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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?

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If you change the function $f$ far from the compact $\bigcup\limits_{t\in[0,1]}\Phi_t(\overline D^2)$ the conditions seem to remain valid. However, the conclusion $\max(f)-\min(f)\ge\pi$ can be easily violated. Probably, you forgot to write some details here. – Sasha Anan'in Jan 5 '14 at 21:50
Sorry. I don't see why max(f)-min(f) can be less than pi if you change the function outside the compact set....Oh, perhaps I should say that f is assumed to be bounded. – user44651 Jan 5 '14 at 22:46
It was my mistake as I have read it as $\le\pi$. I am sorry. – Sasha Anan'in Jan 5 '14 at 22:52
I don't understand what you are looking for. Do you want an elementary proof of this result in this special case, or are you looking to understand what happens when you take one of the proofs of an energy-capacity inequality and specialize it to $\mathbb{R}^2$? – Sam Lisi Jan 7 '14 at 12:39
I thinks I want an elementary of this in the special case of R^{2} – user44651 Jan 7 '14 at 17:37
up vote 2 down vote accepted

(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.

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Excellent! I got the idea. Thank you so much. – user44651 Jan 10 '14 at 22:14

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