Here is a problem that I have encountered in PDE book several times. But I have never seen a proof of it. I will be very grateful if someone could give me a proof.

Question: Let $B$ be the unit ball in $\mathbb{R}^n$, $f$ a non-negative function in $H_0^1(B)$, prove that there exists a sequence of non-negative functions $\varphi_k\in C_c^\infty(B)$ such that $\varphi_k\rightarrow f$ in $H_0^1(B)$.

Edit: What if we replace $B$ by a general domain $\Omega$?

Edit II: Thanks to Hans's idea (which should work for any star-shaped domain), if the boundary of $\Omega$ suitably good (for example, it admits finite covering of star-shaped open sets), then using partition of unity we should be able to construct the desired approximation.

Edit III: If I didn't make any mistake. L.C. Evans's Partial Differential Equations (First Edition) page 260 gives a proof for $C^1$ domain. Although he was actually proving something else, the key ingredient works in our situation!

  • $\begingroup$ An idea: Make $f$ zero outside the ball of radius $a<1$ (use an appropiate cut-off function) and apply the usual regularization trick, in fact the standard mollifiers give you a sequence with the desired properties, let $a\to 1$. I'm assuming that you mean $H^1=H_0^1$. $\endgroup$ – Jose27 Oct 29 '11 at 1:52
  • $\begingroup$ @Jose27: Yes, it should be $H_0^1(B)$. Thanks. I have tried this idea, but when I cutoff $f$ at radius $a$, a large gradient (of the cutoff function) shows up, and I don't know how to control it. $\endgroup$ – Syang Chen Oct 29 '11 at 2:07
  • $\begingroup$ You're right, and I don't think this issue is easy to solve. Another idea: If one could prove that $u\mapsto |u|$ is a continous function from $H_0^1$ to itself, we could reduce the issue to the case in which $f$ has compact support in $B$, and in this case the standard mollifiers give an answer. (I haven't been able to prove the first claim however) $\endgroup$ – Jose27 Oct 29 '11 at 7:09
  • $\begingroup$ Good point. The map $u\mapsto |u|$ is indeed continuous at $0$. But it's hard to see whether it's continuous at other points.. $\endgroup$ – Syang Chen Oct 29 '11 at 16:36
  • $\begingroup$ If I'm interpreting this correctly, exercise 18 in chapter 5 of Evans' PDE book implies that $|D_iu|=D_i|u|$, and from the triangle inequality, the claim would follow. $\endgroup$ – Jose27 Oct 29 '11 at 17:39

Let $\psi$ be a standard non-negative positive mollifier, $C^\infty$, supported on the unit ball in $\mathbb{R}^n$, with $\int_{\mathbb{R}^n} \psi = 1$. Define $\psi_k(x) = k^n \psi(kx)$ as usual. Extend $f$ trivially outside the unit ball and define $\tilde f_k(x) = f(kx/(k-1))$ for all $x$, for $K > 1$. This function is now supported on the ball with radius $1 - k^{-1}$ and is non-negative and in $H^1_0$. Show that $\tilde f_k \to f$ in $H^1_0$. (It is enough to show weak convergence in $L^2$ and convergence of the norm, since we are working in a Hilbert space.)

Then define
$$ \varphi_k(x) = \int_{\mathbb{R}^n} \psi_k(x-y)\tilde f_k(y) dy $$ as usual. Then $\varphi_k$ is $C^\infty$, supported in the unit ball, and non-negative. Apply a triangle inequality argument to show that $\varphi_k \to f$ in $H^1_0$.

  • $\begingroup$ Nice! Is it possible to generalize the proof to general domain $\Omega$? When I asked the question, I was actually expecting a proof that works for general domain. $\endgroup$ – Syang Chen Nov 1 '11 at 18:14
  • $\begingroup$ Thx for the bounty :) You can probably generalize this to the case of a general compact domain with smooth ($C^1$ or better) boundary. Let $\Omega_s = \{ x \in \Omega | dist(x, \partial \Omega) \ge s\}$. For sufficiently small $s$ say $0 \le s \le s_0$, there is a diffeomorphism $\Phi_s$ from $\Omega_s$ onto $\Omega$ that is the identity on $\Omega_{2s_0}$. First approximate $f \in H_0^1$ by $f_k(x) = f(\Phi_{k^{-1}}(x))$. Then $ supp \, f_k \subset \Omega_{k^{-1}}$ and $f_k = f$ on $\Omega_{2s_0}$. Then approximate $f_k$ as before. I have to admit that this approach is a bit cumbersome. $\endgroup$ – Hans Engler Nov 5 '11 at 13:19
  • $\begingroup$ Yes, this is another option. But the analysis will be a bit tough (at least for me).. $\endgroup$ – Syang Chen Nov 5 '11 at 16:53
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    $\begingroup$ Alternative (if you have some semigroup theory): Let $S_t$ be the heat semigroup on this domain with 0 Dirichlet boundary data. Set $\varphi_k = S_{k^{-1}} f$. Done. $\endgroup$ – Hans Engler Nov 5 '11 at 17:26
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    $\begingroup$ Just add $\frac{1}{k}$ to the functions that were constructed above. $\endgroup$ – Hans Engler Jun 25 '15 at 21:28

Are you getting mixed up with the definitions?

Let $\Omega\subset\mathbb{R}^n$ be open. $H^{1}_0(\Omega)=H^{1,2}_0(\Omega)$ is $\mathbf{defined}$ to be the closure of $C^{\infty}_c(\Omega)$ in the $W^{1,2}(\Omega)$ norm, which is the norm that $H^{1,2}_0(\Omega)$ inherits.

Hence, $f\in H^{1,2}_0(\Omega)$ if and only if there exists a sequence $(\varphi_n)$ in $C^{\infty}_c(\Omega)$ such that $\varphi_n\longrightarrow f$ in $H^{1,2}_0(\Omega)$, by definition.

I.e. every function in $H^{1,2}_0(\Omega)$ is either a function in $C^{\infty}_c(\Omega)$, or is the limit of some sequence in $C^{\infty}_c(\Omega)$, under the $W^{1,2}(\Omega)$ norm.

There is nothing to prove.

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    $\begingroup$ I want the approximating sequence to be nonnegative, that's the point. $\endgroup$ – Syang Chen Nov 5 '11 at 3:48

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