Why can't a $C^1$-class mapping with nonzero derivative fill a square?

Question

Let $f: [0,1] \rightarrow \mathbb R^2$ be of class $C^1$ with $f'(t)\neq (0,0)$. Why can't $f$ be a Peano-type curve, i.e. $f(I) \neq I\times I$?

Answer 1

There's a higher-dimensional version of the Mean Value Theorem which tells you that $\|f(s)-f(t)\| \le M|s-t|$, where $M = \max\limits_{u\in [s,t]} \|f'(u)\|$. Using this you can prove that you can cover the image of $f$ with rectangles of arbitrarily small total area.

Let me know how you progress.

Answer 2

If $f : [0,1] \to \mathbb{R}^2$ is $C^1$ then $f$ is lipschitz, so $$\dim_H f([0,1]) \leq \dim_H [0,1]=1< 2 = \dim_H [0,1]^2,$$

where $\dim_H$ denotes the Hausdorff dimension.