Complete metric on $\mathcal C([0,1],\mathbb S^1)$

Question

On $\mathcal C([0,1],\mathbb R^n)$ we have for example a norm, namely $\sup_{s\in [0,1]}\|f(s)-g(s)\|_{\mathbb R^n}$ that make $\mathcal C([0,1],\mathbb R^n)$ complete. Is there such a metric on $\mathcal C([0,1],\mathbb S^1)$, the space of continuous function that map $[0,1]$ on the unit circle ? My idea would be to consider $g$ being the tensor metric of $\mathbb S^1$, and thus, to consider $\sup_{s\in [0,1]}\|f(s)-g(s)\|_g$ where $\|f(t)\|_g:= g(f(t),f(t))$. But in somehow, I have the impression that it doesn't make so much sense, since depending on which chart we are, the quantity $\sup_{0\leq s\leq 1}\|f(s)\|_g$ may be not well defined... Any idea ?

Answer 1

The $\sup$-metric (let us call it $d$) still works: Let $(f_n)_{n \in \mathbb{N}} \subseteq \mathcal{C}([0, 1], \mathbb{S}^1)$ be a Cauchy-Sequence in the $\sup$-metric. Since $\mathcal{C}([0, 1], \mathbb{R}^n)$ is complete, there exists some $f \in \mathcal{C}([0, 1], \mathbb{R}^n)$ such that $$ \sup_{s \in [0, 1]} \lVert f_n(s) - f(s) \rVert = d(f_n, f) \overset{n \rightarrow \infty}{\longrightarrow} 0. $$ Therefore $f_n$ converges pointwise, which implies that $\lVert f_n(x) \rVert$ converges to $\lVert f(x) \rVert$ for all $x \in [0, 1]$. By assumption, for arbitrary $x \in [0, 1]$ we have $$ 1 = \lVert f_n(x) \rVert $$ and as $n \rightarrow \infty$ $$ 1 = \lVert f(x) \rVert. $$ So, $f \in \mathcal{C}([0, 1], \mathbb{S}^1)$ and we are done.

In general: If $(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $K \subseteq X$ is compact, then $\mathcal{C}(K, Y)$ is complete w.r.t. to the metric $\displaystyle d(f, g) := \sup_{x \in K} d_Y(f(x), g(x))$ iff $Y$ is complete.