In a problem we are asked to show that there existe a unique function $\alpha:\mathbb{C}\rightarrow\mathbb{R}$ such that :
- $\alpha(x)=x$ for all real $x\geq0$
- $\alpha(zw)=\alpha(z)\alpha(w)$ for all $z,w\in\mathbb{C}$
- $\alpha$ is bounded on the unit circle $C(0,1)$
My question is : considering that the characterization of the absolute value is only hinted in the title, how would you tackle this problem ? Would you just prove that $\alpha(z) = |z|$ ?
Indeed, $\alpha (z) = |z|$ does work, so let's set about proving that this is unique.
Consider any $e^{i \theta}$. We claim that $\alpha(e^{i\theta}) = \pm 1$. If not, then note $\alpha(e^{i n \theta}) = \alpha(e^{i \theta})^n$ for all integers; taking a very positive or negative $n$ will yield $\alpha(e^{i \theta})^n$ unbounded and hence $\alpha(e^{i n \theta})$, contradicting the boundedness condition.
Note that $\alpha(e^{i \theta}) = -1$ is impossible, since $\alpha(e^{i\theta/2})^2 = \alpha(e^{i \theta}) = -1$, but a real cannot square to $-1$.
So we conclude that $\alpha(e^{i \theta}) = 1$ for all $\theta$, and hence $\alpha(re^{i\theta}) = \alpha(r) \alpha(e^{i \theta}) = r$ for all reals $r \geq 0$, giving the desired answer (by taking $r = |z|$, $\theta = \arg(z)$).