Fdtxjr

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 $f:[a,b] to mathbbR$ 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 $lceil fracb-aepsilon rceil$ squares, so we have to compute $limlimits_epsilon to 0 fraclog lceil fracb-aepsilon rceillog frac1epsilon$.

We have $lceil fracb-aepsilon rceil ge fracb-aepsilon implies log lceil fracb-aepsilon rceil ge log fracb-aepsilon$. Thus, we get $fraclog lceil fracb-aepsilon rceillog frac1epsilon ge fraclog(b-a)-log(epsilon)-log(epsilon)$ 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 $f:[a,b] to mathbbR$. Could you check my conclusions? (note the proof verification tag). I wonder if this can be generalized to higher dimensions.