Homeomorphism from $S^1$ to square differentiable?

Question

It is commonly known that $f:S^1\to \{(x_1,x_2)\in\mathbb{R}^2:\max \{|x_1|,|x_2|\}=1 \},\, f(x_1,x_2)=\frac{1}{\max \{|x_1|,|x_2|\}}(x_1,x_2)$ is a homeomorphism from $S^1$ to the unit square. My question is if $f$ is even differentiable and if not if there is a differentiable injective map from $S^1$ to the unit square.

Answer 1

When you ask whether $f$ is differentiable, I presume that you mean differentiable as a function $f : S^1 \to \mathbb R^2$: otherwise the question does not make sense, because the boundary of the square is not a smooth submanifold of $\mathbb R^2$, and hence has no natural smooth structure to which one can apply the question.

The answer to the presumptive question is no: if $f : S^1 \to \mathbb R^2$ were differentiable then, letting $g(x) = \sqrt{1-x^2}$ for $-1<x<1$, the composition $f \circ g(x)$ would be differentiable, but I'm sure you can show that composition is not differentiable at $x=\sqrt{2}$.

But there is indeed a differentiable injection $f : S^1 \to \mathbb R^2$ whose image is the boundary of the unit square, in fact even a smooth injection. The tool you need is a differentiable bijection $$g : [-1,1] \to [-1,1] $$ whose one-sided derivative vanishes at each endpoint. It's not too hard to construct such a beast, and even worse beasts are constructed in analysis textbooks, for example a smooth bijection whose one-sided derivatives of all orders vanish at each endpoint.

You can then use this function $g$ to construct the desired $f$. For example, on the quarter circle of $S^1$ where $-\frac{\sqrt{2}}{2} \le x \le \frac{\sqrt{2}}{2}$ and $y \ge \frac{\sqrt{2}}{2}$ you can define $$f(x,y) = (g(\sqrt{2} \, x),1) $$ Using similar formulas on the other four quarter circles, you can then check that the whole function is differentiable at the four points of $S^1$ where the quarter circles meet in pairs (and it is clearly differentiable at all other points of $S^1$).