Could anyone tell me?
The No. of Continuous function satisfying the condition $$xf(y)+yf(x)=(x+y)f(x)f(y)$$ is
$1,2,3,$ or none of them?
or give me hints please.
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3I can think of 2: $f(x)=0$ and $f(x)=1$. They are not interesting though. – Daryl Aug 29 '13 at 08:07
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1The trick usually in solving these equations is to plug in some interesting values. Otherwise you can't do much. – Patrick Da Silva Aug 29 '13 at 08:23
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Related: Find all $f:\mathbb R\to\mathbb R$ such that $\forall x,y\in\mathbb R$ the given equality holds: $xf(y)+yf(x)=(x+y)f(x)f(y)$. – Martin Sleziak Dec 04 '18 at 18:39
1 Answers
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Assuming your functions are from $\mathbb{R} \to \mathbb{R}$ $$ f(1)+f(1) = (1+1)f(1)f(1) \Rightarrow f(1) = 0 \text{ or } 1 $$ If $f(1) = 1$, then for any $x\neq 0$ $$ x + f(x) = (x+1)f(x) \Rightarrow f(x) = 1 $$ By continuity, $f\equiv 1$ Similarly, if $f(1) = 0$, then $f\equiv 0$.
Hence, the answer is 2.
Prahlad Vaidyanathan
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