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Let $f$ : $\mathbb{S^1}$ $\to$ $\mathbb{R}$ continuous such that $\forall$ $z,z'$ $\in$ $\mathbb{S^1}$, $f(zz')=f(z)+f(z')$

Let $K=\{{e^{i\theta},\frac{\theta}{2\pi} \in \mathbb{Q}}\}$, show that f is trivial on $K$ i.e, $f(z)=0$ $\forall z \in K$

Also what topological property $K$ enjoys inside $\mathbb{S^1}$ $?$ I dont know how to start the proof and i need some hints please !!

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  • $\begingroup$ It should be clear that no non-trivial subgroup of $\mathbb R$ with addition is compact. Thus, the image of a continuous homomorphism from a compact group must be trivial. $\endgroup$ Oct 12 '16 at 20:11
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Let $z=1$, then $f(1z')=f(1)+f(z')$ and so $f(1)=0$. So, you know at least one element is always mapped to $0$. That's a good starting point for you I think.

Next, try looking at the case that $z=z'$. What about $z'=z^2$, $z'=z^n$?

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  • $\begingroup$ Ok.. then $f(z^n)=nf(z)$, also $f(0)=f(0)+f(0) then f(0)=0$ and by this i can conclude that $\forall$ $z$,$f(z)=0 $ since $f(0.z)=f(0)+f(z)=f(0)$ so what is the point of the problem considering the set $K$ ? $\endgroup$
    – outlaw
    Oct 12 '16 at 16:32
  • $\begingroup$ Be careful. Does the circle contain 0? $\endgroup$
    – Dan Rust
    Oct 12 '16 at 16:54
  • $\begingroup$ @RamyTanios: $0\notin\Bbb S^1$ hence $f(0)$ is not defined. $\endgroup$
    – Piquito
    Oct 12 '16 at 16:56
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    $\begingroup$ Well, if $x = \theta/2\pi =p/q$ is rational, then what is $q(\theta/2\pi)$? How about $\exp(i\theta)^q$? $\endgroup$
    – Dan Rust
    Oct 12 '16 at 17:25
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    $\begingroup$ Small nitpick: $e^{ki2\pi}=1$ for all integers $k$, not $(-1)^k$. Other than that, you're reasoning is right! $\endgroup$
    – Dan Rust
    Oct 13 '16 at 14:07
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$$\begin{cases}e^{\pm i\pi}=\cos(\pm\pi)+i\sin(\pm\pi)\pi=-1\in K\text{ because } \frac{\pm\pi}{2\pi}\in\Bbb Q\\e^{0i}=\cos 0+i\sin 0=1\in K \text{ because } \frac{0}{2\pi}\in\Bbb Q\end{cases}$$ Besides $f(\pm z)=f(\pm1)+f(z)\Rightarrow f(\pm1)=0$.

From $$\pi=\frac{\pi}{2}+\frac{\pi}{2}\\\frac{\pi}{2}=\frac{\pi}{4}+\frac{\pi}{4}\\\frac{\pi}{4}=\frac{\pi}{8}+\frac{\pi}{8}\\......\\......\\\frac{\pi}{2^n}=\frac{\pi}{2^{n+1}}+\frac{\pi}{2^{n+1}}$$ we get easily $$f(e^{i\frac{\pi}{2}})= f(e^{i\frac{\pi}{4}})= f(e^{i\frac{\pi}{8}})=….= f(e^{i\frac{\pi}{2^n}})=0\text { for all } n$$ Furthermore $f(e^{i\frac{\pi}{2}})=f(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2})=f(i)=0$.

You can finish now with $f(z)=0$ thinking a little more.

On the other hand $K$ is easily seen to be a subgroup of the group multiplicative $\Bbb S^1$. Because of $z\in K\iff rz\in K$ for all non-zero rational $r$ one can prove that $K$ is dense in $\Bbb S^1$.

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