2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 Jun 24 '16 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 Jun 24 '16 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.