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The spherical harmonics of degree $k$ in $n$ dimensions are the restriction to the sphere $\mathbb S^{n-1}$ of harmonic polynomials homogeneous of degree $k$ in $n$ variables. It is a classical fact of analysis that an Hilbertian basis of $L^2(\mathbb S^{n-1})$ can be made with spherical harmonics and that the Laplace operator on $L^2(\mathbb S^{n-1})$ can be written as $$ -\Delta_{\mathbb S^{n-1}}=\sum_{k\ge 0}k(k+n-2)\mathbb P_k, \qquad \text{Id}_{L^2(\mathbb S^{n-1})}=\sum_{k\ge 0}\mathbb P_k, $$ where $\mathbb P_k$ is the orthogonal projection onto the space of spherical harmonics of degree $k$.

My Question : The proofs that I know of the second identity above are not so elementary and I would like to know if there is a simple proof of the fact that the orthogonal (in $L^2(\mathbb S^{n-1})$) of the space of all spherical harmonics is reduced to $\{0\}$.

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    $\begingroup$ To fix a threshold for "elementarity", can you mention an elementary proof that the trigonometric system is complete on $L^2(\mathbb S^1)$? (By this I mean that its $L^2$ orthogonal is ${0}$) $\endgroup$
    – Giuseppe Negro
    Commented Oct 11, 2017 at 13:11

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In other words, you want to prove that spherical harmonics are dense in $L^2(S)$. By a standard argument it is enough to show that they are dense in $C(S)$ (continuous functions). Or even dense in the space of smooth functions. For every continuous function on the sphere, we can solve Dirichlet problem (this is elementary: there is an explicit formula due to Poisson). So we obtain a harmonic function in the ball. Expand it to a series of homogeneous harmonic polynomials (again this is an explicit series: just expand the kernel of the Poisson formula and integrate term-by-term). Taking a partial sum of this series we obtain a harmonic polynomial approximating our $L^2(S)$ function.

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    $\begingroup$ The space $P$ of all polynomials on ${\mathbb R} ^n$ can be written as a direct sum $$P=\sum _{m=0}^{\infty} Q^m H$$ where $H$ is the space of harmonic polynomials and $Q=\sum x_i^2$ is the standard quadratic form on $\mathbb R ^n$. It follows, upon restriction to the unit sphere, that all polynomials on $S^{n-1}$ are sums of harmonic polynomials. Then by Stone Weierstrass, Harmonic polynomials are dense in the space of continuous functions on the sphere. $\endgroup$
    – Venkataramana
    Commented Oct 11, 2017 at 14:34
  • $\begingroup$ Thanks for your comment: How do you get that identity? $\endgroup$
    – Bazin
    Commented Oct 13, 2017 at 11:59
  • $\begingroup$ @Venkataramana Thanks for your comment: How do you get that identity? $\endgroup$
    – Bazin
    Commented Oct 29, 2017 at 15:55
  • $\begingroup$ @Bazin, this identity, while simple and algebraic, is a little involved to explain. BY Zariski density, we may deal with $V=\mathbb C ^n$ instead of ${\mathbb R }^n$. Consider the quadric $Z$ defined by $Q=0$ in the projective space ${\mathbb P}({\mathbb C}^n)$. The $m$-th symmetric power $sym ^m (V^*)$ is irreducible for $GL_n(\mathbb C)$, and restricts non-trivially to the quadric $Z$; the quadric $Z=G/P$ where $G=SO(n)$ and $P$ a suitable parabolic. Then (by the Borel-Weil theorem), the restriction of $sym ^m (V^*)$ to $Z$ is irreducible. This is the space of Harmonic degree $m$ polys . $\endgroup$
    – Venkataramana
    Commented Oct 30, 2017 at 14:38
  • $\begingroup$ For a reference to this, you may look at Goodman and Wallach book. $\endgroup$
    – Venkataramana
    Commented Oct 30, 2017 at 14:40

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