A consequence of the Lipschitz inequality

65 Views Asked by At

Is the following true?

If $F : \mathbb R\to\mathbb R$ is Lipschitz then there exists $L'$ such that for every $x \in\mathbb R$ $$|F(x)| \le L'(1+|x|)$$

1

There are 1 best solutions below

0
On

Note that $$|f(x)|\leqslant |f(x)-f(0)| +|f(0)| \leqslant|x|\cdot\sup_{u<v}\frac{|f(v)-f(u) |}{v-u} +|f(0)| \leqslant L'(1+|x|),$$ where $$L':=\max\left\{\sup_{u<v}\frac{|f(v)-f(u) |}{v-u},|f(0)| \right\}.$$