Morse stability with explicit parameter dependence

We first derive auxiliary metric estimates using only the product constant $\kappa$. For a specified segment $[u,v]$ and a point $z$, choose its point $m$ at distance $(z\mathrm{\mid }v{)}_u$ from $u$. This parameter lies between zero and $d(u,v)$. Put $P=(u\mathrm{\mid }v{)}_z$. Expanding products shows that $d(u,z)+d(v,m)=d(v,z)+d(u,m)=d(u,v)+P$. The four-point inequality therefore gives $d(z,m)\le P+2\kappa$. Every point $w$ on $[u,v]$ satisfies $P\le d(z,w)$, by adding the two triangle inequalities through $w$. Also, for any $w\in [u,v]$, $d(z,w)\le \max \{d(z,u),d(z,v)\}+\kappa$: if $w$ is within $\kappa$ of an endpoint this follows by the triangle inequality; otherwise the four-point inequality gives $d(z,w)\le \max \{d(z,u)-d(w,u),d(z,v)-d(w,v)\}+2\kappa$, which gives the claim since both sublengths exceed $\kappa$.

Let $Y$ be nonempty and $Q$-quasiconvex as in F4, and suppose $p,r\in Y$ are closest points of $x,y$ respectively. For any $z\in Y$, choose its quasiconvexity segment $[p,z]$ and its point $u$ at parameter $t=(x\mathrm{\mid }z{)}_p$. Write $A=d(x,p)$ and $P=d(p,z)$; from the definition of $t$ we have $A+P=d(x,z)+2t$. Applying the product inequality at basepoint $x$ with bridge $u$ gives $(p\mathrm{\mid }z{)}_x\ge \min \{(p\mathrm{\mid }u{)}_x,(u\mathrm{\mid }z{)}_x\}-\kappa$, and the products expand to $(p\mathrm{\mid }z{)}_x=A-t$ and $(p\mathrm{\mid }u{)}_x=(u\mathrm{\mid }z{)}_x=(A+d(x,u)-t)\mathrm{/}2$, since $d(p,u)=t$. Hence $d(x,u)\le d(x,p)-t+2\kappa$. Points of $Y$ occur arbitrarily close to distance $Q$ from $u$, hence closest-point minimality implies $t\le Q+2\kappa$. Thus $d(x,p)+d(p,z)\le d(x,z)+2Q+4\kappa$. With $A=d(x,p)$, $B=d(y,r)$ and $s=d(p,r)$, adding the estimates for $z=r$ and $z=p$ gives $d(x,r)+d(y,p)\ge A+B+2s-4Q-8\kappa$. The four-point inequality bounds this by $\max \{A+B,d(x,y)+s\}+2\kappa$. If $s>2Q+5\kappa$, the first maximum entry cannot suffice; therefore $s\le \max \{2Q+5\kappa ,d(x,y)-A-B+4Q+10\kappa \}\mathrm{.}$ The first radial estimate with $z=r$ also gives $s\le d(x,r)-A+2Q+4\kappa$, since $B\le d(y,p)\le d(x,y)+A$. If $B\le A$ then $d(x,r)\le d(x,y)+B\le d(x,y)+A$; if $B>A$ then the same radial estimate with the roles of $x,p$ and $y,r$ interchanged gives $s\le d(y,p)-B+2Q+4\kappa \le d(x,y)+A-B+2Q+4\kappa$. Hence in every case the prime projection bound $s\le d(x,y)+2Q+4\kappa$ holds.

For a specified segment $H$, put $V_R=\{x:d(x,H)\le R\}$, $R\ge 0$. Closest points on $H$ exist by F4. If $p$ projects $x$ onto $H$ and $d(x,H)\ge R$, the radius-$R$ point $q$ on $[p,x]$ projects $x$ onto $V_R$: the triangle inequality gives $d(x,y)\ge d(x,H)-R$ for $y\in V_R$, and $q$ attains equality. The same argument and F4 show that $q$ projects every later point of $[p,x]$ onto $V_R$. Moreover $V_R$ is $8\kappa$-quasiconvex. Join two points of $V_R$ to closest points on $H$ by radial segments of length at most $R$. The radial segments and intervening part of $H$ lie in $V_R$. Splitting the resulting quadrilateral by a diagonal and applying F2 twice puts every point of any fourth side within $8\kappa$ of these three segments, with arbitrarily small witness errors. Hence that side lies in the closed $8\kappa$-neighbourhood of $V_R$. This includes $R=0$, where the sharper quasiconvexity constant zero is available.

