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Aug 12, 2016 at 5:22 vote accept John
Aug 11, 2016 at 16:40 comment added Christian Remling @D G: What I was trying to say is that $f=x$ on $[-1,1]$, and outside this interval I define $f$ in such a way that $f\in C_0^{\infty}$.
Aug 11, 2016 at 10:58 comment added D G I agree with the example by Christian Remling, only pointing out that $f \in C^\infty$, since $f(\pm 1) = \pm 1 \ne 0$.
Aug 10, 2016 at 17:08 history answered Christian Remling CC BY-SA 3.0