exponential function, lie group homomorphism

Question

Let $f: \mathbb{R} \to \mathbb{C}^*$ be a continuous map satisfying for all $x, y \in \mathbb{R}$:

1. $f(x + y) = f(x)f(y)$.
2. $f(x) = 1$ for all $t = 2\pi n, n \in \mathbb{Z}$.

Show that there exists $c \in \mathbb{C}^*$ and $\zeta \in \mathbb{C}$ so that $f(x) = ce^{x\zeta}$ for all $x$.

I know how to show that all $f$ that just satisfy the first condition are of the form $x \mapsto e^{iax}$ for some $a \in \mathbb{R}$ via covering spaces. But I am not sure on how to incorporate the second condition.

Answer 1

By the first condition, we have that $f(a/b) = f(1/b)^a$ and $f(1/b)^b = f(1)$ for any $a/b \in \mathbb{Q}$. Let $f(1) = ce^{i\theta}$. Then $f(1/b) = c^{1/b}e^{i(\theta + 2\pi k)/b}$ where $k$ is an integer depending on $b$. It follows $f(a/b) = e^{a/b}e^{i(\theta + 2\pi k)a/b}$. By continuity of $f$ and density of rationals, it follows that $k$ is locally constant. Since $\mathbb{R}$ is connected, this means $k$ is constant. Thus $f(a/b) = (ce^{i(\theta + 2\pi k)})^{a/b}$ for all rationals, and density of rationals again gives $f(t) = e^{\zeta t}$ for all $t$, where $\zeta = i(\theta + 2\pi k)$.

EDIT: Whoops, this is not entirely correct. With regards to the statement, "by continuity of $\psi$ and density of rationals, it follows that $k$ is locally constant," all we know is that for fixed $s$ the value is constant... what we have is not enough! Exercise left to the reader to rectify this oversight.

Answer 2

First, I think the $ c $ in your claim is dispensable. Use condition 1. to get $ f(x)f(y) = f(x+y) = c e^{x \zeta} e^{y \zeta} = f(x) \frac{f(y)}{c} $. So actually, $ c = 1 $.

Second, plugging in your condintion 2., you get $ 1 = f(2 \pi n) = e^{2 \pi n \Re(\zeta) } e^{2 \pi n i \Im(\zeta)} $ and taking absolute values, we see $ 1 = e^{2 \pi n \Re(\zeta)}, \ \forall n \in \mathbb{Z} $. We conclude $ \Re(\zeta) = 0 $. So you see, your solution suits perfectly.

Third, I don't know anything about covering spaces, but there is a pretty easy way to see of what form the $ f $'s have to be. One sees immediately $ f(0)=1 $ and $ f(n) = f(1)^{n}, \ \forall n \in \mathbb{Z} $. A next step shows $ f(\frac{p}{q}) = f(1)^{\frac{p}{q}}, \ \forall p \in \mathbb{Z}, q \in \mathbb{N} $. By continuity of the $ f $'s you get the statement for reals: $ f(x)=f(1)^{x}, \ \forall x \in \mathbb{R} $. Now $ f(1) \in \mathbb{C}^{*} $, so we just write it as $ f(1) = e^{\alpha}e^{i \beta} $ for $ \alpha, \beta \in \mathbb{R} $. Together we get that $ f(x) = e^{(\alpha + i \beta) x} $ The same arument as in the second paragraph shows $ \alpha = 0 $. After all, $ f(x) = e^{i \beta x} $ for $ \beta \in \mathbb{R} $.

The second point gives you even more, the fact that $ e^{2 \pi n i \beta} = 1, \ \forall n \in \mathbb{Z} $ shows $ n \beta \in \mathbb{Z} $.