= Solution
Assume $\omega\ne0$. With $h=0$, the <method of characteristics> gives $f(t,\Phi(t,0,z))=f_0(z)$. Hence
$$
\operatorname{supp}f_t=\Phi(t,0,\operatorname{supp}f_0).
$$
Let $K=\operatorname{supp}f_0$ and $E_*=\max_KH_\omega$; if $f_0=0$, the conclusion is immediate. By the <Hamiltonian energy balance>, every image point of $K$ remains in the <energy> sublevel
$$
\mathcal K=\{(x,v):|v|^2+\omega^2|x|^2\leq2E_*\}.
$$
This is a fixed compact <ellipsoid> in <phase space>. Therefore \b[the <support> bound is uniform in time]:
$$
\boxed{\operatorname{supp}f_t\subseteq\mathcal K,\qquad
|v|\leq\sqrt{2E_*},\quad |x|\leq\frac{\sqrt{2E_*}}{|\omega|}.}
$$
This <uniform support bound from a coercive conserved energy> requires no explicit solution of the curves.
The nonzero-frequency qualification is necessary. At $\omega=0$ the <energy> does not control $x$, and the equation is <free transport equation>. Choose a smooth <compactly supported> $f_0$ that is nonzero at $(x_0,v_0)$ with $v_0\ne0$. Its transported value at $(x_0+tv_0,v_0)$ stays nonzero, so the union of the supports is unbounded. Each individual <support> is compact, but there is no fixed <compact support> for all times.
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