Star-shaped approximations of a Hamiltonian, agreeing with it to first order along two periodic orbits, for which these orbits and all their iterates are non-degenerate

Let $S=H^{-1}(1)\subset\mathbb{R}^4$ be a strictly star-shaped level and let $P=(x_P,T_P)$, $Q=(x_Q,T_Q)$ be periodic orbits of $X_H$ on $S$. Then there is a sequence of Hamiltonians $H_k:\mathbb{R}^4\to\mathbb{R}$ such that

1. every level $H_k^{-1}(1)$ is strictly star-shaped;
2. every periodic orbit $R$ of $X_{H_k}$ on $H_k^{-1}(1)$ whose image lies in $x_P(\mathbb{R})\cup x_Q(\mathbb{R})$ (that is, $P$, $Q$ and all their iterates, with any time shift) is non-degenerate;
3. $H_k=H$ and $dH_k=dH$ at every point of $x_P(\mathbb{R})\cup x_Q(\mathbb{R})$;
4. $H_k\to H$ in $C^\infty_{\mathrm{loc}}$.

$H_k\to H$ in $C^\infty_{\mathrm{loc}}$ means: for every $m\in\mathbb{N}$ and every compact $K\subset\mathbb{R}^4$, $D^mH_k\to D^mH$ uniformly on $K$. A periodic orbit $R=(x,T)$ on $H^{-1}(1)$ (prime or not) is non-degenerate when, for its linearized flow $Y$, the linearized return map on $\xi=TS/\mathbb{R}X_{H}$, written in the global frame $(Z_1,Z_2)$ as the matrix $\varphi(1)$ of linearizedXiPath, does not have $1$ as an eigenvalue. A Hamiltonian is non-degenerate when all its periodic orbits on the level $1$ are.

By (3), $X_{H_k}=X_H$ along both orbits, so $P$ and $Q$ stay periodic orbits of every $X_{H_k}$. Nothing is asserted about the other periodic orbits of $X_{H_k}$. In contact language: $H_k^{-1}(1)$ carries $g_k\lambda$ with $g_k\to1$ in $C^\infty$, $g_k\equiv1$ and $dg_k\equiv0$ on the two orbits, and the two orbits are non-degenerate for $g_k\lambda$.