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find all functions $f:\mathbb{R^{*}}\to \mathbb{R}$ such that $$f(y^2f(x)+x^2+y)=x(f(y))^2+f(y)+x^2,\;\forall x,y\in \mathbb{R^{*}}$$ ($\mathbb{R^{*}}=\{x\in\mathbb{R},x\ne 0\})$

Babymath
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    Why? Where does this come from? – Gerry Myerson Mar 08 '14 at 07:14
  • got it, let me tex. i think it's $t \stackrel{f}{\mapsto} t^{2}$ – Max Mar 08 '14 at 08:05
  • @Max It can't be that, as you can check by substituting it into the function equation to see that it would require $\left(y^2 x^2 + x^2 + y\right)^2 = x y^4 + y^2 + x^2$ – David H Mar 08 '14 at 08:54
  • okay, either this means i made a mistake or that means each function fulfilling this criteria cannot be continously differentiable in $0$. can you check if you find any mistakes in my answer? – Max Mar 08 '14 at 09:01

2 Answers2

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does anyone find the mistake? (there must be one) remark: i'm assuming $f$ to be continuously extendable in $0$

I'm assuming $f$ to be once continously differentiable on it's connected domain in $\mathbb{R}$ and $0$ is contained in this domain.

$ \forall x,y : \frac{f\left(y^{2}f\left(x\right)+x^{2}+y\right)-f\left(y\right)}{x}=f\left(y\right)^{2}+x $

$ \underbrace{\Rightarrow}_{y\rightarrow 0} \forall x: \frac{f\left(x^{2}\right)-f\left(0\right)}{x}=f\left(0\right)^{2}+x $

this gives

$ \underbrace{\Rightarrow}_{x\rightarrow 0} 2f'\left(0\right)=f\left(0\right)^{2} $

and

$ \forall x:f\left(x^{2}\right)-f\left(0\right)=xf\left(0\right)^{2}+x^{2} $

$ \underbrace{\Rightarrow}_{\partial_{x}} \forall x: 2f'\left(x\right)=f\left(0\right)^{2}+2x $

from there we see firstly, $f$ is parabolic (and also twice differentiable)

secondly

$ \underbrace{\Rightarrow}_{x\rightarrow 0} f\left(0\right)=0 \underbrace{\Rightarrow}_{2f'\left(0\right)=f\left(0\right)^{2}} f'\left(0\right)=0 $

and thirdly by applying $\partial_{x}$ again

$ 2f''\left(x\right)=2 $

thus $f$ is parabolic and we have $f'\left(0\right)=0$, $f\left(0\right)=0$ and $f''\left(0\right)=1$.

this means $f\left(t\right)=\frac{1}{2}t^{2}$

Max
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The correct solution to the functional equation that doesn't exclude $x=0$ from its domain, that is $$f(y^2f(x)+x^2+y)=x(f(y))^2+f(y)+x^2,\;\forall x,y\in \mathbb{R},$$

can be found as follows.

Letting $y=0$, the functional equation reduces to:

$$f(x^2)=x(f(0))^2+f(0)+x^2$$

Set $f_0=f(0)$, and substitute $x=\text{sgn}{(u)}\sqrt{u}$. Then $u=x^2$, and

$$f(u) = u + f_0 + f_0^2 \text{sgn}{(u)}\sqrt{u}$$.

Letting $u=0$,

$$f_0=f_0+f_0^2\\ \implies f_0^2=0\\ \implies f_0=0.$$

Hence, the general solution is simply the identity map, $f(x)=x$. It's a simple matter to check that this satisfies the functional relation. On the LHS we have,

$$f(y^2f(x)+x^2+y)=f(y^2x+x^2+y)\\ =y^2x+x^2+y.$$

And on the RHS we have,

$$x(f(y))^2+f(y)+x^2=xy^2+y+x^2.$$

David H
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