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Apr 17, 2013 at 14:57 history edited Julián Aguirre CC BY-SA 3.0
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Apr 6, 2013 at 18:14 comment added Ben McKay You want $\phi_{\varepsilon}(x)=\varepsilon^{-n}\phi(x/\varepsilon)$ to get a mollifier.
Jul 3, 2010 at 22:14 comment added Julián Aguirre I have included some more detail, but I have also realizaed that it is not clear at all that $\lim_{\varepsilon\to0}G_\varepsilon(t)=0$.
Jul 3, 2010 at 22:12 history edited Julián Aguirre CC BY-SA 2.5
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Jul 3, 2010 at 21:42 history edited Julián Aguirre CC BY-SA 2.5
Included more detail
Jul 2, 2010 at 5:23 comment added Vince Hi Julian, thanks for replying. I followed most of what you wrote, except I'm unclear about the last step. I am guessing you meant $g_\varepsilon$ in place of $g$? Could you please provide some insight as to how the inequality $\omega_\alpha(g_\varepsilon) \leq \int \phi(t)G_\varepsilon(t) dt$ holds? I ask because I thought I had a counterexample in mind in the boundary case of $\alpha = 1$, and it didn't seem like there was a dependence on the choice of $\alpha$, unless it appears in that particular inequality. Of course, I could be mistaken.
Jul 1, 2010 at 18:02 history answered Julián Aguirre CC BY-SA 2.5