Positive-measure families of invariant Lagrangian tori for nondegenerate real analytic Hamiltonians: persistence for $\epsilon \le c{\alpha }^2$

Put $\sigma :=\min ({\sigma }_0,s_0\mathrm{/}32)$, so $3{\sigma }_0\mathrm{/}4\le \sigma \le {\sigma }_0$, and put $E:=2\sqrt{\epsilon M_2}\mathrm{/}(\alpha {\sigma }^{\mu })$. Choose a compact complex neighborhood of $C^{\ast }$ inside ${\mathrm{\Omega }}^{\mathrm{\prime }}$ on which $b$ has bounded derivative and ${\mathrm{dist}}_{\mathrm{\infty }}(b(\lambda ),K)<r_0\mathrm{/}2$; compactness and $b(C^{\ast })\subset K$ permit this. Reduce $c$ so that $r\le r_0\mathrm{/}4$, $h_0+A{B}_0$ is smaller than the width of this neighborhood, and $E$ meets the initial smallness requirements of [F1]-[F3]. The parameter-sequence lemma allows $K_0=O(1+\mathrm{\mid }\log E\mathrm{\mid })$ and $h_0=\alpha \mathrm{/}(2K_0^{\tau +1})$. Since $E_0\le (4\mathrm{/}3{)}^{\mu }E$ and $E(1+\mathrm{\mid }\log E\mathrm{\mid }{)}^{\tau +1}\to 0$ as $E\downarrow 0$, $c$ can also be chosen so every small Lipschitz bound used below is below its fixed positive threshold. Finally reduce $c$ so $C_{n,\tau }(\mathrm{diam}{C}^{\ast }{)}^{n-1}\alpha <\mathrm{\mid }{C}^{\ast }\mathrm{\mid }\mathrm{/}2$ whenever $0<\alpha <c$.

Let $\mathrm{\Lambda }:={\mathrm{\Omega }}_{\alpha ,\tau }$. By [F6], $\mathrm{\Lambda }$ is a closed subset of the compact cube $C^{\ast }$, hence compact. For every $\delta >0$, the resonant part $C^{\ast }\setminus \mathrm{\Lambda }$ is contained in the resonant part of the translated open cube obtained by enlarging each side by $2\delta$; [F5] bounds its measure by the slab constant with half-width $R+\delta$. Letting $\delta \downarrow 0$ gives $\mathrm{\mid }{C}^{\ast }\setminus \mathrm{\Lambda }\mathrm{\mid }\le C_{n,\tau }(\mathrm{diam}{C}^{\ast }{)}^{n-1}\alpha \mathrm{.}$ Since $C^{\ast }$ and $\mathrm{\Lambda }$ are measurable, subtraction from $\mathrm{\mid }{C}^{\ast }\mathrm{\mid }$ gives the lower bound in clause 1; choose $c$ so this lower bound is positive. Every label is in the regular frequency patch; differentiating $Dh(b(\nu ))=\nu$ gives $D^{2}h(b(\nu ))Db(\nu )=I_n$, so $Db(\nu )$ is invertible. On a sufficiently small neighborhood of each frequency, $b$ is lower-Lipschitz by [F6]. The frequency cube is covered by finitely many smaller closed cubes $Q_1,\dots ,Q_N$ whose slightly larger convex neighborhoods have this property and lie in ${\mathrm{\Omega }}^{\mathrm{\prime }}$. At least one, say $Q$, has $\mathrm{\mid }\mathrm{\Lambda }\cap Q\mathrm{\mid }>0$, because the finite cover covers the positive-measure set $\mathrm{\Lambda }$. Choose the constants in step 1.1 so ${\chi }_{\mathrm{\infty }}(\mathrm{\Lambda }\cap Q)$ remains in that larger neighborhood.

Define the physical family map on $\mathrm{\Lambda }\times {\mathbb{T}}^n$ by $\mathcal{P}(\nu ,\vartheta ):=(b({\chi }_{\mathrm{\infty }}(\nu ))+a(\vartheta ,\nu ),\vartheta +w(\vartheta ,\nu ))\mathrm{.}$ For fixed $\nu$, the translation $I\mapsto b({\chi }_{\mathrm{\infty }}(\nu ))+I$ is canonical. Since $H_{\mathrm{p}\mathrm{a}\mathrm{r}}(I,\vartheta ,{\chi }_{\mathrm{\infty }}(\nu ))=H(b({\chi }_{\mathrm{\infty }}(\nu ))+I,\vartheta )$ by [F4], [F3] transports the invariance identity and vanishing Lagrangian pullback to $H$. The reality and embedding statements of [F3] make each descended ${\widehat{\mathrm{\Psi }}}_{\nu }^{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$ an embedded real torus. The geometric choice in step 1.1 keeps the image inside the actual domain of $H$. This proves clause 2.

