# $L^2$-Oscillation

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 $\max\limits_{[0,1]}f-\min\limits_{[0,1]}f$?

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1. Edit your inequality inserting $dxdy$, please. 2. Inequality holds for every $\varepsilon$ or fixed? –  M. Strochyk Oct 4 '12 at 16:51

## 2 Answers

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.

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$\def\abs#1{\left|#1\right|}$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 $\max f_n - \min f_n = n$.

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I dont think you can say that $|f|^{2}$ is small.you 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