# Star-shaped domain and approximation by smooth functions

Let $$1 \leq p < \infty$$ and let $$\Omega \subset \mathbb{R}^n$$ be a domain. Suppose $$\Omega$$ is star-shaped in the sense that there exists $$x_0 \in \Omega$$, such that for every $$x \in \Omega$$, the line segment connecting $$x$$ and $$x_0$$ stays in $$\Omega$$. Let $$u \in W^{1,p}(\Omega)$$, and for $$\lambda > 0$$ define $$\Omega^{\lambda} = \{x : x/ \lambda \in \Omega \}$$ and $$u^{\lambda}(x) =u(x/ \lambda)$$ for any $$x \in \Omega^{\lambda}$$.

Using these definitions, one may easily show that $$u^{\lambda} \in W^{1,p}(\Omega^{\lambda})$$. Next, I aim to show that $$C^{\infty}(\bar{\Omega)}$$ is dense in $$W^{1,p}(\Omega)$$, by showing $$u$$ can be approximated by functions in $$C^{\infty}(\bar{\Omega)}$$. The idea is to suitably mollify $$u^{\lambda}$$ with $$\lambda$$ close to $$1$$ in order to subsequently approximate $$u$$. However, I do not know how to proceed and even if I did, how this would lead to the solution. What confuses me is that the introduction of a mollifier would introduce another variable whose limit we have to take, and I also do not see how this approach would lead to working with $$\bar{\Omega}$$.

To elaborate, if we mollify with a smooth function $$\rho \in C_{c}^{\infty}(B_1(0))$$ with compact support in the unit ball, we'd introduce a new convolution $$u_{\lambda}^{\epsilon} := \rho_{\epsilon} * u_{\lambda}$$, then this is only well-defined on the set: $$$$\Omega_{\epsilon}^{\lambda} = \{ y \in \Omega_{\lambda} | B_{\epsilon}(y) \in \Omega^{\lambda}\}.$$$$ So although these approximations are infinitely smooth and can be shown to converge in $$L^{p}_{loc}(\Omega^{\lambda})$$ to $$u^{\lambda}$$ as $$\epsilon \to 0$$ for $$\lambda$$ fixed, I don't see how to handle the limit $$\lambda \to 1$$ and how one would get the closure of $$\Omega$$ involved.

I am not necessarily looking for a full solution - any hints or intuition would be very appreciated as well. Thanks!

## 1 Answer

I just remembered that this question was never answered and that I was working with a wrong formulation of the problem. The proper definition of star-like here is that the segment $$(xx_0]$$ is in $$\Omega$$, for any $$x \in \overline{\Omega}$$ (the original formulation of star-like is a bit too constraining). Then a sketch of a solution would go as follows:

WLOG suppose that $$x_0 =0$$, so the domain is star-like about the origin. By continuity of the dilation in $$L^p$$, we can obtain via routine calculations: $$$$||u - u^{\lambda}||_{W^{1,p}(\Omega)} \to 0 \text{ as \lambda \to 1.}$$$$

The next idea is to observe that the star-shapedness means that $$\overline{\Omega} \subset \Omega^{\lambda}$$ is a compact subset for $$\lambda >1$$, then for all $$\epsilon \ll 1$$ sufficiently small, we can take $$u^{\lambda} * \rho_{\epsilon} \in C^{\infty}(\overline{\Omega})$$. Moreover: $$$$||u^{\lambda}* \rho_{\epsilon} - u^{\lambda}||_{W^{1,p}(\Omega)} \to 0 \text{ as \epsilon \to 0.}$$$$

So we're pretty close to finished, but the issue is that we need to rigorously reconcile the two limits of $$\epsilon$$ and $$\lambda$$. So given a fixed $$\delta > 0$$, we take $$\lambda_{\delta}$$ close to $$1$$ so that: $$$$||u^{\lambda_{\delta}} - u||_{W^{1,p}(\Omega)} < \frac{\delta}{2}.$$$$ Then, select $$\epsilon = \epsilon(\delta)$$ sufficiently small, so that: $$$$||u^{\lambda_{\delta}} * \rho_{\epsilon} - u^{\lambda_{\delta}}||_{W^{1,p}(\Omega)} < \frac{\delta}{2},$$$$ and so that $$u^{\lambda_{\delta}} * \rho_{\epsilon} \in C^{\infty}(\overline{\Omega})$$. So by the triangle inequality, $$$$||u^{\lambda_{\delta}} * \rho_{\epsilon} - u||_{W^{1,p}(\Omega)} < \delta,$$$$ so $$C^{\infty} (\overline{\Omega})$$ is dense in $$W^{1,p}(\Omega)$$.