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

Solution 1:

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.

Solution 2:

If you enforce continuity on $f(x)$ (continuity even at a single point should do), then the only function which satisfies this is an exponential function of the form $f(x) = e^{ax}$.

You need to enforce continuity to extend this from rationals to reals (as you would expect).

If you have $f: \mathbb{Q} \rightarrow \mathbb{R}$, then $f(x) = e^{ax}$ is the only function.

(The proof for this is analogous to proving $f(x) = cx$ when $f(x+y) = f(x) + f(y)$)

In fact one of the definitions of $f(x) = a^x$ is as follows. If there exists a function $f(x)$ satisfying the following properties,

1. $f(x) : \mathbb{R} \rightarrow \mathbb{R}$

2. $f(x+y) = f(x) \times f(y)$

3. $f(x) \text{ is continuous at-least at one point in } \mathbb{R}$

4. $f(1) = a$

Then, $f(x) = a^x$.

Note: $e^x$ could be defined like above with $f(1) = e$, where $\displaystyle e = \lim_{n \rightarrow \infty} \left( 1+\frac{1}{n} \right)^n$. This is one of the many definitions for $e^x$.

EDIT:

Without continuity you could define a function as follows:

We know that along all rationals $f(x) = e^{ax}$ where $x \in \mathbb{Q}$. Now lets define $f(x)$ over $x \in \mathbb{R} \backslash \{ \mathbb{Q} \}$.

For this, consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. This is an infinite-dimensional vector space and the axiom of choice guarantees the existence of a basis for this space.

Now consider a basis for this space. Let the set of basis be $\{h_{\alpha}\}$ i.e. every real number can now be written as $r = \displaystyle \sum_{i=1}^{N(r)} q_i h_{\alpha_i}$, where $q_i \in \mathbb{Q}$, $N(r) \in \mathbb{N}$, and $h_{\alpha_i} \in \{h_{\alpha}\}$.

Now you have complete freedom to decide what the value of $f$ should take on each element in the set $\{ h_{\alpha} \}$.