I am interested in an algebraic approach to the following theorem:
Theorem. Consider the sphere $S^{n-1} \subseteq \Bbb{R}^n$ and for each $k=0,1,2,\ldots$, the space $H^k$ consisting of homogeneous harmonic polynomials of degree $k$, restricted to $S^{n-1}$. Then $\bigoplus_{k=0}^\infty H^k$ (the set of all linear combinations of elements of $H^k$) is dense in $L^2(S^{n-1})$.
I am familiar with an analytic derivation of this fact (as described at the bottom). However, it relies on some heavy functional machinery (diagonalization in Hilbert spaces, Lioville theorem and singularity removal for harmonic functions), so I am looking for a more elementary, algebraic argument instead - just for my own satisfaction. If one accepts the fact that polynomials are dense in $L^2(S^{n-1})$ (which is much easier to prove than the machinery mentioned earlier), the rest is purely algebraic (comparing all polynomials with harmonic polynomials). However, I found it hard to complete this strategy.
My problem is the following: give an elementary algebraic proof for either one of the following statements:
Fact 1. If $u(x)$ is a non-zero homogeneous harmonic polynomial, then $|x|^{2j} u(x)$ is not harmonic for $j = 1,2,\ldots$.
Fact 2. If $u(x)$ is a non-zero homogeneous polynomial, then $|x|^2 u(x)$ is not harmonic.
I do have a proof for these (see below), but it doesn't satisfy me, because it employs some analytic tools. However, the statements only concern explicit linear maps between finite-dimensional spaces ($u(x) \mapsto |x|^2 u(x)$ and $u(x) \mapsto \Delta u(x)$) and should have an elementary justification.
Context. To provide further motivation, let me outline both approaches:
Algebraic approach. $\newcommand{\oP}{\overline{P}}\newcommand{\oH}{\overline{H}}$ Denote the space of homogeneous degree $k$ polynomials $\oP^k$ and let $\oH^k \subseteq \oP^k$ be the harmonic ones. It's easy to show that $\Delta \colon \oP^k \to \oP^{k-2}$ is onto, which means that $\dim \oH^k = \dim \oP^k - \dim \oP^{k-2}$. Together with either Fact 1 or 2 holds, this implies that $\oP^k = \bigoplus_{j \ge 0} |x|^{2j} \oH^{k-2j}$. Taking restrictions to $S^{n-1}$, we thus have $$ \bigoplus_{k = 0}^m P^k = \bigoplus_{k = 0}^m H^k \qquad \text{for each } m = 0,1,2,\ldots, $$ and so density of polynomials in $L^2(S^{n-1})$ finishes the proof.
Analytic approach. We can show that the Poisson equation on the sphere $-\Delta u = f$ is solvable, and that the solution operator $(-\Delta)^{-1}$ is a self-adjoint compact operator on $L^2(S^{n-1})$. Thus, it is diagonalizable, and $L^2$ is spanned by its eigenfunctions. If we have $-\Delta u = \lambda u$, then the extension $v(rx) = r^\mu u(x)$ is harmonic in $\Bbb{R}^n \setminus \{ 0 \}$ for a suitable choice of $\mu$. One can remove the singularity (so that $v$ is harmonic in $\Bbb{R}^n$) and use the Lioville theorem to show that $v$ is a (homogeneous) polynomial.
Analytic solution of my problems. Let's take Fact 1, since Fact 2 is equivalent. If $u,v$ are two harmonic polynomials of different orders, they are Laplace eigenfunctions with different eigenvalues, and in consequence they are orthogonal in $L^2(S^{n-1})$ (by integration by parts). This means that also $|x|^{2j} u(x)$ is orthogonal to all harmonic polynomials of degree $2j+\deg u$, and it cannot be one of them. [Maybe one can translate this argument into an elementary one?]
Proposition. If $u(x)$ and $|x|^2 u(x)$ are both harmonic (say on the whole space $\mathbf{R}^n$ for simplicity), then $u$ is identically zero.
Analytic proof. The difference $(1-|x|^2) \, u(x)$ is harmonic, and zero on the boundary of the unit ball; hence it's zero inside the ball by the maximum principle. So $u(x)$ is zero inside the ball as well, since the nonzero factor $1-|x|^2$ can be cancelled. And harmonic functions are real-analytic, so a harmonic function which is zero on an open set is zero everywhere.
Algebraic proof, for polynomials. A polynomial is harmonic iff all its homogeneous parts are, so assume that $u$ is a homogeneous polynomial of degree $m$. Then Euler's theorem for homogeneous functions says that $x \cdot \nabla u(x) = m \, u(x)$, and the rule for the Laplacian of a product gives $$ \begin{aligned} 0 & = \Delta ( \, |x|^2 u(x) \,) \\ & = (\Delta |x|^2) \, u(x) + 2 \nabla |x|^2 \cdot \nabla u(x) + |x|^2 \Delta u(x) \\ & = 2n \, u(x) + 4x \cdot \nabla u(x) + |x|^2 \Delta u(x) \\ & = 2n \, u(x) + 4m \, u(x) + 0 \\ & = (2n + 4m) \, u(x) , \end{aligned} $$ i.e., $u(x) = 0$.
Corollary. If $u(x)$ is harmonic and not identically zero, then $|x|^2 u(x)$ cannot be harmonic.