Example 16.14.274EA4AAF7D98

Example 16.14.2.74EA4AAF7D98 Let $n\geq 2$ and $x_0\in X$. Choose holomorphic coordinates $z$ centered at $x_0$ in which $h_{i\bar j}(0)=\delta _{ij}$, which is possible after a complex-linear change of coordinates, and choose $0<r_0<\mathrm e^{-1}$ such that the coordinates are defined on a neighborhood of $\{ |z|\leq r_0\} $. On $0<|z|<1$ put

and $\phi (0)=0$. Write $r=|z|$. Since $\partial _{z_ i}\log r=\bar z_ i/(2r^2)$ and $\partial _{\bar z_ j}(r^2\log r)=z_ j(\log r+\frac12)$, we have

For $0<r<r_0$ we have $\log r<-1$, hence $|\log r+\frac12|\leq |\log r|$, and these formulas give $|\nabla \ell |\leq C/(r|\log r|)$ and $|D^2\ell |\leq C/(r^2|\log r|)$ for the real gradient and Hessian, with a constant $C$ depending only on $n$. Moreover $|P|\leq r^2$, $|\nabla P|\leq 2r$, the real Hessian of $P$ is constant, $P_{1\bar1}=1$, $P_{2\bar2}=-1$, and all other $P_{i\bar j}$ vanish. By the Leibniz rule, on $0<r<r_0$,

In particular $\phi $ is continuously differentiable on $\{ |z|<r_0\} $, with $\nabla \phi (0)=0$, since $|\phi |\leq r^2\ell $ and $\nabla \phi \to 0$ as $z\to 0$; and its pointwise second derivatives off the origin are bounded by $C(1+\ell )\leq C(1-\log |z|)$, hence locally integrable.

These pointwise derivatives are the distributional second derivatives of $\phi $. Indeed, let $\eta $ be a smooth function on $\mathbb R$ with $\eta =0$ on $(-\infty ,1]$ and $\eta =1$ on $[2,\infty )$, put $\eta _\varepsilon (z)=\eta (|z|/\varepsilon )$, and let $\psi $ be a test function on $\{ |z|<r_0\} $. Since $\phi $ is $C^1$, we have $\int \phi \, \partial _ a\partial _ b\psi =-\int \partial _ a\phi \, \partial _ b\psi $ for real coordinate derivatives $\partial _ a,\partial _ b$. Splitting $\psi =\eta _\varepsilon \psi +(1-\eta _\varepsilon )\psi $ and integrating by parts once more in the first term, where $\phi $ is smooth, we get

The last integral is bounded by $\sup _{|z|\leq 2\varepsilon }|\nabla \phi |\cdot C_\psi (\varepsilon ^{-1}+1)\varepsilon ^{2n}$, which tends to zero, and the first tends to $\int \partial _ b\partial _ a\phi \, \psi $ by dominated convergence.

Choose a smooth function $\chi $ on $X$ with support in $\{ |z|<r_0\} $ that equals $1$ on $\{ |z|\leq r_0/2\} $, and put $f=\chi \phi $, extended by zero. Then $f$ is continuous. On $\{ |z|<r_0/2\} $, (16.14.1) and the estimates above give

because $h^{1\bar1}-h^{2\bar2}$ is smooth and vanishes at the origin; the right side is bounded, since $r\ell \to 0$ as $r\to 0$, and elsewhere $f$ is smooth. Hence $f$ has bounded metric Laplacian. On the other hand $f_{1\bar1}=\phi _{1\bar1}\geq \ell -C/|\log r|$ near $x_0$, which is unbounded. Thus the complex Hessian of $f$ is not locally bounded.