Interpolation between $L_1^0$ and $L_2^0$ Let $L_p^0$ be the mean zero functions in $L_p(G)$, where, say, $G$ is an infinite compact group endowed with normalized Haar measure. Suppose that $T$ is a bounded linear operator on $L_1$ that maps $L_1^0$ into itself and $L_2$ into $L_2$.  Suppose that $\|T\|_{L_1 \to L_1} =1$ and $\|T\|_{L_2^0\to L_2^0} <1$.  
Q1: If $1<p<2$, must $\|T\|_{L_p^0\to L_p^0} < 1$?
Q2: What if, in addition,  $T$ is given by convolution with respect to a probability measure on $G$?
Several papers claim that Q2 has an affirmative answer in such a way that they seem to imply that Q1 also has an affirmative answer.  (The claim is expressed by something like “By the Riesz convexity theorem”, but I do not see that it follows from either the theorem or the standard proof of the theorem.)
 A: This is a partial answer to Q2 that has nothing to do with interpolation. It concerns the case of finitely supported measures. 
Here I am not trying to give general criteria for $\|T\|<1$, as this is a hard question, only to relate this "spectral gap" property for different $p$'s.
Remark: Currently I am still unsure about the general case (measures which are not finitely supported). This is incompatible with a comment I made to the original post, which was too optimistic. However using the method Mikael indicates in a comment to this post one can say something, namely that the property of having $\|T^n\|<1$ for some $n$ holds for every $p\in (1,2)$.
Let $G$ be a compact group, $\mu$ a finitely supported probability measure on $G$. Let $T_p:L^0_p(G)\to L^0_p(G)$ be the operator given by left convolution with $\mu$.
Claim: Either for every $1<p<\infty$, $\|T_p\|=1$ or for every $1<p<\infty$, $\|T_p\|<1$.
Let $\Gamma<G$ be the (countable) group generated by the support of $\mu$.
We assume as we may that $\Gamma$ is dense in $G$, otherwise $\|T_p\|=1$ for all $p$. It follows that there are no $\Gamma$-invariant vectors. By strict convexity of the norm we conclude that there are no $T_p$-invariant vectors. Moreover, by the uniform convexity of the norm it is not hard to see the equivalence (for a given $p$): $T_p$ has almost invariant vectors iff $\Gamma$ has almost invariant vectors.
The former is equivalent to $\|T_p\|=1$. However the latter is independent of $p\in (1,\infty)$. 
Indeed,
for every $p,q$ we can define the (non-linear) Mazur map $L_p\to L_q$, $f\mapsto \text{sgn}(f)|f|^{p/q}$ which has two nice properties: it is uniformly continuous on the sphere and it commutes with isometries, eg with $\Gamma$ (this is for $p\neq 2$, by Banach-Lamperti).
For more details see section 4 in http://arxiv.org/pdf/math/0506361v2.pdf.
It follows that the statement $\|T_p\|=1$ is independent of $p$ which proves the claim.
