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Jan 2, 2019 at 5:24 comment added Yemon Choi Re your last comment, you probably need to look up unordered summation -- and then you will find that a series converges as an unordered sum iff it is absolutely convergent, which of course need not happen for the Fourier series of a continuous function on ${\bf T}$. So you will probably need to pick some kind of sequence of finite subsets that exhausts $k$, if you want an analogue of Carleson's theorem
Jan 2, 2019 at 5:02 comment added D_S For $k$ a global field (which is countable), and $\mathbb A$ the adele group of $k$, the quotient group $\mathbb A/k$ is compact, and its character group identifies with $k$. So in the Fourier expansion, I would want the infinite sum over $k$ to be taken in any order, since there is no predetermined order on $k$.
Jan 2, 2019 at 4:56 comment added Yemon Choi D_S, I was gently trying to point out that it's not even clear fo the particular case of ${\bf T}^2$. However, perhaps for the kinds of groups you ask about in your final line, one can say more? I am not familiar enough with Fourier analysis on those kinds of groups to hazard a guess
Jan 2, 2019 at 4:53 comment added D_S I see. Maybe the statement of my question is unclear for general $G$.
Jan 2, 2019 at 4:49 comment added Yemon Choi A hint at some of the potential problems with making your question more precise can be found in the remarks on the Wikipedia page for Carleson's theorem: en.wikipedia.org/wiki/Carleson%27s_theorem
Jan 2, 2019 at 4:48 comment added Yemon Choi How are you performing your summation for a general $I$? That is: your infinite sum must be interpreted as some kind of limit, but how do you wish to do it? This is a problem even if $G={\bf T}^2$, if I recall correctly
Jan 2, 2019 at 4:43 history asked D_S CC BY-SA 4.0