Hölder continuity of a function from $[0,1]$ to $[0,1]^2$

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
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), and i didn't know about the tag [homework]..(should i put it anyway?). 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

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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