The angle component of the parameter invariance equation is $Dh(b(\lambda )+a)+F_p(b(\lambda )+a,\vartheta +w)=\nu +D_{\vartheta }w\nu ,\lambda ={\chi }_{\mathrm{\infty }}(\nu )\mathrm{.}$ Average over $\vartheta \in {\mathbb{T}}^n$. Each component of the periodic $w$ has mean-zero derivatives, hence $\nu -\lambda =\mathrm{mean}(Dh(b(\lambda )+a)-Dh(b(\lambda )))+\mathrm{mean}{F}_p(b(\lambda )+a,\vartheta +w)\mathrm{.}$ Polarization of the Hessian quadratic-form bound gives $\mathrm{\parallel }{D}^{2}h{\mathrm{\parallel }}_{\mathrm{\infty }\to \mathrm{\infty }}\le n{M}_2$ on the real tube. The composition bounds [F2] and Cauchy's estimate on action discs of radius $r_0\mathrm{/}4$ give $\mathrm{\mid }a\mathrm{\mid }\le ArE$ and $\mathrm{\mid }{F}_p\mathrm{\mid }\le 4\epsilon \mathrm{/}{r}_0$. Therefore $\mathrm{\mid }\lambda -\nu \mathrm{\mid }\le n{M}_{2}ArE+4\epsilon \mathrm{/}{r}_0\mathrm{.}$ With $\mathrm{\mid }Db\mathrm{\mid }\le M_b$ on the chosen compact neighborhood, $\mathrm{\mid }b(\lambda )+a-b(\nu )\mathrm{\mid }\le ArE+M_b(n{M}_{2}ArE+4\epsilon \mathrm{/}{r}_0)\mathrm{.}$ Since $\epsilon \mathrm{/}(rE)=\alpha {\sigma }^{\mu }\mathrm{/}2\le 1\mathrm{/}2$, this is bounded by a fixed multiple of $rE$. Also $\mathrm{\mid }w\mathrm{\mid }\le A\sigma E$. The comparisons $E\le (4\mathrm{/}3{)}^{\mu }{E}_0$ and $\sigma \le {\sigma }_0$ give clause 3 after enlarging $C_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$.

Fix $\alpha >0$. On a lifted physical phase chart define the invertible linear output scaling $S_{\alpha }(p,\theta ):=(p\mathrm{/}\alpha ,\theta ),{\mathcal{P}}_{\alpha }:=S_{\alpha }\circ \mathcal{P}\mathrm{.}$ Use the output norm $\mathrm{\parallel }(u,v){\mathrm{\parallel }}_{\ast }:=\mathrm{\mid }u{\mathrm{\mid }}_{\mathrm{\infty }}+\mathrm{\mid }v{\mathrm{\mid }}_{\mathrm{\infty }}$ and the input distance $d_{\alpha }((\nu ,\vartheta ),({\nu }^{\mathrm{\prime }},{\vartheta }^{\mathrm{\prime }})):=\mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}\mathrm{/}\alpha +\mathrm{\mid }\vartheta -{\vartheta }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}\mathrm{.}$ We prove a lower-Lipschitz estimate for ${\mathcal{P}}_{\alpha }$ from this input distance to $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\ast }$ on $(\mathrm{\Lambda }\cap Q)\times A$, where $A$ is a fixed closed angular cube of positive volume lying strictly inside one angle chart. Shrink $c$ so the angular correction has supremum less than the distance from $A$ to the boundary of a larger angle chart. Then both angular outputs have unique lifts in that chart. The composition estimates [F2], with constants enlarged to absorb the finite dimensional norm conversions, give for $X:=\mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}\mathrm{/}\alpha$ and $Y:=\mathrm{\mid }\vartheta -{\vartheta }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}$, $\frac{\mathrm{\mid }a(\vartheta ,\nu )-a({\vartheta }^{\mathrm{\prime }},{\nu }^{\mathrm{\prime }}){\mathrm{\mid }}_{\mathrm{\infty }}}{\alpha }\le C(\frac{rE}{\alpha \sigma }Y+\frac{rE}{h_0}X),$ $\mathrm{\mid }w(\vartheta ,\nu )-w({\vartheta }^{\mathrm{\prime }},{\nu }^{\mathrm{\prime }}){\mathrm{\mid }}_{\mathrm{\infty }}\le C(EY+\frac{\alpha \sigma E}{h_0}X)\mathrm{.}$ The frequency map ${\chi }_{\mathrm{\infty }}$ has $\mathrm{Lip}({\chi }_{\mathrm{\infty }}-\mathrm{i}\mathrm{d})\le C_L{B}_0\mathrm{/}{h}_0$, with $B_0=C\alpha E{\sigma }^{\mu }$. Our choice of $c$ makes this less than $1\mathrm{/}2$; hence $\mathrm{\mid }{\chi }_{\mathrm{\infty }}(\nu )-{\chi }_{\mathrm{\infty }}({\nu }^{\mathrm{\prime }}){\mathrm{\mid }}_{\mathrm{\infty }}\ge \mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}\mathrm{/}2$. Take $m>0$ to be the minimum of the finitely many lower-Lipschitz constants for $b$ on the larger frequency cubes chosen in step 2.1; these cubes and constants depend only on the fixed analytic data. In particular it is a lower-Lipschitz constant on the selected cube. Thus the scaled base map ${\mathcal{B}}_{\alpha }(\nu ,\vartheta ):=(b({\chi }_{\mathrm{\infty }}(\nu ))\mathrm{/}\alpha ,\vartheta )$ satisfies $\mathrm{\parallel }{\mathcal{B}}_{\alpha }(\nu ,\vartheta )-{\mathcal{B}}_{\alpha }({\nu }^{\mathrm{\prime }},{\vartheta }^{\mathrm{\prime }}){\mathrm{\parallel }}_{\ast }\ge (m\mathrm{/}2)X+Y\ge m_{\ast }(X+Y),m_{\ast }:=\min (m\mathrm{/}2,1)\mathrm{.}$ The two error bounds above show that the scaled perturbation $(a\mathrm{/}\alpha ,w)$ has Lipschitz constant at most $C(E+\frac{rE}{\alpha \sigma }+\frac{rE}{h_0}+\frac{\alpha \sigma E}{h_0})\mathrm{.}$ Here $r\mathrm{/}\alpha \le \sqrt{c\mathrm{/}{M}_2}$, $h_0=\alpha \mathrm{/}(2K_0^{\tau +1})$, and $E{K}_0^{\tau +1}\to 0$ as $E\downarrow 0$ because $K_0=O(1+\mathrm{\mid }\log E\mathrm{\mid })$. Reducing $c$ makes this Lipschitz constant smaller than $m_{\ast }\mathrm{/}2$. Subtracting the error difference from the base difference proves the explicit scaled inequality $\mathrm{\parallel }{\mathcal{P}}_{\alpha }(\nu ,\vartheta )-{\mathcal{P}}_{\alpha }({\nu }^{\mathrm{\prime }},{\vartheta }^{\mathrm{\prime }}){\mathrm{\parallel }}_{\ast }\ge (m_{\ast }\mathrm{/}2)d_{\alpha }((\nu ,\vartheta ),({\nu }^{\mathrm{\prime }},{\vartheta }^{\mathrm{\prime }}))\mathrm{.}$ The unscaled physical output has an equivalent norm for each fixed $\alpha >0$, but no uniform unscaled lower-Lipschitz constant as $\alpha \downarrow 0$ is asserted.

