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Jun 21, 2011 at 22:01 comment added Tapio Rajala Averaging on spheres is equivalent to estimating by the radial direction. What argument comes after that can be chosen in many ways. I simply put one argument which I found to be the simplest to verify by one line of calculations.
Jun 21, 2011 at 10:10 vote accept Tom Leness
Jun 21, 2011 at 6:56 comment added Andrew The reasoning simplify somewhat if to take an extention (for constant function 1) which depends on $r$ only? Such an extention can be obtained from an arbitrary one by averaging on shperes. Put $r_0=1$, $r_1=1+\varepsilon$. By stretching arguments (to annulus $1<r<2$) there seems to exisit an extention $f$ with estimates $\|\partial ^l f\|_{L_p(\mathbb R^n)}\le C \varepsilon^{(n-l)/p}$, $|l|\le k$. So norms of some of the derivatives yet tends to zero then $\varepsilon\to+0$.
Jun 20, 2011 at 12:48 history answered Tapio Rajala CC BY-SA 3.0