Let $g: \mathbf R \to \mathbf R$ be a function which is not identically zero and which satisfies the equation $$ g(x+y)=g(x)g(y) \quad\text{for all } x,y \in \mathbf{R}. $$ Show that $g(x)\gt0$ for all $x \in \mathbf{R}$.
A non-zero function satisfying $g(x+y) = g(x)g(y)$ must be positive everywhere
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functional-equations
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3Look at $g(x)=g(x/2+x/2)$. – 2012-05-05
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5$g(x) = g(x/2)^2 \geq 0$. Suppose $g(x) = 0$ then $g(y+x) = 0$ for all $y$. – 2012-05-05
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0thanks. but how can one that g(x)<0 as wel? to complete the proof? – 2012-05-05
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0I don't understand: the whole point is that $g(x)$ is **never** negative. – 2012-05-05
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0sorry, i ment since we've already shown that _g(x)_ is never zero, how can one show that _g(x)_ is also never nevative? – 2012-05-05
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0@Sikhanyiso Brain and theo didn't show it was never zero, they showed it was always greater than or equal to zero. – 2012-05-05
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0@Ragib: Actually, t.b.'s hint shows both. – 2012-05-05
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0@RagibZaman, in Theo's answer, by assuming that g(x)=0,showing that g(x+y)=0,it's to prove that g(x) should not be zero, so as to satisfy the proposed condition that g is not identically zero – 2012-05-05
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0Yes. And both answers show that $g(x)\ge 0$ for all $x\in\Bbb R$. Put the two together, and you have the desired result. – 2012-05-05