The following elementary real facts justify the extrema and numerical steps. A Lipschitz function on a compact interval is bounded by its value at one endpoint plus or minus its Lipschitz constant times the interval length, and attains its infimum: bisect, keep the left half when it has the same infimum and otherwise the right half; nested left endpoints have a supremum, and the Lipschitz bound times the interval length tends to zero, forcing attainment there. Infima exist by the real completeness convention in F1. A continuous real function with opposite weak signs at the ends has a zero by the same interval bisection retaining opposite endpoint signs; continuity at the limiting point proves its value is zero. Finally F6's positive series terms give $\exp (t)\ge 1+t$ for $t\ge 0$. Put $q_0=693\mathrm{/}1000$. Exact rational arithmetic gives ${\sum }_{j=0}^6\frac{q_0^j}{j!}+\frac{q_0^7}{7!}\frac{1}{1-q_0\mathrm{/}8}<2,{\sum }_{j=0}^{16}\frac{((459\mathrm{/}50)q_0{)}^j}{j!}>579.$ In the first inequality the omitted exponential tail has successive ratios at most $q_0\mathrm{/}8$ and is bounded by the displayed geometric tail. Thus $\exp (q_0)<2$, so $\log 2>q_0$. Monotonicity and the second inequality yield $(\log 2)\exp ((459\mathrm{/}50)\log 2)>q_0\cdot 579=401247\mathrm{/}1000>400$, and consequently $\frac{3200\exp (-(459\mathrm{/}50)\log 2)}{\log 2}<8.$ All these are rational finite-sum comparisons and geometric tail bounds, not rounded numerical estimates.

Fix $\rho >\kappa$. Here is the required small-gap assertion. Suppose $f:[u,v]\to X$ is continuous, $p(t)$ is any supplied closest-point selection in a $Q$-quasiconvex set, and $4\rho +2Q\le h\le d(p(u),p(v))$. There is $w\in [u,v]$ with $h-4\rho -2Q\le d(p(u),p(w))\le h$ and with $d(p(u),p(t))\le h$ for every $t\in [u,w]$. Write $J(t)=d(p(u),p(t))$. If $J\le h$ everywhere, use $w=v$. Otherwise let $t_0=\inf \{t:J(t)>h\}$. The prime projection bound of step 1.2 and continuity of $f$ at $u$ make $J(t)<h$ on some initial interval, because $4\kappa +2Q<h$; hence $t_0>u$. All $t<t_0$ have $J(t)\le h$. By continuity at $t_0$, choose $w<t_0$ and $t^{\mathrm{\prime }}\ge t_0$ with $J(t^{\mathrm{\prime }})>h$, both close enough to $t_0$ that $2d(f(w),f(t^{\mathrm{\prime }}))<4(\rho -\kappa )$. Such bad $t^{\mathrm{\prime }}$ occur arbitrarily close to the infimum. The prime projection bound of step 1.2 gives $d(p(w),p(t^{\mathrm{\prime }}))\le d(f(w),f(t^{\mathrm{\prime }}))+2Q+4\kappa <2(\rho -\kappa )+2Q+4\kappa =2\rho +2Q+2\kappa \le 4\rho +2Q$, so $J(w)>h-4\rho -2Q$. This proves the assertion without continuity of $p$. Reversing the real parameter gives the corresponding assertion from the right endpoint.

