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Recently I learned that the distribution of zeroes of a continuous real-valued function in a closed interval can be such that they might be uncountable. An example was given by Daniel Fischer who answered my question:

Let $C \subset [0,1]$ be a Cantor set (could be the standard middle-thirds one, could be a fat Cantor set), and

$$f(t) = \operatorname{dist}(t,C) = \inf \{ \lvert t-c\rvert : c \in > C\}.$$

Then $f\colon [0,1] \to [0,1]$ is a continuous function with $C = > f^{-1}(0)$, and $f$ is not constant on any nondegenerate interval. Since Cantor sets are perfect, $f^{-1}(0)$ has uncountably many ($2^{\aleph_0}$) accumulation points.

However, it recently occurred to me that because a trigonometric function can have at most countably many zeroes in the real line, the fourier series of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has uncountably many zeroes can't possibly converge to $f$, as a fourier series is a countable sum of sines/cosines which have countably many zeroes.

I think my argument is complete but surprisingly simple as I wasn't taught this while learning analysis. Can there be a hole in my argument?

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A (non-trivial) trigonometric polynomial can have only countably many zeros, but a trigonometric series can have uncountably many, effectively by your example: If $C$ is a Cantor set constructed from $[-\pi, \pi]$, and if $f$ is the periodic extension of a smooth function in $[-\pi, \pi]$ whose zero set is $C$ (every closed subset of $\mathbf{R}$ is the zero set of a smooth function), then the Fourier series of $f$ converges pointwise everywhere but the zero set is the uncountable set $C + 2\pi \mathbf{Z}$.

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Andrew D. Hwang
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