# Box dimension of the graph of an increasing function

This Hausdorff dimension of the graph of an increasing function shows that:

Let $$f$$ be a continuous, strictly increasing function from $$[0,1]$$ to itself with $$f(0)=0, f(1)=1$$. Then $$dim_H \; G = 1$$ where $$G$$ is the graph of $$f$$.

I have at hand the Casino function, described as follows in Massopoust's Interpolation and Approximation with Splines and Fractals:

Let $$X = [0,1] \times \mathbb{R}$$, $$N = 4$$ and $$Y = \{(x_v,y_v):0 = x_0 < \ldots x_N = 1, 0 = y_0 < \ldots < y_N = 1\}$$. Define an IFS by $$f_i(x,y) = \begin{pmatrix} x_i-x_{i-1} & 0 \\ 0 & y_i - y_{i-1} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} x_{i-1} \\ y_{i-1} \end{pmatrix}$$ for $$i = 1, \ldots, N$$.

The associated RB operator $$T$$ is contractive and its unique fixed point is called a Casino function $$c:[0,1] \to [0,1]$$. These functions are monotone increasing and therfore $$dim_H \; graph(c) = \dim_B \; graph(c) = 1$$.

I was wondering how can I show that $$dim_B \; graph(c) = 1$$ and whether there is a general argument establishing:

Let $$f$$ be a continuous, strictly increasing function from $$[0,1]$$ to itself with $$f(0)=0, f(1)=1$$. Then $$dim_B \; G = 1$$ where $$G$$ is the graph of $$f$$.

I don't find an argument stablishing $$dim_B \; G \le 1$$.

• Pietro Majer's argument that you cited shows the upper box dimension is $1,$ since graphs of real-valued Lipschitz functions defined on an interval (of positive length) have upper box dimension $1.$ See, for example, Corollary 11.2(a) [using $s=1$] on p. 147 of Falconer's 1990 book Fractal Geometry. You might also want to look at p. 120 (beginning of Section 10.4) and Section 12.4 (pp. 148-150) of Tricot's 1995 book Curves and Fractal Dimension, and this paper. – Dave L Renfro Mar 17 at 19:10
• @DaveLRenfro You should post your comment as an answer. – Piotr Hajlasz Mar 17 at 19:38
• @Piotr Hajlasz: OK, I've posted a slightly revised version of my comment. I figured you might have a lot more to say about this (I just cited some references, and I'm not all that sure whether much better places to read about this topic exist), which is why I didn't bother answering. – Dave L Renfro Mar 17 at 20:04

Pietro Majer's argument that you cited actually shows that the upper box dimension is $$1,$$ and hence the lower box, the upper and lower packing, and the Hausdorff dimensions are all equal to $$1.$$ Also, graphs of real-valued Lipschitz functions defined on an interval (of positive length) have upper box dimension $$1,$$ and as Pietro Majer also pointed out there in a comment, the graph of a strictly increasing continuous function is geometrically congruent to the graph of a Lipschitz function.

Regarding graphs of Lipschitz functions, see Corollary 11.2(a) [using $$s=1$$] on p. 147 of Falconer's 1990 book Fractal Geometry. Regarding graphs of monotone and bounded variation functions, see p. 120 (beginning of Section 10.4) and Section 12.4 (pp. 148-150) of Tricot's 1995 book Curves and Fractal Dimension and this paper.

Theorem If $$\gamma:[a,b]\to (X,d)$$ is an injective rectifiable curve and $$\Gamma=f([a,b])$$, then $$\mathcal{H}^1(\Gamma)=L(\gamma).$$

This theorem applies in our context since we have a strictly increasing function and:

Theorem Every rectifiable curve $$\gamma:[a,b]\to (X,d)$$ can be reparametrized as a $$1$$-Lipschitz curve.

Our curve is rectifiable, that is is length is finite, choosing the sum distance:

$$L(\gamma) = \sup\left\{\sum_{i=1}^{n-1}d(\gamma(t_i),\gamma(t_{i+1}))\right\} = \sup\left\{\sum_{i=1}^{n-1}(t_{i+1}-t_i) + (\gamma(t_{i+1})-\gamma(t_i)) \right\} = 2$$

So the theorem applies and we get $$0 < \mathcal{H}^1 < \infty$$ which implies $$\dim_H(graph(f)) = 1$$.

As it was pointed out in the accepted answer, Falconer's Fractal Geometry contains in Corollary 11.2 a result which can be adapted in our context to:

Let $$f:[0,1] \to \mathbb{R}$$ a Lipschitz function, then $$\dim_H(graph(f)) \le \dim_B(graph(f)) \le 1$$

So in conclusion we get $$dim_H(graph(f)) = dim_B(graph(f)) = 1$$. In summary we have obtained that:

If $$\gamma:[a,b]\to (X,d)$$ is an injective rectifiable curve and $$\Gamma=f([a,b])$$, then $$dim_H(graph(f)) = dim_B(graph(f)) = 1$$.