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

Question

Consider $B \subset \mathbb{R}^N = \mathbb{R}^{N_1} \times \mathbb{R}^{N_2}$ the unit ball and consider $$ H^1_{0, x,y}(B) = \{u \in H^1_0(B) : u(x,y) = u(R(s), S(y)), \forall R \in O(N_1), S \in O_2(N_2)\}. $$ It is very common in papers the authors just sai that $H^1_{0,x,y}(B) = \overline{C^\infty_{0,x,y}(B)}^{H^1_0(B)}$, without more comments, where $$ C^\infty_{0,x,y}(B) = \{u \in C^\infty_{0}(B) : u(x_1,y_2) = u(x_2,y_2) \text{ if }|x_1| = |x_2| \text{ and } |y_1| = |y_2|\}. $$ I am trying to find a precise proof of this. Here there is a discussion for the radial case, but I couldn't adapt the proof for this case.

Answer 1

In this question and in Dense subspace of space of radial functions in $H^1_0(\Omega)$ the question is about the density of compactly supported smooth function, having a certain symmetry, in the Sobolev space subject to the same symmetry. Let's deal with this question once and for all. Namely, let's answer the slightly more general question:

Let $G\subseteq \mathrm{GL}_n(\mathbb{R})$ be a compact group, $\Omega\subseteq \mathbb{R}^n$ be an open set which is invariant under $G$ (i.e. for all $R\in G$ holds $R(\Omega)=\Omega$). Then $$ C_{c,G}^\infty(\Omega) =\{ u\in C_c^\infty(\Omega) \ \vert \ \forall R\in G : u = u\circ R \} $$ is dense in $$ H_{0,G}^1(\Omega) = \{ u\in H_0^1(\Omega) \ \vert \ \forall R\in G : u = u\circ R \}.$$

Strategy: We know that $C_c^\infty(\Omega)$ is dense in $H_0^1(\Omega)$. "Symmetrize" the approximating sequence and check that the symmetrized sequence also converges to our function. The key concept we will use is the so-called Haar measure.

Haar measure: By abstract nonesense (see for example here https://personal.math.ubc.ca/~feldman/m421/haar.pdf for a neat introduction, or any textbook on Haar measures), every compact topological group $G$ admits a measure $\mu_G$ on the Borel sigma-algebra of $G$ which is invariant with respect to $G$. That means that for every $R\in G$ and every Bore-measurable set $A\subseteq G$ we have $$ \mu_G(R(A))=\mu_G(A)= \mu_G(\{ aR \ : \ a\in A\}, $$ i.e. the measure is invariant under multiplication by elements in $G$ both from the left and from the right.

How to symmetrize: Now, if we are given $u\in C_{c}^\infty(\Omega)$, how can we produce an element in $C_{c,G}^\infty(\Omega)$? Let's start with an easier example. Let's say we want to produce a function which is invariant under permutation of the coordinates, that is for all permutations $\tau\in S_n$ we have $$ u(x_{\tau(1)}, \dots, x_{\tau(n)})=u(x_1, \dots, x_n). $$ In linear algebra we have learnt that we can symmetrize in the following way $$ (\mathrm{Sym}\, u)(x_1, \dots, x_n)=\frac{1}{n!}\sum_{\sigma\in S_n} u(x_{\sigma(1)}, \dots, x_{\sigma(n)}). $$ Indeed, if $\tau\in S_n$, then we have $$ (\mathrm{Sym}\, u)(x_{\tau(1)}, \dots, x_{\tau(n)})= \frac{1}{n!}\sum_{\sigma\in S_n} u(x_{\sigma(\tau(1))}, \dots, x_{\sigma(\tau(n))})$$ and if we make the change of variables $\omega = \sigma\circ \tau$, then we see that $\mathrm{Sym}\, u$ is invariant under permutation. Furthermore, if $u$ was symmetric with respect to permutation, then $\mathrm{Sym}\, u = u$ as $$ (\mathrm{Sym}\, u)(x_1, \dots, x_n)=\frac{1}{n!}\sum_{\sigma\in S_n} u(x_{\sigma(1)}, \dots, x_{\sigma(n)}) = \frac{1}{n!} \sum_{\sigma\in S_n} u(x_1, \dots, x_n) = u(x_1, \dots, x_n). $$

