I'm currently trying to specialise a rather general variational inequality to known simple examples to check if my assumptions on the problem are plausible. While doing this, I stepped over the following question:
Given a function $u$ in some Sobolev space.
- (say, for instance, $u\in W^{1,2}_0(\Omega)$, $\Omega \subset \mathbb R^n$ bounded with smooth boundary)
Is it possible to approximate $u$ in norm by a sequence $(\varphi_n)$ of smoother functions
- (say, $\varphi_n\in C^\infty_0(\Omega)$, or just $\varphi_n\in C_0(\Omega)$, or $\varphi_n \in W^{1,p}_0(\Omega)$, where $p$ is given)
where all $\varphi_n$ are one-sided order-bounded by $u$.
- (that is, $\varphi_n(x) \geq u(x)$ for a.$\,$e. $x\in\Omega$)
Of course, if $u=0$, the standard approximation $(\varphi_n)$ by convolution is non-negative. But in the general case, the sign of $\varphi_n-u$ may change.
I found a result that goes in my direction in Function Spaces and Potential Theory by Adams and Hedberg, chapter 3.4 named "One-sided Approximation". There, $\Omega = \mathbb R^d$, and $-$ for the special case $u\leq 0$ a.$\,$e. $-$ there is at least some approximating sequence $(\varphi_n)\subset L^\infty$ with $\varphi_n \geq u$ a.$\,$e.
It would be very interesting to know if this is the best one can achieve. Thank you in advance.
A function $u \in W^{1,2}(\Omega)$ might be unbounded for $n \ge 2$. Hence, it is not possible by continuous functions.
In the case $p > 2$, I think one can develop an argument along the following lines: There is a function $u \in W^{1,2}(\Omega)$, which has a singularity "which is stronger as the space $W^{1,p}(\Omega)$ allows". Hence, the one-sided approximation is not possible in $W^{1,p}(\Omega)$.