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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 $f(x) - f(x-1) \leq f^\prime (x)=200 x$?

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4 Answers 4

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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 $x$.

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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.

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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 (!).

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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.

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