Proof. Let $e(t)={\rm e}^{\pi J_{2n}t}X$, where $X\in V^{n,k}_0$ and $|X|=1$. For any $x=y+\lambda e$, where $y\in E^{-}\oplus E^{0}$ and $\lambda >0$, it holds that

\[ \mathfrak{a}(x)\le\frac{1}{2}\|\lambda e\|^{2}_E= \frac{ \pi }{2}\lambda^{2} \]

and

\[ \int_0^{1}\langle x(t),{\rm e}^{\pi J_{2n}t}X\rangle \,{{\rm d}}t= \int_0^{1}\langle \lambda \,{\rm e}^{\pi J_{2n}t}X,{\rm e}^{\pi J_{2n}t}X\rangle \,{{\rm d}}t=\lambda. \]

It follows that

\[ \mathfrak{a}(x)\le\frac{\pi }{2}\left(\int_0^{1} \langle x(t),{\rm e}^{\pi J_{2n}t}X\rangle \,{{\rm d}}t\right)^{2}\le\frac{\pi}{2}\int_0^{1}|x(t)|^{2} \,{{\rm d}}t. \]

This and proposition 3.2 lead to

\[ \inf_{x\in\gamma(S^{+})}\Phi_H(x) \le\sup_{x\in E^{-}\oplus E^{0}\oplus\mathbb{R}_+e} \Phi_H(x)\le \sup_{z\in\mathbb{R}^{2n}}\left\{\frac{\pi }{2}|z|^{2}-H(z)\right\} \quad\forall\ \gamma\in\Gamma_{n,k}, \]

and hence (3.3) is proved.

Proof. We assume $k< n$. For a constant $\varepsilon >0$ define $\gamma _\varepsilon \in \Gamma _{n,k}$ by $\gamma _\varepsilon (x)=z_0+\varepsilon x\ \forall \ x\in E$. We claim that

(3.6)\begin{equation} \inf_{y\in\gamma_\varepsilon(S^{+})}\Phi_H(y)>0 \end{equation}

for sufficiently small $\varepsilon$. Since

(3.7)\begin{equation} \Phi_H(z_0+x)=\frac{1}{2}\|x\|^{2}_{E}-\int_0^{1}H(z_0+x)\,{{\rm d}}t\quad\forall\ x\in E^{+}, \end{equation}

it suffices to prove that

(3.8)\begin{equation} \lim_{\|x\|_E\rightarrow 0}\frac{\int_0^{1} H(z_0+x)\,{{\rm d}}t}{\|x\|^{2}_E}=0. \end{equation}

Otherwise, suppose there exists a sequence $(x_j)\subset E$ and $d>0$ satisfying

(3.9)\begin{equation} \|x_j\|_E\rightarrow 0 \quad \hbox{and} \quad \frac{\int_0^{1} H(z_0+x_j)\,{{\rm d}}t}{\|x_j\|^{2}_E} \geq d>0\quad\forall\ j. \end{equation}

Let $y_j={x_j}/{\|x_j\|_E}$ and hence $\|y_j\|_E=1$. Then lemma 2.1 implies that $(y_j)$ has a convergent subsequence in $L^{2}$. By a standard result in $L^{p}$ theory, we have $w\in L^{2}$ and a subsequence of $(y_j)$, still denoted by $(y_j)$, such that $y_j(t)\rightarrow y(t)$ a.e. on $(0,1)$ and that $|y_j(t)|\leq w(t)$ a.e. on $(0,1)$ for each $j$. It follows from (3.5) that

\begin{align*} & \frac{H(z_0+x_j(t))}{\|x_j\|_E^{2}}\leq C_1 \frac{|x_j(t)|^{2}}{\|x_j\|_E^{2}} =C_1|y_j(t)|^{2}\le C_1w(t)^{2},\quad \hbox{a.e. on}\ (0,1),\ \forall\ j,\\ & \frac{H(z_0+x_j(t))}{\|x_j\|_E^{2}}\leq C_2\frac{|x_j(t)|^{3}}{\|x_j\|_E^{2}} =C_2|x_j(t)|\cdot |y_j(t)|^{2} \le C_2|x_j(t)|w(t)^{2},\quad \hbox{a.e. on}\ (0,1),\enspace \forall j. \end{align*}

The first claim in (3.9) implies that $(x_j)$ has a subsequence such that $x_{j_l}(t)\rightarrow 0$, a.e. in $(0,1)$. Hence the Lebesgue dominated convergence theorem leads to

\[ \int_0^{1} \frac{H(z_0+x_{j_l}(t))}{\|x_{j_l}\|_E^{2}} {{\rm d}}t\rightarrow 0. \]

This contradicts the second claim in (3.9).

For any fixed $\mathbb {R}^{n,k}$-admissible $H\in C^{2}(\mathbb {R}^{2n},\mathbb {R}_{\ge 0})$, pick some $z_0\in \mathbb {R}^{n,k}\cap {\rm Int}(H^{-1}(0))$. Since (H1) implies that $H$ vanishes near $z_0$, by (H2) and the Taylor expansion of $H$ at $z_0\in \mathbb {R}^{2n}$, we have constants $C_1>0$ and $C_2>0$ such that $H$ satisfies (3.5).

Proof. Since $c^{n,k}(H)$ is a finite positive number by proposition 3.6, the first claim follows from the arguments above proposition 3.5.

