A strictly monotone, continuous path with maximal length

Question

If $f:[0,1]\to [0,1]$ is an increasing continuous function such that $f(0)=0$ and $f(1)=1$, then the length of the graph $\{(t,f(t))|t\in [0,1]\}$ is bounded above by $2$. This is easily proven by considering an arbitrary partition $$0=t_0<t_1<\cdots < t_n=1$$ and noting that $$\sum_{k=1}^n\sqrt{(t_k-t_{k-1})^2+(f(t_k)-f(t_{k-1}))^2}\leq\sum_{k=1}^n|t_k-t_{k-1}|+|f(t_k)-f(t_{k-1})|=1+f(1)-f(0)=2$$

A somewhat less trivial argument shows that for the Cantor function there is an equality: the length of the graph is exactly equal to $2$. The Cantor function is not strictly monotone. In fact, it is constant on every open interval that was removed in the process of constructing the Cantor set.

By considering graphs of the form above, it is clear that for every $\varepsilon>0$ there exists a strictly monotone continuous function such that $f(0)=0$ and $f(1)=1$, whose graph has length larger than $2-\varepsilon$.

Question: Is there a strictly monotone continuous function with $f(0)=0$ and $f(1)=1$ whose graph has length $2$?

Answer 1

If a real-valued monotone function $f$ on $[a,b]$, where $a<b$ satisfies $$\text{length}(\text{graph}(f))=|b-a|+|f(b)-f(a)|,$$ call $f$ a function with the longest graph or a long function on $[a,b]$. If furthermore $f$ is continuous and strictly monotone, call $f$ extremely-monotone.

The question asks whether there are extremely-monotone functions from $[0,1]$ onto itself.

This answer will present several such functions.

An elementary example

Let $f_0(t)=t$ for $t\in[0,1]$. The graph of $f_0$ is the line segment that connects $P_{0,0}=(0,0)$ and $P_{0,1}=(1,1)$ with slope $1=\frac1{2^0}$.

Suppose we have defined $f_k(t)$ that is the piecewise linear function whose graph connects points $P_{2^{2k},0}=(0,0), P_{2^{2k},1},\cdots, P_{2^{2k},2^{2k}}=(1,1)$ such that the slopes of all line segments from left to right alternate between $\frac1{2^k}$ and $2^k$. Furthermore, the increase of both $t$ and $f_k(t)$ on each line segment is $\le\frac1{2^k}$ (where the equality happens only when $k=0$, a fact we do not need though).

The curve in black, purple, orange and green is the graph of $f_0, f_1, f_2, f_3$ respectively.

Let $P_{2^{2(k+1)},2^{2(k+1)}}=P_{2^{2k},2^{2k}}=(1,1)$. For all $0\le i<2^{2k}$,

• let $P_{2^{2(k+1)},4i}=P_{2^{2k},i}\,$,
• let $P_{2^{2(k+1)},4i+2}$ be the midpoint between $P_{2^{2k},i}$ and $P_{2^{2k},i+1}\,$,
• let $P_{2^{2(k+1)},4i+1}$ be the intersection of the line through $P_{2^{2(k+1)},4i}$ with slope $\frac1{2^{k+1}}$ and the line through $P_{2^{2(k+1)},4i+2}$ with slope $2^{k+1}$.
• let $P_{2^{2(k+1)},4i+3}$ be the intersection of the line through $P_{2^{2(k+1)},4i+2}$ with slope $\frac1{2^{k+1}}$ and the line through $P_{2^{2(k+1)},4i+4}$ with slope $2^{k+1}$.

Define $f_{k+1}(t)$ as the piecewise linear function whose graph connects points $P_{2^{2(k+1)}, 0}$, $P_{2^{2(k+1)}, 1}$, $\cdots$, $P_{2^{2(k+1)}, 2^{2(k+1)}}$. The slopes of all line segments of the graph of $f_{k+1}(t)$ from left to right alternate between $\frac1{2^{k+1}}$ and $2^{k+1}$. Furthermore, the increase of both $t$ and $f_{k+1}(t)$ on each line segment is $<\frac1{2^{k+1}}$.

Define $f(t)=\lim_{k\to\infty}f_k(t)$. Since $f_{k+1}(t)\le f_k(t)\le f_{k+1}+\frac1{2^{k+1}}$ for all $k$, $f(t)$ is well-defined and continuous. Since $f_k(t)$ is increasing for all $k$, $f(t)$ is non-decreasing.

Since all endpoints of all line segments of $(t, f_k(t))$ are also the endpoints of some line segments of $(t, f_{k+1}(t))$, $\ (t,f(t))$ passes all the endpoints of all line segments of $(t, f_k(t))$ for all $k$.

• In particular, $f(0)=0$, $f(1)=1$ and $$\text{length}(\text{graph}(f))\ge\text{length}(\text{graph}(f_k))=2\sqrt{1-\frac{2^{k+1}}{(1+2^k)^2}},$$ which goes to $2$ when $k$ goes to $\infty$. Hence $\text{length}(\text{graph}(f))=2$.
• Since $f_k(t)$ is strictly increasing at the endpoints $P_{2^{2k},0}=(0,0), P_{2^{2k},1},\cdots, P_{2^{2k},2^{2k}}=(1,1)$, where the $t$-coordinates of each adjacent pair of points $\le\frac1{2^k}$, which goes to $0$ when $k\to\infty$, $f(t)$ is strictly increasing everywhere.

Two advanced examples

Claim: Let $f$ be a real-valued function on $[a,b]$, $a<b$. $f$ is extremely-monotone iff it is a strictly-monotone continuous function whose derivative is $0$ almost everywhere.
Proof: This follows from Lebesgue's theorem that says a monotone function is differentiable almost everywhere.

A non-constant continuous function is called singular if its derivative is $0$ almost everywhere. So what we are looking for is none other than a singular strictly-monotone function.

This post has constructed a strictly-increasing singular function on $\Bbb R$. So did this post. From such a function we can get a strictly-increasing singular function from $[0,1]$ onto itself by simple linear transformation.

These examples are deemed advanced because it is not very elementary to prove either the claim above or the fact that the derivative is $0$ almost everywhere.

On the other hand, it is straightforward to prove the sum of two long functions is still a long function. Hence, the example function that comes from the sum of countably many shrunken Cantor functions can be proved to be long easily. It can be considered an elementary example as well.

Thanks to comments on the question by Martin R and Dave L. Renfro that guide me to the ideas in this answer. There are a wealth of papers, articles, posts on singular functions and singular strictly-monotone functions.