Negative partial-homogeneity levels trap all sufficiently close forcings

We first prove a scalar gluing fact, without inferring absolute continuity of a composition with a merely continuous function. Let $I=[u,v]$, let $C\subseteq I$ be closed, let $h:I\to \mathbb{R}$ be continuous, and let $Q:I\to \mathbb{R}$ be absolutely continuous with $h=Q$ on $C$. Suppose that $Q^{\mathrm{\prime }}\le -\kappa$ a.e. on $C$, for some $\kappa >0$, and that $h(t)+\kappa t$ is nonincreasing on every component interval of $I\setminus C$, including its endpoints by continuity. Put $U=h+\kappa t$ and $Z=Q+\kappa t$. If $x<y$ both belong to $C$, the complementary open intervals $({\alpha }_j,{\beta }_j)$ of $C$ inside $(x,y)$ have endpoints in $C$. Absolute continuity of $Z$ and countable additivity of its integrable derivative give $Z(y)-Z(x)={\int }_{C\cap [x,y]}{Z}^{\mathrm{\prime }}(t)dt+{\sum }_j{\int }_{{\alpha }_j}^{{\beta }_j}{Z}^{\mathrm{\prime }}(t)dt$. The series is absolutely convergent because the intervals are disjoint and $Z^{\mathrm{\prime }}\in L^1$. Each gap integral equals $Z({\beta }_j)-Z({\alpha }_j)=U({\beta }_j)-U({\alpha }_j)\le 0$, and the integral over $C$ is nonpositive. Thus $U(y)\le U(x)$. For arbitrary $x<y$, if $[x,y]$ misses $C$ they lie in one gap. Otherwise let $x_{\ast }=\min (C\cap [x,y])$ and $y_{\ast }=\max (C\cap [x,y])$; monotonicity on the two possible end gaps and the preceding inequality yield $U(y)\le U(y_{\ast })\le U(x_{\ast })\le U(x)$. Hence $h(y)-h(x)\le -\kappa (y-x)$ on all of $I$. This explicitly excludes positive singular variation on the closed contact set; no absolute continuity of $h$ has been assumed or concluded.

The bounded alternative admits a direct trap. Choose a regular value $a\in (a_0,0)$ of $H$, and choose $b\in (a,0)$. If this interval lies outside the range, an empty level set causes no difficulty: on each connected component of $\mathrm{\Theta }$, the path cannot cross a missing level by continuity. Otherwise compactness of $\mathrm{\Theta }$, the Neumann condition, and regularity give $\beta >0$ with $\mathrm{\mid }\mathrm{\nabla }H\mathrm{\mid }\ge \beta$ on a thin neighborhood of $\{H=a\}$. Since $\mathrm{\parallel }\mathrm{\nabla }{H}_f-\mathrm{\nabla }H{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }d\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f-f_0\mathrm{\parallel }$, choose $ϵ$ so that $⟨\mathrm{\nabla }H,\mathrm{\nabla }{H}_f⟩\ge {\beta }^2\mathrm{/}2$ on that neighborhood whenever $\mathrm{\parallel }f-f_0\mathrm{\parallel }\le ϵ$. Along a path in the Statement, $H(\theta {)}^{\mathrm{\prime }}=-\rho ⟨\mathrm{\nabla }H,\mathrm{\nabla }{H}_f⟩<0$ a.e. while its value is in the thin collar. An outward first crossing of $a$ is therefore impossible: any crossing would contain a subinterval with values in the collar increasing from a lower to a higher value, contrary to the integrated derivative inequality. A start with $H\le a$ consequently remains there. The argument also applies at the intersection with $\mathrm{\partial }\mathrm{\Theta }$, since the gradients are tangent there by Neumann and the path is constrained to $\mathrm{\Theta }$. This proves the first assertion in the bounded alternative.

For the rest of the proof take $\mathrm{\Theta }={\mathbb{R}}^q$. Write $r=\mathrm{\mid }\theta \mathrm{\mid }$ and $\omega =\theta \mathrm{/}r$ when $r>0$. Choose a smooth radial diffeomorphism $\mathrm{\Psi }$ from ${\mathbb{R}}^q$ to the open unit ball whose radial coordinate equals $u=1-1\mathrm{/}r$ for $r\ge 2$, patching it with a smooth strictly increasing radial coordinate near the origin. Its radial coordinate is chosen linear near zero so that $\mathrm{\Psi }$ and its inverse are smooth there. Let $M$ be the closed unit ball. Choose a positive smooth radial function $k(\theta )$ with $k=r^2$ for $r\ge 2$. The pushforward by $\mathrm{\Psi }$ of $-k\mathrm{\nabla }{H}_f$ has, in the outer coordinates $(u,\omega )$, the components $(-{\mathrm{\partial }}_r{H}_f(r\omega ),-{\mathrm{\nabla }}_{\mathbb{S}}{H}_f(r\omega ))$. The sphere derivative is zero when $q=1$, where the sphere consists of two points. For fixed $f$, its second component converges uniformly by assumption. Its first converges uniformly to zero: along each ray $r\mapsto H_f(r\omega )$ converges uniformly in $\omega$, and its derivative is uniformly Lipschitz in $r$. If its derivative had magnitude at least $\delta >0$ at arbitrarily large radii, its Lipschitz bound would keep the same sign with magnitude at least $\delta \mathrm{/}2$ on a radial interval of a fixed positive length, forcing a fixed nonzero change of $H_f$ there; this contradicts uniform convergence. When that Lipschitz bound is zero, convergence makes the derivative identically zero. Thus the pushed-forward field extends continuously to $M$ with boundary components $(0,-{\mathrm{\nabla }}_{\mathbb{S}}{H}_f^{\mathrm{\infty }})$. Denote this continuous Euclidean field on $M$ by $F_f$, and denote by $\widehat{H}$ the continuous extension of $H\circ {\mathrm{\Psi }}^{-1}$ with boundary value $H^{\mathrm{\infty }}$.

