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Let $C(\mathbb{R})$ be the space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with the compact-open topology, and the associated Borel $\sigma$-algebra. Consider the function $p$ from $C(\mathbb{R})$ to $\mathbb{R}_{\geq 0} \cup \{\infty\}$ that maps a continuous function to its period, with the convention that non-periodic functions get mapped to $\infty$. Is the function $p$ a measurable function on $C(\mathbb{R})$? The only way I know of constructing measurable functions is to realize them as iterated lim, limsup, or liminf of a sequence of continuous functions. It's not clear here what continuous functions approximate the period in any reasonable manner.

Questions:

  1. Is there a way to approximate the period of $f \in C(\mathbb{R})$ using a continuous map from $C(\mathbb{R})$ to $\mathbb{R}$, which on taking appropriate limits, converges to the described function $p$?

  2. Is there some other way of showing that the map $p$ is measurable?

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    $\begingroup$ One way of extracting the period of a function is taking the Fourier transform, see for example this signal processing post. You get a Dirac comb, the width of which is inversely proportional to the period. This operation is continuous, because the compact-open convergence implies convergence as a tempered distribution. However, I don't know if this is useful for you. $\endgroup$ – Giuseppe Negro Oct 29 '19 at 13:40
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    $\begingroup$ Is it clear that $p$ is well-defined? To me it is not completely obvious that an arbitrary continuous function is either non-periodic or has a unique period. $\endgroup$ – B K Oct 30 '19 at 0:57
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    $\begingroup$ @BK: $\inf \{ t \in \mathbb{R}^+ | (\forall x \in \mathbb{R}) f(x+t) = f(x) \}$ $\endgroup$ – R.. GitHub STOP HELPING ICE Oct 30 '19 at 1:57
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Isn't the set $p^{-1}([0, T_0])$ closed for every finite $T_0$? Suppose that $f_n$ has period $T_n \leqslant T_0$ and it converges locally uniformly to $f$. By passing to a subsequence, we may assume that $T_n$ has a limit $T$. Uniform convergence of $f_n$ on $[x, x + T_0]$ implies that $$f(x + T) = \lim f_n(x + T_n) = \lim f_n(x) = f(x),$$ and hence $f$ has period at most $T$.

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  • $\begingroup$ Thanks! I guess I was just looking at the problem the wrong way. And from the fact that $p$ is upper semicontinuous, it shouldn't be too hard to find explicitly find continuous functions that approximate the period as well. $\endgroup$ – sayantankhan Oct 29 '19 at 13:18

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