# convergence of the mean curvature under $L^\infty$ norm

Suppose that I have a Jordan curve $$J$$ parametrized by the function $$\phi$$. Consider a sequence of parametric functions $$\phi_n$$ parametrizing a sequence of Jordan curves $$J_n$$, and denote by $$H$$ and $$H_n$$ the mean curvature of $$J$$ and $$J_n$$, respectively. Suppose also that these Jordan curves are contained in a large open bounded ball $$B \subset \mathbb{R}^2$$. Is it enough to assume the convergence of $$\phi_n$$ to $$\phi$$ in the $$C^1$$ topology to prove that $$\|H_n - H\|_{L^{\infty}(B)} \to 0$$? I think that with the additional assumptions that the parametric functions satisfy the conditions that their derivatives (up to second order) are bounded, then the convergence follows. Are these the minimum requirements for the said convergence to hold? Any lead or reference dealing with this kind of problem is very much welcome. (If $$\phi_n$$ converges to $$\phi$$ in $$C^2([0,1);\mathbb{R}^2)$$, I think there is no problem with the desired convergence.)

The hypotheses you impose are not strong enough to ensure convergence of the curvature.

This is ultimately a local problem, so to find a counterexample we work with graphs of functions. Let $$(f_n \mid n \in \mathbf{N})$$ be a sequence of functions in $$C^2(-1,1)$$ say with $$\sup \lvert f_n \rvert + \lvert f_n' \rvert \to 0 \text{ and} \sup \lvert f_n'' \rvert \leq 2,$$ but $$\lvert f_n''(0) \rvert \geq 1$$ for all $$n$$.

The (signed) curvature of the graph of $$f_n$$ is $$k_n = (1 + (f_n')^2)^{-3/2} f_n''.$$ By construction we have that for large enough $$n$$, $$\lvert k_n(0) \rvert \geq 1/2.$$

Edit. In fact the following is true: if $$\gamma_n \to \gamma \text{ in C^1} \text{ and }k_n \to k \text{ in L^\infty}$$ then $$\gamma_n \to \gamma \text{ in C^2}.$$

This is essentially for the same reason as above: locally the curves can be written as graphs over a common line. After rotating and translating the graphs, we are reduced to the case where $$f_n,f \in C^2(-1,1)$$ and $$f_n \to f \text{ in C^1}.$$

But then also $$f_n'' = (1 + (f_n')^2)^{3/2} k_n \to (1 + (f')^2)^{3/2} k = f'' \text{ in L^\infty}.$$ The convergence $$\gamma_n \to \gamma$$ in $$C^2$$ follows by piecing together these local arguments.

• Thank you for your comment Leo. My apologies, but can you explain how does the last inequality violates the desired convergence? Does this mean that I need to require the parametrizing functions to be Lipschitz in addition to the convergence of $\phi_n$ to $\phi$ in the C^2 topology? Jul 8 at 12:13
• I'm not sure I understand the question - if the curves are $C^2$ then they must also be Lipschitz. Jul 8 at 12:20
• Sorry, my mistake. I mean, does it need to be that the second-derivatives of the parametrizing functions are (uniformly) Lipschitz? Jul 8 at 12:23
• @JullienneFranz: I think Leo is answering your explicitly stated question, which only assume that $\phi_n$ converges in the $C^1$ topology. (The demonstrated counterexample converges in $C^1$ but not necessarily $C^2$.) I think the summary of the answer is just that "mean curvature is defined based on the second derivative, and so convergence in $C^1$ can never be enough to control it". Jul 8 at 13:41
• @JullienneFranz I second Willie's comment. I've added more details - I hope that will clarify some elements that I'd left open. Does that answer your questions? Jul 8 at 14:55