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Let $T:S^1\to C^\ast$ be a group theoretic homomorphism from the circle to the non-zero complex numbers. Presumably it is true that if $T$ is $L^2$, then it is continuous. Is there a simple proof, or a pointer to the literature?

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    $\begingroup$ Even better: consider the composite with the exponential map, $\exp\colon \mathbb{R} \to S^1$, and then $T$ is continuous if and only if $T\circ \exp$ is continuous. So then you just need to consider a homomorphism $S\colon \mathbb{R} \to \mathbb{C}$ with $S(0)=0$ and $S(x+y) = S(x)+S(y)$. Possibly $S$ measurable implies it is continuous, I have vague memories of such a result for homomorphisms $\mathbb{R} \to \mathbb{R}$. $\endgroup$ – David Roberts Nov 6 at 9:22
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    $\begingroup$ @DavidRoberts There is indeed such a result for homomorphisms $\mathbb R\to\mathbb R$, and hence also $\mathbb R\to\mathbb C$. However, the homomorphisms at hand map to $\mathbb C^*$ and it's not clear to me how we could reduce it (we can't take a logarithm since it is multivalued). Do you have an idea how to make that work? $\endgroup$ – Wojowu Nov 6 at 9:24
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    $\begingroup$ Aha, found it: golem.ph.utexas.edu/category/2010/12/… (see this Tricki entry, Example 3). Apparently one only needs to be bounded on a set of positive measure $\endgroup$ – David Roberts Nov 6 at 9:25
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    $\begingroup$ @Wojowu I didn't quite see that, garn. I would be tempted to lift through the complex exponential, but without continuity then extra work is needed to ensure nothing drastic happens when this is done. In any case, it is possible the argument at the Tricki could work for multiplicative case directly, with some tweaks. $\endgroup$ – David Roberts Nov 6 at 9:29
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    $\begingroup$ To be $L^2$ requires being measurable, and measurability is enough (this is done at this MathSE question: Every measurable homomorphism from $\mathbb{R}^n$ to $\mathbb{C}^*$ is exponential. The $L^2$ assumption plays no role. $\endgroup$ – YCor Nov 6 at 13:01

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