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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.

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1 Answer 1

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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 $W^{1,p}(\Omega)$.

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  • $\begingroup$ Yes, you're right; for general unbounded functions there is really no hope. But I think there is some hope in the direction outlined above, namely $u\leq 0$. (In the book they approximated $u$ "from inside" with functions $\varphi_n$ fulfilling $|\varphi_n| \leq |u|$ and $\varphi_n u\geq 0$. $\endgroup$
    – Dreipunkt
    Commented Jun 24, 2016 at 11:03
  • $\begingroup$ You could take any smooth approximation $\varphi_n$ and consider $\max(u, \varphi_n)$. This is pointwise greater than $u$, has at least the same regularity of $u$, and converges in $W^{1,2}$. Again, I think that you cannot get better regularity in the general case. $\endgroup$
    – gerw
    Commented Jun 24, 2016 at 12:57

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