Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Take $f(x)=e^x$, so $f(x+x_0)-f(x) = e^x(e^{x_0}-1)$ which is unbounded. There isn't a whole lot you can say about this difference in general - what sort of thing are you looking to do? – Alon Amit Apr 14 '11 at 18:09
@Alon: Note the later comment "I forget to say that $f'$ is bounded". I.e., assume $0\leq f'(x)\lt M$ for some fixed $M$. – Arturo Magidin Apr 14 '11 at 18:13
@Arturo: well that does change things quite a bit... – Alon Amit Apr 14 '11 at 18:15

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, $$|f(x+x_0) - f(x)|\leq M|x_0|.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.