When $H$ has bounded derivatives of second order, (2.4) is satisfied naturally. Then $\nabla \Phi _H$ satisfies the global Lipschitz condition, and thus has a unique global flow $\mathbb {R}\times E\to E: (u,x)\mapsto \varphi _u(x)$ satisfying lemma 2.6, that is, $\varphi _u(x)={\rm e}^{-u}x^{-}+x^{0}+{\rm e}^{u}x^{+}+\widetilde {K}(u,x)$, where $\widetilde {K}:\mathbb {R}\times E\rightarrow E$ is continuous and maps bounded sets into precompact sets. For a set $F=\gamma (S^{+})\in \mathcal {F}_{n,k}$ with $\gamma \in \Gamma _{n,k}$, we have $\alpha :=\inf (\Phi _H|_{\gamma (S^{+})})>0$ by the definition of $\mathcal {F}_{n,k}$. Let $\rho :\mathbb {R}\rightarrow [0,1]$ be a smooth function such that $\rho (s)=0$ for $s\le 0$ and $\rho (s)=1$ for $s\ge \alpha$. Define a vector field $V$ on $E$ by

\[ V(x)=x^{+}-x^{-}-\rho(\Phi_H(x))\nabla \mathfrak{b}(x). \]

Clearly $V$ is locally Lipschitz and has linear growth. These imply that $V$ has a unique global flow, denoted by $\Upsilon _u$. Moreover, it is obvious that $\Upsilon _u$ has the same property as $\varphi _u$ described in lemma 2.6. For $x\in E^{-}\oplus E^{0}$, we have $\Phi _H(x)\le 0$ and hence $V(x)=-x^{-}$, which implies that $\Upsilon _u(E^{-}\oplus E^{0})=E^{-}\oplus E^{0}$ and $\Upsilon _u(E\setminus E^{-}\oplus E^{0})=E\setminus E^{-}\oplus E^{0}$ since $\Upsilon _u$ is a homeomorphism for each $u\in \mathbb {R}$. Therefore, $\Upsilon _u\in \Gamma _{n,k}$ for all $u\in \mathbb {R}$.

Note that $V|_{\Phi _H^{-1}([\alpha,\infty ])}=\nabla \Phi _H(x)$. For each $u\ge 0$ we have $\Upsilon _u(x)=\varphi _u(x)$ for any $x\in \Phi _H^{-1}([\alpha,\infty ])$, and especially $\Upsilon _u(F)=\varphi _u(F)$, that is, $(\Upsilon _u\circ \gamma )(S^{+})=\varphi _u(F)$. Since $\Gamma _{n,k}$ is closed for the composition operation and

\[ \inf(\Phi_H|_{(\Upsilon_u\circ\gamma)(S^{+})})=\inf(\Phi_H|_{\varphi_u(F)})\ge\inf(\Phi_H|_F)>0, \]

we obtain $\varphi _u(F)\in \mathcal {F}_{n,k}$, that is, $\mathcal {F}_{n,k}$ is positively invariant under the flow $\varphi _u$ of $\nabla \Phi _H$.

Proof. The functional $\mathfrak {b}$ is differentiable. Its gradient $\nabla \mathfrak {b}$ is compact and satisfies a global Lipschitz condition on $E$. Since $\nabla \Phi _{H}(x)=x^{+}-x^{-}-\nabla \mathfrak {b}(x)$ for any $x\in E$, we have

(3.11)\begin{equation} x_j^{+}-x_j^{-}-\nabla \mathfrak{b}(x_j)\rightarrow 0. \end{equation}

Case 1. $(x_j)$ is bounded in $E$. Then $(x^{0}_j)$ is a bounded sequence in the space $\mathbb {R}^{n,k}$ of finite dimension. Hence $(x^{0}_j)$ has a convergent subsequence. Moreover, since $\nabla \mathfrak {b}$ is compact, $(\nabla \mathfrak {b}(x_j))$ has a convergent subsequence, and so both $(x_j^{+})$ and $(x_j^{-})$ have convergent subsequences in $E$. It follows that $(x_j)$ has a convergent subsequence.Case 2. $(x_j)$ is unbounded in $E$. Without loss of generality, we may assume $\lim _{j\rightarrow +\infty }\|x_j\|_E=+\infty$. For $z_0\in \mathbb {R}^{n,k}$ defined as in (H2), let

\[ y_j=\frac{x_j}{\|x_j\|_E}-\frac{1}{2a}z_0. \]

Then $|y_j^{0}|\le \|y_j\|_E\le 1+|{z_0}/{2a}|$, and (3.11) implies

(3.12)\begin{equation} y_j^{+}-y_j^{-}-\jmath^{{\ast}}\left(\frac{ \nabla H(x_j)}{\|x_j\|_E}\right)\rightarrow 0. \end{equation}

Also by (H2) there exist constants $C_1$ and $C_2$ such that

\[ \left\|\frac{\nabla H(x_j)}{\|x_j\|_E}\right\|_{L^{2}}^{2}\leq \frac{8a^{2} \|x_j\|_{L^{2}}^{2}+C_1 }{\|x_k\|_E^{2}}\leq C_2 \]

that is, $(\nabla H(x_j)/\|x_j\|_E)$ is bounded in $L^{2}$. Hence the sequence $\jmath ^{\ast }( \nabla H(x_j)/\|x_j\|_E)$ is compact. (3.12) implies that $(y_j)$ has a convergent subsequence in $E$. Without loss of generality, we may assume that $y_{j}\rightarrow y$ in $E$. Since (H2) implies that $H(z)=Q(z):=a|z|^{2}+ \langle z, z_0\rangle + b$ for $|z|$ sufficiently large, there exists a constant $C>0$ such that $|\nabla H(z)-\nabla Q(z)|\leq C$ for all $z\in \mathbb {R}^{2n}$. It follows that as $j\to \infty$,

\begin{align*} \left\|\frac{\nabla H(x_j)}{\|x_j\|_{E}}-\nabla Q(y)\right\|_{L^{2}} & \leq \left\|\frac{\nabla H(x_j)}{\|x_j\|_{E}}-\nabla Q(y_j)\right\|_{L^{2}}+ \left\|\nabla Q(y_j)-\nabla Q(y)\right\|_{L^{2}}\\ & \leq \left\|\frac{\nabla H(x_j)-\nabla Q(x_j)}{\|x_j\|_E}\right\|_{L^{2}}+ \frac{|z_0|}{\|x_j\|_E}+2a\|y_j-y\|_{L^{2}}\\ & \leq \frac{C}{\|x_j\|_E}+\frac{|z_0|}{\|x_j\|_E}+2a\|y_j-y\|_{L^{2}}\rightarrow 0. \end{align*}