The fields $F_f$ satisfy $\mathrm{\parallel }{F}_f-F_{f_0}{\mathrm{\parallel }}_{C^0(M)}\le C\mathrm{\parallel }f-f_0\mathrm{\parallel }$ for a finite $C$. On bounded radial intervals this follows from bounded $d\phi$ and smoothness of the coordinate change. On the outer region the radial component difference is at most $\mathrm{\parallel }d\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f-f_0\mathrm{\parallel }$. For the sphere component, consider $T_r:\mathcal{H}\to C^1({\mathbb{S}}^{q-1})$, $T_{r}f(\omega )=⟨f,\phi (r\omega )⟩$, for $r\ge 2$. Each $T_r$ is a bounded linear operator. For every fixed $f$, its assumed convergence as $r\to \mathrm{\infty }$ makes its $C^1$ norms bounded at large radii; ordinary $C^1$ regularity on compact radial intervals supplies the remaining bound. Hence [F1] gives ${\sup }_{r\ge 2}\mathrm{\parallel }{T}_r\mathrm{\parallel }<\mathrm{\infty }$. This proves the required sphere bound uniformly in $r$ and, by passing to the pointwise limits, also on the boundary. Smooth changes between these outer coordinates and the Euclidean ball have bounded differentials on the compact outer collar, proving the claimed estimate on $M$. No convergence in operator norm is asserted.

Choose $a\in (a_0,0)$ regular both for $H$ and for $H^{\mathrm{\infty }}$. To justify the simultaneous choice, regular values of $H^{\mathrm{\infty }}$ form an open set, since its critical values form a compact set; density, with empty level sets counted as regular, supplies a nonempty open interval within $(a_0,0)$ consisting of its regular values. Density for $H$ supplies a regular value in that interval. The same reasoning applies if the interval is outside either range. If $\widehat{H}$ has no values in a neighborhood of $a$, its continuous paths cannot cross that neighborhood, and the desired trap follows immediately. Otherwise choose $d>0$ and $\kappa >0$, with $a_0<a-2d<a+2d<0$, so that the following estimates hold on the band $\mathcal{B}=\{a-2d\le \widehat{H}\le a+2d\}$. At interior points the directional derivative of $\widehat{H}$ along $F_0:=F_{f_0}$ is $-k\mathrm{\mid }\mathrm{\nabla }H{\mathrm{\mid }}^2\le -\kappa$; at boundary points the derivative of $H^{\mathrm{\infty }}$ along its boundary field is $-\mathrm{\mid }{\mathrm{\nabla }}_{\mathbb{S}}{H}^{\mathrm{\infty }}{\mathrm{\mid }}^2\le -\kappa$. Indeed, near the boundary, uniform $C^1$ sphere convergence and regularity of $a$ for $H^{\mathrm{\infty }}$ bound $\mathrm{\mid }{\mathrm{\nabla }}_{\mathbb{S}}H(r\cdot )\mathrm{\mid }$ below whenever $H(r\omega )$ lies in this band, and $k=r^2$ cancels the Euclidean angular factor $r^{-2}$. The remaining region is compact; regularity of $a$ for $H$ and a smaller band give a positive lower bound there. If the boundary portion of the band is empty, only the compact interior argument is needed.

Every absolutely continuous solution $\gamma :[u,v]\to M$ of ${\gamma }^{\mathrm{\prime }}=F_0(\gamma )$ whose image lies in $\mathcal{B}$ obeys $\widehat{H}(\gamma (v))-\widehat{H}(\gamma (u))\le -\kappa (v-u)$, even if it touches the compactification boundary on a complicated closed time set. Its speed is bounded by $\mathrm{\parallel }{F}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$, so it is Lipschitz. Put $C=\{t:\gamma (t)\in \mathrm{\partial }M\}$, which is closed. Choose a $C^1$ function $Q_0$ on $M$ equal to $H^{\mathrm{\infty }}$ on the boundary: on a collar use its radial extension multiplied by a smooth cutoff that vanishes before the collar's inner edge, and extend by zero. This is $C^1$ because $H^{\mathrm{\infty }}$ is $C^1$. The function $Q=Q_0\circ \gamma$ is absolutely continuous. At a.e. $t\in C$ its derivative equals the boundary directional derivative in step 5.1, since the radial component of $F_0$ is zero there; hence $Q^{\mathrm{\prime }}\le -\kappa$ a.e. on $C$. The continuous function $h=\widehat{H}\circ \gamma$ agrees with $Q$ on $C$. On any interior gap it is locally absolutely continuous by smooth coordinates and the chain rule, and $h^{\mathrm{\prime }}\le -\kappa$. Integrating on compact subintervals of a gap and taking endpoint limits proves that $h+\kappa t$ is nonincreasing on the whole gap. Apply the scalar gluing argument of step 1.1 to obtain the asserted estimate. In particular this argument does not infer absolute continuity from continuity of $\widehat{H}$, and does not presume uniqueness for the extended ODE at the boundary.

