1018

设函数 ${\displaystyle f(x)}$ 对一切实数 ${\displaystyle x,y}$ 满足 ${\displaystyle f(xy)=x^{2}f(y)+y^{2}f(x)-(xy{)}^2}$, 且 ${\displaystyle |f(x)-x^2|\le 1}$, 则函数 ${\displaystyle f(x)=}$ _____.

首先代入 ${\displaystyle x=y=0}$, 立即得 ${\displaystyle f(0)=0}$.

若${\displaystyle x,y\ne 0}$, 对条件式同除${\displaystyle (xy{)}^2}$, 有$$\begin{array}{r}\frac{f(xy)}{(xy{)}^2}-1=\frac{f(x)}{x^2}-1+\frac{f(y)}{y^2}-1,\end{array}$$
因此记${\displaystyle g(x)={\displaystyle \frac{f(x)}{x^2}-1}}$, 有$$g(xy)=g(x)+g(y).$$
此时不等式条件等价于$$\begin{array}{r}{x}^2|g(x)|\le 1\Rightarrow |g(x)|\le \frac{1}{x^2}(x\ne 0).\end{array}$$

对 ${\displaystyle g(x)}$, 代入 ${\displaystyle x=y=0}$, 又有 ${\displaystyle g(0)=0}$; 代入 ${\displaystyle x=y=1}$, 有 ${\displaystyle g(1)=0}$. 再令 ${\displaystyle x=y}$, 有 $$g(x^2)=2g(x)\Rightarrow g(x^3)=g(x)+g(x^2)=3g(x)\Rightarrow g(x^n)=ng(x).$$

• 如果存在 ${\displaystyle x_0,|x_0|>1}$ 使 ${\displaystyle g(x_0)\ne 0}$, 则$$\begin{array}{r}n|g(x_0)|=|g(x_0^n)|\le \frac{1}{x_0^n}\Rightarrow |g(x_0)|\le \frac{1}{n{x}_0^n}\to 0(n\to \mathrm{\infty }),\end{array}$$
  矛盾；
• 如果存在 ${\displaystyle x_0,0<|x_0|<1}$ 使 ${\displaystyle g(x_0)\ne 0}$, 则$$\begin{array}{r}0=g(1)=g(x_0)+g\left(\frac{1}{x_0}\right)\Rightarrow g\left(\frac{1}{x_0}\right)=-g(x_0)\ne 0,\end{array}$$
  化归到上一种情形，再次矛盾.

因此，只有 ${\displaystyle g(x)\equiv 0}$, 即 ${\displaystyle f(x)=x^2,x\ne 0}$. 再结合 ${\displaystyle f(0)=0}$, 可得 ${\displaystyle f(x)=x^2,\mathrm{\forall }x\in \mathbb{R}}$.