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Dec 17, 2015 at 9:39 comment added Piero D'Ancona Just by rescaling. The norm of $f_a$ is a power of $a$ times the norm of $f$. When $a\to0$ the behaviour is the one written above
Dec 16, 2015 at 22:03 comment added Dee Sorry, now that I've written it down, I don't see how sobolev norm of $f_a$ is calculated?
Dec 16, 2015 at 15:14 comment added Piero D'Ancona Take a test function $f(x)\ge0$, $0<s<-1+n/p$, and let $f_a(x)=a^{-s}f(x/a)$ for $a>0$. The Sobolev norm of $f_a$ is $\simeq a^{-s-1+n/p}\to0$ as $a\to0$, while the sup norm of $f_a$ tends to $\infty$. Thus we can modify locally the sign of a function by adding or subtracting a perturbation $f_a$ arbitrarily small in the Sobolev norm.
Dec 16, 2015 at 13:51 review First posts
Dec 16, 2015 at 13:54
Dec 16, 2015 at 13:42 history asked Dee CC BY-SA 3.0