Fix ${\tau }_{\ast }>0$ so small that every path in $M$ with speed at most $B:=1+\mathrm{\parallel }{F}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ changes the value of $\widehat{H}$ by strictly less than $d$ on an interval of length at most ${\tau }_{\ast }$. Such a choice follows from uniform continuity of $\widehat{H}$ on compact $M$, because displacement is at most $B{\tau }_{\ast }$. There exists $e>0$, with $e\le 1$, such that every absolutely continuous curve $\gamma :[0,{\tau }_{\ast }]\to M$ staying in $\mathcal{B}$ and satisfying $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}-F_0(\gamma )\mathrm{\mid }\le e$ a.e. obeys $\widehat{H}(\gamma ({\tau }_{\ast }))-\widehat{H}(\gamma (0))\le -\kappa {\tau }_{\ast }\mathrm{/}2$. Otherwise choose curves violating this assertion with errors $e_n\downarrow 0$. Their speeds are bounded by $B$, and [F2] gives a uniformly convergent subsequence. Continuity of $F_0$ and the integral equations show that its limit solves ${\gamma }^{\mathrm{\prime }}=F_0(\gamma )$; the limit stays in the closed band. Step 6.1 gives an endpoint drop at least $\kappa {\tau }_{\ast }$, while uniform convergence and continuity of $\widehat{H}$ give a drop at most $\kappa {\tau }_{\ast }\mathrm{/}2$. This contradiction proves the claim.

Put $b=a+d$. Any absolutely continuous curve satisfying $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}-F_0(\gamma )\mathrm{\mid }\le e$ a.e. and starting with $\widehat{H}(\gamma (0))\le a$ cannot reach $\widehat{H}=b$. For otherwise consider its first time $v$ at level $b$ and its last preceding time $u$ at level $a$. Both exist by continuity, and its values on $[u,v]$ lie in $[a,b]\subset \mathcal{B}$. The speed bound and the choice of ${\tau }_{\ast }$ force $v-u>{\tau }_{\ast }$. Apply step 7.1 to the final interval $[v-{\tau }_{\ast },v]$. Its endpoint value must be strictly below its starting value, contradicting that $v$ is the first hit of the upper level $b$. Thus the curve remains strictly below $b$. Choose $ϵ$ small enough that step 4.1 gives $\mathrm{\parallel }{F}_f-F_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le e$ for $\mathrm{\parallel }f-f_0\mathrm{\parallel }\le ϵ$. The preceding conclusion then holds for every forcing path in that neighborhood, including Borel time-dependent forcing.

For a physical path in the Statement define $\tau (t)={\int }_{t_0}^t\rho (v)\mathrm{/}k(\theta (v))dv$. On each finite interval its derivative is positive and continuous, bounded away from zero, because $\rho$ is positive continuous and the physical path has compact image. Its inverse is consequently continuously differentiable. By change of variable, $\gamma (\tau )=\mathrm{\Psi }(\theta (t(\tau )))$ is absolutely continuous and solves ${\gamma }^{\mathrm{\prime }}=F_{f(t(\tau ))}(\gamma )$ a.e. Step 8.1 applies and proves $H(\theta (t))<b$ whenever its initial value is at most $a$. The same argument works on every finite physical interval, regardless of whether normalized time has a finite limit on an infinite horizon. This proves the first assertion in the unbounded alternative.

In either domain alternative, reverse a path satisfying the opposite-sign equation on $[t_0,t_1]$: the path $\tilde{\theta }(u)=\theta (t_0+t_1-u)$ satisfies the favorable equation with $\tilde{f}(u)=f(t_0+t_1-u)$ and $\tilde{\rho }(u)=\rho (t_0+t_1-u)$. These retain the same Borel, continuity, positivity and closeness conditions. The first assertion therefore proves the backward assertion with exactly the same constants. Finally $\mathrm{\mid }{H}_f-H\mathrm{\mid }\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f-f_0\mathrm{\parallel }$, so decrease $ϵ$ until this bound is at most $-b\mathrm{/}2$; then $H<b$ implies $H_f<b\mathrm{/}2<0$. The numbers $a,b,ϵ$ were chosen only from the fixed feature, forcing center and negative level, completing the proof.