1)Let $\displaystyle f(x)$ be a continous function on $\displaystyle \mathbb{R}$ solve the functional equation: $\displaystyle f(x+y)=f(x)f(y)$.

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- Nov 18th 2007, 04:59 PMThePerfectHackerProblem 41
1)Let $\displaystyle f(x)$ be a continous function on $\displaystyle \mathbb{R}$ solve the functional equation: $\displaystyle f(x+y)=f(x)f(y)$.

- Nov 18th 2007, 05:26 PMangel.white
- Nov 18th 2007, 05:33 PMThePerfectHacker
- Nov 18th 2007, 05:50 PMangel.white
- Nov 20th 2007, 05:21 AMmath sucks
Let $\displaystyle n$ be a positive integer. Then

$\displaystyle f(1)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)= f\left(\frac{1}{n}\right)^n$

So

$\displaystyle f\left(\frac{1}{n}\right)=f(1)^{\frac{1}{n}}$

Now let $\displaystyle \frac{m}{n}$ be a positive rational number. Then

$\displaystyle f\left(\frac{m}{n}\right)=f\left(\frac{1}{n}+\ldot s+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^m=f (1)^{\frac{m}{n}}$

Now let $\displaystyle x$ be a positive real number. By continuity,

$\displaystyle f(x)=\lim_{z\rightarrow x}f(z)=f(1)^x$

where the last result is established by our definition of $\displaystyle f$ over the rational numbers.

Assume $\displaystyle f(1)\neq0$, then it is trivial that $\displaystyle f(0)=1$. Thus, for any positive real number $\displaystyle y$,

$\displaystyle f(-y)=\frac{f(-y)f(y)}{f(y)}=\frac{f(0)}{f(y)}=f(1)^{-y}$

Therefore, $\displaystyle f(x)=f(1)^x$ for all $\displaystyle x$ or assume $\displaystyle f(1)=0$, then it is trivial that $\displaystyle f(x)=0$ for all $\displaystyle x$ - Nov 20th 2007, 11:22 AMmath sucks
? I don't understand. How did you think that?

Maybe this will clarify,

$\displaystyle f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1 }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n$

The second equality is straight from the problem. - Nov 21st 2007, 08:42 AMtopsquark
- May 26th 2009, 08:39 PMRebesques
[cheater] What of discontinuous f? (Angry)

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