# Existence of a curve of finite length on the image of an embedding which is Sobolev

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?

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.
• 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? Sep 29 at 6:12