Example for an integral, rectifiable varifold with unbounded first variation

Question

I'm just looking for an example of an integral, rectifiable varifold, which has no locally bounded first variation.

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Recapitulation

for every $m$-rectifiable varifold $\mu$ exists a $m$-rectifiable set $E$ in $\mathbb R^n$, meaning $E=E_0 \cup \bigcup_{k\in\mathbb N} E_k$ with $\mathcal H^m(E_0)=0$ and $E_k\subseteq F_k$ for some $\mathcal C^1$-manifolds $F_k$ of dimension $m$, and a non-negativ function $\theta\in L^1_{\text{loc}}(\mathcal H^m|_E)$ such that $\mu=\theta \mathcal H^m|_E$. This is a characterisation of $m$-recitifiable varifolds. The first variation $\delta\mu$ of a varifold $\mu$ is for $\eta\in\mathcal C^1_c(\mathbb R^n;\mathbb R^n)$ given by $$\delta\mu(\eta)=\int div_\mu\eta\,d\mu,$$ where $ div_\mu(\eta)(x) = \sum_{i=1}^n \tau_i^T(x)\cdot D\eta(x)\cdot \tau_i(x)$ where $\tau_i(x)$ is a orthogonal basis of the tangentspace of $\mu$ in $x$, which coinsidence $\mu$-almost everywhere with $T_xF_i$ for $x\in E_i\subseteq F_i$ as above. So $div_\mu\eta(x)$ is just the divergence in the manifold $F_i$, with $x\in E_i\subseteq F_i$.

We say $\mu$ has an locally bounded first variation, if for all $\Omega'\subseteq \Omega$ there exists $c(\Omega')<\infty$ such that $$ \delta\mu(\eta) \le C(\Omega',\Omega) \Vert \eta\Vert_{L^\infty(\Omega)} \qquad\forall\;\eta\in\mathcal C^1_c(\Omega'). $$ See for more explanation for example http://eom.springer.de/G/g130040.htm.

For a $\mathcal C^2$-manifold $M$ in $\mathbb R^n$ with mean curvature $H_M$ the first variation is $$ \delta M(\eta)=-\int_M H_M \cdot \eta \,dvol_M -\int_{\partial M} \tau_0 \cdot \eta \,dvol_{\partial M} \qquad\forall\;\eta\in\mathcal C_c^1(\mathbb R^n)$$ with the inner normal $\tau_0\in T_xM\cap(T_x\partial M)^\bot$ and where the mean curvature is the trace of the second fundamental form $A$ by the meaning of $H_M(x)=\sum_{i=1}^m A_x(\tau_i,\tau_i)$ in the normal space of $M$. As obviouse in this case the first variation is locally bounded.

Answer 1

Something is strange here: it seems like for the sawtooth curve (the Lipschitz curve that goes up and down with slope $1$), the first variation is just the sum of $\delta$-measures at the turning points times the unit bisector vectors, so we can have fixed length and arbitrarily large first variation (just make turning points more and more dense), which can be now trivially turned into an example of finite area and infinite first variation: take more and more rigged closed sawtooth curves around infinitely many circles with finite some of radii contained in a compact domain. Am I missing something?

Answer 2

There are several ways to construct a rectifiable varifold which has no locally bounded first variation. To start let me rewrite the first variation for the m-varifold $v = V(E,\theta)$ by $$ \delta\mu(\eta) = \int_E \mathrm{div}_E(\eta)\,\theta d\mathcal{H}^m $$ Supposing for a moment $E$ is a smooth $C^2$ manifold with boundary and $\theta$ a smooth function, we obtain: $$ \delta\mu(\eta) = \int_E [\mathbf{H_E}\cdot \eta\,\theta + \nabla_E\theta \cdot \eta] d\mathcal{H}^{m} + \int_{\partial E} \eta\cdot \tau_0 \,d\mathcal{H}^{m-1} $$ In particular when $E = \{(x,0):\, x\in [a,b]\}$ and $\Omega=\mathbb{R}^2$ we have $$ \delta\mu(\eta) = \int_E \nabla_E\theta \cdot \eta \,d\mathcal{H}^{1} + \eta(a)\cdot (-1,0) + \eta(b) \cdot (1,0) $$ This formula suggests two easy ways of constructing a rectifiable varifold as the one you are looking for:

• take as $\theta$ a function (of one variable) whose derivative is not a Radon-measure, that is a function that is not in $BV((a,b))$, in which case the "first term explodes"

• glue together infinitely many (possibly disjoint) intervals so that "0-dimenstional length of the boundary explodes".