This implies that $\jmath ^{\ast }(\nabla H(x_k)/\|x_k\|_{E})$ tends to $\jmath ^{\ast }(\nabla Q(y))$ in $E$, and thus we arrive at

\[ y^{+}-y^{-}-\jmath^{{\ast}}(\nabla Q (y))=0\quad\hbox{and}\quad \left\|y+\frac{z_0}{2a} \right\|_E=1. \]

Then $y$ is smooth and satisfies

\[ \dot{y}=J_{2n}\nabla Q(y)\quad\hbox{and}\quad y(1)\sim y(0),\enspace y(0),\ y(1)\in\mathbb{R}^{n,k}. \]

Clearly $y(t)$ is given by

\[ y(t)+\frac{1}{2a}z_0={\rm e}^{2aJ_{2n}t}\left(y(0)+\frac{1}{2a}z_0\right). \]

Since $\|y+({1}/{2a})z_0\|_{E}=1$ implies that $y+({1}/{2a})z_0$ is nonconstant, using the boundary condition satisfied by $y$ and the assumption that $z_0\in \mathbb {R}^{n,k}$, we deduce that $2a\in m\mathbb {N}\pi$. This gives rise to a contradiction because $H$ is strong non-resonant.

Proof. The method is similar to that of [Reference Jin and Lu26, lemma 3.5]. For clearness we give it in details. By lemma 2.4, $\Phi _H$ is $C^{1,1}$. Lemma 2.5 implies that all critical points of $\Phi _{H}$ sit in $C^{2n+2}_{n,k}([0,1],\mathbb {R}^{2n})$. Thus the restriction of $\Phi _{H}$ to $C^{1}_{n,k}([0,1],\mathbb {R}^{2n})$, denoted by $\hat \Phi _{H}$, and $\Phi _{H}$ have the same critical value sets. As in the proof of [Reference Jin and Lu24, claim 4.4] we can deduce that $\hat \Phi _{H}$ is of class $C^{2n+1}$.

Let $P_0$ and $P_1$ be the orthogonal projections of $\mathbb {R}^{2n}$ to the spaces $V_0^{n,k}$ and $V_1^{2k}$ in (1.1) and (1.2), respectively. Take a smooth $g:[0,1]\rightarrow [0,1]$ such that $g$ equals $1$ (resp. $0$) near $0$ (resp. $1$). Denote by $\phi ^{t}$ the flow of $X_{H}$. Since $X_{H}$ is $C^{2n+1}$, we have a $C^{2n+1}$ map

\[ \psi:[0,1]\times\mathbb{R}^{n,k}\rightarrow \mathbb{R}^{2n},\quad (t,z)\mapsto g(t)\phi^{t}(z)+(1-g(t))\phi^{t-1}(P_0\phi^{1}(z)+P_1z). \]

For any $z\in \mathbb {R}^{n,k}$, since $\psi (0,z)=\phi ^{0}(z)=z$ and $\psi (1,z)=P_0\phi ^{1}(z)+P_1z$, we have

\[ \psi(1,z),\quad \psi(0,z)\in\mathbb{R}^{n,k}\quad\hbox{and}\quad \psi(1,z)\sim\psi(0,z). \]

These and [Reference Jin and Lu24, corollary B.2] show that $\psi$ gives rise to a $C^{2n}$ map

\[ \Omega:\mathbb{R}^{n,k}\to C^{1}_{n,k}([0,1],\mathbb{R}^{2n}),\quad z\mapsto \psi({\cdot},z). \]

Hence $\Phi _{H}\circ \Omega : \mathbb {R}^{n,k}\to \mathbb {R}$ is of class $C^{2n}$. By Sard's theorem we deduce that the critical value sets of $\Phi _{H}\circ \Omega$ is nowhere dense (since $\dim \mathbb {R}^{n,k}<2n$).

Let $z\in \mathbb {R}^{n,k}$ be such that $\phi ^{1}(z)\in \mathbb {R}^{n,k}$ and $\phi ^{1}(z)\sim z$. Then $P_0\phi ^{1}(z)-P_0z=\phi ^{1}(z)-z$ and therefore $P_0\phi ^{1}(z)+P_1z=\phi ^{1}(z)$, which implies $\psi (t,z)=\phi ^{t}(z) \forall \ t\in [0,1]$.

For a critical point $y$ of $\Phi _H$, that is, $y\in C^{2n+2}_{n,k}([0,1],\mathbb {R}^{2n})$ and solves $\dot {y}=J_{2n}\nabla H(y)=X_H(y)$, with $z_y:=y(0)\in \mathbb {R}^{n,k}$ we have $y(t)=\phi ^{t}(z_y)\ \forall \ t\in [0,1]$, which implies that $\phi ^{1}(z_y)\in \mathbb {R}^{n,k}$, $\phi ^{1}(z_y)\sim z_y$ and therefore $y=\psi (\cdot,z_y)=\Omega (z_y)$. Hence $z_y$ is a critical point of $\Phi _{H}\circ \Omega$ and $\Phi _{H}\circ \Omega (z_y)=\Phi _{H}(y)$. Thus the critical value set of $\Phi _{H}$ is contained in that of $\Phi _{H}\circ \Omega$. The desired claim is obtained.

