Convolution mollification of $H^s$ functions uniformly in the unit ball of this sobolev space

Question

Let $\phi$ be a nonnegative $C_c^\infty(B(0,1))$ function, where $B(0,1)\in \mathbb R^n$ is the unit ball, and $\int \phi =1$. Let $\phi_{\epsilon}(x) =\epsilon^{-n} \phi(x/\epsilon).$ For any $L^2$ function $u$, we can get a $C^\infty$ approximation by taking $u\ast \phi_\epsilon.$ The problem I am focusing on here is: do we have $ \lim_{\epsilon \to 0} \|u\ast \phi_\epsilon - u\|_{L^2(\mathbb R^n)} =0$ uniformly in $\{u\in H^s(\mathbb R^n): \|u\|_{H^s} \leq 1\} ?$

Clearly, this is not true for $s< 1/2,$ since for this range of $s,$ the indicator function $\chi$ of $B(0,1)$ lies in $H^s.$ This can be used to construct a example for how uniform convergence fails, due to the ineffectiveness of mollification against jump discontinuities.

Also, clearly this is true for $s\geq 1,$ because the mollification error can be easily estimated by the (weak) gradient of $u.$

What about the range of value in the middle, $s\in [1/2,1)?$ do we have uniform convergence for these $s?$

Answer 1

This works for all $s>0$. If you take Fourier transforms (and write $\widehat{\varphi}=\psi$), then you are asking if $$ \lim_{\epsilon\to 0}\sup_{\|(1+|t|^s)\widehat{u}\|=1}\|\widehat{u}(\psi(\epsilon t)-1)\| =0 . $$ That's the same as asking if $T_{\epsilon}\to 0$ in operator norm on $L^2$, where $$ (T_{\epsilon}v)(t)= \frac{\psi(\epsilon t)-1}{1+|t|^s} v(t). $$ The norm $M(\epsilon)=\|T_{\epsilon}\|$ of a multiplication operator is given by $$ M(\epsilon)=\sup_{t\in\mathbb R} \left| \frac{\psi(\epsilon t)-1}{1+|t|^s} \right| . $$ By considering separately $|t|<\epsilon^{-1/2}$ and $|t|\ge\epsilon^{-1/2}$ (say), we see that $M(\epsilon)\to 0$, for any $s>0$.