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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?

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As a first step: note that if either $1<p\leq \infty$ 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). – user16299 Nov 2 '11 at 3:22
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 thus – Philip Brooker Nov 3 '11 at 22:51
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. – Philip Brooker Nov 3 '11 at 22:58
@Philip-Brooker: why don't you write that up as an answer? (I had missed the trick with the Rademachers) – user16299 Nov 4 '11 at 4:22

(Turning my comments into an answer, as per Yemon's suggestion).

The case where $\mu$ is purely atomic is essentially the only case where a Pitt-type theorem holds for operators $L_p(\mu)\longrightarrow L_q(\mu)$, where $1\leq q,p<\infty$. Indeed, let us consider, say, the case where $\mu$ is Lebesgue measure on $[0,1]$ (all other non-purely atomic cases should reduce to this one). By the assertions of Yemon's first comment, one wants to show that there is a noncompact operator $L_p(\mu)\longrightarrow L_q(\mu)$. To this end I suggest use the fact that for $1\leq r<\infty$ the closed linear hull $R_r$ in $L_r(\mu)$ of the sequence of Rademacher functions on $[0,1]$ is isomorphic to $\ell_2$, and $R_r$ is complemented in $L_r(\mu)$ if and only if $1<r<\infty$ (a good reference for this result is Albiac and Kalton's book Topics in Banach space theory; there it is stated as Proposition 6.4.2). The time to treat the cases $p=1$ and $p>1$ separately has now arrived.

For $p=1$, we let $X$ be a complemented subspace of $L_1(\mu)$ isomorphic to $\ell_1$ and let $P$ be a projection of $L_1(\mu)$ onto $X$. Since every separable Banach space is the range of a continuous linear operator from $\ell_1$, there exists a surjective operator $T:X \longrightarrow L_q(\mu)$. The map $x\mapsto T(Px)$ is a surjective (hence noncompact) linear operator of $L_1(\mu)$ onto $L_q(\mu)$.

For $p>1$, let $Q$ be a projection of $L_p(\mu)$ onto $R_p$ and let $U: R_p\longrightarrow$ be an isomorphic embedding onto $R_q$. The map $x\mapsto U(Qx)$ is a linear operator of $L_p(\mu)$ onto the infinite dimensional closed subspace $R_q$ of $L_q(\mu)$; this operator is, in particular, noncompact.

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