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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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    $\begingroup$ Look at $g(x)=g(x/2+x/2)$. $\endgroup$ – Brian M. Scott May 5 '12 at 7:41
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    $\begingroup$ $g(x) = g(x/2)^2 \geq 0$. Suppose $g(x) = 0$ then $g(y+x) = 0$ for all $y$. $\endgroup$ – t.b. May 5 '12 at 7:42
  • $\begingroup$ thanks. but how can one that g(x)<0 as wel? to complete the proof? $\endgroup$ – Sikhanyiso May 5 '12 at 8:00
  • $\begingroup$ I don't understand: the whole point is that $g(x)$ is never negative. $\endgroup$ – Brian M. Scott May 5 '12 at 8:02
  • $\begingroup$ 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? $\endgroup$ – Sikhanyiso May 5 '12 at 8:08
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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 $x \in \mathbf{R}$.

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