Proof. By the assumptions each $H\circ \psi _s$ is also $\mathbb {R}^{n,k}$-admissible and strong nonresonant. Hence $c(H\circ \psi _s)$ is a positive critical value for each $s$. Let $x\in E$ be a critical point of $\Phi _{H\circ \psi _s}$ with critical value $c(H\circ \psi _s)$. Then $x\in C^{2n+2}_{n,k}([0,1],\mathbb {R}^{2n})$ and solves $\dot {x}=J_{2n}\nabla (H\circ \psi _s)(x)=X_{H\circ \psi _s}(x)$. Let $y_s=\psi _s\circ x$. Then $y_s\in C^{2n+2}_{n,k}([0,1],\mathbb {R}^{2n})$ and satisfies

\[ \dot{y}_s(t)=({\rm d}\psi_s(x(t))\dot{x}(t)= ({\rm d}\psi_s(x(t))X_{H\circ\psi_s}(x(t))= X_{H}(\psi_s(x(t))=J_{2n}\nabla H(y_s(t)) \]

since ${\rm d}\psi _s(z)X_H(z)=X_{H}(\psi _s(z))$ for any $z\in \mathbb {R}^{2n}$ by [Reference Hofer and Zehnder20, p. 9]. Therefore $y_s$ is a critical point of $\Phi _H$ on $E$. We claim that

(3.14)\begin{equation} \Phi_H(y_s)=\Phi_{H\circ\psi_s}(x). \end{equation}

Clearly, it suffices to prove the following equality:

(3.15)\begin{equation} A(y_s)=\frac{1}{2}\int^{1}_0 \langle -J_{2n}\dot{y}_s, y_s\rangle \,{{\rm d}}t= \frac{1}{2}\int_0^{1}\langle -J_{2n}\dot{x},x\rangle \,{{\rm d}}t=A(x). \end{equation}

Extend $x$ into a piecewise $C^{2n+2}$-smooth loop $x^{\ast }:[0, 2]\to \mathbb {R}^{2n}$ by setting $x^{\ast }(t)=(2-t)x(1)+(t-1)x(0)$ for any $1\le t\le 2$. We get a piecewise $C^{2n+2}$-smooth loop extending of $y_s$, $y_s^{\ast }=\psi _s(x^{\ast })$. Clearly, we can extend $x^{\ast }$ into a piecewise $C^{2n+2}$-smooth $u:D^{2}\to \mathbb {R}^{2n}$, where $D^{2}$ is a closed disc bounded by $\partial D^{2}\equiv [0,2]/\{0,2\}$. Then $\psi _s\circ u:D^{2}\to \mathbb {R}^{2n}$ is piecewise $C^{2n+2}$-smooth and $\psi _s\circ u|_{\partial D^{2}}=y_s^{\ast }$. Stokes theorem yields

\begin{align*} & \frac{1}{2}\int_0^{2}\langle -J_{2n}\dot{x}^{{\ast}},x^{{\ast}}\rangle \,{{\rm d}}t=\int_{D^{2}}u^{{\ast}}\omega_0,\\ & \frac{1}{2}\int^{2}_0 \langle -J_{2n}\dot{y}^{{\ast}}_s, y^{{\ast}}_s\rangle \,{{\rm d}}t= \int_{D^{2}}(\psi_s\circ u)^{{\ast}}\omega_0=\int_{D^{2}}u^{{\ast}}\omega_0. \end{align*}

Moreover, for any $t\in [1,2]$ we have $\dot {x}^{\ast }(t)=x(0)-x(1)\in V_0^{n,k}$ and $x^{\ast }(t)\in \mathbb {R}^{n,k}$, and therefore $\langle -J_{2n}\dot {x}^{\ast }(t), x^{\ast }(t)\rangle =0$ because $\mathbb {R}^{2n}$ has the orthogonal decomposition $\mathbb {R}^{2n}=J_{2n}V^{n,k}_0\oplus \mathbb {R}^{n,k}$. Then (3.15) follows from these.

Since $s\mapsto c^{n,k}(H\circ \psi _s)$ is continuous by proposition 3.5, and a critical point $x$ of $\Phi _{H\circ \psi _s}$ with critical value $c(H\circ \psi _s)$ yields a critical point $y_s$ of $\Phi _H$ on $E$ satisfying (3.14), we deduce that each $c(H\circ \psi _s)$ is also a critical value of $\Phi _H$. Lemma 3.10 shows that $s\mapsto c^{n,k}(H\circ \psi _s)$ must be constant.

Finally, let $\psi _s(z)=z+ w_s$. It is clear that $H\circ \psi _s$ is $\mathbb {R}^{n,k}$-admissible and strong nonresonant. Thus $c(H\circ \psi _s)$ is a positive critical value. If $x\in E$ is a critical point of $\Phi _{H\circ \psi _s}$ with critical value $c(H\circ \psi _s)$, then $y_s:=\psi _s\circ x$ is a critical point of $\Phi _H$ on $E$ and (3.14) holds. Hence $s\mapsto c^{n,k}(H(\cdot + w_s))$ is constant.

