Firstly, Nf is define for whole R^n, but f is defined in a bounded domain(my previous comment just means this).

Secondly, if x is strictly away from \Omega, then Nf is differential obviously. The convolution has no singularity near x.

Last, I think the convergence can be made uniformly on R^n. Take a larger domain than \Omega and D_{x}w_\epsilon will converge to DNf. Outside that domain, you have control obviously. I think that works, but I did not really write it down.

]]>btw is your source from Gilbarg and Trudinger? I ask because there’s a very similar derivation in that book and they also conclude uniform convergence in compact subsets of R^n. It’s a rather minor issue in the context of this proof imo and doesn’t change the result but still just curious…

]]>Did I miss something or is it a gap?

Thanks.

]]>Did I miss something or is it a gap?

Thanks. ]]>