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I am trying to show for a 1d map, $r$, where every number involved is in $\Bbb{R}$ that if: $$r(x_1+x_2)=r(x_1)+r(x_2)\tag{*}$$ then $$r(x)=\alpha x$$ for some $\alpha\in \Bbb{R}$. How do I show this? My attempt is given below:

Let $x_2=\frac{p}{q} x_1$ where $p, q \in \Bbb{Z}$ then: $$r(x_2)=r\left( p \times \frac{1}{q}x_1\right)$$ then from (*) applied $p$ times we have: $$r(x_2)=p\times r\left(\frac{1}{q} x_1\right)$$ but since: $$r(x_1)=r\left( q\times\frac{1}{q} x_1 \right)=q\times r\left( \frac{1}{q} x_1 \right)$$ we have: $$ r\left( \frac{1}{q} x_1 \right)=\frac{1}{q} r(x_1)$$ and thus: $$r(x_2)=\frac{p}{q} \times r(x_1)$$ This means that if we define $\alpha$ so $r(x_1)=\alpha x_1$ then:$$r(x_2)=\frac{p}{q} \alpha x_1=\alpha x_2$$ I am now stuck on proving it when $x_2=\beta x_1$ where $\beta$ is irrational. Is it enough to say that it has to work since rationals are dense in $\Bbb{R}$? Either way can you explain or present another valid proof.

Quantum spaghettification
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    Cauchy's equation again, sigh. According to Jagger and Richards, you can't always get what you want. – kimchi lover Jan 06 '18 at 12:50
  • @kimchilover Opps I didn't realize this had a name, thanks. I will take a look at the wiki page. – Quantum spaghettification Jan 06 '18 at 12:52
  • Is it possible that the problem you're working on also stated that $r$ is continuous? – John Hughes Jan 06 '18 at 12:53
  • The wikpiedia page on the subject is surprisingly good, but the short answer is that there aren't any unexpected solutions with some reasonable assumptions (e.g., continuity, although that can be weakened significantly). In the opposite direction, it's easy to construct a pathological solution from a Hamel basis. – anomaly Jan 06 '18 at 12:54
  • @JohnHughes As far as know, no. The problem I am working on is finding the 1d irreducible representations of $U(1)$, I have that we require $R(e^{i\theta})=e^{i r(\theta)}$ which from the definition of reps means that I need (*) to hold. As far as I am aware there is nothing in the definition of a representation that says that $R$ (or $r$) must be continuous. – Quantum spaghettification Jan 06 '18 at 12:57
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    See https://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – Mark Bennet Jan 06 '18 at 12:58
  • How about measurability? – kimchi lover Jan 06 '18 at 13:17

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If you're talking about a representation of $U(1)$ as a Lie Group, the following (copied from the Wikipedia page on representations) suggests that smoothness is relevant (as well it might be, since it's essential to preserve Lie-group-ness):

Let us first discuss representations acting on finite-dimensional vector spaces over a field ${\displaystyle \mathbb {C} }$. (Occasionally representations over the field of real numbers are also considered.) A representation of a Lie group $G$ on a finite-dimensional vector space $V $over ${\displaystyle \mathbb {C} }$ is a smooth group homomorphism

${\displaystyle \psi :G\rightarrow \mathrm {GL} (V)}$,

where ${\displaystyle \mathrm {GL} (V)}$ is the group of all invertible linear transformations of ${\displaystyle V}$. For n-dimensional $V$, the group ${\displaystyle \mathrm {GL} (V)}$ is identified with the ${\displaystyle \mathrm {GL} (n;\mathbb {C} )}$, the group of ${\displaystyle n\times n}$ invertible matrices. Smoothness of the map should be regarded as a technicality, in that any homomorphism ${\displaystyle \psi }$ that is continuous will automatically be smooth.

[boldface in the quote above is mine -- jfh ]

Once you have smoothness, then the theorem you're looking for becomes straightforward (and is shown in the links given in the comments).

John Hughes
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