Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be multiplicative. Is it exponential? For function $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies $f\left(x+y\right)=f\left(x\right)f\left(y\right)$
and is not the zero-function I can prove that $f\left(1\right)>0$
and $f\left(x\right)=f\left(1\right)^{x}$ for each $x\in\mathbb{Q}$.
Is there a way to prove that for $x\in\mathbb{R}$?
This question has been marked to be a duplicate of the question whether $f(xy)=f(x)f(y)$ leads to $f(x)=x^p$ for some $p$. I disagree on that. Both questions are answered by means of construction of a function $g$ that suffices $g(x+y)=g(x)+g(y)$. In this question: $g(x)=\log f(x)$ and in the other $g(x)=\log f(e^x)$. So the answers are alike, but both questions definitely have another startpoint.
 A: No, because if $f$ is any of your functions, you may take any additive function $g : \mathbb{R} \to \mathbb{R}$ (that is, a function such that $g(x+y) = g(x) + g(y)$), and $f \circ g$ will still satisfy your assumption, as $f \circ g (x + y) = f(g(x+y)) = f(g(x) + g(y)) = f(g(x)) f(g(y)) = f \circ g(x) f \circ g(y)$.
And there are plenty such $g$, see under Hamel basis.
A: Continuity (or continuity in some point or measurability) is required. See Cauchy's functional equation. Your problem is reducible to this.
A: If $f(x_0)=0$ for some $x_0$ then $f(x)=f((x-x_0)+x_0)=f(x-x_0)f(x_0)=0$ for all $x$. As the zero function is excluded,  $f(x)\ne 0$ for all $x$ and in fact $f(x)=f(\frac x2)^2>0$. Therefore we can define $g(x)=\ln f(x)$ and find the functional equation $$ g(x+y)=g(x)+g(y)$$ for $g$. As has been pointed out, this (i.e. Cauchy's) functional equation has precisely the affine linear functions as continuous solutions, but also many wild and non-continuous solutions. Each solution $f$ is of the form $f(x)=e^{g(x)}$ for such a $g$.
