Let $\gamma: [0, 1] \to \mathbb{R}^2$ be a curve (continuous, at least piecewise differentiable?) with end points $\gamma(0) = (0, 0)$, $\gamma(1) = (1, 1)$.
I'm interested under which conditions I can bound the length of $\gamma$ by the side lengths of the triangle {(0, 0), (0, 1), (1, 1)}. The lower length bound of the diagonal $\sqrt 2$ is obvious, so the question is about in what case the length of the curve has an upper bound of 2 (the length of the other two sides).
I think a sufficient (but not necessary1) condition is that the curve is monotonous (in both coordinates). From this it follows that the range of both coordinates is $[0, 1]$ and we can express the curve as the graph of a function $f: [0, 1] \to [0, 1]$ which is also monotonous. Further, we can assume w.l.o.g. that the function is contained in the triangle given above (i.e. $f(x) \ge x$). (However, I'm not sure if this is useful)
To wit: If $f$ does not fulfill this, mirror any parts inside the corresponding lower triangle over the diagonal, creating a new function $g$ which is still monotonous and has the same length as $f$.
Now I don't know how to proceed in using the monotonicity to prove that the length of the function $f$ is bounded by the triangle sides, i.e. less than 2.
The question can probably be phrased / solved in a much more general way. It somehow reminds me of Bézier curves and their properties w.r.t to convex hulls, but I'm not sure how to this relates.
1 There are obviously non-monotonous curves with the given end-points but outside of $[0, 1]^2$ that have length less than 2. What would a necessary and sufficient condition look like?
