1. 设$f: mathbb{R}^{n} o mathbb{R}^{n}$且$f=(f_{1},f_{2},cdots,f_{n})$.$|Jf(x)|leq frac{1}{2},f(x)in C^{1}(mathbb{R}^{n})$. 证明: $g(x)=x+f(x)$是一一映射.
证明: 首先证明$g: mathbb{R}^{n} o mathbb{R}^{n}$为单射,设 $g(x_{1})=g(x_{2}),x_{1},x_{2}in mathbb{R}^{n}$, 那么有
$$|x_{1}-x_{2}|=|f(x_{1})-f(x_{2})|leq max|Jf(x)|cdot |x_{1}-x_{2}|leq frac{1}{2}|x_{1}-x_{2}|$$
从而得
$$|x_{1}-x_{2}|=0, x_{1}=x_{2}$$
满射: 即证明 $g(x)=x+f(x)=y$在$mathbb{R}^{n}$中有解，即证明$g$（局部）可逆
$$|Jg(x)|=|I+Jf(x)|geq |I|-|Jf(x)|geq frac{1}{2}$$
由逆映射定理知$g$可逆. 从而 $g$为满射.