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In the context of $L^2$ space, it is usually stated that any square-integrable function can be expanded as a linear combination of Spherical Harmonics: $$ f(\theta,\varphi)=\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell f_\ell^m \, Y_\ell^m(\theta,\varphi)\tag 2 $$ where $Y_\ell^m( \theta , \varphi )$ are the Laplace spherical harmonics.

The context here is important because this equality holds only in the sense of the $L^2$-norm:

$$\lim_{N\to\infty} \int_0^{2\pi}\int_0^\pi \left|f(\theta,\varphi)-\sum_{\ell=0}^N \sum_{m=- \ell}^\ell f_\ell^m Y_\ell^m(\theta,\varphi)\right|^2\sin\theta\, d\theta \,d\varphi = 0.$$

Do we also have pointwise convergence almost everywhere?

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    $\begingroup$ @CarloBeenakker: But of course the convergence can't be uniform for a general $f\in L^2(S)$. $\endgroup$
    – Christian Remling
    Commented May 12, 2020 at 18:01
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    $\begingroup$ @Coltrane8: The question itself looks very reasonable to me, though I suspect that this is answered somewhere in the literature (or recognized as an open problem). $\endgroup$
    – Christian Remling
    Commented May 12, 2020 at 22:11
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    $\begingroup$ This 2002 paper indicates that the question was open at the time: projecteuclid.org/download/pdf_1/euclid.pcma/1416322431 $\endgroup$
    – Christian Remling
    Commented May 12, 2020 at 22:20
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    $\begingroup$ @ChristianRemling I feel that your last comment should be the answer (non-pdf link). (Also, I believe there are counterexamples to simpler generalizations of Carleson’s theorem?) $\endgroup$
    – Francois Ziegler
    Commented May 12, 2020 at 23:15
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    $\begingroup$ @ChristianRemling I had in mind Fourier series on $(\mathbf R/\mathbf Z)^n$ for $n>1$ where I believe subtle “summation method” problems already arise (but I am far from an expert). $\endgroup$
    – Francois Ziegler
    Commented May 12, 2020 at 23:29

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A fundamental issue in appraising (uniform!?! especially...) pointwise convergence of eigenfunction expansions is having a reasonable estimate on the sup-norms of the eigenfunctions, as a function of the eigenvalue.

For Fourier series, the sup-norms are all $1$, and we can easily overlook this convenience.

For spherical harmonics, it is "well known" (see Stein-Weiss, and also some of my own notes...) that there is a polynomial bound for the sup-norm versus $L^2$ norm in terms of the eigenvalue. This does admit some abstraction in situations where a compact group acts transitively on the physical space on which we consider the functions.

Even for mildly general Sturm-Liouville situations, I myself do not know of a general approach to (uniform?) pointwise convergence, ... but I wish I did, and my ignorance surely indicates very little about the state of the art. :)

EDIT: that polynomial bound does also enable a reasonable "Sobolev space" set-up here, which gives tighter estimates than just saying "oh, it converges distributionally" (which, indeed, is a good remark).

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Suppose, for a moment, that the spherical harmonics expansion for a function $g\in L^2$ is everywhere convergent to $g$.

Now change the function from $g$ to $f$ by changing the values of $g$ on some nonempty set of measure $0$. Then $f\in L^2$ and the Fourier coefficients $f^m_l$ of $f$ will be the same as $g^m_l$, and hence the expansion for $f$ will be the same as that for $g$. However, now the expansion for $f\in L^2$ will not be everywhere convergent to $f$.

Members of $L^2$ are actually, not functions, but classes of functions differing only on a set of measure $0$. So, it makes no sense to even talk about everywhere convergence to a member of $L^2$ in general.

However, if $f$ is smooth enough, then the corresponding expansion will converge to $f$ everywhere. For instance, by Theorem 1 on page 9 in the paper Kalf - On the Expansion of a Function in Terms of Spherical Harmonics in Arbitrary Dimensions referenced in the comment by Carlo Beenakker, if $f\in C^1$, then the corresponding expansion will converge to $f$ uniformly and hence everywhere.

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    $\begingroup$ So how to deal with series sums term by term or derivation, for series expansions of classes of functions? I mean what is the formal development for the properties in analogy with usual functions? $\endgroup$
    – Coltrane8
    Commented May 12, 2020 at 18:47
  • $\begingroup$ Since $\lim_n(a_n+b_n)=\lim_n a_n+\lim_n b_n$ (provided that the latter two limits exist and, say, are finite) and $(f+g)^m_l=f^m_l+g^m_l$, the expansion for $f+g$ can be obtained by the term-wise addition of the expansions for $f$ and $g$ and, moreover, the expansion for $f+g$ will converge to $f+g$ if (and in the same sense as) the expansions for $f$ and $g$ converge to $f$ and $g$, respectively. $\endgroup$
    – Iosif Pinelis
    Commented May 12, 2020 at 19:00
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    $\begingroup$ Moreover I'm now still a bit confused, how can I now talk about convergence uniformly if the orthogonal projection doesn't distinguish by function differing for a set of measure $0$ ? Is that the case that the functions in the class of $f$ are not in $C^1$? $\endgroup$
    – Coltrane8
    Commented May 12, 2020 at 19:51
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    $\begingroup$ I'm sorry. I'll rephrase: 1) The result stated above, where the expansion converge uniformly is valid for spherical harmonics. There is a more general theorem that says something like that for any expansion on any complete orthonormal set of $L^2$? (Say if I have another function defined on $R$ bounded and continuous, or $C^1$, and I expand it on the set of Hermite polinomials. Can I say it converges uniformly. I'm looking for this kind of generalization) $\endgroup$
    – Coltrane8
    Commented May 13, 2020 at 10:00
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    $\begingroup$ @Coltrane8 You do have the (perhaps weak) consolation that if you work in the generalised context of the space of distributions on the sphere then you have everything you might desire (except for pointwise convergence)--the series always convergences in the distributional sense, you can freely manipulate the terms, you can differentiate term by term, .... The price you have to pay is that the corresponding convergence notion is much weaker than any of the classical ones for functions. Only you can know if it is too high. $\endgroup$
    – bathalf15320
    Commented Jun 7, 2021 at 3:28

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