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Let $B \subset \mathbf{R}^{n+1}$ be the unit ball, and $M \subset B$ be a minimal hypersurface. By this we mean that $M$ is an embedded $n$-dimensional submanifold with vanishing mean curvature. We allow for the closure $\overline{M}$ of $M$ in $B$ to not be embedded, write $\mathrm{sing} \, M = \overline{M} \setminus M$ and call this the singular set of $M$. However this is assumed to be small enough for $M$ to be stationary in $B$: compactly supported deformations $X \in C_c^1(B;\mathbf{R}^{n+1})$ do not change the area of $M$ up to first order. For example one might take $n \geq 2$ and consider a surface $M$ embedded outside the origin with $\mathrm{sing} \, M = \{ 0 \}$.

Question 1. Are there conditions that allow the extension of $M$ to a globally defined immersed minimal hypersurface in $\mathbf{R}^{n+1}$? That is, when is there a minimal hypersurface $\tilde{M}$ in $\mathbf{R}^{n+1}$ (immersed away from a small singular set) with $\tilde{M} \cap B = M$?

Let me make some remarks summarising my own conclusions.

  • The Cauchy--Kovalevskaya theorem could be relevant, but I am not sure whether this can be used to construct a globally defined extension. Moreover, one would have to worry about pieces coming together and meeting tangentially.
  • This is not a purely PDE-theoretic question. If one considers the case where $M$ is the graph of a smooth function $u$ defined on $\mathbf{R}^n \cap B$---this function satisfies the (quasi-linear) minimal surface equation---then it is not hard to see that $u$ can in general not be extended to a globally defined function $\tilde{u}: \mathbf{R}^n \to \mathbf{R}$. The Bernstein theorem is one way to see this, but also simple examples can be constructed using a suitable portion of the catenoid.
  • One can use the unique continuation property of minimal surfaces against the question, by taking $M$ to be a portion of a known surface. For example, by taking $M$ to be an embedded portion of an immersed minimal surface $\tilde{M}$ one can see that one cannot hope for a globally defined and embedded minimal extension. Moreover, if one chooses $M$ to be portion of a singly-periodic Scherk surface one sees that $\tilde{M}$ may have unbounded area growth: $\lim_{R \to \infty} \mathcal{H}^n(\tilde{M} \cap B_R)/R^n = \infty$.

I am especially interested in the case where $M$ is one of the surfaces constructed by Caffarelli--Hardt--Simon. These are defined in $B$, embedded outside the origin, where they are prescribed to be tangent to a given minimal cone $\mathbf{C}$.

Question 2. How does the answer change if $M$ is one of those surfaces? Is there $\tilde{M}$ extending $M$, perhaps even with bounded area growth, that is with a constant $C > 0$ so that $\mathcal{H}^n(\tilde{M} \cap B_R) \leq C R^n$ for all radii $R > 1$?

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    $\begingroup$ What does "globally defined" mean? A proper immersion? If so then one can use Nadirashvilli's examples to see that one can't even ensure this. In any event I doubt there is any checkable condition. One sometimes sees real analyticity up to the boundary to extend a little bit, but that is far from what you are asking. $\endgroup$
    – RBega2
    Dec 19, 2020 at 17:07
  • $\begingroup$ @RBega2 Thank you for the comment; I had forgotten about Nadirashvili's surfaces. I was a bit vague in my statement, and avoided the term 'proper immersion' because I was worried that the small singular set might make this fail on a technicality. I wanted to avoid using too many terms from GMT, but to be precise 'globally defined' would mean something like the following: the surface $\tilde{M}$ has locally finite area densities, and defines a stationary varifold in $\mathbf{R}^{n+1}$. I guess the local finiteness would work like a GMT-analog of properness. $\endgroup$
    – Leo Moos
    Dec 19, 2020 at 17:24
  • $\begingroup$ @RBega2 In any case you're right the Nadirashvili example is troublesome, I'm happy to concede. By rescaling and Allard regularity you can have $M$ arbitrarily close to the plane which does not admit 'global' extensions. However I'll leave the question up, in the hope that somebody might comment on Question 2 in particular. $\endgroup$
    – Leo Moos
    Dec 19, 2020 at 17:31

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