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.)



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


