# What can we say about functions satisfying $f(a + b) = f(a)f(b)$ for all $a,b\in \mathbb{R}$? [duplicate]

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Is there a name for such kind of function?

I am investigating functions satisfying the exponentiation identity $f(a + b) = f(a)f(b)$ for all $a,b\in \Bbb R$. This is satisfied by $f(x) = \exp(A x)$ but what other functions are possible, if any? What properties must such functions have?

This is what I have figured out so far - I am interested in any other properties that are required.

1. $f(0) \in \{0,1\}$
Proof: $f(0) = f(0 + 0) = f(0)f(0) \implies$ either $f(0) = 0$ or dividing by $f(0)$ gives $f(0) = 1$

2. $f(x)$ is strictly positive everywhere or zero everywhere
Proof: If $f(0) = 0$, then $f(x) = f(x+0) = f(x)f(0) = f(x) \cdot 0 = 0$ for all $x$

If $f(0) = 1$ then first we show that $f(x)$ is non zero for all $x$: $$1 = f(0) = f(x - x) = f(x)f(-x)$$ which implies that both $f(x)$ and $f(-x)$ can not be zero and this is true for all $x$.
$$f(x) = f(x/2 + x/2) = f(x/2)^2$$ but we already know that $f(x/2)$ is non zero, hence its square is strictly positive.

Corollary: $f(-x) = 1/f(x)$ if $f(x)$ is non-zero.

For the rest of this question lets assume the non-trivial case where $f(x)$ is strictly positive.

1. $f(p/q) = f(1)^{p/q}$ for all $p/q \in \Bbb Q$
Proof: We have $$f(1/q) ^ q = \underbrace{f(1/q)\cdots f(1/q)}_{q \text{ times }} = f(q/q) = f(1)$$ so taking the $q^{\text{th}}$ root, $f(1/q) = f(1)^{1/q}$. $$f(px) = \underbrace{f(x)\cdots f(x)}_{p \text{ times }} = f(x)^p$$ Combining these we get the result.

Setting $A = \log(f(1))$ we see that $f(x) = \exp(A_t x)$ where $x = p/q \in \Bbb Q$. But what about $x \in \Bbb R - \Bbb Q$ (irrational values of $x$)?

1. If $f$ is required to be continuous, then by the density of $\Bbb Q$ in $\Bbb R$, $f(x) = \exp(A_t x)$ everywhere. But if $f$ is not required to be continuous then I think I can define $f(x) = \exp(A_t x)$ where $x$ in $\Bbb Q$ where $t$ in some coset $\Bbb R / \Bbb Q$ and $t \Bbb Q = {t + q \text{ where } q\in \Bbb Q}$ and $A_t$ is different for each coset. This makes for quite an interesting function. (FYI The cosets of $\Bbb R / \Bbb Q$ are discussed in this question What do the cosets of $\mathbb{R} / \mathbb{Q}$ look like?)

I am pretty sure in this case that $f(x)$ is not Lesbesque integrable. I am not sure what else we can tell about $f(x)$...

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## marked as duplicate by Nate Eldredge, Marvis, The Chaz 2.0, Eric NaslundJun 2 '12 at 17:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

To get $\Bbb R$ and $\Bbb Q$, use \Bbb R and \Bbb Q. –  MJD Jun 2 '12 at 16:48
–  user17762 Jun 2 '12 at 16:58
Specifically, @JonasMeyer has given an excellent answer to the same question here (math.stackexchange.com/questions/22069/…). –  user17762 Jun 2 '12 at 16:59

## 1 Answer

Let $B$ be a basis of $\Bbb R$ over $\Bbb Q$. Then $f$ is uniquely determined by its value on this basis.

If $x \in \Bbb R$ is written

$$x=\sum_{i=1}^n c_i x_i \,;\, c_i \in \Bbb Q , x_i \in B (*)$$ then

$$f(x)= \prod_{i=1}^n f(x_i)^{c_i} \,.$$

Conversely if you pick any function $g: B \to [0, \infty)$ you can define

$$f(x)= \prod_{i=1}^n g(x_i)^{c_i} \,.$$ where x is as in $(*)$ and get a function which has your desired properties.

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The OP mentions this in his post ;) –  N. S. Jun 2 '12 at 17:05