calculus - Is there a name for function with the exponential property $f(x+y)=f(x) cdot f(y)$?

I was wondering if there is a name for a function that satisfies the conditions

$f:\mathbb{R} \to \mathbb{R}$ and $f(x+y)=f(x) \cdot f(y)$?

Thanks and regards!

Answer

If $f(x_0)= 0$ for some $x_0\in\mathbb{R}$, then for all $x\in\mathbb{R}$, $f(x)=f(x_0+(x-x_0))=f(x_0)\cdot f(x-x_0)=0$. Therefore, either $f$ is identically $0$ or never $0$. If $f$ is not $0$, then it is a homomorphism from the group $\mathbb{R}$ with addition to the group $\mathbb{R}\setminus\{0\}$ with multiplication. If $f(x)<0$ for some $x$, then $f(\frac{x}{2})^2\lt 0$, which is impossible, so $f$ is actually a homomorphism into the positive real numbers with multiplication. By composing with the isomorphism $\log:(0,\infty)\to\mathbb{R}$, such $f$ can be analyzed by first analyzing all additive maps on $\mathbb{R}$. Assuming continuity, these all have the form $x\mapsto cx$ for some $c\in \mathbb{R}$, and hence $f(x)=\exp(cx)$. Assuming the axiom of choice, there are discontinuous additive functions on $\mathbb{R}$ that can be constructed using a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, and thus there are also discontinuous homomorphisms from $\mathbb{R}$ to $(0,\infty)$.

So for an actual answer to the question: Yes, they are called (the zero map or) homomorphisms from the additive group of real numbers to the multiplicative group of positive real numbers.