Suppose that we have an embedding $f:\mathbb{R}^2\to\mathbb{R}^3$ which belongs in the Sobolev space $W^{1,p}_{loc}(\mathbb{R^2},\mathbb{R}^3)$ for some $p>2$. Is it true then that for any two points in $f(\Bbb R^2)$ there exists a curve of finite length connecting them?


1 Answer 1


Without loss of generality, we can assume that the two points are $(0,0)$ and $(1,0)$. We then define the family of curves $\gamma_t(s) = (s, ts(1-s))$ and note that one has $$ \int_0^1\int_0^1|\nabla f(\gamma_t(s))|^p s(1-s)\,ds\,dt = \int_0^1\int_0^{x(1-x)} |\nabla f(x,y)|^p\,dy\,dx \le \|\nabla f\|_p^p < \infty\;. $$ It follows that there exists some $t \in [0,1]$ such that $\int_0^1|\nabla f(\gamma_t(s))|^p s(1-s)\,ds < \infty$. Since $p > 2$, Hölder's inequality then yields a constant $C$ such that $$ \int_0^1 |\nabla f(\gamma_t(s))|\,ds \le C \Big(\int_0^1|\nabla f(\gamma_t(s))|^p s(1-s)\,ds\Big)^{1/p} < \infty, $$ which implies the claim.

  • $\begingroup$ Could you elaborate a bit on why the integral $\int_0^1|\nabla f(\gamma_t(s))|ds$ shows that the length of $\gamma_t$ is finite? We do not know that $f$ is absolutely continuous on $\gamma_t$ so how is that integral related to the length? $\endgroup$
    – Mad Max
    Sep 29 at 6:12

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