Can an arbitary Jordan curve be approximated by a smooth Jordan curve?

Question

Given a Jordan curve $\gamma_1 \colon [a,b] \to \mathbb{C}$ and an $\epsilon > 0$, can we find a continuously differentiable Jordan curve, $\gamma_2 \colon [a,b] \to \mathbb{C}$, such that $ |\gamma_1(t) - \gamma_2(t) | < \epsilon$ for all $ t \in [a,b]$?

Answer 1

Up to reparametrization, yes:

By the Jordan curve theorem and Caratheodory's conformal mapping theorem, we can find a homeomorphism $\varphi$ from $\gamma_1\cup \text{int}(\gamma_1)$ to the closed unit disc such that its restriction to $\text{int}(\gamma_1)$ is a conformal map. Now, let $0<r<1,\hat\gamma_r(t)=re^{i2\pi t}$, and note that $\varphi^{-1}(\hat\gamma_1(t))$ is a reparametrization of $\gamma_1$ . Since $\varphi^{-1}$ is a continuous function on a compact set, it is uniformly continuous. Thus we can find $\delta(\varepsilon)$ such that $|z_1-z_2|<\delta\rightarrow |\varphi(z_1)-\varphi(z_2)|<\varepsilon$, and so if we take $r$ such that $1-r<\delta$, we get that $$\forall_{t\in[0,1]}|\varphi^{-1}(\hat\gamma_1(t))-\varphi^{-1}(\hat\gamma_r(t))|<\varepsilon$$ Since $\varphi^{-1}$ is a conformal map on the open unit disc, $\tilde\gamma_2(t):=\varphi^{-1}(\hat\gamma_r(t))$ is a smooth jordan curve, approximating the reparametrization of $\gamma_1$ as desired.

To get the result withouth reparametrization, note that the reparametrization $f$ is monotone continuous, and as such can be uniformly approximated by monotone bernstein polynomials $f_n$ (see this thread on MO) with fixed endpoints. Since $\tilde\gamma_2(t)$ is uniformly continous, $\tilde\gamma_2(f_n(t))\to \tilde\gamma_2(f(t))$ uniformly. It is easy to see that $\gamma_2(t):=\tilde\gamma_2(f_n(t)))$ is a smooth jordan curve satisfying our requirement.