Fundamental proof of the baby case of Hofer's theorem about displacement energy In 1990, Hofer proved that the displacement energy of a standard ball in $C^{n}$ equals it's Gromov area.
Here is the baby case:  Consider a smooth bounded function $f:R^{2}\rightarrow R$. Consider the vector field $(-\frac{\partial f}{\partial y},\frac{\partial f}{\partial x})$, which is the rotation of the gradient field. Now consider the differmorphism $\Phi_{t}:R^{2}\rightarrow R^{2}$ generated by this vector field, with $\Phi_{0}=$identity. Suppose that $\Phi_{1}(D^{2})\cap D^{2}=\emptyset$, where $D^{2}$ is the unit ball in $R^{2}$. Then we have:
$\text{max}(f)-\text{min}(f)\geq \pi$.
My question is: since this problem seems very easy to understand, is there a fundamental proof for this? 
 A: (To match the standard definitions, you want $f$ or at least $df$ to have compact support. This doesn't really matter for this simple computation. I just need the Hamiltonian flow to be complete.)
Let $\gamma \colon [0,1] \to \mathbb{R}^2$ be a smooth path. Then, define a map 
$\bar \gamma \colon [0,1]\times[0,1] \to \mathbb{R}^2$ by $\bar\gamma(s,t) = \Phi_t( \gamma(s))$, where $\Phi_t$ is the time $t$ flow of the Hamiltonian vector field of $f$. The area swept out by $\bar \gamma$ is obtained by integrating 
\[
\int \bar\gamma^*\omega = \int \omega( T\Phi_t(\gamma(s))\cdot \dot \gamma(s),
X_f( \Phi_t (\gamma(s))) ds\wedge dt = \int_0^1 \int_0^1 df(\dot \gamma) ds dt = \int_0^1 df( \dot \gamma(s)) ds = f(\gamma(1)) - f(\gamma(0)).
\]
(I have taken the convention that $\omega(X_f, \cdot) = - df \cdot$. This matches your sign convention also. In simplifying this, I have used the time-independence of $f$.)
Take $p$ and $q$ to be maximum and minimum of the function $f$ and $\gamma$ a path connecting them that is to "one side" of the disk (but very close). If the time 1 map of $\Phi$ displaces the disk, it will move the path to the "other side" of the disk. The points $p$ and $q$ are kept fixed by the flow, since they are critical points, so the image of the rectangle is now a region that contains the disk. Its area is thus at least $\pi$, but the above computation shows this is the difference between max and min.
There are clearly some points in here that could use a few extra details to nail the ideas down completely -- in particular, this "one side" and the "other side" are compelling in pictures, but could use a little more care.
