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Jun 18, 2016 at 15:19 vote accept Bill Johnson
Jun 17, 2016 at 6:35 history edited Uri Bader CC BY-SA 3.0
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Jun 15, 2016 at 10:41 comment added Bill Johnson Thanks very much, Uri and Mikael! I'll not accept Uri's answer right away to see if someone can answer Q1, but ya'll have answered the question that was important for Gideon and me.
Jun 15, 2016 at 9:56 comment added Uri Bader Thanks for the reminder, Mikael. It is indeed a better argument as it gives more info on general measures and on Q1.
Jun 15, 2016 at 9:44 comment added Mikael de la Salle One can also use interpolation instead of Mazur maps (this is something that Uri and I discussed a long time ago): indeed, since the natural projection on $L_p^0$ has norm $\leq 2$, interpolation leads to the inequality $\|T_{p_\theta}\| \leq 2 \|T_{p_0}\|^{1-\theta} \|T_{p_1}\|^\theta $. This implies that the property $\exists n, \|T_p^n \|<1$ does not depend on $1<p<\infty$, and by the same uniform convexity argument in Uri's answer the property $\|T_p^n\|<1$ is equivalent to the representation having no almost invariant vectors, and in particular does not depend on $n$.
Jun 15, 2016 at 9:44 comment added Uri Bader To clarify the role of "finite support" in the above answer let me make the following comment: the above works whenever $\mu$ could be identified as a pushed measure under a continuous homomorphism $\Gamma\to G$, where $\Gamma$ is a compactly generated locally compact group (eg, f.g as in the above answer). Then we can see $L_p^0(G)$ as a $\Gamma$ space and relate the spectral gap of $T$ with a.i of $\Gamma$. Currently I am unsure about general $\mu$. Some of my comments up there were premature and not enough well thought of (sorry).
Jun 15, 2016 at 9:30 history answered Uri Bader CC BY-SA 3.0