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

Let $f(x)$ be a positive function on $[0,\infty)$ such that $f(x) \leq 100 x^2$. I want to bound $f(x) - f(x-1)$ from above. Of course, we have $$f(x) - f(x-1) \leq f(x) \leq 100 x^2.$$ This is not good for me though. I need a bound which is linear (or at worst linear-times-root) in $x$.

Is there an inequality of the form $f(x) - f(x-1) \leq f^\prime (x)=200 x$?

share|cite|improve this question
up vote 1 down vote accepted

There is no such bound. Let $c$ be a real number, and let $$f(x)=\begin{cases} 0&\text{if $x<c$}\\100x^2&\text{if $x\geq c$}.\end{cases}$$ Then for $x\in [c,c+1)$ the inequality $f(x)-f(x-1)\leq 100x^2$ is the best bound possible. So we cannot make a better bound for a general function satisfying $f(x)\leq 100x^2$ for all $x$.

share|cite|improve this answer

Suppose $f(x) = 100x^2\sin^2(\pi x/2)$. Then $f(x) = 0$ when $x$ is an even integer and $f(x) = 100x^2$ when $x$ is an odd integer. So $f(x)-f(x-1)\ge 100(x-1)^2$, with equality when $x$ is even.

share|cite|improve this answer

There is no such bound. Let $f(2)=0$, and $f(x)=100x^2$ for other $x$. Surely $f(3)-f(2)=900\le100x^2$, but equality holds (!).

share|cite|improve this answer

For $2^n \le x<2^{n+1}$, let $f(x)=100(2^{n})^2$. There is an enormous jump from $f(2^{n+1}-1)$ to $f(2^{n+1})$. So even if we assume that $f$ is non-decreasing, we can have jumps of size comparable to $100x^2$. At the cost of complicating the description, we can modify the above $f(x)$ to make it strictly increasing.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.