In the book Interpolation and Approximation with Splines and Fractals by Prof. Massopoust one studies the following theorem with regard two the box dimension of affine fractal interpolation functions:
Suppose that $f:[a,b] \to \mathbb{R}$ is an affine fractal interpolation funtion defined as above and $G$ its graph. If $\sum_{i = 1}^{N} |\lambda_i| > 1$ and if the interpolation set $Y$ is not collinear, then the box dimension of $G$ is given by the formula $\dim_B \; G = 1 + \log_N(\sum_{i = 1}^N|\lambda_i|)$
But more precisely, I'm interested in the following argument, which is used during the proof:
Given that f is a continuous function, then $\dim_B G \ge 1$
However, I cannot find a justification for this. Any hints?
You can find any missing piece of information in these lecture notes.
My try
To cover the graph of the function in the interval $[a,b]$ we need more than $\lceil \frac{b-a}{\epsilon} \rceil$ squares, so we have to compute $\lim\limits_{\epsilon \to 0} \frac{\log \lceil \frac{b-a}{\epsilon} \rceil}{\log \frac{1}{\epsilon}}$.
We have $\lceil \frac{b-a}{\epsilon} \rceil \ge \frac{b-a}{\epsilon} \implies \log \lceil \frac{b-a}{\epsilon} \rceil \ge \log \frac{b-a}{\epsilon}$. Thus, we get $\frac{\log \lceil \frac{b-a}{\epsilon} \rceil}{\log \frac{1}{\epsilon}} \ge \frac{\log(b-a)-\log(\epsilon)}{-\log(\epsilon)}$ and taking limits $dim_B \; G \ge 1$.
But in fact, I didn't need the hypothesis that the function was continuous. So I guess my conclusion extends to any function $f:[a,b] \to \mathbb{R}$. Could you check my conclusions? (note the proof verification tag). I wonder if this can be generalized to higher dimensions.