# A uniform limit of bi-Lipschitz embedding which preserves total length

I'm trying to verify part of a claim from Freedman-He-Wang's paper on Mobius knot energy, in Prop. 8.3. My question is less about knot energies, and more about the preservation of length in a uniform limit. To be specific, suppose $$\gamma_i: [0,1] \to \mathbb{R}^3$$ is a sequence of rectifiable simple closed curves, each parametrized with respect to arc length and with total length $$1$$. Furthermore, suppose for each $$\varepsilon > 0$$, there is some $$\delta > 0$$ such that each $$\gamma_i$$ is $$(1 + \varepsilon)$$ bi-Lipschitz on subarcs of length $$\leq \delta$$. (This property is a consequence of Lemma 1.2, for knots with finite Mobius energy). Suppose $$\gamma_i$$ converges to a curve $$\gamma$$ uniformly. Must it be the case that $$\gamma$$ also has total length $$1$$?

Showing $$L(\gamma) \leq 1$$ is a straightforward application of Fatou's lemma. However, the uniform limit of curves could lose length. For instance, consider a sequence $$\Gamma_i$$ of loops of length $$1$$ each contained in concentric spheres of radius $$\frac{1}{i}$$. Then the uniform limit is a single point with zero length. This problem doesn't occur under our assumptions, but I'm having trouble showing the arc-length doesn't decrease in the limit.

That is a beautiful paper. Regarding the question: Given $$\epsilon>0$$, consider the corresponding $$\delta$$, and take $$n>1/\delta$$. Then each $$\gamma_i$$ is $$(1+\epsilon)$$-bilipschitz on $$[\frac{k-1}n,\frac{k}{n}]$$. In particular, $$\forall i \ge 1, \quad \forall k=1,2\ldots, n , \quad \text{we have} \quad \Big|\gamma_i(\frac{k}{n})-\gamma_i(\frac{k-1}n)\Big| \ge \frac1{n(1+\epsilon)} \,.$$ Passing to the limit, $$\forall k=1,2\ldots, n , \quad \text{we have} \quad \Big|\gamma (\frac{k}{n})-\gamma (\frac{k-1}n)\Big| \ge \frac1{n(1+\epsilon)} \,.$$ We infer that the length $$L(\gamma)$$ is at least $$\sum_{k=1}^n \Big|\gamma (\frac{k}{n})-\gamma (\frac{k-1}n)\Big| \ge \frac1{ 1+\epsilon } \,.$$ Since $$\epsilon>0$$ is arbitrary, we conclude that $$L(\gamma)=1$$.