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Consider the unit sphere $S_p^{n-1}$ of an $L^p$ normin $\mathbb{R}^n.$ The question is: what is the expected value of the $L^q$ norm on $S_p^{n-1}?$ Since (I assume) this is intractable in closed form, what are the asymptotics in $q, n$ (for $p$ fixed)?

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  • $\begingroup$ A naive question: what's your choice of measure on $S_p^{n-1}$ when $p\neq 2$? Is it just the image of the uniform measure on the Euclidean sphere under the natural homeo from that sphere to the $L^p$-version? $\endgroup$ – Yemon Choi Sep 9 '14 at 0:15
  • $\begingroup$ @YemonChoi Yes, it is the induced measure, though I have to admit that I was secretly thinking $p=2$ - that case might well be much easier. $\endgroup$ – Igor Rivin Sep 9 '14 at 0:26
  • $\begingroup$ In that case, doesn't it reduce to the $p=2$ case by scaling? (It is 1am here so forgive me if I have missed something obvious) $\endgroup$ – Yemon Choi Sep 9 '14 at 0:27
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    $\begingroup$ Also, for the $p=2$ case I am tempted to make a first guess (not a proper derivation of the correct asymptotic) by looking at a Gaussian vector with i.i.d. entries that are $N(0,n^{-1/2})$, this is "mostly" concentrated on the unit sphere and the expected $L^q$-norm would seem to have some closed form that allows for decent estimates asymptotic in $n$. But as I said this may be a case of working out the asymptotics for something different from what was intended. $\endgroup$ – Yemon Choi Sep 9 '14 at 0:34
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    $\begingroup$ Two papers that might be relevant are: G. Schechtman and J. Zinn, On the volume of the intersection of two $L^n_p$ balls, Proc. A.M.S. 110 (1990), 217–224. G. Schechtman and M. Schmuckenschlager, Another remark on the volume of the intersection of two $L^n_p$ balls, GAFA Seminar 89/90, Lecture Notes in Math., Vol 1469, 174–178, Springer (1991). $\endgroup$ – Bill Johnson Sep 9 '14 at 9:21
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For $p=2$: up to multiplicative universal constants, the average $M$ of $\|\cdot\|_q$ over $S^{n-1}$ is equal to

  1. $M \simeq \sqrt{q} \cdot n^{1/q-1/2}$ when $1 \leq q \leq \log n$,
  2. $M \simeq \sqrt{\log n}/\sqrt{n}$ for $q \geq \log n$.

This can be checked most easily after switching to a Gaussian integral as Yemon mentions. For the lower bound in 1, it may be useful to consider using concentration of measure. If it is a matter of reference, it can probably be extracted from Chapter 5.4 in Milman-Schechtman, "Asymptotic theory of finite-dimensional normed spaces". Indeed, the value of this average is closely related to the dimension of almost Euclidean sections of the space $\ell_q^n$.

Edit: let me add more detail. First, (this is true for any norm of $\mathbb{R}^n$, just by rotational invariance of the Gaussian measure $\gamma_n$), we have $$ M = \frac{1}{\alpha_n} \int_{\mathbb{R}^n} \|x\|_q \, \mathrm{d} \gamma_n(x), $$ where $$\alpha_n = \int_{\mathbb{R}^n} \|x\|_2 \, \mathrm{d} \gamma_n(x) $$ is a constant very close to $\sqrt{n}$. Now write $$ M \leq \frac{1}{\alpha_n} \left(\int_{\mathbb{R}^n} \|x\|^q_q \, \mathrm{d} \gamma_n(x) \right)^{1/q} \simeq \sqrt{q} \cdot n^{1/q-1/2} $$ (use the fact the $L^q$ norm of a standard Gaussian variable is or order $\sqrt{q}$). This upper bound is sharp when $q \leq \log n$, this follows from concentration of measure. Finally for $q \geq \log n$, the norms $\|\cdot\|_q$ and $\|\cdot\|_{\infty}$ are equivalent, and the question reduces to estimating the expected maximum of $n$ i.i.d. standard Gaussian variables.

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  • $\begingroup$ It would be nice to have more of a derivation of this nice result. $\endgroup$ – Igor Rivin Sep 10 '14 at 0:02
  • $\begingroup$ @Christian : start from the Gaussian integral and use polar integration (& homogeneity of the norm) to produce an integral over the sphere $\endgroup$ – Guillaume Aubrun Sep 11 '14 at 16:51
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Igor, pending finding the original or the version in W+W, here are my notes, let me stick to $\mathbb R^3;$ evidently there were also a bunch of lower case Greek letters that i set to 1 here: Given $x,y,z > 0$ and $$ x^p + y^q + z^r < 1, $$ we get $$ \int x^{a-1} y^{b-1} z^{c-1} dx dy dz $$ as $$ \frac{ \Gamma\left( \frac{a}{p} \right) \Gamma\left( \frac{b}{q} \right) \Gamma\left( \frac{c}{r} \right) }{\Gamma\left( 1 + \frac{a}{p} + \frac{b}{q}+ \frac{c}{r}\right) } $$

Dirichlet, Über eine neue Methode zur Bestimmung vielfacher Integrale, original 1839

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  • $\begingroup$ Very cool (though not obviously useful here, since the norms are not polynomials [this does, of course, give the answer for $\|x\|_q^q.$, which comes out quite nicely] $\endgroup$ – Igor Rivin Sep 9 '14 at 1:53
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I don't know if you are very interested in this, but one special case where we can work out the average explicitly is $q=1$. In this case, $$ \frac{1}{S_{n-1}}\int_{S_{n-1}}\|x\|_1\, d\sigma(x) = \frac{n}{S_{n-1}}\int_{S_{n-1}} |x_1|\, d\sigma(x) = \frac{2n S_{n-2}}{S_{n-1}} \int_0^1 x (1-x^2)^{(n-3)/2}\, dx\\ =\frac{2nS_{n-2}}{(n-1)S_{n-1}} = \frac{2n}{n-1}\, \frac{\Gamma(\frac{n-1}{2})}{\pi^{1/2}\Gamma(\frac{n}{2}-1)}. $$ Here, I write $S_d$ for both the $d$-dimensional unit sphere and its surface area. (It so happens I did the same calculation earlier on this site, in a different context.)

If $n=2k+2$, say, then this equals $$ \frac{4k+4}{2k+1}\, \frac{k(2k)!}{4^k(k!)^2} \sim \frac{2}{\pi^{1/2}}\, k^{1/2} \sim \sqrt{\frac{2}{\pi}}\, n^{1/2}, $$ by Stirling's formula to obtain the asymptotics.

(This asymptotic behavior is of course consistent with the intuition suggested by Yemon in the comments.)

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  • $\begingroup$ That is a nice calculation, there is a more general computation (which is related to @WillJagy's thing) written up by Gerry Folland in "How to integrate a polynomial over a sphere", in the Monthly a while ago [I just found this]. $\endgroup$ – Igor Rivin Sep 9 '14 at 1:59

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