If $f(x)$ agrees with $e^x$ on the integers, and $f(x)f(y)=f(x+y)$, then is $f(x)=e^x$?

Question

Suppose that the function $f:\mathbf R\to\mathbf R$ satisfies $f(n)=e^n$, where $n\in \mathbf Z$, and $f(x)f(y)=f(x+y)$ for all $x,y\in\mathbf R$. Must it be the case that $f=\exp$?

What I have tried so far:

If we can prove that $\lim_{h\to0}f(h)=1$, then it follows that $$\lim_{h\to0}f(a+h)=\lim_{h\to0}f(a)f(h)=f(a)\lim_{h\to0}f(h)=f(a) \, ,$$ and so $f$ is continuous. With the extra hypothesis of $f$ being continuous, it can then be shown that $f=\exp$. So if the statement is true, then the problem reduces to proving that $\lim_{h\to0}f(h)=1$. If the statement is false, then there is a function $f$ satisfying the hypotheses of the question, but not satisfying $\lim_{h\to0}f(h)=1$.

Answer 1

There exist pathological (specifically non-measurable) functions $g$ such that $g(x+y)=g(x)g(y)$ for all $x,y$. For such a function $g$ we have $g(n)=a^{n}, n \in \mathbb Z$ for some $a$. Let $f(x)=(\frac e a)^{x}g(x)$. The $f$ satisfies the hypothesis of your question. So some continuity assumption is essential.

Answer 2

Must it be the case that $f=\exp$?

No.

From the functional equation you can conclude that $f=\exp$ on $\Bbb Q$ and that $f(0)=1=e^0$. But it's completely fine to have $$f(y)=2^y \quad\text{where}\ y\in \sqrt{2}\Bbb Q$$

The functional equation implies the definition on the additive group $G=(\Bbb Q+\sqrt{2}\Bbb Q,+)$ $$ f(x+\sqrt{2}y) = e^x2^y\quad\text{ where }\quad x,y\in\Bbb Q$$

The condition you are missing is that $f$ must be continuous to conclude for all of $\Bbb R$. Without continuity, your limits are not defined. Or at least it must be continuous on some open interval.

Answer 3

Let $\{b_i\}_{i\in I}$ be a basis for $\mathbb{R}$ over $\mathbb{Q}$ which contains $1$, and let $\{c_i\}_{i\in I}$ be any collection of real numbers. Then $$f\left(\sum_{i\in I}\lambda_i b_i\right):=\prod_{i\in I}e^{c_i\lambda_i}$$ satisfies the functional equation. Set the $c_i$ corresponding to $b_i=1$ to $1$, and take some other $c_j$ and set it to anything but $1$, say $2$. Then $f$ agrees with $\exp$ on $\mathbb{Q}$, but not on $\mathbb{R}$.