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Jan 19, 2022 at 21:50 comment added Student Aah, of course. Thanks!
Jan 19, 2022 at 21:09 comment added Fedor Petrov $|f(x)g(x)-f(y)g(y)|\leqslant A(x,y)+B(x,y)$, where $A=|f(x)|\cdot |g(x)-g(y)|$, $B=|g(y)|\cdot |f(x)-f(y)|$. Now use the triange inequality bound $\|A\|+\|B\|\geqslant \|A+B\|$, where the norm is taken in the weighted $L^2(\mathbb{R}^n\times \mathbb{R}^n,1/|x-y|^{n+2s})$.
Jan 19, 2022 at 20:54 history asked Student CC BY-SA 4.0