$L^2$-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 $\max\limits_{[0,1]}f-\min\limits_{[0,1]}f$?

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.
