 Solution

Let $\varphi \in {\mathrm{Ham}}_c({\mathbb{R}}^{2n})$ be nonidentity and choose $x$ with $\varphi (x)\ne x$. Choose disjoint open neighbourhoods $U,V$ of $x,\varphi (x)$, respectively. Continuity gives a small Euclidean ball $B=B(x,r_0)$ with $B\subset U$ and $\varphi (B)\subset V$. Therefore $\varphi (B)\cap B=\varnothing$.

Inside this ball construct an explicit closed rational Lagrangian submanifold. For $0<a<r_0/\sqrt{n}$, take the product Lagrangian torusIts distance from $x$ is $\sqrt{n}a<r_0$. Each coordinate circle is one-dimensional, so the pullback of $d{q}_j\wedge d{p}_j$ to it is zero; summing proves that this $n$-dimensional product is a Lagrangian submanifold. It is compact without boundary.

On the $j$th fundamental circle, $q_j=a\cos t$, $p_j=a\sin t$, and the primitive in part (a) restricts to $\frac{1}{2}{a}^{2}dt$. Its integral is $\pi a^2$. Translation of the torus adds only an exact constant-coefficient one-form to the primitive, so it does not change the periods. Since $H_1(T^n;\mathbb{Z})$ is generated by these circles,The given displacement-energy bound applies to this explicitly constructed torus. Because $L\subset B$, the map $\varphi$ displaces $L$, and henceThis is Hofer nondegeneracy from rational Lagrangian displacement. For two different maps $\varphi ,\psi$, apply it to ${\varphi }^{-1}\psi \ne \mathrm{id}$ and use the distance definition. Therefore ${\rho }_{\infty }(\varphi ,\psi )>0$. The equivalent oscillation Hofer metric is nondegenerate as well.

Only the stated energy inequality was used; its conclusion was not used to presume positive energy of an arbitrary open set. The rational torus and its positive period have been constructed inside the particular displaced ball.