Dense subspace of space of radial functions in $H^1_0(\Omega)$

Question

Consider $\Omega\subset \mathbb{R}^N$ a bounded domain, which is radial. By definition, the space of radial functions in $H^1_0(\Omega)$ is $$ H^1_{0, rad}(\Omega) = \{u \in H^1_0(\Omega) : u = u \circ R, \forall R \in O(N)\}. $$ It is very common in papers the authors just sai that $H^1_{0,rad}(\Omega) = \overline{C^\infty_{0,rad}(\Omega)}^{H^1_0(\Omega)}$, without more comments, where $$ C^\infty_{0,rad}(\Omega) = \{u \in C^\infty_{0}(\Omega) : u \text{ is radial}\}. $$ I am trying to find a precise proof of this. If we replace $\Omega$ by $\mathbb{R}^N$ I know how to prove using standad molifications. The hard part for me in the case of a bounded domain it is to conclude that the molification of a function in $H^1_{0,rad}(\Omega)$ has its support in $\Omega$. Any help with this or just a reference with this fact will be appreciated.

Answer 1

In the other answer it is noted that one can redo the entire construction as in the non-radial case by choosing the mollifier and the cut-off function radially. Another approach is to directly utilize the fact that we already know the characterization of $H_0^1(\Omega)$.

Let $u\in H_{0,\mathrm{rad}}^1(\Omega)$. Then in particular we know that $u\in H_0^1(\Omega)$ and $u$ is radial. As $u\in H_0^1(\Omega)$ we know that there exists a sequence $(u_n)_{n\in \mathbb{N}} \subseteq C_c^\infty(\Omega)$ such that $\Vert u-u_n\Vert_{H^1(\Omega)} \rightarrow 0$. The issue here is of course that the $u_n$ need not be radial. Thus, we define $$ v_n(x) = \int_{O(N)} u_n(R x) dR $$ where $dR$ is the Haar measure on $O(N)$. One readily checks (change of variables and the fact that $dR$ is $O(N)$-invariant) that $v_n\in C_c^\infty(\Omega)$ is radial.

Next we are going to use the fact that $\Vert u_n - u\Vert_{H^1(\Omega)}\rightarrow 0$ to conclude $\Vert v_n-u\Vert_{H^1(\Omega)} \rightarrow 0$. First we going to compute some gradients. Namely, if $R\in O(N), g\in C^1(\Omega)$ and $x\in \Omega, v\in \mathbb{R}^N$, then we have $$ \langle \nabla (g\circ R)(x), v\rangle = D(g\circ R)(x)[v] = Df(Rx) [Rv] = \langle (\nabla g)(Rx), Rv \rangle = \langle R^{-1} (\nabla g)(Rx), v \rangle.$$ Hence, we have $$ \nabla (g\circ R)(x) = R^{-1} (\nabla g)(Rx). $$ With this at hand, we can do the estimates. Using Minkowski's integral inequality we get \begin{align*} \left( \int_\Omega \vert \nabla(v_n(x)-u(x)) \vert^2 dx \right)^{1/2} &= \left( \int_\Omega \vert \int_{O(N)} \nabla[(u_n\circ R)(x)-(u\circ R)(x)] dR \vert^2 dx \right)^{1/2} \\ &\leq \int_{O(N)} \left( \int_\Omega \vert \nabla [(u_n\circ R)(x)-(u\circ R)(x) \vert^2 dx \right)^{1/2} dR \\ &=\int_{O(N)} \left( \int_\Omega \vert R^{-1}[(\nabla u_n)(Rx)-(\nabla u)(Rx)] \vert^2 dx \right)^{1/2} dR \\ &=\int_{O(N)} \left( \int_\Omega \vert (\nabla u_n)(Rx)-(\nabla u)(Rx) \vert^2 dx \right)^{1/2} dR \\ &=\int_{O(N)} \left( \int_\Omega \vert (\nabla u_n)(y)-(\nabla u)(y) \vert^2 dy \right)^{1/2} dR \\ &= \Vert \nabla u_n - \nabla u\Vert_{L^2(\Omega)} \int_{O(N)} dR \\ &=\Vert \nabla u_n - \nabla u\Vert_{L^2(\Omega)}. \end{align*} Similar computation yields $\Vert v_n - u \Vert_{L^2(\Omega)}\leq \Vert u_n-u\Vert_{L^2(\Omega)}$. Hence, we get $$ \Vert v_n-u\Vert_{H^1(\Omega)} \leq \Vert u_n - u \Vert_{H^1(\Omega)} \rightarrow 0. $$ Hence, $u\in \overline{C_{0,\mathrm{rad}}^\infty(\Omega)}^{H_0^1(\Omega)}$.

