Is any continuous group homomorphism from R to C* an exponential map? Consider $\mathbb{R}$ to be an additive topological group, and $\mathbb{C}^{\ast}$ to be a multiplicative topological group.
Is the following statement true?
If so, then how can one prove it?
Statement: any continuous group homomorphism $\mathbb{R} \to \mathbb{C}^{\ast}$ is always of the form $exp(\lambda x)$, $\lambda \in \mathbb{C}^{\ast}$.
 A: Yes. First, decompose $\mathbf{C}^\times$ as $\mathbf{R}_{>0}\times\mathbf{T}$, where $\mathbf{T}=\{z\in\mathbf{C}^\times:\vert z\vert=1\}$. We have $\mathbf{R}_{>0}\cong\mathbf{R}$ via the logarithm, and any continuous homomorphism $\mathbf{R}\rightarrow\mathbf{R}$ is necessarily $\mathbf{R}$-linear, and hence given by $x\mapsto sx$ for some (unique) $s\in\mathbf{R}$. The set of continuous homomorphisms $\mathbf{R}\rightarrow\mathbf{T}$ is the Pontryagin dual of $\mathbf{R}$, which is isomorphic to $\mathbf{R}$ via $s\mapsto(x\mapsto\exp(ixs))$. Combining these facts, one finds that the continuous homomorphisms $\mathbf{R}\rightarrow\mathbf{C}^\times$ are of the form $x\mapsto \exp(zx)$ for $z\in\mathbf{C}$.
A: It is certainly true that the only $differentiable$ homomorphisms are exponential maps. To see this let $f$ be a differentiable homomorphism, and note that $f(0) = 1$. Write $f'(0) = a$, so
$\lim_{t \to 0} (f(t) - 1)/t = a$. Now we compute $f'(x)$ for arbitrary $x$. We have
$$
f'(x) = \lim_{t \to 0} \frac{f(x+t) - f(x)}{t},
$$
and $f(x+t) = f(x)f(t)$ since $f$ is a homomorphism. This yields
$$
f'(x) = \lim_{t \to 0} f(x)\frac{f(t) - 1}{t} = a f(x).
$$
The unique solution to this differential equation with initial condition $f(0) = 1$ is
$f(x) = \exp(ax)$.

Added later:${\ \ }$ I now have a complete proof using only first-year calculus techniques. This does not rely on my previous argument.  Define
$F(x) = \int_0^x f(t) dt$, and note that $F(x) \ne 0$ for
$x \ne 0$ and $F'(x) = f(x)$ by the fundamental theorem of calculus. For a fixed constant $y$, make the change of variables $u = t + y$. This yields
$$
F(x) = \int_y^{x+y} f(u-y) du = \frac1{f(y)}\int_y^{x+y} f(u)du = 
\frac1{f(y)} (F(x+y) - F(y)).
$$
This yields
$$
f(y)F(x) + F(y) = F(x+y) = f(x)F(y) + F(x),
$$
where the second equality follows by interchanging $x$ and $y$. Then
$F(x)(f(y) - 1) = F(y)(f(x) - 1)$ for all $x,y$.
We can assume $f$ is not the constant $1$, so choose $a$ with $f(a) \ne 1$ and write $k = F(a)/(f(a) - 1)$, so $k \ne 0$ . Then $F(x) = k(f(x) - 1)$, and we have $f(x)= F(x)/k + 1$. Since $F$ is differentiable, so is $f$, and differentiation yields $f'(x) = F'(x)/k = f(x)/k$. Then $f(x) = \exp(x/k)$, and we are done.