Let us now try to do the same thing for a general, finite, group $G$. For the symmetric group we have averaged over all possible permutations (the $1/(n!)$ was important to ensure that functions which were symmetric from the start would remain unchanged). If the group was finite, we could define the symmetrized function as $$ (\mathrm{Sym_G} u)(x) = \frac{1}{\vert G\vert} \sum_{R\in G} u(Rx) $$ and we would have the same properties as before. Namely, $(\mathrm{Sym_G} u)(Rx)=(\mathrm{Sym_G} u)(x)$ for all $R\in G$ (i.e. it is $G$-invariant) and if $u$ is $G$-invariant, then $\mathrm{Sym_G} u =u$.

If we want to do this for a general compact group, we need to replace the sum by an integral. The overarching theme was that stuff needs to be $G$-invariant and this is exactly what the Haar measure gives us. So, we define $$ (\mathrm{Sym_G} u)(x)= \int_G u(Rx) \mu_G(dR). $$ As $\mu_G$ is $G$-invariant, we get by the general change of variables formula Is there a change of variables formula for a measure theoretic integral that does not use the Lebesgue measure that $\mathrm{Sym_G} u$ is $G$-invariant and that $\mathrm{Sym_G} u=u$ if $u$ was $G$-invariant to start with.

The approximating sequence: Let $u\in H_{0,G}^1(\Omega)$. Then in particular we know that $u\in H_0^1(\Omega)$ and $u$ is $G$-invariant. As $u\in H_0^1(\Omega)$ we know that there exists a sequence $(u_m)_{m\in \mathbb{N}} \subseteq C_c^\infty(\Omega)$ such that $\Vert u-u_m\Vert_{H^1(\Omega)} \rightarrow 0$. The issue here is of course that the $u_m$ need not be $G$-invariant. Thus, we define $$ v_m(x) = \int_{G} u_m(R x) \mu_G(dR) $$ where $\mu_G$ is the Haar measure on $G$. As explained before $v_m$ is $G$-invariant. We now need to check that $v_m\in C_c^\infty(\Omega)$ (i.e. $v_m$ has compact support and $v_m$ is smooth). Firstly, we check that $v_m$ has compact support. For this we define $$ F_m: G\times \mathrm{supp}(u_m) \rightarrow \mathbb{R}^n, F_m(R,x)=Rx. $$ Clearly, $\mathrm{supp}(v_m) \subseteq \mathrm{Im}(F_m)$. However, because $F_m$ is continuous and $G\times \mathrm{supp}(u_n)$ is compact (as Cartesian product of compact sets is compact) we get that the image of $F_m$ is compact.

Next we check that $v_m$ is smooth. For this it helps to note that by assumption $G\subseteq \mathrm{GL}_n(\mathbb{R})$ is compact. By compactness of $G$ we know that there exists a constant $C>0$ such that for $R=(R_{ij})_{1\leq i,j \leq n} \in \mathrm{GL}_n(\mathbb{R})$ holds $\vert R_{ij}\vert \leq C$. Thus, we get $$ \Vert \partial_j (u_m \circ R)\Vert_{L^\infty(\Omega)} = \Vert \sum_{1\leq i\leq n} R_{ij} (\partial_i u_m)\circ R \Vert_{L^\infty(\Omega)} \leq Cn \max_{i=1, \dots, n} \Vert \partial_i u_m \Vert_{L^\infty(\Omega)}<\infty $$ as $u_m\in C_c^\infty(\Omega)$. As $v_m$ has compact support, we can differentiate under the integral sign (Differentiation under (measure theoretical) integral sign) and conclude that $v_m$ is differentiable. The very same argument works for higher derivatives. Hence, we get that $v_m\in C_{c,G}^\infty(\Omega)$.

