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.
Answer
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.
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