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 $\mathbb R^3$?



The limit of a sequence of embedded minimal disks in $\mathbb{R}^3$
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](http://www.ugr.es/~jperez/papers/monograph-book2.pdf)).




---


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](http://www.ams.org/mathscinet-getitem?mr=2153399). 


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](http://www.ams.org/mathscinet-getitem?mr=2153399) paper.