Proof. We only prove (iii). By (ii) it suffices to prove that for each given $H\in \mathscr {E}_{n,k}(\mathbb {R}^{2n},B)$ there exists $G\in \mathscr {E}^{l}_{n,k}(\mathbb {R}^{2n},B)$ such that $G\ge H$. We may assume that $H(z)=a|z|^{2}+ \langle z, z_0\rangle + b$ outside a larger closed ball $\overline {B^{2n}(R_1)}$, where $a>\pi$ and $a\notin \pi \mathbb {N}$ for $k=n$, and $a>\pi /2$ and $a\notin \pi \mathbb {N}/2$ for $0\le k< n$. Let $U_\epsilon (B)$ be the $\epsilon$-neighbourhood of $B$. We can also assume that $H$ vanishes in $U_{2\epsilon }(B)$. Since $\overline {B^{2n}(R_1)}$ is compact, we may find numbers $a'>a$, $b'$ such that $a'\notin \pi \mathbb {N}$ for $k=n$, $a'\notin \pi \mathbb {N}/2$ for $0\le k< n$, and $a'|z|^{2}+ \langle z, z_0\rangle + b'\ge H(z)$ for all $z\in \mathbb {R}^{2n}$. Take a smooth function $f:\mathbb {R}^{2n}\to \mathbb {R}_{\ge 0}$ such that it equals to zero in $U_{\epsilon }(B)$ and $1$ outside $U_{2\epsilon }(B)$. Define $G(z):=f(z)(a'|z|^{2}+ \langle z, z_0\rangle + b')$ for $z\in \mathbb {R}^{2n}$. Then $G\ge H$ and $G\in \mathscr {E}^{\infty }_{n,k}(\mathbb {R}^{2n},B)$.

Proof of proposition 1.1. Proposition 3.5(i)–(iii) lead to the first three claims. Let us prove (iv). We may assume that $B$ is bounded. By (3.17) we have a sequence $(H_j)\subset \mathscr {F}_{n,k}(\mathbb {R}^{2n},B)$ such that $c^{n,k}(H_j)\to c^{n,k}(B)$. Note that $H_j(\cdot -w)\in \mathscr {F}_{n,k}(\mathbb {R}^{2n},B+w)$ for each $j$. Hence

\[ c^{n,k}(B+w)\le\inf_j c^{n,k}(H_j({\cdot}{-}w))= \inf_j c^{n,k}(H_j)=c^{n,k}(B) \]

by the final claim in proposition 3.11. The same reasoning leads to $c^{n,k}(B)=c^{n,k}(B+w+(-w))\le c^{n,k}(B+w)$ and so $c^{n,k}(B+w)=c^{n,k}(B)$.

Proof. Note that

\[ \mathscr{E}_{n,k}(\mathbb{R}^{2n}, A)\cap C^{\infty}(\mathbb{R}^{2n},\mathbb{R}_{\ge 0})\to \mathscr{E}_{n,k}(\mathbb{R}^{2n}, \psi_s(A))\!\cap\! C^{\infty}(\mathbb{R}^{2n},\mathbb{R}_{\ge 0}),\quad H\!\mapsto\! H\circ\psi_s^{{-}1} \]

is a one-to-one correspondence. Then

\begin{align*} c^{n,k}(\psi_s(A))& = \inf\{c^{n,k}(G)\,|\, G\in \mathscr{E}_{n,k} (\mathbb{R}^{2n}, \psi_s(A))\cap C^{\infty}(\mathbb{R}^{2n},\mathbb{R}_{\ge 0})\}\\ & = \inf\{c^{n,k}(H\circ\psi_s^{{-}1})\,|\, H\in \mathscr{E}_{n,k} (\mathbb{R}^{2n}, A)\cap C^{\infty}(\mathbb{R}^{2n},\mathbb{R}_{\ge 0})\}\\ & = \inf\{c^{n,k}(H)\,|\, H\in \mathscr{E}_{n,k}(\mathbb{R}^{2n}, A) \cap C^{\infty}(\mathbb{R}^{2n},\mathbb{R}_{\ge 0})\}=c^{n,k}(A). \end{align*}

Here the third equality comes from proposition 3.11.

Proof of theorem 1.2. We may assume that $B$ is bounded, and complete the proof in two steps.

Step 1. Prove $c^{n,k}(\Phi (B))=c^{n,k}(B)$ for every $\Phi \in {\rm Sp}(2n,k)$. Take a smooth path $[0, 1]\ni t\mapsto \Phi _t\in {\rm Sp}(2n,k)$ such that $\Phi _0=I_{2n}$ and $\Phi _1=\Phi$. We have a smooth function $[0,1]\times \mathbb {R}^{2n}\ni (t,z)\mapsto G_t(z)\in \mathbb {R}^{2n}$ such that the path $\Phi _t$ is generated by $X_{G_t}$ and that $G_t(z)=0\ \forall \ z\in \mathbb {R}^{n,k}$ (see step 2 below). Since $\cup _{t\in [0, 1]}\Phi _t(\overline {B})$ is compact, it can be contained in a ball $B^{2n}(0, R)$ for some $R>0$. Take a smooth cut function $\rho :\mathbb {R}^{2n}\to [0, 1]$ such that $\rho =1$ on $B^{2n}(0, 2R)$ and $\rho =0$ outside $B^{2n}(0, 3R)$. Define a smooth function $\tilde {G}:[0,1]\times \mathbb {R}^{2n}\to \mathbb {R}$ by $\tilde {G}(t,z)=\rho (z)G_t(z)$ for $(t,z)\in [0,1]\times \mathbb {R}^{2n}$. Denote by $\psi _t$ the Hamiltonian path generated by $\tilde {G}$ in ${\rm Ham}^{c}(\mathbb {R}^{2n},\omega _0)$. Then $\psi _t(z)=\Phi _t(z)$ for all $(t,z)\in [0,1]\times B^{2n}(0, R)$. Moreover each $\psi _t$ restricts to the identity on $\mathbb {R}^{n,k}$ because $\tilde {G}(t,z)=\rho (z)G_t(z)=0$ for all $(t,z)\in [0,1]\times \mathbb {R}^{n,k}$. Hence we obtain $c^{n,k}(\Phi (B))=c^{n,k}(\Phi _1(B))=c^{n,k}(B)$ by proposition 3.16.

