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Nov 21, 2016 at 5:12 history edited Alexey Ustinov CC BY-SA 3.0
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Nov 21, 2016 at 5:01 comment added Alexey Ustinov @fedja Yes, nice argument!
Nov 20, 2016 at 19:26 comment added fedja Certainly not: 3 shifted Fejer kernels (triangles on the Fourier side) dominate the rectangle with the same base, so you can get $3$ instead of $4$ at no cost even if you replace $\le N$ with $<N$ on the RHS (assuming $N\ge 1$, of course). With such replacement $3$ gets sharp but I don't know if without it you can actually get $2$.
Nov 20, 2016 at 4:43 history edited Alexey Ustinov
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Nov 20, 2016 at 3:27 history asked Alexey Ustinov CC BY-SA 3.0