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I'm trying to prove that if $g: [0,1] \longrightarrow [0,1]^2$ is an $\alpha$-Hölder continuous mapping whose image is the entire square $[0, 1]^2$ then $\alpha \leq 1/2$. I wouldn't know where to start, surely the surjectivity is essential to prove the result but i don't know how..

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1. It is considered rude here to use the imperative. 2. What have you tried so far? – Brad Apr 1 '12 at 22:03
Although you've been at the site for about 2months, I wanted to remind you about a few things: In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("Find", "Show") to be rude when asking for help; please consider rewriting your post. – Arturo Magidin Apr 1 '12 at 22:09
I'm sorry, I ri-edited the question..I didn't mean it in a rude way, i just translated from my language the sentence! Anyway this is an exercise i found on the internet (that's why i don't know how to start). I will keep it in mind for the next time! – balestrav Apr 1 '12 at 22:47
Probably I would use Hausdorff dimension to investigate this. – GEdgar Apr 2 '12 at 0:27
I just posted a full answer (with no use of Hausdorff dimension) and deleted it. I will undelete it as soon as the accept rate of the OP becomes more, well... acceptable. – Did Apr 2 '12 at 8:57
up vote 4 down vote accepted

Assume $g$ is Hölder continuous with exponent $a$, hence $\|g(x)-g(y)\|\leqslant C|x-y|^a$ for every $x$ and $y$ in the interval $[0,1]$. In particular all the images of an interval of length $1/n$ are at distance at most $C/n^a$ from each other, hence the area these occupy in the square $[0,1]^2$ is at most $\pi C^2/n^{2a}$. Since $n$ intervals of length $1/n$ cover $[0,1]$, the image of $[0,1]$ covers at most an area $\pi C^2n^{1-2a}$ in the square. If $a\gt1/2$, this goes to zero when $n\to\infty$, which is a contradiction to the hypothesis that the image of $g$ is the whole square. Thus, $a\leqslant1/2$.

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