I understand the other way around, where if a function is exponential then it will satisfy the equality f(a+b)=f(a)f(b). But is every function that satisfies that equality always exponential?
Answer
First see that f(0) is either 0 or 1. If f(0)=0, then for all x∈R, f(x)=f(0)f(x)=0. In this case f(x)=0 a constant function.
Let's assume f(0)=1. See that for positive integer n, we have f(nx)=f(x)n which means f(n)=f(1)n. Also see that:
f(1)=f(n1n)=f(1n)n⟹f(1n)=f(1)1/n.
Therefore for all positive rational numbers:
f(mn)=f(1)m/n.
If the function is continuous, then f(x)=f(1)x for all positive x. For negative x see that:
f(0)=f(x)f(−x)⟹f(x)=1f(−x).
So in general f(x)=ax for some a>0.
Without continuity, consider the relation: xRy if xy∈Q (quotient group R/Q). This relation forms an equivalence class and partitions R to sets with leaders z. In each partition the function is exponential with base f(z).
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