Prove that there exists a function $f : [0,1]\to [0,1]$ that is continuous and nondecreasing so that the length of the graph of $f$ is at least $2$. The length of the graph of $f$ is the supremum, over all partitions $0=x_0 < x_1<\cdots < x_n = 1$, of the sum $\sum_{i=1}^n \sqrt{(\Delta_i x)^2 + (\Delta_i y)^2}$ where $\Delta_i x = x_i - x_{i-1}$ and $\Delta_i y = f(x_i) - f(x_{i-1})$.
I think the Cantor function $f:[0,1]\to [0,1]$ satisfies the requirement. It is defined as follows. Let $C$ be the Cantor set (equivalently $C$ is the set of numbers that can be written in base $3$ as $0.a_1 a_2\cdots, a_i \in \{0,2\}\,\forall i$). For $x\in C,$ write $x$ in base 3 as $x=0.a_1a_2\cdots, a_i \in \{0,2\}\,\forall i$, and define $f(x) = 0.\frac{a_1}2 \frac{a_2}2\cdots$.
How can one show that for $x,y\in C, x < y, f(x) = f(y)$ only if $x$ and $y$ are of the form $x=0.a_1\cdots a_k 022\cdots, y = 0.a_1\cdots a_k 20000$?
Equivalently, the open interval $(x,y)$ is one of the open middle thirds deleted in the construction of $C$.
$f$ is nondecreasing because if $x = 0.a_1a_2\cdots < y = 0.b_1b_2\cdots $ then there exists a smallest $i$ so that $a_i < b_i$. Then $f(x) = 0.\frac{a_1}2\cdots \frac{a_{i-1}}2 0 \frac{a_{i+1}}2\cdots, f(y) = 0.\frac{a_1}2 \cdots \frac{a_{i-1}}2 1\frac{b_{i+1}}2\cdots $, and so the largest $f(x)$ can be is $0.\frac{a_1}2\cdots \frac{a_{i-1}}2 0 1 1\cdots = 0.\frac{a_1}2\cdots \frac{a_{i-1}}2 1 \leq f(y).$
To show the function is continuous, one can use the Weierstrass M-test with $f_k(0.a_1a_2\cdots) = \frac{a_k}{2^{k+1}}$ for each $k \ge 1$ and $f(x) = \sum_{i=1}^\infty f_k(x)$ (since the convergence is uniform by the Weierstrass M test and the uniform limit of continuous functions is continuous).
Finally, why is it that when we use the partition $0=x_0 < x_1<\cdots < x_n = 1,$ where the $x_i$ are the $2^{k+1}$ endpoints of the $2^k$ closed intervals in $C_k,$ we get $\sum_{i=1}^n \sqrt{(\Delta_i x)^2 + (\Delta_i y)^2} \ge 1 + (\frac{1}3 + \frac{2}9 +\cdots + \frac{2^{k-1}}{3^k}) = 2 - (\frac{2}3)^k \to 2$ as $k\to \infty$?
Perhaps it would be useful to use the fact that the intervals in $C_k$ are of size $\frac{1}{3^k}$ and result from repeatedly removing the open middle thirds of all the intervals in $C_{k-1}$, etc.? I think this is equivalent to saying that the $k$th bit in the ternary expansion of each element of $C_k$ that uses only the digits 0 and 2 must be 2.
