11
$\begingroup$

The famous Pitt's theorem asserts that if $p>q$ then each bounded operator $T\colon \ell^p\to\ell^q$ is compact. Since $\ell^p$ and $\ell^q$ are incomparable ($p\neq q$, $p,q\geq 1$), each operator $T\colon \ell^p\to\ell^q$ is strictly singular anyway.

Now I want to ask about $L^p(\mu)$ (put any assumptions on $\mu$ as you wish). Under what conditions on $p$ and $q$ each weakly compact operator $T\colon L^p(\mu)\to L^q(\mu)$ is compact?

  • 0
    As a first step: note that if either $1 or $1\leq q < \infty$ then every linear operator $L^p\to L^q$ is weakly compact. This is clear when either $p$ or $q$ lies in $(1,\infty)$, because then either the domain or range is reflexive; and the only other case is $p=\infty$, $q=1$ when my claim follows from Grothendieck's theorem (one can also prove it without G's thm, but I don't know the argument without looking it up).2011-11-02
  • 1
    The case where $\mu$ is purely atomic is essentially the only case where a Pitt-type theorem holds for operators between. Consider, say, $\mu$ to be Lebesgue measure on $[0,1]$ (you should be able to reduce to this case if $\mu$ is not purely atomic); by the assertions of Yemon's comment, one wants to show in most cases that there is a noncompact operator from $L_p(\mu)\longrightarrow L_q(\mu)$. For $p\in [1,\infty)$, $L_p(\mu)$ contains a subspace isomorphic to $\ell_2$s (spanned by the Rademacher functions). For $p,q \in (1,\infty)$, this copy of $\ell_2$ is complemented and you can thus2011-11-03
  • 0
    find an operator from $L_p$ to $L_q$ whose range is the complemented copy of $\ell_2$ in $L_q$; this operator is of course noncompact. For the case $p=1$, note that every separable Banach space $X$ is the image of a bounded linear operator from $L_1(\mu)$, and such a surjection is non compact whenever $X$ is infinite dimensional.2011-11-03
  • 1
    @Philip-Brooker: why don't you write that up as an answer? (I had missed the trick with the Rademachers)2011-11-04

1 Answers 1