TITLE: Box dimension of the graph of an increasing function
QUESTION [7 upvotes]: 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 $dim_B \; G \le 1$.

REPLY [3 votes]: 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  $dim_H(graph(f)) = dim_B(graph(f)) = 1$.