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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 $f$?
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    $\begingroup$ 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. $\endgroup$
    – Paul Siegel
    Commented Apr 9, 2014 at 2:33
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    $\begingroup$ How are you ordering the summation? Note that one must specify an ordering before it makes sense to talk about pointwise convergence. $\endgroup$
    – Mark Lewko
    Commented Apr 9, 2014 at 3:01
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    $\begingroup$ 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 $\endgroup$
    – Yemon Choi
    Commented Apr 9, 2014 at 4:17
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    $\begingroup$ @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). $\endgroup$
    – Mark Lewko
    Commented Apr 9, 2014 at 17:26
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    $\begingroup$ 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). $\endgroup$
    – Asaf
    Commented Apr 14, 2014 at 14:52

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  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.

  3. Again you want to work with smooth functions, which for profinite groups means it factors through a finite quotient. So the statement would be trivial.

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  • $\begingroup$ @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). $\endgroup$
    – Asaf
    Commented Apr 14, 2014 at 14:42
  • $\begingroup$ Marc, could you, please, advise reading about the case of $C^\infty$? $\endgroup$
    – Sergei Akbarov
    Commented May 19, 2014 at 18:19
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    $\begingroup$ Harish-Chandra (1966), "Discrete series for semisimple Lie groups. II. Explicit determination of the characters", $\endgroup$
    – Marc Palm
    Commented May 20, 2014 at 17:46
  • $\begingroup$ If you use the framework of Hilbert's 5th problem, you can generalize it to the locally compact version, see chapter 6 of my Phd Thesis. $\endgroup$
    – Marc Palm
    Commented May 20, 2014 at 17:47

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