We now prove the continuous estimate, assuming more precisely that $f:[a,b]\to X$ is Lipschitz and is a $(\lambda ,C)$-quasi-geodesic, $C\ge 0$. Fix $z\in [a,b]$ and $\rho >\kappa$. Set $L=18\rho$, $D=55\rho$, $\alpha =3\mathrm{/}25$, and $\beta =\frac{\alpha \log 2}{5(4+(L+2\rho )\mathrm{/}D)\rho \lambda },A_{\ast }=4\sqrt{2}\lambda \exp (-\frac{(1-\alpha )D\log 2}{5\rho }),B_{\ast }=\frac{L+4\rho }{L-13\rho }\frac{A_{\ast }}{\beta }\mathrm{.}$ All denominators and these constants are positive. Put $T_{\ast }={\lambda }^2(D+3L\mathrm{/}2+\rho +11C\mathrm{/}2)-2\rho$ and $h_0=\rho \mathrm{/}(4\lambda )$. We shall prove, by induction on a nonnegative integer $n$, for every $a\le u\le z\le v\le b$ with $v-u\le n{h}_0$, that $(f(u)\mathrm{\mid }f(v){)}_{f(z)}\le T_{\ast }+B_{\ast }(1-\exp (-\beta (v-u)))\mathrm{.}$ For $n=0$, $u=v=z$ and the left side is zero, whereas $T_{\ast }\ge 81\rho \ge 0$. The constants are independent of $u,v,n$.

Assume the induction bound for $n$, and let $v-u\le (n+1)h_0$. Write $Z=f(z)$, $U=f(u)$, $V=f(v)$, $P=(U\mathrm{\mid }V{)}_Z$. If $P\le L$, the bound holds since $T_{\ast }\ge 81\rho >L$ and the exponential contribution is nonnegative. Suppose $P>L$. Choose any segment $[U,V]$ and its point $m$ at parameter $(Z\mathrm{\mid }V{)}_U$, as in step 1.1. Then $P\le d(Z,m)\le P+2\kappa <P+2\rho$. On a specified segment $H=[Z,m]$, let $\pi$ have distance $P$ from $Z$; thus $d(\pi ,m)\le 2\rho$. Using F4 and AC, select closest points $p(t)\in H$ for all $t\in [a,b]$. In particular $p(z)=Z$. Closest-point minimality gives $d(U,Z)\le d(U,p(u))+d(p(u),Z)\le d(U,m)+d(p(u),Z)$; since $d(U,Z)-d(U,m)=P$, we have $d(p(u),Z)\ge P$. Thus $p(u)$ lies between $\pi$ and $m$. The same computation holds for $p(v)$.

Apply step 2.1 on $[u,z]$ with $Q=0$ and threshold $L+d(\pi ,p(u))$. It is at least $4\rho$ and is less than $d(p(u),Z)$ since $L<P$. Obtain $y_{-}\in [u,z]$ with $L+d(\pi ,p(u))-4\rho \le d(p(u),p(y_{-}))\le L+d(\pi ,p(u)),$ and this upper bound holds on all of $[u,y_{-}]$. From the right endpoint similarly obtain $y_{+}\in [z,v]$. Measure coordinates on $H$ from $m$. The initial projections have coordinates in $[0,d(m,\pi )]$, and the upper bounds imply that all projections in these two outer intervals have coordinates in $[0,L+d(m,\pi )]$. Hence their pairwise distance is at most $L+2\rho$, and their individual distances from $\pi$ are at most $L$ (since $d(m,\pi )\le 2\rho <L$). The function $t\mapsto d(f(t),H)$ is Lipschitz, by the triangle inequality and Lipschitzness of $f$. Step 1.4 gives minima $d_{-},d_{+}$ attained at $c_{-}\in [u,y_{-}]$, $c_{+}\in [y_{+},v]$.

For $r\in [u,y_{-}]$ and $s\in [y_{+},v]$, the lower quasi-geodesic inequality and the route through their projections give $s-r\le \lambda (d(f(r),H)+d(f(s),H)+L+2\rho +C)\mathrm{.}$ We also have the product recurrence $P\le (f(r)\mathrm{\mid }f(s){)}_Z+L+4\rho \mathrm{.}$ Indeed F4 gives $(f(r)\mathrm{\mid }p(r){)}_Z\ge d(Z,p(r))-2\kappa \ge P-L-2\kappa$, and the same holds on the right. The product of $p(r),p(s)$ at $Z$ is the smaller of their radial distances, at least $P-L$. Applying the product inequality twice through these two projections gives $(f(r)\mathrm{\mid }f(s){)}_Z\ge P-L-4\kappa \ge P-L-4\rho$.

