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It has been asked ~here~ how to understand the mapping of a period function onto its Fourier coefficients, which has led me to look into Pontryagin duality. It's fascinating stuff, except I don't buy it.

My problem with this is, suppose you have a function $\phi(x)$ that is continuous on a finite interval $[0,1]$ with Dirichlet boundary conditions for simplicity. It has the Fourier expansion $\phi(x) = \sum c_n \sin(n \pi x) $. These $c_n$'s constitute the same information as $\phi(x)$: knowledge of either one implies knowledge of both. Yet, when you think of $\phi(x)$ and the sequence $\{c_n\}$ as vectors in infinite-dimensional spaces, one of them appears to be uncountably infinite while the other is countably infinite.

This reminds me of the motivation for having different cardinalities, namely, the mapping of integers onto rationals on a finite interval. How is it possible that a countable set ( $\{c_n\}$ ) could be used as a stand-in for $\phi(x)$?

The Wikipedia article on Pontryagin duality contains a clue: "[A] group $G$ and its dual group $\widehat{G}$ are not in general isomorphic, but their group algebras are." I am coming from physics/engineering background, though, so I am not even sure if this a relevant statement.

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Beepboop bebop
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    The secret is that this countable set is dense (in the right topology). E.g., suppose you knew that $\phi(x)$ is continuous on $[0, 1]$, with Dirichlet boundary conditions, and suppose we knew the values of $\phi(x)$ only on the (countably many) rational points $x$ in $[0, 1]$. This would give us all of $\phi(x)$. – avs Aug 12 '16 at 23:24
  • @avs can I ask you to spell this out a bit more? I have investigated some of the math, and the term dense comes up, but I am not able to connect the dots still. Maybe I'm just being dense... – Beepboop bebop Aug 16 '16 at 17:52
  • "$A$ dense in $B$" means that if we choose an arbitrary point $b \in B$, then every neighborhood of $b$ will contain a point from $A$. In short, every point in $b$ can be approximated with any tolerance by points from $A$. Equivalently, and more importantly for us, this means that for each $b \in B$ there exists a sequence of points in $A$ that converges to $b$.

    If $b$ an irrational in $[0, 1]$, then there exists a sequence $x_{n}$ of rationals converging to $b$, and since $f$ is continuous, $f(b) = \lim_{n} f(x_{n})$.

     – avs Aug 16 '16 at 18:00
  • Thus, from knowing the values $f(x)$ only at the rational $x$, we can use limit passages and the continuity of $f$ to restore $f$ on the irrational points $x \in [0, 1]$. – avs Aug 16 '16 at 18:01

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