# How to show that If $f(x+y)=f(x)f(y)$ then $f(x)\geq 0$ [duplicate]

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 $f(x)\geq 0$

• Is $f$ real-valued? – Eclipse Sun May 15 '16 at 5:47
• Yes! $f$ real-valued function – Young May 15 '16 at 5:47
• It is not necessary to assume continuity. – André Nicolas May 15 '16 at 5:50
• Why not $a^{x+y}=a^x\cdot a^{y}$? – Aditya Dev May 15 '16 at 5:56
• – Martin Sleziak May 15 '16 at 5:57

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