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continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$

Let $g: \mathbf{R} \to \mathbf{R}$ be a function which is not identically zero and which satisfies the functional equation $g(x+y)=g(x)g(y)$

Suppose $a>0$, show that there exists a unique continuous function satisfying the above, such that $g(1)=a$.

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marked as duplicate by t.b., Martin Sleziak, The Chaz 2.0, lhf, J.D. May 7 '12 at 16:12

This question was marked as an exact duplicate of an existing question.

Uhm - homework? – user20266 May 7 '12 at 14:52
Patrova = Sikhanyiso? – t.b. May 7 '12 at 15:31
Also, almost verbatim the same question was completely answered here. – t.b. May 7 '12 at 15:32

Note that $g(n)=g(1)^n=a^n$, and $g\left(\frac{1}{n}\right)^n=g(1)=a>0$ so that $g\left(\frac{1}{n}\right)=a^{\frac{1}{n}}$. Putting these together you get that $g(q)=a^q$ for every $q\in\mathbb{Q}$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and $a^x$ and $g(x)$ are continuous and agree on $\mathbb{Q}$ they must be equal.

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