Step 2. Prove $c^{n,k}(\phi (B))=c^{n,k}(B)$ in case $w_0=0$. Let $\Phi =({\rm d}\phi (0))^{-1}$. Since $c^{n,k}(\Phi \circ \phi (B))=c^{n,k}(\phi (B))$ by step 1, and $\Phi \circ \phi (w)=w\ \forall \ w\in \mathbb {R}^{n,k}$, replacing $\Phi \circ \phi$ by $\phi$, we may assume ${\rm d}\phi (0)={\rm id}_{\mathbb {R}^{2n}}$. Define a continuous path in ${\rm Symp}(\mathbb {R}^{2n},\omega _0)$,

(3.19)\begin{equation} \varphi_t(z)=\left\{\begin{array}{@{}ll} z & \hbox{if}\ t\le 0,\\ \dfrac{1}{t}\phi(tz) & \hbox{if}\ t> 0, \end{array}\right. \end{equation}

which is smooth except possibly at $t=0$. As in [Reference Schlenk36, proposition A.1] we can smoothen it with a smooth function $\eta :\mathbb {R}\to \mathbb {R}$ defined by

(3.20)\begin{equation} \eta(t)=\left\{\begin{array}{@{}ll} 0 & \hbox{if}\ t\le 0,\\ { e}^{2}\,{ e}^{{-}2/t} & \hbox{if}\ t> 0, \end{array}\right. \end{equation}

where $e$ is the Euler number. Namely, defining $\phi _t(z):=\varphi _{\eta (t)}(z)$ for $z\in \mathbb {R}^{2n}$ and $t\in \mathbb {R}$, we get a smooth path $\mathbb {R}\ni t\mapsto \phi _t\in {\rm Symp}(\mathbb {R}^{2n},\omega _0)$ such that

(3.21)\begin{equation} \phi_0={\rm id}_{\mathbb{R}^{2n}},\quad \phi_1=\phi,\quad \phi_t(z)=z,\quad \forall\ z\in\mathbb{R}^{n,k},\enspace \forall\ t\in\mathbb{R}. \end{equation}

Define $X_t(z)=(({{{\rm d}}}/{{{\rm d}}t})\phi _t)(\phi _t^{-1}(z))$ and

(3.22)\begin{equation} H_t(z)=\int^{z}_0 i_{X_t}\omega_0, \end{equation}

where the integral is along any piecewise smooth curve from $0$ to $z$ in $\mathbb {R}^{2n}$. Then $\mathbb {R}\times \mathbb {R}^{2n}\ni (t,z)\mapsto H_t(z)\in \mathbb {R}$ is smooth and $X_t=X_{H_t}$. By the final condition in (3.21), for each $(t,z)\in \mathbb {R}\times \mathbb {R}^{n,k}$ we have $X_t(z)=0$ and therefore $H_t(z)=0$. As in step 1, we can assume that $\cup _{t\in [0, 1]}\phi _t(\overline {B})$ is contained a ball $B^{2n}(0, R)$. Take a smooth cut function $\rho :\mathbb {R}^{2n}\to [0, 1]$ as above, and define a smooth function $\tilde {H}:[0,1]\times \mathbb {R}^{2n}\to \mathbb {R}$ by $\tilde {H}(t,z)=\rho (z)H_t(z)$ for $(t,z)\in [0,1]\times \mathbb {R}^{2n}$. Then the Hamiltonian path $\psi _t$ generated by $\tilde {H}$ in ${\rm Ham}^{c}(\mathbb {R}^{2n},\omega _0)$ satisfies

\[ \psi_t(z)=\phi_t(z),\quad \forall\ (t,z)\in [0,1]\times B^{2n}(0, R)\quad\hbox{and}\quad \psi_t(z)=z,\quad \forall\ (t,z)\in [0,1]\times \mathbb{R}^{n,k}. \]

It follows from proposition 3.16 that $c^{n,k}(\phi (B))=c^{n,k}(\psi _1(B))=c^{n,k}(B)$ as above.

Step 3. Prove $c^{n,k}(\phi (B))=c^{n,k}(B)$ in case $w_0\ne 0$. Define $\varphi (w)=\phi (w+w_0)$ for $w\in \mathbb {R}^{2n}$. Then ${\rm d}\varphi (0)={\rm d}\phi (w_0)\in {\rm Sp}(2n,k)$ and $\varphi (w)=\phi (w+w_0)=w \forall \ w\in \mathbb {R}^{n,k}$. By step 2 we arrive at $c^{n,k}(\varphi (B-w_0))=c^{n,k}(B-w_0)$. The desired equality follows because $\phi (B)=\varphi (B-w_0)$ and $c^{n,k}(B-w_0)= c^{n,k}(B)$ by proposition 1.1.

Proof of corollary 1.3. As discussed above the proof is reduced to the case $w_0=0$. Moreover we can assume that both sets $A$ and $U$ are bounded and that $U$ is also star-shaped with respect to the origin $0\in \mathbb {R}^{2n}$.

