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Can a sequence $f:\mathbb{Z}\to\mathbb{C}$ be in $\ell^1$ but not $\ell^2$? (any one counter example will suffice)

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Are you sure you mean $\Bbb R\to\Bbb C$? That wouldn't be a sequence.... – Cameron Buie Oct 31 '12 at 18:21
@CameronBuie Sorry, $\mathbb{Z}\to\mathbb{C}$ – apk Oct 31 '12 at 18:40
Sure, take $f(n) = 1/n$. More generally, you can find a sequence in ${\ell}^a$ but not in ${\ell}^b$ for any $a > b \geq 1$. Do you see why? – Jonah Sinick Oct 31 '12 at 18:52
@JonahSinick That example is in $\ell^2$ but not in $\ell^1$. The OP wants to know whether it can be the other way around. – user38355 Oct 31 '12 at 19:16
@ brom - Hah, yes, you're right, I misread. – Jonah Sinick Oct 31 '12 at 22:16
up vote 8 down vote accepted

$\ell_p$ is just $L_p(\mu)$ where $\mu$ is counting measure on $\mathbb N$, and it's a general fact of $L_p$-spaces that if the measure doesn't admit sets of arbitrarily small positive measure, then for $p \leq q$ we have the inclusion $L_p \subseteq L_q$.

Of course for $\ell_1$ and $\ell_2$, this is gross overkill since it's obvious that each $(x_n)$ in $\ell_1$ has to have $x_n \to 0$ as $n\to\infty$, so in particular $|x_n|<1$ for all $n$ greater than or equal to some $N$ and thus $|x_n|^2 < |x_n| < 1$ for $n \geq N$, so $(x_n) \in \ell_2$.

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I guess the same could be said for $L^1$ then? i.e., if $g:\mathbb{R}\to\mathbb{C}$ is in $L^1$, then $g$ is also in $L^2$? – apk Oct 31 '12 at 18:42
"if the measure doesn't admit sets of arbitrarily small positive measure" so no, $L^1(\mathbb{R})$ is not a subset of $L^2(\mathbb{R})$. I can even give you a specific example, take $f(x) = \sqrt{|x|}^{-1}\cdot \chi_{[-1,1]}$. This has a Riemann integrable singularity(with finite integral) at the origin, but it's square doesn't have a finite Lebesgue integral. – John Stalfos Oct 31 '12 at 18:51
@kahen Could I ask a reference for this general fact?>if the measure doesn't admit.....$L_p\subset L_q$. I guess I have seen only the inclusion of $L_p$'s with finite measure using Holder, or the inclusion $l_p$ in a separate discussion, and I find this fact interesting. – Oct 22 '15 at 5:01

This really is a long comment to answer this query, but I think that it is unlikely to fit in the space of the single comment. For the standard Lebesgue spaces on $\mathbb{R}$ or $\mathbb{R}^n$, there are essentially two important features for a generic function $f\in L^p$. The two features are how $f$ behaves

  1. at the very large scale; that is to say, how $f$ decays as $|x|\to\infty$
  2. at the very small scale; that is to say, near a point $x_0$ where $|f(x_0)| = \infty$ (interpreted suitably since $f$ is an equivalence class of functions defined up to sets of measure zero), how does $f$ blow-up.

That these two features are the important ones can be seen by the fact that

Finite linear combinations of characteristic functions of bounded intervals (i.e. step functions of compact support) are always in $L^p$ for every $1\leq p \leq \infty$.

Using with hind-sight the result that step functions are dense in $L^p$, we see that the some of the differences between the various $L^p$ spaces can be captured by what infinite linear combinations of characteristic functions are admitted in each of the $L^p$ spaces. And the two places where this infinity really comes into play are when we drop the requirement of compact support and when we drop the requirement of the bound on the (essential) supremum of the function.

The rough heuristic is that for a function in $L^p$:

  1. The smaller the $p$, the faster it decays as $|x|\to\infty$.
  2. The larger the $p$, the better it behaves in the small scale (the "blowup" is confined to smaller and smaller regions).

This heuristic plays an important role in interpolation theory: if we can control the small-scale behaviour in a large $p$ norm, and the decay behaviour in a small $q$ norm, then the function itself can be guaranteed to be in $L^s$ for every $q \leq s \leq p$. These types of argument are often used in classical Fourier analysis (such as Calderon-Zygmund theory).

In the special case of $\ell^p$ spaces, because the underlying space is countable and discrete, there is no such thing as "small scale" (which is what kahen refers to as "sets of arbitrarily small positive measure"). Hence the very small scale differences do not manifest between the $\ell^p$ spaces, and we can capture the differences just in the decay behaviour.

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Great answer, I love it. It's very intuitive and gives the feeling of how $L^p$ behaves. +1 – Patrick Da Silva Nov 1 '12 at 9:15

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