1
$\begingroup$

Given a continuous and increasing function $f$ on $\mathbb R$, and a given point $x_{0}\in \mathbb R$, what we can say about $ f(x+x_{0})-f(x)$ for all $x\in \mathbb R$? Do we have a bound for this difference? I forget to say that f' is bounded on $\mathbb R$. Does this change anything!

  • 0
    @Arturo: well that does change things quite a bit...2011-04-14

1 Answers 1

6

If we assume that f' is bounded (and in particular, it exists at all points), then 0\leq f'(x)\leq M for some $M\gt 0$ ($0\leq f'(x)$ because $f$ is increasing).

By the Mean Value Theorem, for every $a\lt b$ there exists $c$ such that f'(c) = \frac{f(b)-f(a)}{b-a} with $a\leq c\leq b$, so f'(c)(b-a) = f(b)-f(a). Therefore, we have f(b)-f(a) = f'(c)(b-a) \leq M(b-a). Thus, if $x_0\gt 0$, then $f(x+x_0) - f(x) \leq Mx_0$ and if $x_0\lt 0$ then $f(x) - f(x+x_0) \leq M(-x_0)$. In summary, $|f(x+x_0) - f(x)|\leq M|x_0|.$