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Aug 23, 2014 at 2:48 comment added Terry Tao Generally the situation with $p \leq 1$ is quite pathological, particularly the non-convex, non-locally integrable case $p<1$. Things are better if one uses Hardy spaces instead of Lebesgue spaces though.
Aug 23, 2014 at 1:05 comment added timur @TerryTao: What happens for $0<p\leq1$? Is there a Littlewood-Paley theory in this case?
Aug 21, 2014 at 0:27 comment added timur @Dunham: That is how DeVore and Popov proved their Besov space interpolation results. I think the idea goes back to Nikolsky's school. What do you have in mind as easier approaches?
Aug 20, 2014 at 20:28 comment added Dunham We know the equivalence between interpolation spaces and approximation spaces, so you could try to directly show the equivalence between the approximation space norms and Besov norms for band-limited approximation. However, it seems that there are easier approaches to this problem.
Aug 20, 2014 at 6:51 comment added timur @TerryTao: It seems so, and it seems not as complicated as I initially thought. But probably it would be a nonstandard application of CZ theory (The spaces I have are approximation spaces related to multiresolution analysis defined by piecewise polynomials on triangulations). Thank you for your input!
Aug 20, 2014 at 5:35 comment added Terry Tao Ah, I see now. (I had misread the Wikipedia entry.) In that case, one probably has no choice but to use some Calderon-Zygmund theory (as is for instance implicit in the statement that $L^p = F^0_{p2}$) as otherwise one doesn't seem to be able to handle all the varying indices simultaneously.
Aug 20, 2014 at 3:54 comment added timur @TerryTao: Thanks for looking this up. Actually I have the book and the theorem is identical to what is on Wikipedia. If you look at the conditions on the indices, they require $t=p$ because we already fixed the indices of other two spaces. The theorem in Bergh and Lofstrom follows from another earlier theorem on interpolation between Banach valued sequence spaces, and if the indices are general one would get Besov spaces (with 4 indices) based on Lorentz spaces $L_{p,q}$.
Aug 20, 2014 at 3:27 comment added Terry Tao I don't believe t=p is necessary. Bergh and Lofstrom (for instance) should have details; I don't have access to it here, but the Wikipedia page en.wikipedia.org/wiki/… asserts that the required interpolation result is in Theorem 6.4.5 of that text.
Aug 20, 2014 at 2:33 comment added timur I mean, we will have $[B^0_{p,1},B^s_{q,q}]_{\theta,r}\subset[L_{p},B^s_{q,q}]_{\theta,r}\subset[B^{0}_{p,\infty},B^s_{q,q}]_{\theta,r}$ but (from what I understand) the spaces on the two sides are not Besov spaces.
Aug 20, 2014 at 1:10 comment added timur @TerryTao: I think in order for $[B^0_{p,t},B^s_{q,q}]_{\theta,r}=B^\alpha_{r,r}$ to happen we must have $t=p$.
Aug 20, 2014 at 0:43 comment added Terry Tao I haven't checked this carefully, but perhaps one could simply use the inclusions $B^0_{p,1} \subset L_p \subset B^0_{p,\infty}$ together with Besov space interpolation?
Aug 20, 2014 at 0:25 comment added timur @Dunham: Reiteration approach seems to require writing $L_p$ as an interpolation space between Besov spaces, which I think is not possible (By interpolation you cannot get out of the Besov scale).
Aug 20, 2014 at 0:19 history edited timur CC BY-SA 3.0
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Aug 19, 2014 at 22:45 history edited timur CC BY-SA 3.0
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Aug 19, 2014 at 17:14 answer added Joonas Ilmavirta timeline score: 2
Aug 19, 2014 at 17:11 answer added Dunham timeline score: 2
Aug 19, 2014 at 16:37 comment added Dunham I suspect that this can be derived using the reiteration theorem for interpolation.
Aug 19, 2014 at 16:10 history asked timur CC BY-SA 3.0