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Im going through some notes for fractal geometry and the following exercise is stated:

"Let $f:[0,1]\rightarrow{}\mathbb{R}$ be a continuously differentiable function with $f(0)\neq{}f(1)$. Show that the box dimension of the image of $f$, namely $f([0,1])=\{f(x):0\le{}x\le{}1\}$ is $1$."

Firstly, my attempt:

Since f is differentiable on $[0,1]$ it is bounded on $[0,1]$ and hence has a maximum and minimum . The intermediate value theorem tells us that the image is also an interval, the result follows (simple calculation).

Secondly:

Is the question stated wrong? could it be in fact to calculate the box dimension of the graph $\{(x,f(x):0\le{x}\le{1})\}\subset\mathbb{R}^2$? Even if it's not an error, is the box dimension of this determinable? Is it dependant on the function?

Mark McClure
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kam
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    Both sets you refer to have box dimension 1, though that assertion is certainly is easier to prove for one than the other. I don't know that the problem is a typo, though; it might be meant to be an easy problem. Also, the use of term "image" agrees with the set theoretic statement $f([0,1])={f(x):0\le{}x\le{}1}$. Your second set is commonly called the graph of the function. – Mark McClure May 22 '20 at 19:58
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    The first one requires only a continuous function. For the second one, you will use the fact that it is continuously differentiable. – GEdgar May 22 '20 at 20:02

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