If both $d_{-},d_{+}\le D+4C$, put $s_0=d(f(c_{-}),f(c_{+}))$. The sum of the two upper quasi-geodesic bounds via $z$ gives $2(f(c_{-})\mathrm{\mid }f(c_{+}){)}_Z\le \lambda (c_{+}-c_{-})+2C-s_0\mathrm{.}$ If $s_0\ge 12\rho$, step 5.1 gives $c_{+}-c_{-}\le \lambda (2D+L+2\rho +9C)$, hence $(f(c_{-})\mathrm{\mid }f(c_{+}){)}_Z\le {\lambda }^2(D+L\mathrm{/}2+\rho +11C\mathrm{/}2)-6\rho$. If $s_0\le 12\rho$, the direct lower quasi-geodesic bound gives $c_{+}-c_{-}\le \lambda (12\rho +C)$, so the same product is at most ${\lambda }^2(6\rho +3C\mathrm{/}2)$, which is bounded by that preceding expression because $D=55\rho$, $L=18\rho$ and $\lambda \ge 1$. Adding the recurrence loss $L+4\rho \le {\lambda }^{2}L+4\rho$ in step 5.1 proves $P\le T_{\ast }$ and thus the required induction bound.

In the remaining case one of $d_{-},d_{+}$ is at least $D+4C$ and is the larger. First suppose $d=d_{-}\ge D+4C$ and $d_{+}\le d$. By AC choose, for each parameter $t$, a radial segment from $p(t)$ to $f(t)$. For integers $k\ge 0$ define $R_k=(2^k-1)d$, $V_k=V_{R_k}$ from step 1.3, and let $q_k(t)$ be the point at radius $\min \{R_k,d(f(t),H)\}$ on that radial segment. Set $Q_0=0$ and $Q_k=8\rho$ for $k\ge 1$. Step 1.3 makes $V_k$ $Q_k$-quasiconvex; when $d(f(t),H)\ge R_k$, $q_k(t)$ is a closest point to $V_k$. We use the following finite stopping construction. At stage $k$ there is $x\in [u,y_{-}]$ such that $d(f(t),H)\ge (2^{k+1}-1)d(u\le t\le x),d(q_k(u),q_k(x))\ge L-4\rho +7Q_k\mathrm{.}$ The initial stage $k=0$ uses $x=y_{-}$ by its projection lower bound in step 4.1 and the definition of the minimum $d$.

Given a stage, apply step 2.1 to $V_k$ on $[u,x]$ with threshold $9\rho +4Q_k$. This lies between $4\rho +2Q_k$ and $L-4\rho +7Q_k$. Obtain $w\in [u,x]$ with $5\rho +2Q_k\le d(q_k(u),q_k(w))\le 9\rho +4Q_k,$ and the upper bound holds for every parameter in $[u,w]$. If some $t\in [u,w]$ satisfies $d(f(t),H)\le (2^{k+2}-1)d$, choose such a $t$ and call it $r$. This is the stopping case, handled next. Otherwise every parameter in $[u,w]$ has distance strictly greater than that threshold; the next-stage construction below applies with $x=w$.

In the stopping case $u\le r\le w\le x\le y_{-}$. For $t\in [r,x]$, step 7.1 and radial projection give $d(f(t),V_k)\ge 2^{k}d$. Step 5.1 and $d(f(r),H)\le (4\cdot 2^k-1)d$ give $c_{+}-r\le \lambda 2^k(4+(L+2\rho )\mathrm{/}D)d\mathrm{.}$ Here $L+C+2\rho \le ((L+2\rho )\mathrm{/}D)d$, since $d\ge D+4C$ and $4(L+2\rho )\ge D$, and $d_{+}\le d$. Also $2^{k}d-C\mathrm{/}2-Q_k\ge (1-\alpha )D+\alpha 2^{k}d\mathrm{.}$ For $k=0$ this is $(1-\alpha )(d-D)\ge C\mathrm{/}2$, which follows from $d-D\ge 4C$. For $k\ge 1$, $(1-\alpha )(2^{k}d-D)\ge (1-\alpha )(D+8C)\ge 8\rho +C\mathrm{/}2$, proving the other case.

