Say in $R^3$, is there a sequence of smooth graphs $f_n$ over some plane P, such that the graphs as submanifolds in $R^3$ converge in the sense of varifold (as Radon measures on $R^3 \times Gr(2,3)$ ) to a limit V, which is multiplicity 2 of the plane P? Seems impossible but is there a proof?
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3$\begingroup$ It's possible to converge to a multiplicity 2 plane which is perpendicular to P though. $\endgroup$– SamCommented Jun 13, 2016 at 20:15
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I think that the following argument rules out getting $2[P]$:
Suppose that $v_n : = graph (f_n) \rightharpoonup 2[P]$ as varifolds. Consider the open set $U = \{(x,P') \in \mathbb{R}^3\times Gr(2,3) : |x| < 1, d(P,P') <\epsilon\}$. Then, a standard result for weak convergence of measures implies that $$ \liminf_{n\to\infty} v_n(U) \geq 2[P](U) \qquad (= 2 |B_1(0)|). $$ However, it is easy to bound $v_n(U) \leq |B_1(0)| + O(\epsilon)$ as $\epsilon\to0$, since $v_n(U)$ is the $\mathcal{H}^2$ measure of the set of points in $graph(f_n) \cap B_1(0)$ whose normal vectors differ from that of $P$ by a distance $O(\epsilon)$ (the precise distance we choose on $Gr(2,3)$ is irrelevant). This is a contradiction.
answered Jun 14, 2016 at 12:46
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1$\begingroup$ Thanks. Seems can use this argument to rule out any multiplicity 2 plane not perpendicular to P, for the perpendicular ones we need a gradient bound for the graphs to rule out that. $\endgroup$– SamCommented Jun 14, 2016 at 18:33
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$\begingroup$ But the $\mathcal{H}^2$ measure of the graph is getting close to $2|B_1(0)|$. And more and more of it could have normal vector far from the normal to $P$. For example, isn't it fairly clear you can converge to $\sqrt{2}[P]$ by zig-zagging? $\endgroup$– ThompsonCommented Oct 20, 2016 at 18:17
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1$\begingroup$ @Thompson, The measures might converge but the associated varifolds do not converge as varifolds $\endgroup$ Commented Oct 20, 2016 at 18:38