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This question is inspired by How kinky can a Jordan curve get?

What is the least upper bound for the Hausdorff dimension of the graph of a real-valued, continuous function on an interval? Is the least upper bound attained by some function?

It may be noted that the area (2-dimensional Hausdorff measure) of a function graph is zero. However, this does not rule out the possible existence of a function graph of dimension two.

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>>It may be noted that the area (2-dimensional Hausdorff measure) of a function graph is zero. Really? I don't know how to prove it if the function is non-measurable. –  Catcat Mar 25 at 0:07
@Catcat I had continuous functions in mind, as indicated in the second paragraph. –  Harald Hanche-Olsen Mar 25 at 12:59

2 Answers 2

up vote 7 down vote accepted

The answer is 2.

Besicovitch and Ursell, Sets of fractional dimensions (V): On dimensional numbers of some continuous curves. J. London Math. Soc. 12 (1937) 18–25. doi:10.1112/jlms/s1-12.45.18

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Here is one example with Hausdorff dimension 2.

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Thank you. Have you checked the proof yourself? I ask because that journal (Chaos, Solitons and Fractals) has a somewhat dubious reputation (see Physics World, january 2009 and Nature, November 2008; also zeit.de/2009/03/N-El-Naschie?page=1, arstechnica.com/science/news/2008/11/…), scienceblogs.com/pontiff/2008/11/…) –  Harald Hanche-Olsen May 8 '10 at 21:45
To follow up on my own comment, at least the paper looks serious. So I think it's probably okay. –  Harald Hanche-Olsen May 8 '10 at 21:50
I haven't checked it carefully. Thanks for the articles, I had never heard of this before. –  Gjergji Zaimi May 8 '10 at 22:16
2 is the right answer, but of course 2007 is not the earliest example –  Gerald Edgar May 8 '10 at 23:58
Once you get arbitrarily close to 2, just use disjoint intervals and put graphs on them with dimensions $\gt 2-1/n$ –  Gerald Edgar May 9 '10 at 0:31

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