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so let $T=\tau(M,\theta,\xi)$ a rectifiable current in $U\subset\mathbb{R}^{n+k}$, then $T$ has mean curvature vector $H$ if for every $X\in C^1_c(U\setminus\partial T,\mathbb{R}^{n+k})$ the variation $$\int \operatorname{div}_M X \; d\mu_T = \int X \cdot H \;d\mu_T\quad (\ast)$$ holds.

There are different integrands one can consider to get such currents. My question is: If one calculates the first variation and get an expression like $$\Big|\int \operatorname{div}_M X\; d\mu_T\Big| \le \lambda \int |X\wedge \xi| \; d\mu_T$$

why follows the existance of a vector $H$ s.t. $|H|\le\lambda$, $H\in (T_xM)^\perp$ and the variation $(\ast)$ holds.

Thanks a lot!

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  • $\begingroup$ What do you mean by $|H|\leq \lambda$ ? Is that an $L^\infty$ bound ? Your condition will probably imply the existence of a vector valued measure that would act as the mean curvature vector, but I am not sure under which condition this would give a genuine vector valued map defined on the support of $T$. $\endgroup$
    – Thomas Richard
    Commented May 21, 2016 at 11:45
  • $\begingroup$ I'm not sure. This is basically from Frank Duzaar's and Klaus Steffen's Paper "$\lambda$ Minimizing Currents". They write: There exist a vector field $H$ on $\mathbb{R}^{n+k}$, s.t. $H$ is $\mu_T$ measurable and the conditions I wrote above. So maybe $|H|$ means $\sup_{x\in\mathbb{R}^{n+k}} \sqrt{H_1(x)^2+\ldots+H_{n+k}(x)^2}$ ?? $\endgroup$
    – leander
    Commented May 22, 2016 at 7:44

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