MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Do there exist, either in the literature or in folklore, theorems that characterize some particular $\ell^p$ space(s) ($p\not=1,2,\infty$)?

Such a theorem should reveal the particular space(s) as somehow idiosyncratic, in the sense that no obvious modification of the characterization works for general $\ell^p$ spaces.

Thus it would not be interesting here to learn, say, that $\ell^3$ alone has a dual isomorphic to $\ell^{3/2}$; obviously this just specializes a general fact from the theory of all the $\ell^p$'s.

share|cite|improve this question
Are you expecting a "best" answer or are you looking for a bunch of good answers? In the former case, I think this question might deserve to be made "community wiki", but I am open to persuasion. – Yemon Choi Dec 13 '10 at 7:18
Are you expecting a "best" answer or are you looking for a bunch of good answers? I think I'll see how the response unfolds for a day or so, and then decide whether to opt for "community wiki." For example, a priori I couldn't rule out as an answer of "No, and here's why you shouldn't expect such..." Even positive answers don't rule this out, since later remarks might reveal such as special cases of general results after all. – David Feldman Dec 13 '10 at 7:44
David: fair enough. I haven't thought seriously about the question yet. – Yemon Choi Dec 13 '10 at 8:55
This is too cheap to qualify as more than a comment, but $p={\frac{1+\sqrt{5}}{2}}$ is the only case where $\ell^p$ is dual to $\ell^{p^2}$. – Aaron Meyerowitz Dec 13 '10 at 8:56
@Aaron Perhaps if there were some natural abstract connection between $\ell^p$ and $\ell^{p^2}$, you'd have a winner. – David Feldman Dec 13 '10 at 9:02
  1. Littlewood's $4/3$-inequality singles out $\ell^{4/3}$.

    Namely, given a real valued array $\hat{a}=(\hat a_{m,n}:(m,n)\in\mathbb N^2)$, the norm $\|\hat a\|_{\ell_p}$ is finite for all $\hat a$ such that $$\sup \left\{\left|\sum\limits_{m\in\mathcal M,n\in\mathcal N}\hat a_{m,n}x_my_n\right|:x_m,y_n\in[-1,1],\mathcal M,\mathcal N\mbox{ are finite}\right\} < \infty$$ if and only if $p\geq 4/3$.

  2. The second example is somewhat tangential to the question but I find it worth mentioning. It is concerned with the peculiar asymptotics of $L^4$-norms of the Hermite functions (see, e.g., Lectures on Hermite and Laguerre expansions by Thangavelu, Lemma 1.5.2).

    Proposition. As $n\to\infty$ the Hermite functions satisfy the estimates $$\|h_n\|_{p}\sim\begin{cases} n^{\frac{1}{2p}-\frac{1}{4}}, & 1\leq p< \infty, \\\ \\\ n^{-\frac{1}{8}}\log n, & p=4, \\\ \\\ n^{-\frac{1}{6p}-\frac{1}{12}}, & 4 < p\leq \infty. \end{cases} $$ Here $a_n\sim b_n$ means $a_n=O(b_n)$ and $b_n=O(a_n)$.

share|cite|improve this answer
Interesting, and close to what I want, but this doesn't so much single out the space $\ell^{4/3}$ as single out the number $4/3$. If it could be combined with a condition characterizing $\ell^p$'s for $p\leq 4/3$, that would be perfect. – David Feldman Dec 13 '10 at 7:51
Also, if you use the higher dimensional version by Bohnenblust and Hille, you can also similarly characterize all numbers $p = \frac{2m}{m+1}$ for $m\in \mathbb{N}$. (The inequality is $$ \|\hat{a}\|_{\ell^p(\mathbb{N}^m)} \leq 2^{(m-1)/2} \|\hat{a}\| $$ where on the RHS is $\sup \hat{a}(x_1,\ldots,x_m)$ with each $x_i\in \ell^\infty(\mathbb{N})$.) – Willie Wong Dec 13 '10 at 12:25
Dear Andrey: I got confused by where you put the if and only if. Sorry. Deleted the irrelevant comments. – Willie Wong Dec 13 '10 at 16:54
Dear Willie, many thanks for your comment. – Andrey Rekalo Dec 13 '10 at 17:15
In fact there is nothing special in Littlewood's inequality. It is a manifistation of general result of Kwapien: any operator from $\mathcal{L}_1$ space to $\mathcal{L}_p$ space is $(r,1)$-summing for $r^{-1}=1-|p^{-1}-2^{-1}|$ and $p>1$. For details see page 208 in Absolutely Summing Operators by Joe Diestel – Norbert May 30 '14 at 0:49

The following theorem is due to Plotkin and Rudin and characterizes $p \neq 2,4,6,\dots.$

Theorem: (Plotkin-Rudin): Let $0< p< \infty$ and $p \neq 2,4,6,\dots$. Let $(\Omega,\mu)$ and $(\Omega',\nu)$ be two probability measure spaces. Let finally $n$ be a positive integer and $f_1, \dots f_n \in L_p(\mu)$, $g_1, \dots g_n \in L_p(\nu)$.

