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 $f:[0,1]\to \mathbb{R}$ be a smooth function such that the following property is satisfied. $$\int\limits_{[0,1]}\int\limits_{[0,1]}|f(x)-f(y)|^2dxdy\leq \varepsilon.$$ What can I most say about $\max\limits_{[0,1]}f-\min\limits_{[0,1]}f$?

share|cite|improve this question
1. Edit your inequality inserting $dxdy$, please. 2. Inequality holds for every $\varepsilon$ or fixed? – M. Strochyk Oct 4 '12 at 16:51
up vote 3 down vote accepted

If you choose $f_n(x) = x^n$, with $n$ a positive integer, a quick computation shows that $\int_{[0,1]} \int_{[0,1]} |f_n(x)-f_n(y)|^2 \, dx dy = \frac{2n^2}{(n+1)^2(2n+1)}$. Furthermore, $\max_{x\in [0,1]} f_n(x) - \min_{x\in [0,1]} f_n(x) = 1$ for all $n$.

Hence the integral can be made arbitrarily small, yet the range is 1. So, roughly speaking, not much can be said about the range given integral bound information.

share|cite|improve this answer

$\def\abs#1{\left|#1\right|}$Nothing. We have \begin{align*} \int_{[0,1]^2} \abs{f(x) - f(y)}^2\, d(x,y) &\le \int_{[0,1]^2} \abs{f(x)}^2\,d(x,y) + \int_{[0,1]^2} 2\abs{f(x)}\abs{f(y)}\, d(x,y) + \int_{[0,1]^2} \abs{f(y)}^2\, d(x,y)\\ &= \|f\|^2 + 2\|f\|^2 + \|f\|^2\\ &= 4\|f\|^2 \end{align*} That is, your term will be small if $\|f\|^2$ is small. Now let $f_n$ be a smooth, positive function with $\max f_n = n$, $\min f_n = 0$, $\mathrm{supp}\, f_n \subseteq [0, \frac 1{n^3}]$. Then $\|f\|^2 \le n^2 \cdot \frac 1{n^3} = \frac 1n$, but $\max f_n - \min f_n = n$.

share|cite|improve this answer
I dont think you can say that $|f|^{2}$ is can say it may be small, but from the given condition you cant get that implication(atleast not directly) – Phani Raj Oct 4 '12 at 17:15

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.