China Southeastern Mathematical Olympiad

Let $g(x)=f(x)-x^2$, then $$\begin{array}{rl}g(x+1)-g(x)& =f(x+1)-f(x)-(x+1{)}^2+x^2\\ & =(2x+1)-(x^2+2x+1-x^2)\\ & =2x+1-(2x+1)\\ & =0.\end{array}$$ Thus, $g(x)$ is a periodic function with 1 as its period. On the other hand, as is given $|f(x)|\le 1$ when $x\in [0,1]$, so $$|g(x)|=|f(x)-x^2|\le |f(x)|+x^2\le 1+x^2\le 2,$$ when $x\in [0,1]$. Therefore, the periodic function $g(x)$ satisfies $|g(x)|\le 2$ ($x\in \mathbb{R}$), thus arriving at $$|f(x)|=|g(x)+x^2|\le |g(x)|+x^2\le 2+x^2(x\in \mathbb{R}),$$ as desired.