Assume that for all complex numbers $z_1, \dots z_n \in \mathbb C$,

$$\int |1 + z_1 f_1 + \dots z_n f_n |^p d \mu = \int |1 + z_1 g_1 + \dots z_n g_n|^p d \nu.$$

Then $(f_1 ,\dots f_n)$ and $(g_1 ,\dots g_n)$ form two equimeasurable families. This means that the ${\mathbb R}^n$-valued random variables $(f_1 ,\dots f_n)$ and $(g_1 ,\dots g_n)$ have the same distribution.

share|cite|improve this answer
Are there easy counterexamples in the remaining cases? – Qiaochu Yuan Dec 21 '10 at 3:38

Perhaps if you look at applications to other domains and you admit the Lebesgue space $L^p(\mathbb R^n)$... The space $L^n(\mathbb R^n)$ is critical for Navier-Stokes in space dimension $n$. For instance a theorem of T. Kato says that if the initial data is small in $L^3(\mathbb R^3)$, then the Navier-Stokes equation for an incompressible fluid admits a unique solution, global in time. Removing the smallness assumption is worth a million dollars. The exponent $p=3$ is the only one for which such a result holds true.

share|cite|improve this answer

If you are looking for a true characterization of some $\ell^p$ space (an if and only if) I suspect no example satisfies your demands (probably it's only my ignorance :). There are of course inequalities which are known to be true for special values of $p$ and are an open problem for other values, at least for $L^p$ (and a discrete analogue seems reasonable). Best candidate for this kind of inequalities is $p=4$ since the norm is the square of a square and can be represented in a fairly reasonable way using Fourier transform and convolutions. This method allows to prove e.g. Zygmund's inequality $$ \|\sum c_{n}e^{i(n^2t+nx)}\|^2_{L^4(\mathbb{T}^2)}\le C \sum|c_n|^2 $$ for $L^4$ on $\mathbb{T}^2_{t,x}=[0,2\pi]^2$.

share|cite|improve this answer
@Piero. Isn't it a special case of a Strichartz inequality ? – Denis Serre Dec 13 '10 at 8:30
@Denis: yes indeed, it is the 1D Strichartz inequality for the Scroedinger equation, and it was proved by Zygmund around 1950 – Piero D'Ancona Dec 13 '10 at 11:24
P.S. notice that the other Strichartz inequalities are for a $L^p_tL^q_x$ norm with p different from q. – Piero D'Ancona Dec 13 '10 at 11:25

$p$-stability singles out $0 < p \le 2$. Specifically, there is no probability distribution $P$ such that the linear combination $\sum^n a_i X_i$ is distributed as $\|a\|_p Y$, where $X_1 ... X_n$ and $Y$ are random variables distributed according to $P$, if $p$ is not in the range $(0, 2]$.

For $p = 0.5, 1, 2$ these distributions have closed-form expressions.

(note: updated to reflect Gideon Schectman's comment)

share|cite|improve this answer
There are p-stable distributions for all 0<p≤2. I think this is usually attributed to Paul Levy. You can find this in many books. I particularly like the construction in Chung's A course in probability theory. – Gideon Schechtman Dec 13 '10 at 10:05
Ah correct. I'll edit accordingly. – Suresh Venkat Dec 13 '10 at 11:46
A functional analytic version and amplification of this is that if $p<2$, then $\ell_r$ embeds isometrically isomorphically into $L_p$ for $p\le r\le 2$, while if $r$ is not in this range, then $\ell_r$ does not embed even isomorphically into $L_p$. – Bill Johnson Dec 14 '10 at 0:51
what is an 'isomorphic' embedding ? – Suresh Venkat Dec 14 '10 at 11:04
An isomorphic embedding is a linear homeomorphism into. – Bill Johnson Dec 14 '10 at 23:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.