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 $dim_B \; G \le 1$.