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Let $C$ be a $C^1$ hypersurface in $R^n$ and let $u \in C^1(R^n)$. Suppose

$$\nabla u(x) \cdot \eta(x)=|\nabla u| \ \ \forall x\in C$$

where $\eta(x)$ is the normal vector to $C$ at $x$ ($\nabla u$ is parallel to $\eta$ on $C$). Then one can easily show that $h(t)=u(f(t))$ is constant for every $C^1$ curve lying on the surface $C$, and therefore $u$ must be constant on $C$.

My questions is that how much the assumption $u \in C^1$ can be weakened. Can we reach a similar (and possibly weaker) conclusion if $u$ is only assumed to be in $W^{1,1}$ or $BV$? For instance if we only assume $u\in BV$ and $\frac{Du}{|Du|}\cdot \eta =1$, $|Du|$ a.e. on $C$, then can we conclude that $u^{-1}(C)$ is countable in the trace sense? (Assuming that $\frac{Du}{|Du|} \in (L_c^{\infty}(R^2))^n$ has a trace $T\in L^{\infty}(C)$. See http://www.math.northwestern.edu/~gqchen/10-Papers/ChenTorresZiemer.pdf)

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Some thoughts on your problem. If $u$ is so smooth that $Du$ has a well defined trace, e.g., $u\in H^s$ for $s>3/2$, then you could flatten the surface by a $C^1$ change of variables and reduce to the case $S=\{x_1=0\}$. Then the assumption on $Du$ becomes $Du=(D_1u,0,...,0)$ which means the derivatives of the trace are zero a.e. on the coordinate plane $S$, and hence $u$ is constant there. This seems to work fine for Sobolev functions and maybe for other functional spaces.

For $BV$ functions I am a bit skeptical; you have a well defined $L^1$ trace for $u$, but in what sense would you formulate your assumption on $Du$ restricted to $S$?

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  • $\begingroup$ Under certain conditions, one can make sense of the trace $T\in (L^{\infty}(\partial \Omega))^n$ of $\frac{Du}{|Du|}\in (L^{\infty}(\Omega))^n$. So the assumption would be $T\cdot \eta=|T|$. For instance see: math.northwestern.edu/~gqchen/10-Papers/ChenTorresZiemer.pdf $\endgroup$
    – A random mathematician
    Commented Jun 6, 2016 at 0:06
  • $\begingroup$ Interesting. Then you should at least include these conditions, or assume that the trace is defined, in your question $\endgroup$
    – Piero D'Ancona
    Commented Jun 6, 2016 at 5:35
  • $\begingroup$ The question is revised. $\endgroup$
    – A random mathematician
    Commented Jun 8, 2016 at 6:40

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