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Talk:Riesz-Thorin theorem - Wikipedia, the free encyclopedia

Talk:Riesz-Thorin theorem

From Wikipedia, the free encyclopedia

[edit] ambiguous sentence "Its usefulness stems from ..."

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)

Of course they have a more complicated structure :-( Is the new formulation better? Gadykozma 20:09, 29 Oct 2004 (UTC)
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)
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)

BTW, just wondering: In your example I see theorems that link L^p and L^q where 1/p + 1/q = 1. So when p=q you get theorems about L^2 alone. Is this related to L^2 being simpler and/or Hilbert ? (I know I am probably only exposing my utter ignorance of the matter by asking this question ;-) --FvdP 20:28, 29 Oct 2004 (UTC)

Yes, it's almost the same. Lp is Lq's dual and a Hilbert space is a self-dual Banach space. Gadykozma 20:41, 29 Oct 2004 (UTC)

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