2

Find all $f:\mathbb{R} \to \mathbb{R}$ s.t. $$f\big(x^2+f(y)\big)=(x-y)^2f(x+y)$$

Putting $x=y$ yields $$f\big(x^2+f(x)\big)=0\text.\tag1\label1$$

Putting $y=-x$ yields $$f\big(x^2+f(-x)\big)=(2x)^2f(0)\text.\tag2\label2$$

By \eqref{1}, $(2x)^2f(0)=0$ ($\forall x \in \mathbb{R}$) $$\implies \; f(0)=0$$ Hence putting $y=0$ we have $f\big(x^2+f(0)\big)=f(x)x^2$ which from above is $f\big(x^2\big)=f(x)x^2$.

Using this last equation we have $f(a)=af\left(a^{\frac12}\right)=a^{1+\frac12}f\left(a^{\frac14}\right)=a^{1+\frac12+\frac14}f\left(a^{\frac18}\right)= \cdots =a^2f\left(a^0\right)=a^2f(1)$.

So $f(x)=kx^2$ for some real constant $k$.

Conversely we check and see that it works only for $k=0$ (I think).

My question essentially is - in the sort of limit part is that rigorous (I'd love to avoid explicit calculus) and is my final deduction correct?

Thanks for any help.

Mohsen Shahriari
  • 6,718
John Marty
  • 3,650
  • 1
    If you're given that $f(x)$ needs be continuous, then your limit argument works. However that doesn't seem to be part of what you gave as the problem statement. Also, your construction only works for $a>0$. – vadim123 Jul 31 '13 at 05:34
  • Why does $f(x)$ have to be continuous for that to work? – John Marty Jul 31 '13 at 06:21
  • Because the punch line of your method is that $f(1+\epsilon)$ is close to $f(1)$, for small $\epsilon$. If $f$ isn't continuous at $1$ that isn't true. – vadim123 Jul 31 '13 at 14:10
  • Note that another solution is $f(x) = -x^2$, i.e. $k=-1$ works. – Calvin Lin Jul 31 '13 at 15:01

2 Answers2

2

There's a way to prove $f(x)=0$ or $f(x)=-x^2$ without using continuity.
Put $-x$ in the place of $x$ and you'll get: $$f\big((-x)^2+f(y)\big)=(-x-y)^2f(-x+y)$$ $$\therefore(y+x)^2f(y-x)=(y-x)^2f(y+x)$$ Now put $\frac{y+x}{2}$ and $\frac{y-x}{2}$ in the place of $y$ and $x$, respectively, and you'll have: $$y^2f(x)=x^2f(y)$$ $$\therefore f(x)=x^2f(1)$$ Put $k=f(1)$ and check if the answer works: $$k\big(x^2+ky^2\big)^2=(x-y)^2k(x+y)^2=k\big(x^2-y^2\big)^2$$ $$\therefore k(1+k)y^2\big(2x^2+(k-1)y^2\big)=0$$ The above equation holds for $x=1$ and $y=1$, so the only valid values for $k$ are $0$ and $-1$.

Mohsen Shahriari
  • 6,718
0

You obtained tw0 relations: $(1)$ $f(x^{2}+f(x))=0$ and $(2)$ $f(x^{2})=x^{2}f(x)$ , but for complete of solution you need another relation, the functional equation ( or recently relation) show that the function $f$ is even. Also, It's obvious that there is a kind of symmetric relation in functional equation, so we have $$(3)\ \ f(x^{2}+f(y))=f(y^{2}+f(x))$$ The relations $(2)$ and $(3)$ imply that $$(4)\ \ f(f(x))=x^{2}f(x)$$ We know that $f(0)=0$ and $f$ is an even function, so if the function $f$ is a nonzero function, then relation $(4)$ implies that the point $x=0$ is only root of $f$, therefore relation $(1)$ implies that $f(x)=-x^{2}$ and the proof is done.