1
$\begingroup$

Question: Finding $f:R \to R, $$f(x +y^2) \ge (y+1)f(x) $

My idea is: if $f(x)=const, f=0, \forall x \in R$

if $f\neq const$, trying to prove not exit functional

when y=-1 we have $f(x+1)\ge 0 , \forall x \in R$ so $f(x)\ge 0 , \forall x \in R$

when y=1 we have $f(x+1) \ge 2f(x) \ge f(x), \forall x \in R$

and when $y \to {\infty}$, $f \to {\infty}$

I've tried to find a sequense but I have trouble proving it: when $x=x-y^2$ $f(x) \ge (y+1)f(x-y^2) $ , when $y=f(x), $$f(x) \ge (f(x)+1) f(x-f(x)^2)$

since $y \to {\infty}$, $f \to {\infty}$ exit $x_0, f(x) \ge 1, \forall x \ge x_0 $ hence $f(x) \ge (f(x)+1) f(x-f(x)^2)\ge 2 f(x-f(x)^2), \forall x \ge x_0 $

$\forall x \ge x_0 , let:x_1=x, x_{n+1}=x_n-f(x_n)^2$ so $f(x_{n+1}) \le \frac {f(x_n)}{2}\le...\le \frac{f(x_1)}{2^n} \to 0 $ $x_{n+1}=...=x_1+x_1-f(x_1)-f(x_2)-...-f(x_n) \ge 2x_1-f(x_1)(1+...+\frac{1}{2^n}) \ge 2x_1-2f(x_1) \Rightarrow f(x_{n+1}) \ge f(2x_1-2f(x_1))$ hence $f(2x-2f(x))=0,\forall x \ge x_0$

  • 0
    You can try differentiate wrt x and y separately if you have learned differentiation. – 2017-02-22

1 Answers 1

0

Partial result: If $f$ is differentiable, then it is identically $0$. You can prove this by setting $y= \sqrt{h}$, and rearranging to show that unless $f(x)=0$, the limit $f'(x)$ is undefined.

  • 0
    but maybe not exit f'(x) – 2017-02-22
  • 0
    @AnhThư True, which is why that's only a partial result. – 2017-02-22