Answer 2

Let me make your original approach rigorous. Wlog I'll assume that $\Omega=B(0,1)\subseteq \mathbb{R}^n$ (rescaling everything we can always reduce to this case).

Let $u\in H_{0,\mathrm{rad}}^1(\Omega)$. Pick $\varphi\in C_c^\infty(\mathbb{R})$ even such that $\varphi\geq 0,$ $\varphi(0)=1$ and $\varphi^{(k)}(0)=0$ for all $k\in \mathbb{N}_{\geq 1}$ and $\varphi(r)=0$ for $\vert r \vert \geq 1/2$. Define for $0<\varepsilon<1$ $$ \varphi_\varepsilon(r) = \begin{cases} 1,& -1+\varepsilon\leq r \leq 1-\varepsilon,\\ \varphi((r-(1-\varepsilon))/\varepsilon),& r>1-\varepsilon,\\ \varphi((r-(-1+\varepsilon))/\varepsilon),& r<-1+\varepsilon. \end{cases}$$ Finally, we define $$ \chi_\varepsilon(x)=\varphi_\varepsilon(\vert x \vert). $$ Then we have that $\chi_\varepsilon\in C_c^\infty(\mathbb{R}^n)$ is radial with support in $B(0,1-\varepsilon/2)$.

Cutting off the support: We define $v_\varepsilon = \chi_\varepsilon u$. As $\chi_\varepsilon$ is smooth, we get $v_\varepsilon \in H_0^1(\Omega)$. We want to prove that $\lim_{\varepsilon \rightarrow 0^+} \Vert v_\varepsilon -u \Vert_{H^1(\Omega)} =0$. Using $\lim_{\varepsilon \rightarrow 0^+} \chi_\varepsilon =1$ and dominated convergence yields $\Vert v_\varepsilon - u \Vert_{L^2(\Omega)}=0$. Next we need to deal with the derivative. We have \begin{align*} \Vert \nabla (v_\varepsilon -u ) \Vert_{L^2(\Omega)} \leq \Vert (1-\chi_\varepsilon) \nabla u \Vert_{L^2(\Omega)} + \Vert (\nabla \chi_\varepsilon) u \Vert_{L^2(\Omega)}. \end{align*} The first term on the RHS goes to zero again by dominated convergence. For the second term we need to work a bit harder. First of all we note that by construction we have $$ \vert \nabla \chi_\varepsilon(x) \vert \leq \varepsilon^{-1} \left(\sup_{\vert r \vert \leq 1/2} \vert \varphi'(r) \vert\right) 1_{\Omega_\varepsilon}(x) $$ where $\Omega_\varepsilon = B(0,1)\setminus B(0,1-\varepsilon)$.

