 Solution

LetFor any sufficiently small nonzero $\delta$, translationis a symplectic isotopy with ${\varphi }_1(L)\cap L=\varnothing$. Choosing $|\delta |$ arbitrarily small makes the isotopy arbitrarily small in every $C^k$ norm.

This isotopy has nonzero flux homomorphism, represented by $\delta dx$. In contrast, every Hamiltonian isotopy has zero flux. If a Hamiltonian image $L^{\prime }$ were disjoint from $L$, the two homologous essential circles would bound an annulus $A$, and evaluation of the flux on $[L]$ would equal the signed symplectic areawhich is nonzero. This contradicts vanishing Hamiltonian flux, so $L$ is not Hamiltonian displaceable.