# L_p norm balls for 1<p<2 - is it always similar to an L_q norm ball for some q>2?

The L_1 ball in 2D is shaped like a diamond (L_1 is also known as the Manhattan norm). The L_∞ ball is shaped like a square (L_∞ is also known as the supremum norm). They are similar, i.e. have same shape. The L2 ball is shaped like a circle.

Hypothesis: For all p in the interval (1,2), there is q>2 such that the q-ball and the p-ball are similar. One further hypothesis is that this occurs precisely when p,q are Hölder conjugates.

I wasn't sure how to tag this.

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Your initial observation is more or less a coincidence. Look at what happens in 3D. – Qiaochu Yuan Oct 19 '09 at 22:50

$\ell_p^n$ is isometric to $\ell_r^n$ with $1\le p < r\le \infty$ iff $n=1$ or ($n=2$ and $p=1$ and $r=\infty$ and the scalars are real). While known before L. Dor's 1976 paper in the Israel J. Math. 24, 260-268, it can be deduced from Theorem 2.1 in that paper, which classifies when $\ell_s^n$ isometrically embeds into $L_t(0,1)$.

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AKA Dorembus, if I'm remembering well... – Ady Jan 18 '10 at 19:44
Yes, indeed. Who are you, Ady? – Bill Johnson Jan 18 '10 at 19:52

You can use this applet to generate pictures of the curves you're thinking of. Your conjecture seems unlikely because |x|^p + |y|^p = 1 appears to be a smooth curve for p ≥ 2 and have singularities at (1,0), (-1,0), (0,1), (0,-1) for p < 2.

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I agree that graphs look different, but your argument why doesn't work. For any e>0, (1-|x|^(1+e))^(1/(1+e)) is differentiable at x=0. – Richard Dore Oct 20 '09 at 3:09
You're right. The curves looked pointy, so I didn't actually compute the derivative. – Michael Lugo Oct 20 '09 at 12:01

You are looking for isometries between the ell^p and ell^q

Here is a paper describing some methods on how to identify isometries (but the paper is not freely downloadable)

http://www.informaworld.com/smpp/content~content=a746864853~db=all~order=page

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