As $u\in H_0^1(\Omega)$ there exists a sequence $(g_n)_{n\in \mathbb{N}}\subseteq C_c^\infty(\Omega)$ such that $\Vert u -g_n\Vert_{H^1(\Omega)}\rightarrow 0$. For every $g_n$ and every $x\in \Omega\setminus \{0\}$ we have (this is just the fundamental theorem of calculus walking in the radial direction to the boundary of the ball). \begin{align*} (\ast) \quad g_n(x) &= g_n(x/\vert x\vert)- \int_1^{1/\vert x\vert} \partial_s g_n(sx)ds \\ &= g_n(x/\vert x\vert)- \int_1^{1/\vert x\vert} (\nabla g_n)(sx)\cdot x \ ds \\ &= -\int_1^{L} \partial_{x} (\nabla g_n)(sx)\cdot x \ ds \end{align*} for any $L\geq 1/\vert x \vert$ as $g_n$ has compact support in $\Omega$. Using the previous equality and Minkowski's integral inequality we get \begin{align*} \left( \int_{\Omega_\varepsilon} \vert g_n(x)\vert^2 dx \right)^{1/2} &= \left( \int_{\Omega_\varepsilon} \left\vert \int_1^{1/\vert x \vert}\vert (\nabla g_n)(sx) \vert \cdot \vert x \vert ds \right\vert^2 dx \right)^{1/2} \\ &\leq \left( \int_{\Omega_\varepsilon} \left\vert \int_1^{1/(1-\varepsilon)}\vert (\nabla g_n)(sx) \vert \cdot \vert x \vert ds \right\vert^2 dx \right)^{1/2} \\ &\leq \int_1^{1/(1-\varepsilon)} \left(\int_{\Omega_\varepsilon} \vert (\nabla g_n)(sx)\vert^2 dx \right)^{1/2} ds \\ &= \int_1^{1/(1-\varepsilon)} \left(\int_{\Omega_\varepsilon} \vert (\nabla g_n)(sx)\vert^2 dx \right)^{1/2} ds \\ &\leq \int_1^{1/(1-\varepsilon)} s^{-n/2} \left(\int_{\Omega_\varepsilon} \vert (\nabla g_n)(y)\vert^2 dy \right)^{1/2} ds \\ &\leq \Vert g_n \Vert_{H^1(\Omega_\varepsilon)} \left(\frac{1}{1-\varepsilon}-1\right) \\ &= \varepsilon \frac{1}{1-\varepsilon} \Vert g_n \Vert_{H^1(\Omega_\varepsilon)}, \end{align*} where we have used that $(s\Omega_\varepsilon)\cap \Omega \subseteq \Omega_\varepsilon$ for $s\geq 1$ and $\vert x \vert \leq 1$ for $x\in \Omega_\varepsilon$. Letting $n\rightarrow \infty$ we get $$ \Vert u \Vert_{L^2(\Omega_\varepsilon)} \leq \frac{\varepsilon}{1-\varepsilon} \Vert u \Vert_{H^1(\Omega_\varepsilon)}. $$ This yields \begin{align*} \Vert (\nabla \chi_\varepsilon) u\Vert_{L^2(\Omega)} \leq \varepsilon^{-1} \left( \sup_{\vert r \vert \leq 1/2} \vert \varphi'(r) \vert \right) \Vert u \Vert_{L^2(\Omega_\varepsilon)} \leq \left( \sup_{\vert r \vert \leq 1/2} \vert \varphi'(r) \vert \right) \frac{\Vert u \Vert_{H^1(\Omega_\varepsilon)}}{1-\varepsilon}. \end{align*} Thus, in total we get $$ \lim_{\varepsilon \rightarrow 0^+} \Vert v_\varepsilon - u \Vert_{H^1(\Omega)} =0. $$ This means that we can approximate $u$ by radial functions in $H^1(\Omega)$ with support strictly contained in $B(0,1)$.

