Difference between uniform and Hausdorff convergence of curves in $\mathbb{R}^2$

287 Views Asked by At

I'm specifically looking at this in the context of a continuous closed curve $u:\mathbb{S}^1\rightarrow\mathbb{R}^2$ and a sequence of continuous closed curves $\{u_n\}$ converging in some sense to $u$. Hausdorff convergence seems to be the more intuitive notion for me, defined for two curves $u$ and $v$ by \begin{equation} d_H(u,v)=\inf\{\epsilon>0:u\subseteq\mathcal{N}_\epsilon(v),v\subseteq\mathcal{N}_\epsilon(u)\} \end{equation} where $\mathcal{N}_\epsilon$ is an open $\epsilon$-neighbourhood.

In this way, $u$ and $v$ are close together if every point in $u$ is close to some point in $v$ and vice versa. There is no reference to parametrisation here. But what is uniform convergence of $u_n$ to $u$ in this case? Does this require a parametrisation of each $u_n$ so that we can say \begin{equation} \sup_{x\in\mathbb{S}^1}||u(x)-u_n(x)||\rightarrow0 \end{equation} as $n\rightarrow\infty$? Or is there some way of describing uniform convergence without reference to a parametrisation, in a similar vein to Hausdorff convergence? Does one imply the other? Thanks

1

There are 1 best solutions below

0
On

Uniform convergence implies Hausdorff (this is immediate) but not the other way around (even for Jordan curves). For example, let $C$ denote the unit circle. For each $n$ take a Jordan curve $C_n$ within Hausdorff distance $\le 1/n$ from $C$ such that $C_n$ winds along $C$ in the positive direction $n$ times and then backtracks and winds in the negative direction $n$ times before closing.