Example 17.4.1127DF9596E862

Example 17.4.11.27DF9596E862 The energy is not continuous for $L^1$ convergence, even for smooth potentials on $\mathbb P^1$. Let $\omega =\mathrm{dd}^{\mathrm{c}}\log (1+|z|^2)$, which has mass one. For $0<a<1$ put $\varepsilon _ a=\exp (-a^{-2})$ and

These functions extend smoothly across infinity, and $\omega +\mathrm{dd}^{\mathrm{c}}u_ a=(1-a)\omega +a\mathrm{dd}^{\mathrm{c}}\log (|z|^2+\varepsilon _ a)>0$. They tend to zero in $L^1$ as $a\downarrow 0$, since $a\log (|z|^2/(1+|z|^2))\leq u_ a\leq 0$ and the left side is $a$ times a fixed integrable function.

We compute their Dirichlet energies. Write $r=\log |z|^2$ and $F(r)=\log (\mathrm e^ r+\varepsilon )-\log (1+\mathrm e^ r)$ with $\varepsilon =\varepsilon _ a$, so that $u_ a=aF(\log |z|^2)$. For a smooth function $g=G(\log |z|^2)$ we have $\partial g=G'(r)\, \mathrm dz/z$, hence $\mathrm dg\wedge \mathrm d^{\mathrm c}g=\frac{\mathrm i}{2\pi }\partial g\wedge \bar\partial g=\pi ^{-1}G'(r)^2|z|^{-2}\, \mathrm dA$, where $\mathrm dA$ is Lebesgue measure; integration in polar coordinates gives $\int \mathrm dg\wedge \mathrm d^{\mathrm c}g=\int _{-\infty }^\infty G'(r)^2\, \mathrm dr$. Therefore

The bound $O(1)$ is uniform in $a$. Indeed, we have

On $[\log \varepsilon ,0]$ the integral of $\varepsilon \mathrm e^{-r}+\mathrm e^ r$ is at most $2$, and $1\geq F'^2\geq 1-2(1-F')$, so $\int _{\log \varepsilon }^0F'^2\, \mathrm dr=-\log \varepsilon +O(1)$. On the left tail $F'(r)\leq \mathrm e^ r/\varepsilon $, and on the right tail $F'(r)\leq \mathrm e^{-r}$; the corresponding integrals of $F'^2$ are at most $\frac12$. Since $\int _ Xu_ a\, \omega \to 0$, the formula of example 17.3.16 gives $E_\omega (u_ a)\to -\frac12$, whereas $E_\omega (0)=0$.