# Finding $\alpha$ such that $f(\alpha(x+y))=f(x)+f(y)$

Problem taken from the link: http://web.mit.edu/rwbarton/Public/func-eq.pdf I am stating the question here

For which $\alpha$ does there exists a nonconstant function $f: \mathbb{R} \to \mathbb{R}$ such that $f(\alpha(x+y))=f(x)+f(y)$ for all $x,y \in \mathbb{R}$.

Clearly for $\alpha=1$ we see that this case is satisfied, by taking the identity function. But are there other values of $\alpha$ for which this condition is satisfied.

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## 1 Answer

$f(0+0)=f(0) + f(0)$ implies $f(0)=0$. Set $y=0$ to get $f(\alpha x)=f(x)$ so that $f(\alpha (x+y))=f(x+y)$ and the function is additive. If $\alpha \neq 1$ this means $f((\alpha - 1)x)=0$ and hence $f=0$.

This is too easy to be on rwbarton's list, and in fact I cannot find this problem in the file that you linked. Which problem number is it in the list?

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@T: I am sorry! the file which i had downloaded is outdated. Seems, Reid has updated it. The one which i have was dated on June 2005, but the file present now is September 2006. – anonymous Sep 12 '10 at 2:30
Good to know, thanks. – T.. Sep 12 '10 at 5:07
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