Spherical harmonics expansion 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?
 A: 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).
A: 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.
