Hi,
i have a sequence of immersed disc $u\_n: \mathbb{D} \rightarrow \mathbb{R}^3$ which converge to a singular cover of the disc: $z^k$ for $k\geq 2$, moreprecisely $u\_n \rightarrow z^k$ in $C^2(\mathbb{D})$. Of course the Gauss curvature of the image $\Sigma\_n=u\_n(\mathbb{B})$ blows up thanks to Gauss-Bonnet formula : $\int\_{\Sigma\_n}K= (1-k)2\pi + o(1)$. My questions are the following:


1) can the Gauss curvature be bounded form above? i.e the blow-up come only from necks and there is no pinching region... my feeeling is no, since you have to close the surface.


2) Extra bonus: same question with only a convergence in $C^2\_{loc}(\mathbb{D}\setminus \{ 0\})$, here we allow the closing of the surface be made by a big sphere for example.



degenerating surface
When $k$ is odd, there does exist such a family of immersions satisfying Paul's requirements. (When $k$ is even, Vitali Kapovich has shown, using a clever topological argument, that it's not possible to have such a family of degenerating immersions. Please see his answer for the details.)


Set $k=2m+1$, and consider the (complex) $1$-parameter family of maps $u\_t:\mathbb{C}\to\mathbb{R}^3$ given by
$$
u\_t(z) = \bigl(Re(z^{2m+1}-(2m{+}1)t^2z),\ Im(z^{2m+1}+(2m{+}1)t^2z),\ 
 \tfrac{4m+2}{m+1} Re(t z^{m+1})\ \bigr).
$$
These smooth maps converge smoothly to $u\_0$ as $t\to0$,
and $u\_t$ induces the metric 
$$
ds\_t^2 = (2m{+}1)^2\bigl(|z|^{2m}+|t|^2\bigr)^2 |dz|^2.
$$
Thus, $u\_t$ is an immersion for $t\not=0$,
while $u\_0(z) = \bigl(Re(z^{2m+1}),\ Im(z^{2m+1}),\ 0\ \bigr)$.


The family $u\_t$ was constructed using the Weierstrass formula for minimal immersions, so the image $u\_t({\mathbb{C}})\subset \mathbb{R}^3$ is an immersed minimal surface, and, as a result, the Gauss curvature is everywhere non-positive. In fact, for $t\not=0$, the Gauss curvature only vanishes at $z=0$, and then only when $m>1$. In particular, all of these degenerating immersions have curvature bounded above.


