For function f:R→R that satisfies f(x+y)=f(x)f(y)
and is not the zero-function I can prove that f(1)>0
and f(x)=f(1)x for each x∈Q.
Is there a way to prove that for x∈R?
This question has been marked to be a duplicate of the question whether f(xy)=f(x)f(y) leads to f(x)=xp 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)=logf(x) and in the other g(x)=logf(ex). 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:R→R (that is, a function such that g(x+y)=g(x)+g(y)), and f∘g will still satisfy your assumption, as f∘g(x+y)=f(g(x+y))=f(g(x)+g(y))=f(g(x))f(g(y))=f∘g(x)f∘g(y).
And there are plenty such g, see under Hamel basis.
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