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It is a well known result of Alexandrov that the only compact, connected, constant mean curvature surface is the ball. There is a generalized notion of curvature known as generalized mean curvature which makes sense on rectifiable varifolds. In particular if $V$ is a recifiable varifold than the generalized mean curvature is defined as follows:

Let $\|\delta V\|$ denote the first variation of $V$. Then if this measure is absolutely continuous with respect to $\|V\|$ then by the Riesz representation theorem:

$ \delta V(g) = - \int g \cdot H d\|V\|$, for $H$ defined $\|V\|$ a.e.

  • H as defined here is the generalized mean curvature.

Question: Working in the plane, $\mathbb{R}^2$, assume that $\Omega$ is a set of bounded variation so $|\partial \Omega| < C < +\infty$ and define $H$ as above.

  • If $H=$ constant, what does this imply about $\Omega$? Must $\Omega$ be a ball? Are there examples of singular $BV$ sets which satisfy this condition but are not balls?

  • What is the generalized mean curvature of a square?

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I don't know what a rectifiable varifold is, so I can't answer your questions rigorously. However, I doubt very much that there is any pathology in the plane. Also, using the principle that curvature is the rate of change of the angle of the tangent (or normal) with respect to arclength, the (generalized) mean curvature of the square should be equal to zero on the sides and $\pi/4$ times the delta function at the corners. Moreover, if you define the generalized mean curvature as a measure using the variation of arclength, then that's exactly what you'll get. But this is not absolutely continuous with respect to arclength, so the square does not meet the assumptions for your definition of generalized mean curvature.

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