimes d\mu$.

Since ${\phi }_M$ is bounded and continuous, the weak convergence ${\mu }_n\Rightarrow \mu$ gives $\int {\phi }_{M}d{\mu }_n\to \int {\phi }_{M}d\mu$, hence ${\mathrm{lim\hspace{0.17em}inf}}_n{U}^{{\mu }_n}(z)\ge \int {\phi }_{M}d\mu -\log R_z$ for every $M$.

For $M>0$ put $g_M(z,w):=\min \{M,k_R(z,w)\}$; it is continuous on $K\times K$, bounded by $M$, and $I(\nu )+\log R=\iint k_{R}d\nu \otimes d\nu \ge \iint g_{M}d\nu \otimes d\nu$ for every Borel probability $\nu$ on $K$.

Given $\epsilon >0$, step 1.2 provides $h$ with $\mathrm{\mid }{g}_M-h\mathrm{\mid }\le \epsilon$ on $K\times K$ and hence $\mid \iint g_{M}d{\mu }_n\otimes d{\mu }_n-\iint g_{M}d\mu \otimes d\mu \mid \le 2\epsilon +\mid \iint hd{\mu }_n\otimes d{\mu }_n-\iint hd\mu \otimes d\mu \mid$, so step 2.4 makes the left side tend to $0$: $\iint g_{M}d{\mu }_n\otimes d{\mu }_n\to \iint g_{M}d\mu \otimes d\mu$.

As $M\to \mathrm{\infty }$ one has ${\phi }_M\uparrow \log \frac{R_z}{\mathrm{\mid }z-w\mathrm{\mid }}$ pointwise, so [F6] gives $\int {\phi }_{M}d\mu \uparrow \int \log \frac{R_z}{\mathrm{\mid }z-w\mathrm{\mid }}d\mu (w)=\log R_z+U^{\mu }(z)$ in the extended sense; combining with step 3.1 yields $U^{\mu }(z)\le {\mathrm{lim\hspace{0.17em}inf}}_n{U}^{{\mu }_n}(z)$ for every $z\in \mathbb{C}$.

Therefore ${\mathrm{lim\hspace{0.17em}inf}}_n(I({\mu }_n)+\log R)\ge \iint g_{M}d\mu \otimes d\mu$ for every $M$; since $g_M\uparrow k_R$ pointwise, [F6] gives $\iint g_{M}d\mu \otimes d\mu \uparrow \iint k_{R}d\mu \otimes d\mu =I(\mu )+\log R$, so $I(\mu )\le {\mathrm{lim\hspace{0.17em}inf}}_{n}I({\mu }_n)$.

If $\mu (\{z\})>0$, the diagonal value of the kernel gives $U^{\mu }(z)\ge k(z,z)\mu (\{z\})=+\mathrm{\infty }$; for $\mu ={\delta }_z$, replacing $k(z,z)=+\mathrm{\infty }$ by a finite value $c$ changes $U^{\mu }(z)$ from $+\mathrm{\infty }$ to $c$, so the diagonal value cannot be assigned freely for the class of atomic measures.