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.)

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    $\begingroup$ I am aware of a direct argument that explains why the convolution operator (wrt to a generating probability measure) has norm less than 1 on each $L_p^0(G)$, $1<p<\infty$ ("direct"="not relying on interpolation" in this context). I can elaborate on that, but it does not answer the question. $\endgroup$ – Uri Bader Jun 15 '16 at 7:37
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    $\begingroup$ By "generating probability measure" I mean "not supported on a proper closed subgroup". This is necessary, otherwise $\|T\|=1$ on $L^0_p$ for every $p$. $\endgroup$ – Uri Bader Jun 15 '16 at 7:43
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    $\begingroup$ @UriBader: well, you first comment does answer Q2, no? $\endgroup$ – Mikael de la Salle Jun 15 '16 at 8:04
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    $\begingroup$ Bill, in Q1 do you assume that $T$ maps the constant function $1$ to itself? $\endgroup$ – Mikael de la Salle Jun 15 '16 at 8:09
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    $\begingroup$ On the other hand, the answer to Q2 is (positive and) easier for convolution by finitely supported measures. I guess that this will be clear in Uri's forthcoming answer. $\endgroup$ – Mikael de la Salle Jun 15 '16 at 8:58

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.

  • $\begingroup$ 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). $\endgroup$ – Uri Bader Jun 15 '16 at 9:44
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    $\begingroup$ 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 $n$. $\endgroup$ – Mikael de la Salle Jun 15 '16 at 9:44
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    $\begingroup$ Thanks for the reminder, Mikael. It is indeed a better argument as it gives more info on general measures and on Q1. $\endgroup$ – Uri Bader Jun 15 '16 at 9:56
  • $\begingroup$ 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. $\endgroup$ – Bill Johnson Jun 15 '16 at 10:41

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