Talk:Riesz–Thorin theorem

message: The statement of the theorem seems to place undue restriction to the order of the indices. For example I want to choose p0=1, q0=∞, and p1=2, q1=2, but the theorem (as it's written here) seems to want me to ensure p0≤p1, and q0≤q1.
The choice of indices above is used in the motivation section for the Fourier transform, so the theorem should work for this case.  — Preceding unsigned comment added by 86.169.149.174 (talk) 16:48, 8 May 2014 (UTC)
thread_title:  Constraints on the order of the indices? 

message: What exactly does this mean: "Its usefulness stems from the fact that some of these spaces have much simpler structure than others (namely, L2 which is a Hilbert space, L1 and L∞). " ? It it not clear whether the "namely ... L1 and L∞" part applies to "much simpler" or "others". I suppose L1 and L∞ are simpler but from the sentence, I can't tell for sure. For L2, the qualificative "is a Hilbert space" helps disambiguate. Much less so for L1 and L∞. Do L1 and L∞ also have simpler structure than the other Lp ? (Or is it just a matter of having a "simpler" metric ?) --FvdP 19:35, 29 Oct 2004 (UTC)
replies: text: Of course they have a more complicated structure :-( Is the new formulation better? Gadykozma 20:09, 29 Oct 2004 (UTC)
replies: text: Not much (IMO...). If you really mean that L1 and L(inf) are examples of the "other" = "more complex" kind of algebras, it's better to tell it explicitly. If your mention of L1 and L(inf) is unrelated to the previous sentence, why not something like: "The spaces involved are often L2 (which is a Hilbert space), L1 and L^inf."" --FvdP 20:22, 29 Oct 2004 (UTC)

text: No, it would be incorrect to say that they are simpler, they are not. Particularly L^\infty is a non-separable behemoth. Nonetheless, sometimes they are easier to analyze, as in the examples. I guess "simple" is not so simple... Gadykozma 20:41, 29 Oct 2004 (UTC)

text: Yes, it's almost the same. L^p is L^q's dual and a Hilbert space is a self-dual Banach space. Gadykozma 20:41, 29 Oct 2004 (UTC)
thread_title:  ambiguous sentence "Its usefulness stems from ..." 

message: My question is what are the constraints on p and q in the theorem. Can they be 1 or infinity in addition to anything between, or just in between, or 1 but not infinity. This is because the theorem assumes L^p and L^q have a common dense subspace as domain, but on an infinite measure space if p=infinity and q<infinity then no subspace of L^q is dense in L^p, hence the assumptions of the theorem cannot be satisfied, it seems to me. Scineram (talk) 17:50, 19 May 2008 (UTC)
replies: text: Yes the cases 1 and ∞ are included.  To be a bit more formal the operator doesn't need to be defined on a subspace of Lp and Lq, but rather on a space (say D) so that D\cap L^p is dense in Lp and D\cap L^q is dense in Lq.  There are a few mild assumptions about D, it should contain characteristic functions of finite measure sets and should contain truncations. Thenub314 (talk) 09:54, 10 October 2008 (UTC)
thread_title:  Constraints on p and q? 

message: "This theorem bounds the norms of linear maps acting between Lp spaces." - This is true, but not the entirety of the theorem and very misleading. The theorem is actually proved from one linear space, D, of measurable functions on a general measure space to another space of measurable functions on a general measure space as long as D contains all characteristic functions and truncations. Grimtageuk (talk) 14:45, 12 December 2012 (UTC)
thread_title:  Generality of Theorem 

message: I think there is a mistake here. I tried in vain to prove the statement "  is convex " but, in fact, I think is the map  to be convex (see Lemma 2 here --Gim²y (talk) 17:12, 27 July 2014 (UTC)
thread_title:  Mistake log-convexity of <math>L^p</math> norm