Question: Let $q>p \ge 1$ and let $T:L^p[0,1] \to L^q[0,1]$ be a bounded linear operator. Let $i: L^q[0,1] \to L^p[0,1]$ be the inclusion map (which is bounded). Is the composition $i\circ T$ necessarily a compact operator on $L^p$? We can assume $p\in (1,\infty)$ if the uniform convexity of the underlying space helps in some way, though the boundary cases might be interesting themselves.
Context: This was inspired by this question, where the accepted answer covers the case $q=\infty$ (in that case the composition is indeed always compact).
I've been at it for a couple days trying to prove and also to disprove it. It might be trivial with a silly oversight on my part (because I don't know too many explicit examples of bounded operators from $L_p \to L_q$), but on the other hand it could be inherently nontrivial as well, and I was hoping to learn some new funfact about the geometry of Banach spaces in case of the latter.
Ideas: To prove that it is compact, here are a couple of ideas that I played around with. One suggestion is to somehow translate the problem to a statement about $\ell^p$ spaces and then apply Pitt's theorem. Unfortunately there seems to be no clear way to do this, for example any attempt at using a Fourier transform (which boundedly sends $L^p \to \ell^{\frac{p}{p-1}}$ and $\ell^p \to L^{\frac{p}{p-1}}$ for $1< p < 2$) will somehow "go the wrong way." A second idea is to try to find a Banach space $X$ which has the Dunford-Pettis property (e.g. a $L^1(\mu)$ or $C(K)$ space) and which nests in between $L^p$ and $L^q$, i.e., $L^q \subset X \subset L^p$. This would easily show compactness of $i\circ T$, but finding such $X$ seems not too easy. A third idea is to try to mimic the proof of Pitt's theorem adapted to this $L^p$ context. Basically suppose $i\circ T$ was not compact. Then we can find $f_n \in L^p$ with $\|f_n\|_p=1$ and $f_n \to 0$ weakly, and also $\|Tf_n\|_p \ge \delta>0$. Then (after perhaps passing to a subsequence) one may try to show that $T$ is bounded below on the closed linear span of the $f_n$, which would imply that the image of $T$ contains a closed infinite-dimensional subspace of $L^p$. And I'm farily certain that such a subspace cannot be contained in $L^q$ (never mind: this turns out to be false, see answer below). But formalizing these ideas might take some work. A fourth idea is to try to use type-cotype considerations which I don't know much about, but have been powerful in the context of classifying $L_p$ spaces up to isomorphism.
On the other hand, here are some ideas for potential counterexamples if it turns out to be false. For one idea let us identify $L^p[0,1] \simeq L^p(\Bbb R)$ and consider the Fourier transform $\mathcal F:L^p(\Bbb R) \to L^{\frac{p}{p-1}}(\Bbb R)$, where $p<2$. I've shown that $\mathcal F$ is not a compact operator. Hence there is at least a noncompact operator from $L^p[0,1] \to L^{\frac{p}{p-1}}[0,1]$ for $p<2$ (which brings up another question of what if we restrict attention to $q>p^*$, then can we say that $T$ itself is compact?... but perhaps that's for another day). However, the difficult thing is to show that it remains noncompact when we compose it with the inclusion map, and I think that it actually becomes false. I played around with the Hermite functions $\psi_n$ which orthonormally diagonalize $\mathcal F$ on $L^2(\Bbb R)$, and I was able to compute that $\|\psi_n\|_{L^p(\Bbb R)} \sim_n C_p n^{\frac1{3p}-\frac16}$, so while these converge weakly to $0$ in $L^2(\Bbb R)$ they fail to do so in $L^p(\Bbb R)$ for $p<2$, hence they are useless for us. A second idea is to use some of the abstract mappings mentioned in this MO thread and the subsequent comments. Actually this might all be trivial from some proposition in Albiac-Kalton but I can't access the book at the moment.
This might be a counterexample, but take it with some care. It builds on the fact that every $L^p(0,1)$ contains a Hilbert subspace which is complemented, see Remark on p. 143 in Johnson&Lindenstrauss
Let us call this subspace $X_p$. We get an isometric isomorphism $T_p : X_p \to \ell^2$ and a bounded projection $P_p : L^p(0,1) \to X_p$. Now, we consider the operator $$ T = I_{p,q} \circ T_q^{-1} \circ T_p \circ P_p ,$$ where $I_{p,q} : L^q(0,1) \to L^p(0,1)$ is the canonical embedding.
Now, we can make the spaces $X_p$ and $X_q$ a little bit more explicit. In fact, we can choose $X_p = X_q$ being a space generated by independent normally distributed variables $(g_n)_{n \in \mathbb N}$. Then, we should get $$ (T_q^{-1} \circ T_p)(g_n) = c_{p,q} \, g_n$$ for some constant $c_{p,q} > 0$. Hence, $T$ cannot be compact.