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I would appreciate if somebody could help me with the following problem:

Q: Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function and $f(x+y)=f(x)f(y)$. Then $f(x)\geq 0$

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If $f$ vanishes at one point then it vanishes everywhere. So one solution is $f(x) = 0$ for all $x$. Otherwise $f$ must be non-zero everywhere. Now it is easy to see that if $f(x) < 0$ then $f(2x) = f(x)f(x) > 0$ and hence $f$ changes sign and vanishes somewhere by IVT. Hence we must have $f(x) > 0$ for all $x$.

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Hint: Substitute $x=y=a/2$ into the equation.

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  • $\begingroup$ Upvote because continuity is not required. $\endgroup$ – user369210 May 15 '16 at 5:52
  • $\begingroup$ Your answer is better than mine, because it does not use continuity. +1 $\endgroup$ – Paramanand Singh May 15 '16 at 5:57

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