$L^p$ norm means

Question

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)?

Answer 1

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.

Answer 2

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 $$ \frac{ \Gamma\left( \frac{a}{p} \right) \Gamma\left( \frac{b}{q} \right) \Gamma\left( \frac{c}{r} \right) }{\,p\, q\, r \; \; \Gamma\left( 1 + \frac{a}{p} + \frac{b}{q}+ \frac{c}{r}\right) } $$

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

Answer 3

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.)

Answer 4

[This does not give the mean of ${ q - }$norm on ${ S ^{n} _{p} ,}$ but a related quantity]

Let ${ p \in [1, \infty) }.$ By the computation of ${ f _Z (z) }$ above theorem ${ 2 }$ here, the density $${ \begin{align*} f _{Z} (z) &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( u (1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du \\ &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \frac{1}{\Gamma(1 + \frac{1}{p}) ^{n+1}} \frac{1}{p} \Gamma\left(\frac{n+1}{p}\right) \end{align*} }$$

integrates over ${ z \in (S _{p} ^{n}) ^{+} = \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert _{p} < 1 \rbrace }$ to give ${ 1 }.$

Here densities ${ f _1 (t) = \ldots = f _{n+1} (t) = \frac{1}{\Gamma(1 + \frac{1}{p})} e ^{- t ^p} }$ for ${ t \geq 0 }.$

So $${ \int _{ z \in S _{p} ^{n}} \left( \sum _{i=1} ^{n+1} \vert z _ i \vert ^{2(p-1)}\right) ^{-\frac{1}{2}} dA = 2 ^{n+1} \frac{ p \Gamma(1+\frac{1}{p}) ^{n+1}}{\Gamma\left(\frac{n+1}{p}\right)} }$$

that is

$${ \int _{ z \in S _{p} ^{n}} \left( \sum _{i=1} ^{n+1} \vert z _ i \vert ^{2(p-1)}\right) ^{-\frac{1}{2}} dA = 2 ^{n+1} \frac{1}{p ^n} \frac{\Gamma(\frac{1}{p}) ^{n+1}}{\Gamma\left(\frac{n+1}{p}\right)} .}$$

Setting ${ p = 2 }$ gives area of ${ S _2 ^n }$ to be ${ \frac{2 \pi ^{(n+1)/2}}{\Gamma(\frac{n+1}{2}) } }.$

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A better form for above integral:

This is a generalisation of Folland's article on integrating polynomials over ${ S _2 ^n }$ here.

Consider the same expression $${ \begin{align*} f _{Z} (z) &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( u (1- z _{1} ^{p} - \ldots - z _{n} ^{p}) ^{\frac{1}{p}}) u ^{n} \, du \\ &= \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} \int _{0} ^{\infty} f _{1} (z _1 u) \ldots f _{n} (z _{n} u ) f _{n+1} ( z _{n+1} u ) u ^{n} \, du \end{align*} }$$ for density of ${ Z \in (S _p ^n) ^{+} = \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert _{p} < 1 \rbrace }.$

Set densities $${ f _1 (t) = \frac{p}{\Gamma(d _1 /p)} t ^{d _1 - 1} e ^{-t ^p}, \ldots , f _{n+1} (t) = \frac{p}{\Gamma(d _{n+1}/p)} t ^{d _{n+1} - 1} e ^{- t ^p } }$$ for ${ t \geq 0 }$ (where parameters ${ d _1, \ldots, d _{n+1} > 0 }$).
These are Generalised Gamma densities, as in section 3 of the draft here.

Now we get $${ f _{Z} (z) = \left(\sum _{i=1} ^{n+1} z _i ^{2(p-1)} \right) ^{-\frac{1}{2}} z _1 ^{d _1 -1} \ldots z _{n+1} ^{d _{n+1} -1} p ^n \frac{\Gamma\left(\frac{\sum _{i=1} ^{n+1} d _i}{p}\right)}{\Gamma(\frac{d _1}{p}) \ldots \Gamma(\frac{d _{n+1}}{p})} }$$ for ${ z \in (S _p ^n ) ^{+} }.$

Hence $${\boxed{ \int _{z \in S _p ^n} \left(\sum _{i=1} ^{n+1} \vert z _i \vert ^{2(p-1)} \right) ^{-\frac{1}{2}} \vert z _1 \vert ^{d _1 -1} \ldots \vert z _{n+1} \vert ^{d _{n+1} -1} dA = 2 ^{n+1} \frac{1}{p ^n} \frac{\Gamma(\frac{d _1}{p}) \ldots \Gamma(\frac{d _{n+1}}{p})}{\Gamma\left(\frac{\sum _{i=1} ^{n+1} d _i}{p}\right)} } }$$ for parameters ${ d _1, \ldots, d _{n+1} > 0 }.$

Setting ${ p = 2 }$ gives the theorem in Folland's article.

I don't know how to proceed if the term ${ \left(\sum _{i=1} ^{n+1} \vert z _i \vert ^{2(p-1)} \right) ^{-\frac{1}{2}} }$ is removed from the integrand. (The term is ${ 1 }$ when ${ p = 2 }$, so we know how to integrate polynomials on ${ S _2 ^n }$).