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Laplace's spherical harmonics "form a complete set of orthonormal functions and thus form an orthonormal basis of the Hilbert space of square-integrable functions" [1]. I have three related questions about this statement:

(1) I can prove their orthonormality, but how do you prove that they form a complete set?

(2) What does completeness mean for an set with an infinite number of elements?

(3) How does the assertion that the spherical harmonics form an orthonormal basis of the Hilbert space of square-integrable functions follow from their being a complete set of orthornormal functions?

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2) It means that no nonzero vector is orthogonal to all of them; equivalently, it means their span is dense in the $L^2$ norm. 3) The two statements mean the same thing. –  Qiaochu Yuan Jul 1 '11 at 12:53
    
@Qiaochu Yuan: So then in order to prove (1) is it sufficient to prove their orthogonality? I.e. since there is no spherical harmonic that is equal to zero for all values of its angular arguments, and all of the spherical harmonics are mutually orthogonal. –  okj Jul 1 '11 at 13:01
    
@okj: No, that's not the same statement. That statement is also true for any proper subset of them, whereas Qiaochu's isn't (since the ones in the subset are orthogonal to the ones left out). –  joriki Jul 1 '11 at 13:07
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I think the most accessible argument for completeness of spherical harmonics is to start with the Weierstrass approximation theorem, which asserts that on compact subsets $E$ of $R^n$ polynomials are dense in _sup_norm_, meaning the metric on functions $d(f,g)=\sup_{x\in E} |f(x)-g(x)|$. This applies to the sphere $E=S^{n-1}$. At some point in the development (as in the corresponding chapter in Stein-Weiss, for example), it is shown that every homogeneous polynomial $f$ can be written as $f=f_d + r^2f_{d-2} + r^4f_{d-4}+\ldots$ where $f_i$ is harmonic, and $r$ is radius. Thus, restricted to the sphere, every polynomial is pointwise-equal to a harmonic polynomial. Combining these two points, harmonic polynomials are dense in continuous functions on the sphere, with respect to sup-norm.

Depending what definition of "integral" one uses, one shows that continuous functions on the sphere are dense in $L^2$, so harmonic polynomials are dense in $L^2$. Then it is a small exercise to infer completeness.

A somewhat more sophisticated viewpoint has us note that the sphere is a homogeneous space $S^{n-1}=SO(n)/O(n-1)$ of orthogonal groups, so (thinking in terms of Frobenius reciprocity) the representations of $SO(n)$ appear which restrict to the trivial repn on $O(n-1)$, with multiplicity equal to the dimension of $O(n-1)$-fixed vectors, which is found to be $1$ (or $0$). The completeness in this version of the story is part of the general decomposition of $L^2(SO(n))$ (under the compact (Hilbert-Schmidt) operators coming from compactly-supported continuous functions on the group), and, instead, the issue becomes identifying irreducibles as corresponding to spherical harmonics. This is probably best done by using the fact that every irreducible has a highest weight, and, conversely, isomorphism classes are uniquely determined by highest weights. This fancier viewpoint applies more broadly.

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Garrett: So are you saying that if an irreducible representation of a given group can be written in terms of a particular set of orthonormal basis functions, then this fact in and of itself shows that that set of basis functions is complete? –  okj Jul 1 '11 at 14:17
    
Perhaps ... tho' I'm not entirely sure what is implicit in your wording. What is generally true is that $L^2(K/H)$ for a compact group $K$ and closed subgroup $H$ decomposes as a direct sum of irreducibles. If we find an orthonormal basis for each of those irreducible subs, then taking these bases together is complete in $L^2(K/H)$. That is, it does form a Hilbert-space basis, meaning that finite linear combinations are dense (and linear independence...). But/and this is true more generally, that a Hilbert space which is the completion of a direct sum of subspaces $V_i$, with each $V_i$ ... –  paul garrett Jul 1 '11 at 15:49
    
... having an orthonormal basis $e_{ij}$, has an orthonormal (Hilbert-space) basis consisting of all the $e_{ij}$'s. That is, some part of the question's concern may be about creating Hilbert-space bases of (completions of!) direct-sums of Hilbert spaces from the summands' bases... which is a simpler sort of issue. –  paul garrett Jul 1 '11 at 15:51
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