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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 $dim_B \; G \le 1$.

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    $\begingroup$ 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. $\endgroup$ – Dave L Renfro Mar 17 at 19:10
  • $\begingroup$ @DaveLRenfro You should post your comment as an answer. $\endgroup$ – Piotr Hajlasz Mar 17 at 19:38
  • $\begingroup$ @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. $\endgroup$ – Dave L Renfro Mar 17 at 20:04
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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.

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Let me encode the solution of the problem explicitly. From the linked answer we had:

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 $dim_H(graph(f)) = dim_B(graph(f)) = 1$.

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