# Spherical Harmonics Expansion convergence

• I
Gold Member

## Main Question or Discussion Point

In the contex 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.

This expansion holds in the sense of mean-square convergence which is to say that

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

So in general this limit is NOT pointwise? So I can't say that the value at a point of the function equals the value of the expansion at the same point?

If so, why it's usually stated out of the context of the structure of Hilbert space, that a bounded function or a square integrable function on the unit sphere can be expanded with Spherical Harmonics if it's not pointwise? I mean in some context outside the Hilbert space, where I am not interested in their square integral.

Furthermore, if it's not pointwise, but only in the norm, am I allowed to sum term by term two different functions, with two different expansions, like in the quantum scattering problem?

Related Linear and Abstract Algebra News on Phys.org
mathman
L2 convergence does not preclude pointwise convergence. If the function is bounded and continuous,it will converge pointwise and Lp for any ##p\ge 1##.

Coltrane8
Gold Member
L2 convergence does not preclude pointwise convergence. If the function is bounded and continuous,it will converge pointwise and Lp for any ##p\ge 1##.
Ok, where can I find this result/theorem? Because it seems to be pretty strong if for any basis in ##L^2##. I know that for spherical harmonic expansion, you would need the function to be ##C^1##, so it seems just not true to me what you affirmed.

Moreover when you talk of a basis of ##L^2##, you are dealing with class of function. In fact the expansions of two different functions, differing in a set of measure zero, is the same. So there exist only one function in the class of function differing by a set of measure zero, that suffice to converge pointwise?

I bet that in the last case we just talk about convergence almost everywhere.

Last edited:
mathman
Bounded and continuous functions on a set of finite measure belong to all Lp classes (trivially obvious). Also two continuous function which are equal almost everywhere are identical.

Coltrane8
Gold Member
Ok, and this I assume is valid for any ##C^1## or bounded and continuous function expanded on any complete set of orthonormal basis of a Hilbert space. I would like to see the proof and the formal statement of the theorem. May you provide some references were I can look to?

Say I have a function on ##H## (say ##L^2##) with some restriction (bounded, continuous).
I have a complete set of function that form an orthonormal basis of the Hilbert space.
When can I say that the expansion converges outside ##H##?, for example uniformly or pointwise?

mathman
##C^1## is continuous. I don't have a reference. The basis for my statements is my recollection of course material from many years ago.

martinbn