Number of continuous $[0; 1] \to [0; 1]$ functions for given arc length

Question

Just out of pure curiosity ...

Suppose I want to connect the two points $(0,0)$ and $(1,1)$ with the graph of some continuous and differentiable function

$$f : [0; 1] \to [0; 1]$$

and let $s$ be the arc length of that function in $[0; 1]$.

Of course, the function with minimum $s$ that satisfies the above conditions is $f(x) = x$ with $s = \sqrt 2$. So for $s = \sqrt 2$, exactly one matching function can be found.

But what happens to the number of these functions if $s$ increases?

Surely, more functions can be found to match the given arc length - uncountably many more I suppose due to the nature of the real numbers.

But intuitively, I'd think that the number of such functions grows even more the greater $s$ gets, since there is more "space" the graph can use.

So, despite continuum cardinality, are there any means of measuring the number of such functions against $s$ or is it all the same once that minimal way of $f(x) = x$ as been taken?

And would this change if we limited the ways of constructing such functions to e.g. some elementary ones?

Answer 1

There is probably a sophisticated answer to this; let me just mention an unsophisticated one. Several possible discrete analogues of the question you ask exhibit the desired behavior. Consider, for example, the problem of finding a path from $(0, 0)$ to $(n, n)$ in the lattice $\mathbb{Z}^2$ where the only allowable steps are between diagonally adjacent lattice points. Let the length of such a path be the total number of steps. Then, of course, there is only one path of length $n$, which is the obvious diagonal one. However, there are ${n+2 \choose 2} + n$ paths of length $n+2$, already a big leap.

The general answer can be computed as follows. In a path of length $n+2k$ we need to choose $2j$ of the steps to be up-left or down-right (for some $j$ between $0$ and $k$) and we need to choose $n+2k-2j$ of the steps to be up-right or down-left. Of the first set of steps, half need to be of one type and half need to be the other, and of the second set of steps, $n+k-j$ need to be of one type and $k-j$ need to be the other. This gives the total number of paths of length $n+2k$ as

$$\sum_{j=0}^{k} {n+2k \choose 2j} {2j \choose j} {n+2k-2j \choose k-j}.$$

The dominant term when $n$ is large compared to $k$ is the term associated to $j = k$, which is ${n+2k \choose 2k}$ and hence grows asymptotically like $\frac{n^{2k}}{(2k)!}$. So, again when $n$ is large compared to $k$, it is indeed true that the number of allowable paths grows with the length of the path.

Answer 2

Here's my take on an intuitive answer:

Since there are $2^{\aleph_0}$ continuous functions from the unit interval to itself, it is clear that there can be at most that many functions.

Now, given $s > \sqrt{2}$ (which as you well observed the minimal distance), we define $t = s - \sqrt{2}$ and for any $x\in (0,1)$ we can construct a function that has the wanted arc length, simply by "pulling" the value of $f(x)$ smoothly in a way that extends the length of the arc by $t$.

(The construction is very clear when you don't request $f$ to be differentiable, as you can just stick a triangle in the proper length, however the idea should be clear)

So you have at least $|(0,1)| = 2^{\aleph_0}$ functions as such, which means that you have exactly that many.