# Fourier series of functions on compact groups

Let $G$ be a compact, second countable, Hausdorff topological group with the normalized Haar measure $\mu$. From Peter-Weyl's theorem we now that for any $f\in \mathrm{L}^2(G)$ the Fourier series of $f$: $$\sum_{\rho}d_\rho\langle \hat{f}(\rho),\rho(g)\rangle_{HS}$$ converges in $\mathrm{L}^2(G,\mu)$ to $f$. My first question is:

1. Why do people usually write $f(g)=\sum_{\rho}d_\rho\langle \hat{f}(\rho),\rho(g)\rangle_{HS}$? This notation suggest that the series converges pointwise to $f(g)$ which is definitely wrong. Is it a conventional notation or there is reason behind it?
2. What can we say about the pointwise convergent points of the Fourier series?
3. Let $G$ be a profinite group. Is it true that the Fourier series converges pointwise to $f$?
• Almost everywhere convergence for tori follows from Carleson's theorem, but I'm pretty sure it's open beyond that. The reason for the notation is just to emphasize what variable the two sides depend on. – Paul Siegel Apr 9 '14 at 2:33
• How are you ordering the summation? Note that one must specify an ordering before it makes sense to talk about pointwise convergence. – Mark Lewko Apr 9 '14 at 3:01
• Also, if $f$ belongs to the so-called Fourier algebra of $G$ (which will happen if $f$ is sufficiently differentiable and $G$ is a compact Lie group, by results of Sugiura projecteuclid.org/euclid.ojm/1200693127 ) then the convergence of the Fourier series is absolute, so the point astutely raised by @MarkLewko does not arise – Yemon Choi Apr 9 '14 at 4:17
• @Anton, If you interpret the question like that it isn't very interesting. There are certainly $L^2$ functions on the circle without absolutely convergent Fourier series. You could alternately ask for unconditional pointwise convergence (that is a.e. pointwise convergence for every ordering) but this is false for every complete infinite orthonormal system (such as Fourier series on the cirlce, again). – Mark Lewko Apr 9 '14 at 17:26
• One thing worth mentioning, in respect to question $1$ is that one have a structure theorem for compact groups (say by Von-Neumann's theorem, they are inverse limits of linear groups), and in practice, when one deals with a concrete example (say $SO(n)$), those weird matrix coefficients are actually computable and given (usually) by special functions (i.e. spherical harmonics), and then one can verify the convergence directly (for example, if one have a finite dimensional compact Lie group, all it rep. are algebraic, hence those matrix coefficients are smooth). – Asaf Apr 14 '14 at 14:52

1. No, it does not. It implies convergence in the $L^2$-norm.
2. If you assume $C^\infty$ functions, it converges in the $C^\infty$-sense. Check Harish-Chandra Discretes Series Representations II. HC does actually consider non--compact groups $G$ there and decomposes $C^\infty(G)$ as $K-K$-bimdule for $K$ compact.
• @PM, for $2$ it is enough to consider say Holder functions or BV, the usual proof by Dirichlet will hold (replacing the difference operators with the proper convolution operators). – Asaf Apr 14 '14 at 14:42
• Marc, could you, please, advise reading about the case of $C^\infty$? – Sergei Akbarov May 19 '14 at 18:19