Smoothing: Let $\psi \in C_c^\infty(\mathbb{R}^n)$ be a radial and set $\psi_\delta(x)=\delta^{-n} \psi(x/\delta)$. Define $w_{\varepsilon,\delta} = \psi_\delta * v_\varepsilon$. As both $\psi_\delta$ and $v_\varepsilon$ are radial, we get that $w_{\varepsilon, \delta}$ is radial too. Furthermore, we have for $\delta<\varepsilon/2$ that $$\mathrm{supp}(w_{\varepsilon,\delta}) \subseteq \mathrm{supp}(v_\varepsilon)+\overline{B(0,\delta)} \subseteq \overline{B(0,1-\varepsilon/2+\delta)}\subset \Omega.$$ Thus, $w_{\varepsilon, \delta}\in C_{0, \mathrm{rad}}^\infty(\Omega)$. By standard theory, we get $w_{\varepsilon,\delta} = \psi_\delta * v_\varepsilon \rightarrow v_\varepsilon$ in $H^1$ and therefore, we can approximate $v_\varepsilon$ by functions in $C_{0,\mathrm{rad}}^\infty(\Omega)$. Finally, using $$ \Vert u - w_{\varepsilon, \delta} \Vert_{H^1(\Omega)} \leq \Vert u - v_\varepsilon \Vert_{H^1(\Omega)} + \Vert v_\varepsilon - w_{\varepsilon,\delta}\Vert_{H^1(\Omega)} $$ we conclude that $u\in \overline{C_{0,\mathrm{rad}}^\infty(\Omega)}^{H^1(\Omega)}$.

Answer 3

Let $u \in H^1_0(B)$ a radial function. For a given $\delta > 0$, consider $\varphi_\delta$ a standard radial mollifier. Now, let $\overline{u}$ the extension by zero of $u$ to $\mathbb{R}^N$

Lemma 1: Given $\epsilon > 0$, fixed, there exists $v_\epsilon \in C^\infty_0(\mathbb{R}^N)$, radial, and $\delta_\epsilon > 0$ such that $$ \| \overline{u} - v_\epsilon\|_{H^1_0(B)} < \epsilon \quad \text{ and }\quad \text{supp} (v_\epsilon) + \text{supp} (\varphi_\delta) \subset B, \quad \forall\,\, 0 < \delta < \delta_\epsilon. $$ Let $\epsilon > 0$. Notice that $v_\epsilon \in H^1(\mathbb{R}^N)$. Applying Theorem 1, p. 264, from Evans, PDE, 2010 with $U = \mathbb{R}^N$, we define $u_{\delta,\epsilon}= \varphi_\delta \ast v_\epsilon$, for $\delta > 0$ and obtain, for each $\epsilon > 0$, $u_{\delta,\epsilon} \to v_\epsilon$ in $H^1_{\text{loc}}(\mathbb{R}^N)$ as $\delta \to 0$. In particular, $u_{\delta,\epsilon} \to v_\epsilon$ in $H^1(B)$ as $\delta \to 0$. If we take $0 < \delta < \delta_\epsilon$, notice that $\text{supp} u_{\delta,\epsilon} \subset B$ and $u_{\delta,\epsilon} \in C^\infty_0(B)$. Finally, for this fixed $\epsilon > 0$, choose $0 < \ell_\epsilon < \delta_\epsilon$ and see that
$$ \|v_\epsilon - u_{\ell_\epsilon, \epsilon} \|_{H^1_0(B)} < \epsilon. $$ Now, define $w_\epsilon = u_{\ell_\epsilon, \epsilon} = \varphi_{\ell_\epsilon} \ast v_\epsilon$ and notice that $$ \|u - w_\epsilon\|_{H^1_0(B)} = \|\overline{u} - u_{\ell_\epsilon,\epsilon}\|_{H^1_0(B)} \leq \|\overline{u} - v_\epsilon \|_{H^1_0(B)} + \|v_\epsilon - u_{\ell_\epsilon,\epsilon} \|_{H^1_0(B)} < \epsilon + \epsilon = 2\epsilon. $$ Thus $w_\epsilon \to u$ in $H^1_0(B)$ as $\epsilon \to 0$. Of course convolution of radial functions is also radial, so $w_\epsilon \in C^\infty_{0,\text{rad}}(B)$.