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The Riesz Thorin theorem allows us to interpolate between $L^p$ spaces and the usual assumption is that the map $T$ is linear.

What I was wondering about is whether this is because otherwise you do not know what $T$ on the interpolated spaces shall be or whether this is an intrinsic property of the theorem?

That's why I ask: Assume that the measure space is of finite measure and you interpolate a map $T:L^p \rightarrow L^p$ that is bounded but non-linear such that $T\vert_{L^q}: L^q \rightarrow L^q$ is also bounded and $q>p.$

Does this imply that $T$ restricts to a bounded map on any $L^r$ with $r \in (p,q)$ and an estimate on the operator norm is given by the Riesz-Thorin interpolation theorem?

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  • $\begingroup$ I assume you meant $q > p$? With no further assumptions on $T$, you can set $T u = 0$ for $u \in L^q$ and $T u = u_0$ for $u \in L^p \setminus L^q$, where $u_0$ is a fixed element in $L^p$ that does not belong to $L^r$ for any $r > p$. Then $T$ restricted to $L^r$ does not even take values in $L^r$ for $r \in (p, q)$. $\endgroup$ Commented Aug 7, 2018 at 7:47
  • $\begingroup$ Sorry, yes that is what I meant $\endgroup$
    – user127450
    Commented Aug 7, 2018 at 7:49
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    $\begingroup$ This falls under the general topic of nonlinear interpolation which is in general quite nontrivial. There are lots of works on this (Lions/Peetre themselves have done some, Tartar, Cwikel, ...). A recent article with a lot of references (and a positive, albeit restrictive result for your question) would be "Interpolation of Nonlinear Maps" by Kappeler, Savchuk, Shkalikov and Topalov. $\endgroup$
    – Hannes
    Commented Aug 7, 2018 at 8:26
  • $\begingroup$ @Hannes sorry, which Theorem are you referring to? $\endgroup$
    – user127450
    Commented Aug 7, 2018 at 10:38
  • $\begingroup$ Well, Theorem 1 in the paper, no? For the particular $L^p$ situation, there also seems to be an analogous result proven by Elias Stein in his dissertation in 1954---at least that is what the Wikipedia article about Riesz-Thorin tells me :-) $\endgroup$
    – Hannes
    Commented Aug 7, 2018 at 13:02

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