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Let $\Sigma_n,n\ge 1$ be a sequence of embedded minimal disks in $\mathbb{R}^3$ such that:

(1) $0\in\Sigma_n\subset B(0,r_n)$ with $r_n\to\infty$ as $n$ tend to $\infty$,

(2) $\partial\Sigma_n\subset\partial B(0,r_n)$,

(3) $\left|K_{\Sigma_n}(p)\right|\le 1$ for all $p\in\Sigma_n,n\ge 1$ where $K_{\Sigma_n}$ is the Gauss curvature of $\Sigma_n$.

It is standard that if the surfaces $\Sigma_n$ have local uniform area bound then there exists a subsequence of $\Sigma_n$ that converges to a complete minimal surface in $\mathbb R^3$.

In the case that the surfaces $\Sigma_n$ don't have local uniform area bound: does this sequence have a subsequence of $\Sigma_n$ that converges to a complete minimal surface in $\mathbb R^3$?

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  • $\begingroup$ Curvature bound implies volume bound by computing the change of Jacobian. (Standard volume comparison with hyperbolic model.) $\endgroup$
    – Chih-Wei Chen
    Commented Jul 8, 2014 at 16:30
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    $\begingroup$ @Chih-WeiChen, your remark shows that the intrinsic volumes are bounded, but not the extrinsic volumes which is what the OP meant $\endgroup$
    – Otis Chodosh
    Commented Sep 21, 2014 at 17:03
  • $\begingroup$ @OtisChodosh Thanks for your comment! I was not aware that the radius $r_n$ is measured by the extrinsic distance - an embarrassing mistake. $\endgroup$
    – Chih-Wei Chen
    Commented Jul 17, 2019 at 18:21

1 Answer 1

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A first answer is (under much weaker hypothesis than you require)

Answer 1: If $\Sigma_k$ are a sequence of minimal surfaces with $\partial\Sigma_k \subset B(0,r_k)$ with $r_k\to\infty$ and $|K_{\Sigma_k}|\leq C(S)$ for each compact set $S \subset B(0,r_k)$, then we may pass to the limit as an immersion.

Answer 2: If each $\Sigma_k$ are embedded, we may alternatively pass to the limit as a lamination.

Sketch:

Because the surfaces are minimal (so $2K = -|B|^2$), Gauss curvature bounds are equivalent to second fundamental form bounds. Now, it is a well known fact that a sequence of minimal surfaces with second fundamental form bounds has a subsequence which converges as an immersion (basically the Cheeger--Gromov version of convergence of maps).

See http://arxiv.org/pdf/1006.5697v4.pdf for a proof and recall that if a minimal surface has bounded second fundamental form, then all covariant derivatives are pointwise bounded (second fundamental form bounds imply that locally the surface may be written as a graph of bounded size and $C^2$ norm, the graph satisfies the minimal surface equation, so we can get higher derivative bounds from elliptic PDE, which imply bounds on the derivatives of the second fundamental form).

Given embeddedness, it is possible to patch together the immersion limits to obtain a subsequence of the $\Sigma_k$ which converges to a lamination of $\mathbb{R}^3$ (see, e.g., Definition 4.1.4 here).

The drawback to immersion/lamination limits is that they may not be proper. Because you have specified rather strong conditions, it is actually possible to say more here.

Answer 3: If in addition to the assumptions in Answer 2, we assume that the $\Sigma_k$ have uniformly bounded genus, then the limit will be properly embedded.

This is proven in Lemma 1.5 here.

Note that in this case, you can already conclude convergence in the locally graphical sense, as if you had area bounds on the original sequence.

Answer 4: Actually, under the full hypothesis of your question, you can identify the limit: a subsequence will either converge to a plane or a helicoid in the locally graphical sense.

This follows from the main result of this paper.

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