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I need to prove that for a given $a \in \mathbb{R}, a \geq 0$, there exists a unique monotonic function $f : \mathbb{R} \rightarrow \mathbb{R}$ which satisfies $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ and $f(1) = a$. How do I prove uniqueness? I assumed the opposite and proved that it holds for every natural number via mathematical induction. I do not know what to do next.

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mathion
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    What have you tried? Where are you stuck? – Ethan Bolker Oct 11 '23 at 21:10
  • I assumed the opposite and proved that it holds for every natural number via mathematical induction. I do not know what to do next. – mathion Oct 11 '23 at 21:13
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    What can you deduce about the value at $1/2$? At $1/n$? At rationals? You can [edit] the question, but do not use comments to clarify. – Ethan Bolker Oct 11 '23 at 21:16
  • I do not know where to use the assumption that f is monotonic... – mathion Oct 11 '23 at 21:27
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    Hint: find out what the value of the function must be at every integer. hint2: find out what it must be for every rational. Hint3, apply monotonicity and density of rationals. – M W Oct 11 '23 at 21:31
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    Maybe after some simplification you can reduce the problem to Cauchy's functional equation: https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation – C614 Oct 11 '23 at 21:32

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