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Find function $f :\mathbb{R} \to \mathbb{R}$ satisfy that : $$f(1)=1$$ $$f(x+y)=f(x)+f(y)+2xy$$ $$f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4} \hspace{5pt}\forall x \neq 0$$

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2 Answers 2

up vote 5 down vote accepted

Hint: $$(x+y)^2=x^2+y^2+2xy{}$$

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I do to here : let $g(x)=f(x)-x^2$ , at that time : $$g(x+y)=g(x)+g(y)$$ $$g(1)=g(0)=0$$ $$g(\frac{1}{x})=\frac{g(x)}{x^{4}}$$

I proved $g(x)\equiv 0$ with $ x \in \mathbb{Q}$ but proving $g(x) \equiv 0$ with $x \in \mathbb{R}$ is problem for me, can you help me more

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If $f$ is taken to be continuous then by the fact it is $0$ on $\mathbb Q$ it has to be zero everywhere. –  Asaf Karagila Nov 11 '12 at 14:55
We can't say we must prove –  LevanDokite Nov 11 '12 at 15:57
But you are supposed to find a function and prove it has the properties. Simply prove that $f(x)=x^2$ has the wanted properties and you're done. I don't see why you need to do this. –  Asaf Karagila Nov 11 '12 at 15:58
Example we don't have $f(\sqrt{2})=0$ we just have $f(x)=0$ with $\forall x \in Q$ –  LevanDokite Nov 11 '12 at 16:01
I don't understand. You are supposed to find a function with the properties mentioned in your question. So you are supposed to find a function and then prove that the properties hold for it. Simply say "Let us take $f(x)=x^2$, and we will see that the properties hold..." or something like that. Furthermore without assumption of continuity you cannot possible deduce that $f(q)=0$ for all $q\in\mathbb Q$ implies $f(x)=0$ for all $x\in\mathbb R$. Take for example $$h(x)=\begin{cases}0 & x\in\mathbb Q\\1& x\notin\mathbb Q\end{cases}$$ –  Asaf Karagila Nov 11 '12 at 16:09

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