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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: \begin{equation} \Omega_{\epsilon}^{\lambda} = \{ y \in \Omega_{\lambda} | B_{\epsilon}(y) \in \Omega^{\lambda}\}. \end{equation} 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!

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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: \begin{equation} ||u - u^{\lambda}||_{W^{1,p}(\Omega)} \to 0 \text{ as $\lambda \to 1$.} \end{equation}

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: \begin{equation} ||u^{\lambda}* \rho_{\epsilon} - u^{\lambda}||_{W^{1,p}(\Omega)} \to 0 \text{ as $\epsilon \to 0$.} \end{equation}

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: \begin{equation} ||u^{\lambda_{\delta}} - u||_{W^{1,p}(\Omega)} < \frac{\delta}{2}. \end{equation} Then, select $\epsilon = \epsilon(\delta)$ sufficiently small, so that: \begin{equation} ||u^{\lambda_{\delta}} * \rho_{\epsilon} - u^{\lambda_{\delta}}||_{W^{1,p}(\Omega)} < \frac{\delta}{2}, \end{equation} and so that $u^{\lambda_{\delta}} * \rho_{\epsilon} \in C^{\infty}(\overline{\Omega})$. So by the triangle inequality, \begin{equation} ||u^{\lambda_{\delta}} * \rho_{\epsilon} - u||_{W^{1,p}(\Omega)} < \delta, \end{equation} so $C^{\infty} (\overline{\Omega})$ is dense in $W^{1,p}(\Omega)$.

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