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Consider the standard open ball $B_r:=\left\{x ; \left\lvert x \right\rvert \le R \right\}.$

The trace theorem tells us any function in $W^{k,p}(B_r)$ can be restricted to a function $W^{k-1,p}(\partial B_r)$ on the boundary in a continuous linear way such that

$$\left\lvert Tv \right\rvert_{W^{k-1,p}(\partial B_r)} \le C \left\lvert v \right\rvert_{W^{k-1,p}(\partial B_r)}$$ where $v$ is I was wondering whether the scaling of $C$ with $r$ is known? I suspect that $C$ grows (as r) increases like the area of the boundary, i.e. like $R^{n-1}$ where $n$ is the dimension, but I do not know. Perhaps, there are even explicit constants for balls?

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  • $\begingroup$ try the case of the standard trace theorem first. And then consider $w(x)=v(r x)$ for $ |x|<1$... writing out the trace theorem for $ w(x)$ should give you want you want... $\endgroup$ – Math604 Aug 10 '18 at 5:42

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