Consider the unit sphere $S_p^{n1}$ of an $L^p$ normin $\mathbb{R}^n.$ The question is: what is the expected value of the $L^q$ norm on $S_p^{n1}?$ Since (I assume) this is intractable in closed form, what are the asymptotics in $q, n$ (for $p$ fixed)?

$\begingroup$ A naive question: what's your choice of measure on $S_p^{n1}$ 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

2$\begingroup$ 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

2$\begingroup$ 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
For $p=2$: up to multiplicative universal constants, the average $M$ of $\\cdot\_q$ over $S^{n1}$ is equal to
 $M \simeq \sqrt{q} \cdot n^{1/q1/2}$ when $1 \leq q \leq \log n$,
 $M \simeq \sqrt{\log n}/\sqrt{n}$ 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 MilmanSchechtman, "Asymptotic theory of finitedimensional 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 $$ M \leq \frac{1}{\alpha_n} \left(\int_{\mathbb{R}^n} \x\^q_q \, \mathrm{d} \gamma_n(x) \right)^{1/q} \simeq \sqrt{q} \cdot n^{1/q1/2} $$ (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.

$\begingroup$ 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
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 $$ \int x^{a1} y^{b1} z^{c1} dx dy dz $$ 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

$\begingroup$ 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
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, $$ \frac{1}{S_{n1}}\int_{S_{n1}}\x\_1\, d\sigma(x) = \frac{n}{S_{n1}}\int_{S_{n1}} x_1\, d\sigma(x) = \frac{2n S_{n2}}{S_{n1}} \int_0^1 x (1x^2)^{(n3)/2}\, dx\\ =\frac{2nS_{n2}}{(n1)S_{n1}} = \frac{2n}{n1}\, \frac{\Gamma(\frac{n1}{2})}{\pi^{1/2}\Gamma(\frac{n}{2}1)}. $$ 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.)

$\begingroup$ 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