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 $x \in \mathbf{R}$.
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3Look at $g(x)=g(x/2+x/2)$. – Brian M. Scott May 05 '12 at 07:41
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5$g(x) = g(x/2)^2 \geq 0$. Suppose $g(x) = 0$ then $g(y+x) = 0$ for all $y$. – t.b. May 05 '12 at 07:42
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thanks. but how can one that g(x)<0 as wel? to complete the proof? – Sikhanyiso May 05 '12 at 08:00
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I don't understand: the whole point is that $g(x)$ is never negative. – Brian M. Scott May 05 '12 at 08:02
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sorry, i ment since we've already shown that g(x) is never zero, how can one show that g(x) is also never nevative? – Sikhanyiso May 05 '12 at 08:08
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@Sikhanyiso Brain and theo didn't show it was never zero, they showed it was always greater than or equal to zero. – Ragib Zaman May 05 '12 at 08:12
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@Ragib: Actually, t.b.'s hint shows both. – Brian M. Scott May 05 '12 at 08:23
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@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 – Sikhanyiso May 05 '12 at 08:26
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Yes. 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. – Brian M. Scott May 05 '12 at 08:28
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We have $g(x) = g(\tfrac{x}{2} + \tfrac{x}{2}) = g(\tfrac{x}{2})^2 \geq 0$ for all $x \in \mathbf{R}$.
Suppose we have $g(x_0) = 0$ for some $x_0 \in \mathbf{R}$. Then $g(x_0+y) = g(x_0)g(y) = 0$ for all $y \in \mathbf{R}$, hence $g$ must be identically zero. Since you assume that's not the case, there can't be any such $x_0$, thus $g(x) \gt 0$ for all $x \in \mathbf{R}$.
t.b.
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