Next the proof can be completed following [Reference Schlenk36, proposition A.1]. Now $[0,1]\ni t\mapsto \phi _t(\cdot ):=\varphi _{\eta (t)}(\cdot )$ given by (3.19) and (3.20) is a smooth path of symplectic embeddings from $U$ to $\mathbb {R}^{2n}$ with properties

(3.23)\begin{equation} \phi_0={\rm id}_U,\quad \phi_1=\varphi,\quad \phi_t(z)=z,\quad \forall\ z\in\mathbb{R}^{n,k}\cap U,\enspace \forall\ t\in\mathbb{R}. \end{equation}

Thus $X_t(z):=(({{{\rm d}}}/{{{\rm d}}t})\phi _t)(\phi _t^{-1}(z))$ is a symplectic vector field defined on $\phi _t(U)$, and (3.22) (where the integral is along any piecewise smooth curve from $0$ to $z$ in $\phi _t(U)$) defines a smooth function $H_t$ on $\phi _t(U)$ in the present case. Observe that $H:\cup _{t\in [0,1]}(\{t\}\times \phi _t(U))\to \mathbb {R}$ defined by $H(t,z)=H_t(z)$ is smooth and generates the path $\phi _t$. Since $K=\cup _{t\in [0,1]}\{t\}\times \phi _t(\overline {A})$ is a compact subset in $[0,1]\times \mathbb {R}^{2n}$ we can choose a bounded and relative open neighbourhood $W$ of $K$ in $[0,1]\times \mathbb {R}^{2n}$ such that $W\subset \cup _{t\in [0,1]}(\{t\}\times \phi _t(U))$. Take a smooth cut function $\chi :[0,1]\times \mathbb {R}^{2n}\to \mathbb {R}$ such that $\chi |_K=1$ and $\chi$ vanishes outside $W$. Define $\hat {H}:[0,1]\times \mathbb {R}^{2n}\to \mathbb {R}$ by $\hat {H}(t,z)=\chi (t,z)H(t,z)$. It generates a smooth homotopy $\psi _t$ ($t\in [0,1]$) of the identity in ${\rm Ham}^{c}(\mathbb {R}^{2n},\omega _0)$ such that $\psi _t(z)=\phi _t(z)$ for all $(t,z)\in [0,1]\times A$. Moreover, the final condition in (3.21) implies that $\mathbb {R}^{n,k}\cap U\subset \phi _t(U)$ and $X_t(z)=0$ for any $t\in [0,1]$ and $z\in \mathbb {R}^{n,k}\cap U$. Hence for any $(t,z)\in [0,1]\times \mathbb {R}^{n,k}$ we have $\hat {H}(t,z)=\chi (t,z)H(t,z)=0$ and so $\psi _t(z)=z$. Then proposition 3.16 leads to $c^{n,k}(A)= c^{n,k}(\psi _1(A))=c^{n,k}(\phi _1(A))=c^{n,k}(\varphi (A))$.

Proof. For a critical point $x$ of $\Phi _{\overline {H}}$ with positive critical values there holds

\[{-}J_{2n}\dot{x}(t)=\nabla \overline{H}(x(t))=f'(H(x(t)))\nabla H(x(t)),\quad x(1)\sim x(0),\quad x(1), x(0)\in\mathbb{R}^{n,k} \]

and $H(x(t))\equiv c_0$ (a positive constant). Since

\begin{align*} 0<\Phi_{\overline{H}}(x) & = \frac{1}{2}\int^{1}_0\langle J_{2n}x(t),\dot{x}(t)\rangle\,{{\rm d}}t- \int^{1}_0\overline{H}(x(t))\,{{\rm d}}t\\ & = \frac{1}{2}\int^{1}_0\langle x(t),f'(c_0)\nabla H(x(t))\rangle \,{{\rm d}}t-\int^{1}_0f(s)\,{{\rm d}}t\\ & = f'(c_0)c_0-f(c_0), \end{align*}

we deduce $\beta :=f'(c_0)>0$, and so $c_0>1$. Define $y(t)=({1}/{\sqrt {c_0}})x(t/\beta )$ for $0\le t\le \beta$. Then

\[ H(y(t))=1,\quad -J_{2n}\dot{y}=\nabla H(y(t)),\quad y(\beta)\sim y(0),\quad y(\beta), y(0)\in\mathbb{R}^{n,k} \]

and therefore $f'(c_0)=\beta =A(y)\in \Sigma ^{n,k}_{\partial D}$. By the definition of $f$ this implies $f(c_0)<\epsilon$ and so

\[ \Phi_{\overline{H}}(x)=f'(c_0)c_0-f(c_0)> f'(c_0)-\epsilon\ge \min\Sigma^{n,k}_{\partial D}-\epsilon. \]

Proof. The case $k=n$ was proved in lemma 2 of [Reference Sikorav37, § 6.6]. We assume $k< n$ below. Let $S_{n,k}^{+}=\partial B_{n,k}^{+}$ and $\mathscr {E}^{n,k}_{\epsilon /2}(\mathbb {R}^{2n},\partial D)$ be as in (4.3). Replacing $H$ by a greater function we may assume $H\in \mathscr {E}^{n,k}_{\epsilon /2}(\mathbb {R}^{2n},\partial D)$. Since $H=0$ near $\partial {D}$, by the arguments at the end of proof of proposition 3.6, the condition (3.5) may be satisfied with any $z_0\in \mathbb {R}^{n,k}\cap {\rm Int}({H}^{-1}(0))$. Fix such a $z_0\in \mathbb {R}^{n,k}\cap {\rm Int}({H}^{-1}(0))$. It follows that there exists $\alpha >0$ such that

(5.2)\begin{equation} \inf \Phi_{H}|_{(z_0+\alpha S^{+}_{n,k})}>0\quad\hbox{and}\quad \Phi_{H}|_{(z_0+\alpha B^{+}_{n,k})}\ge 0, \end{equation}

