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May 19, 2017 at 16:24 comment added Paata Ivanishvili Thank you. By the way if $f$ is 0-1 indicator of a set $A \subseteq \{-1,1\}^{n}$ then $|\nabla \chi_{A}| = \frac{\sqrt{w_{A}(x)}}{2}$ and Bobkov's inequality gives $C(1)\geq \sqrt{\frac{2}{\pi}}\approx 0.79$. But Hamming balls give $C(1)\leq \frac{2}{\sqrt{\pi}}$ which is worse than $C(1)\leq 1$. So the optimizer cannot be Hamming ball in this particular case
May 19, 2017 at 13:00 history answered Ryan O'Donnell CC BY-SA 3.0