Verifying that the symmetrized sequence approximates $u$: Next we are going to use the fact that $\Vert u_m - u\Vert_{H^1(\Omega)}\rightarrow 0$ to conclude $\Vert v_m-u\Vert_{H^1(\Omega)} \rightarrow 0$. First we going to compute some gradients. Namely, if $R\in G, g\in C^1(\Omega)$ and $x\in \Omega, v\in \mathbb{R}^N$, then we have $$ \langle \nabla (g\circ R)(x), v\rangle = D(g\circ R)(x)[v] = Df(Rx) [Rv] = \langle (\nabla g)(Rx), Rv \rangle = \langle R^T (\nabla g)(Rx), v \rangle.$$ Hence, we have $$ \nabla (g\circ R)(x) = R^T (\nabla g)(Rx). $$ With this at hand, we can do the estimates. Using Minkowski's integral inequality and the fact that $\mathrm{Sym_G}u(x)=u(x)$ we get \begin{align*} \left( \int_\Omega \vert \nabla(v_m(x)-u(x)) \vert^2 dx \right)^{1/2} &= \left( \int_\Omega \vert \mathrm{Sym_G}\, u_m(x) - \mathrm{Sym_G} \, u(x) \vert^2 dx \right)^{1/2} \\ &=\left( \int_\Omega \vert \int_{G} \nabla[(u_m\circ R)(x)-(u\circ R)(x)] \mu_G(dR) \vert^2 dx \right)^{1/2} \\ &\leq \int_{G} \left( \int_\Omega \vert \nabla [(u_m\circ R)(x)-(u\circ R)(x) \vert^2 dx \right)^{1/2} \mu_G(dR) \\ &=\int_{G} \left( \int_\Omega \vert R^T[(\nabla u_m)(Rx)-(\nabla u)(Rx)] \vert^2 dx \right)^{1/2} \mu_G(dR) \\ &\leq \Vert R^T\Vert_\mathrm{op}\int_{G} \left( \int_\Omega \vert (\nabla u_m)(Rx)-(\nabla u)(Rx) \vert^2 dx \right)^{1/2} \mu_G(dR) \\ &=\Vert R^T\Vert_\mathrm{op} \vert \det(R^{-1})\vert\int_{G} \left( \int_\Omega \vert (\nabla u_m)(y)-(\nabla u)(y) \vert^2 dy \right)^{1/2} \mu_G(dR) \\ &= \Vert R^T\Vert_\mathrm{op} \vert \det(R^{-1})\vert \Vert \nabla u_m - \nabla u\Vert_{L^2(\Omega)} \int_{G} \mu_G(dR) \\ &=\Vert R^T\Vert_\mathrm{op} \vert \det(R^{-1})\vert \Vert \nabla u_m - \nabla u\Vert_{L^2(\Omega)}. \end{align*} Note that, as $G$ is compact, we have that $$ \sup_{S\in G} \Vert S^T\Vert_\mathrm{op},\sup_{S\in G} \vert \det(S)\vert <\infty. $$ Similar computation yields $\Vert v_m - u \Vert_{L^2(\Omega)}\leq \left(\sup_{S\in G} \vert \det(S)\vert\right)\Vert u_m-u\Vert_{L^2(\Omega)}$. Hence, we get $$ \Vert v_m-u\Vert_{H^1(\Omega)} \leq \left(\sup_{S\in G} \vert \det(S)\vert\right)\left(1+\sup_{S\in G} \Vert S^T \Vert_\mathrm{op}\right)\Vert u_m - u \Vert_{H^1(\Omega)} \rightarrow 0. $$ Hence, $u\in \overline{C_{c,G}^\infty(\Omega)}^{H_0^1(\Omega)}$.

The special case at hand: In the previous question Dense subspace of space of radial functions in $H^1_0(\Omega)$ we had $G=O(N)$ which is compact (Orthogonal matrices form a compact set). In this case the group $G$ is given by $G=O(n_1)\times O(n_2)$, which is also compact (as it is a cartesian product of compact groups).

More abstract nonesense: The very same proof would work for any linear Lie group action on $\Omega$ if the Lie group in question is compact.