Let $f: [0,1] \to \mathbb R$ and $\Gamma(f)$ its graphic. Show that $\mathcal H^1(\Gamma(f)) \geq 1$.

Question

Problem: Let $f: [0,1] \to \mathbb R$ and $\Gamma(f)$ its graphic. Show that $\mathcal H^1(\Gamma(f)) \geq 1$.

Attempt: Well, if $f \equiv 0$ we get 1. Provided some sort of goodness like $f \in C^1$ we can draw the line between $f(0)$ and $f(1)$ and use triangular inequality since $H^1(\Gamma(f)) = length(\Gamma(f))$. By the way it looks like we don't need any kind of property on $f$. That seems weird to me!

Thanks!

Answer 1

Consider the standard coordiante system $(x,y)$ for $\mathbb{R}^2 \supset \Gamma(f)$.

Let $P$ be the projection onto the $x$-axis, so that for any covering $C = \bigcup_{i} C_i$ of $\Gamma(f)$, we have that $P(C)=\bigcup_i P(C_i)$ is the projection of that covering onto the $x$-axis.

Clearly, $[0,1] \subset P(C)$. Moreover, for any set $S \subset \mathbb{R}^2$ we have that $\text{diam}\big(P(S)\big)\leq \text{diam}(S)$. Can you see how this solves the problem?