The compact set $\mathrm{\Lambda }\cap Q$ has positive measure, and $A$ has positive $n$-dimensional measure, so $\mathcal{E}:=(\mathrm{\Lambda }\cap Q)\times A$ is Borel and has positive $2n$-dimensional measure by [F5]. To see that $\mathcal{K}$ is measurable, represent each torus angle by a point of $C_0:=[0,2\pi {]}^n$. The domain $\mathrm{\Lambda }\times C_0$ is a closed bounded subset of ${\mathbb{R}}^{2n}$, hence compact by [F6]. The map $(\nu ,\vartheta )\mapsto \mathcal{P}(\nu ,\pi (\vartheta ))$ is continuous by the parameter Lipschitz and angular holomorphy bounds in [F2], and has image exactly $\mathcal{K}$, because $\pi (C_0)={\mathbb{T}}^n$. Thus $\mathcal{K}$ is compact in the metric phase space and is Lebesgue-measurable in its local charts. If the lifted physical image $\mathcal{P}(\mathcal{E})$ had measure zero, then ${\mathcal{P}}_{\alpha }(\mathcal{E})=S_{\alpha }(\mathcal{P}(\mathcal{E}))$ would also have measure zero: for this fixed $\alpha >0$, $S_{\alpha }$ and its inverse are linear Lipschitz maps with finite constants. Step 3.1 makes ${\mathcal{P}}_{\alpha }$ injective, and its inverse on ${\mathcal{P}}_{\alpha }(\mathcal{E})$ is Lipschitz in ordinary Euclidean coordinates. Indeed $\mathrm{\mid }\nu -{\nu }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}+\mathrm{\mid }\vartheta -{\vartheta }^{\mathrm{\prime }}{\mathrm{\mid }}_{\mathrm{\infty }}\le \max (\alpha ,1)d_{\alpha }$; combining this with step 3.1 and the fixed finite-dimensional norm comparisons gives a finite inverse Lipschitz constant for each fixed $\alpha$. By the cube-cover property in [F6], that inverse would send the null set ${\mathcal{P}}_{\alpha }(\mathcal{E})$ to the positive-measure set $\mathcal{E}$, a contradiction. The passage back to physical coordinates uses only invertibility of the fixed scaling, not a bound uniform as $\alpha \downarrow 0$. Therefore this image, and hence $\mathcal{K}$, has positive phase-space measure. This is clause 4.