Example 15.19.25.F7F939140E90 On every such $(C,h)$ there are smooth Hermitian metrics $g_\varepsilon \leq h$, for $\varepsilon \downarrow 0$, such that
\[ \begin{aligned} A_\varepsilon :=\operatorname {Area}_{g_\varepsilon }(C) & \asymp \varepsilon ^2\log (1/\varepsilon ),\\ \operatorname {diam}(C,g_\varepsilon ) & \asymp \varepsilon \log (1/\varepsilon ). \end{aligned} \]
Consequently the preceding diameter bound has sharp order under its uniform upper bound. Without that bound, area can tend to zero while diameter stays bounded below by a positive constant.
Proof.
Choose a coordinate disk $\{ |z|<R\} $ with $0<R<1$, on whose closure $h=\lambda (z)|\mathrm{d}z|^2$ with $\lambda >0$ smooth. Choose a smooth cutoff $0\leq \chi \leq 1$, supported in this disk and equal to one for $|z|\leq R/2$, and a constant $c>0$ with $c\leq \min \lambda $. For sufficiently small $\varepsilon >0$, set
\[ g_\varepsilon = \left(\chi \frac{c\varepsilon ^2}{\lambda (z)(\varepsilon ^2+|z|^2)} +(1-\chi )\varepsilon ^2\right)h \]
in the disk, and set $g_\varepsilon =\varepsilon ^2h$ outside it. The cutoff makes this a smooth positive metric, and both coefficients in its convex combination are at most one.
On $|z|\leq R/2$ its metric density is $c\varepsilon ^2/(\varepsilon ^2+|z|^2)$, whose area is
\[ \pi c\varepsilon ^2 \log \left(1+\frac{R^2}{4\varepsilon ^2}\right). \]
The remaining area is $O(\varepsilon ^2)$, proving the asserted area comparison. Any path from the origin to a point outside this smaller disk must first cross its boundary. Radial variation is bounded by Euclidean speed, so that initial portion has length at least
\[ \sqrt c\, \varepsilon \int _0^{R/2} \frac{\mathrm{d}r}{\sqrt{\varepsilon ^2+r^2}} =\sqrt c\, \varepsilon \operatorname {arsinh}(R/(2\varepsilon )) \asymp \varepsilon \log (1/\varepsilon ). \]
This gives the diameter lower bound. The upper bound follows from lemma 15.19.24, since $A_\varepsilon (1+\log (1/A_\varepsilon ))\asymp \varepsilon ^2\log ^2(1/\varepsilon )$. Finally rescale by
\[ \widetilde g_\varepsilon =\frac{g_\varepsilon }{\varepsilon ^2\log ^2(1/\varepsilon )}. \]
Its area is comparable to $1/\log (1/\varepsilon )$ and tends to zero, whereas its diameter is comparable to one. At the origin its ratio to $h$ tends to infinity, so it has no uniform upper bound.
$\square$
Pome
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