(see (3.6)–(3.8) in the proof of proposition 3.6). Define $\gamma _\varepsilon :E_{n,k}\to E_{n,k}$ by $\gamma _\varepsilon (z)=z_0+ \alpha z$. It is easily seen that $\gamma _\varepsilon \in \Gamma _{n,k}$. The first inequality in (5.2) shows that $\gamma _\varepsilon (S^{+}_{n,k})$ belongs to the set $\mathcal {F}_{n,k}=\{\gamma (S^{+}_{n,k})\,|\,\gamma \in \Gamma _{n,k} \text{ and}\ \inf (\Phi _H|_{\gamma (S^{+}_{n,k})})>0\}$ in (3.10). Lemma 3.7 shows that

\[ c^{n,k}(H)=\sup_{F\in\mathcal{F}_{n,k}}\inf_{x\in F}\Phi_H(x), \]

and $\mathcal {F}_{n,k}$ is positively invariant under the flow $\varphi _u$ of $\nabla \Phi _H$. Define $S_u=\varphi _u(z_0+\alpha S^{+}_{n,k})$ and ${\rm d}(H)=\sup _{u\ge 0}\inf (\Phi _{H}|_{S_u})$. It follows from these and (5.2) that

\[ 0<\inf \Phi_{H}|_{S_0}\le {\rm d}(H)\le \sup_{F\in\mathcal{F}_{n,k}}\inf_{x\in F}\Phi_H(x)=c^{n,k}(H)<\infty. \]

Since $\Phi _{H}$ satisfies the (PS) condition by lemma 4.1, ${\rm d}(H)$ is a positive critical value of $\Phi _{H}$, and ${\rm d}(H)\ge c^{n,k}(D)-\epsilon /2$ by lemma 4.2. Moreover, by the definition of $d({H)}$ there exists $r>0$ such that $\Phi _{H}|_{S_r}\ge {\rm d}(H)-\epsilon /2$ and thus

(5.3)\begin{equation} \Phi_{H}|_{S_r}\ge c^{n,k}(D)-\epsilon. \end{equation}

Because $\Phi _{H}$ is nondecreasing along the flow $\varphi _u$, we arrive at

(5.4)\begin{equation} \Phi_{H}|_{S_u}\ge \Phi_{H}|_{S_0}\ge \inf (\Phi_{H}|_{S_0})>0,\quad\forall\ u\ge 0. \end{equation}

Define $\gamma :E_{n,k}\to E_{n,k}$ by $\gamma (x^{+}+x^{0}+x^{-})=\widetilde {\gamma }(x^{+})+x^{0}+x^{-}$, where

\begin{align*} & \widetilde{\gamma}(x)=z_0+ 2(\alpha/\epsilon)x \quad \hbox{if}\ x\in E^{+}_{n,k}\ \hbox{and} \|x\|_{E_{n,k}}\le\frac{1}{2}\epsilon,\\ & \widetilde{\gamma}(x)=\varphi_{r(2\|x\|_{E_{n,k}}-\epsilon)/\epsilon} (z_0+\alpha x/\|x\|_{E_{n,k}})\quad\hbox{if}\ x\in E^{+}_{n,k}\ \hbox{and}\ \frac{1}{2}\epsilon<\|x\|_{E_{n,k}}\le \epsilon,\\ & \widetilde{\gamma}(x)=\varphi_{r}(z_0+\alpha\ x/\|x\|_{E_{n,k}})\quad\hbox{if}\ x\in E^{+}_{n,k}\ \hbox{and}\ \|x\|_{E_{n,k}}>\epsilon. \end{align*}

The first and second lines imply $\gamma (({\epsilon }/{2})B^{+}_{n,k})=(z_0+\alpha B^{+}_{n,k})$ and $\gamma (B^{+}_{n,k}\setminus ({\epsilon }/{2})B^{+}_{n,k})=\bigcup _{0\le u\le r}S_u$, respectively, and so

\[ \gamma(B^{+}_{n,k})=(z_0+\alpha B^{+}_{n,k})\bigcup_{0\le u\le r}S_u; \]

the third line implies $\gamma (B^{+}_{n,k}\setminus \epsilon B^{+}_{n,k})=S_r$. It follows from these, (5.2) and (5.3)–(5.4) that $\gamma$ satisfies (5.1).

Finally, we can also know that $\gamma \in \Gamma _{n,k}$ by considering the homotopy

\[ \gamma_0(x)=2(\alpha/\epsilon)x^{+}+x^{0}+x^{-},\quad \gamma_u(x)=\frac{1}{u}{(\gamma(ux)-z_0)}+ z_0,\quad 0< u\le 1. \]

Proof. Let $H(z)=(j_D(z))^{2}$ for $z\in \mathbb {R}^{n_1}$ and define

\[ E_R=\{(z,z')\in\mathbb{R}^{2n_1}\times\mathbb{R}^{2n_2}\,|\,H(z)+ (|z'|/R)^{2}<1\}. \]

By the definition and the monotonicity of $c^{n,k}$ we have

\[ c^{n,k}(D\times\mathbb{R}^{2n_2})=\sup_R c^{n,k}(E_R). \]

Since the function $\mathbb {R}^{2n_1}\times \mathbb {R}^{2n_2}\ni (z,z')\mapsto G(z,z'):=H(z)+ (|z'|/R)^{2}\in \mathbb {R}$ is convex and of class $C^{1,1}$, $E_R$ is convex and $\mathcal {S}_R=\partial E_R$ is of class $C^{1,1}$. By theorem 1.4 we arrive at

\[ c^{n,k}(E_R)=\min\Sigma_{\mathcal{S}_R}^{n,k}. \]

Let $\lambda$ be a positive number and $u=(x,x'):[0,\lambda ]\rightarrow \mathcal {S}_R$ satisfy

(5.6)\begin{equation} \dot{u}=X_G(u)\quad\hbox{and}\quad u(\lambda),u(0)\in\mathbb{R}^{n,k},\quad u(\lambda)\sim u(0). \end{equation}

Namely, $u$ is a leafwise chord on $\mathcal {S}_R$ for $\mathbb {R}^{n,k}$ with action $\lambda$. Let $k_1=\min \{n_1,k\}$ and $k_2=\max \{k-n_1,0\}$. Clearly, $k_1+k_2=k$, and (5.6) is equivalent to the following