Suppose f is a real-valued function of one variable, and suppose f is of differentiability class C1. My question is, if $\Gamma$ is the graph of f, then must $\dim_H(\Gamma)=1$? If anyone knows of a published proof of the answer, I'd appreciate a reference greatly.
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1$\begingroup$ James: I'm not a specialist, but I doubt mathoverflow is the suitable place to ask this question (see the faq). Have you tried math.stackexchange.com ? $\endgroup$– Thierry ZellAug 21, 2010 at 0:54
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2 Answers
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Let $f \colon I \to \mathbb{R}$. Since $f$ is $C^1$, the graph $\Gamma_f$ is locally bilipschitz to $I$, via the projection. It follows that Hausdorff dimension is the same as that of $I$ (being defined in terms of the metric space structure only), so it is $1$.
Disclaimer: I haven't seen these topics for quite a while, so I may have said something stupid.
answered Aug 20, 2010 at 22:13
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$\begingroup$ A Lipschitz map does not increase dimension. The map $f$ is Lipschitz on any interval where $f'$ is bounded. Since $f'$ is continuous, the real line is a countable union of intervals where $f'$ is bounded. The countable union of sets of dimension at most $d$ also has dimension at most $d$. $\endgroup$ Aug 21, 2010 at 0:46
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$\begingroup$ Thank you, I gave for obvious that $(id, f)$ is bilipschitz where $f'$ is bounded. :-) My doubt was whether bilipschitz was enough to preserve the Hausdorff dimension; it should be more or less obvious, but I hadn't seen the definition for like 6 years, and it took me a moment of thinking to convince me that it is so. $\endgroup$ Aug 21, 2010 at 2:04
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In spirit, the solution is in any calculus book, using the length formula $$L(f) = \int_a^b \sqrt{f'(x)^2+1} dx.$$ You also need the theorem that a continuous function on a closed interval is bounded.
answered Aug 21, 2010 at 0:47