The stage lower bound and the projection cap up to $w$ imply $d(q_k(r),q_k(x))\ge L-13\rho +3Q_k$. Apply F4's exponential contraction to $f{\mathrm{\mid }}_{[r,x]}$ and $V_k$, with lower distance $2^{k}d$. Its hypothesis $2^{k}d\ge 15\rho \mathrm{/}2+Q_k+C\mathrm{/}2$ follows from $d\ge 55\rho +4C$. For $k=0$, use F4's sharper segment-target estimate, with no additive term. For $k\ge 1$, its additive term is $2Q_k+8\rho =3Q_k$. Thus, in both cases, $L-13\rho \le \max \{5\kappa ,4\sqrt{2}\lambda (x-r)\exp (-\frac{(2^{k}d-C\mathrm{/}2-Q_k)\log 2}{5\rho })\}\mathrm{.}$ Since $L-13\rho =5\rho >5\kappa$, the exponential entry is at least $L-13\rho$. Step 9.1 splits its exponent and yields $L-13\rho \le A_{\ast }(x-r)\exp (-\frac{\alpha 2^{k}d\log 2}{5\rho })\mathrm{.}$ The exponentials before splitting are at most one, and $4\sqrt{2}\le 20$, so also $x-r\ge \rho \mathrm{/}(4\lambda )=h_0$.

The first bound in step 9.1 gives $\exp (-\alpha 2^{k}d\log 2\mathrm{/}(5\rho ))\le \exp (-\beta (c_{+}-r))$. Use $t\le \exp (t)-1$ for $t=\beta (x-r)\ge 0$ from step 1.4 and exponential addition to obtain $\beta (x-r)\exp (-\frac{\alpha 2^{k}d\log 2}{5\rho })\le \exp (-\beta (c_{+}-x))-\exp (-\beta (c_{+}-r))\le \exp (-\beta (c_{+}-x))-\exp (-\beta (v-u))\mathrm{.}$ Combining this with step 10.1 and the definition of $B_{\ast }$ yields $L+4\rho \le B_{\ast }(\exp (-\beta (c_{+}-x))-\exp (-\beta (v-u)))$. Moreover $c_{+}-x+h_0\le c_{+}-r\le v-u\le (n+1)h_0$, so $c_{+}-x\le n{h}_0$. The outer induction hypothesis applies to $[x,c_{+}]$, which contains $z$. Adding its bound to the recurrence in step 5.1 cancels the two intermediate exponentials exactly and gives $P\le T_{\ast }+B_{\ast }(1-\exp (-\beta (v-u)))$. This proves the required induction step whenever the stopping case occurs.

If there is no stopping point in step 8.1, every $t\in [u,w]$ has $d(f(t),H)>(2^{k+2}-1)d$. Both $q_{k+1}(u)$ and $q_{k+1}(w)$ therefore exist at their prescribed radii; their distances to $q_k(u),q_k(w)$ respectively equal $2^{k}d$, and step 1.3 makes those earlier radial points their closest points in $V_k$. Apply the contraction inequality of step 1.2 to these two new points. Because the old projection gap is at least $5\rho +2Q_k>5\kappa +2Q_k$, its second maximum entry must apply. Hence $d(q_{k+1}(u),q_{k+1}(w))\ge 2^{k+1}d-5\rho -2Q_k\ge L-4\rho +7Q_{k+1}\mathrm{.}$ For the last inequality it suffices to use $2^{k+1}d\ge 2D=110\rho$, $Q_k\le 8\rho$, $Q_{k+1}=8\rho$ and $L=18\rho$: the left side is at least $89\rho$ and the required right side is $70\rho$. Thus $x=w$ gives the next stage. Choose in advance a finite integer $k_0$ with $(2^{k_0+1}-1)d>d(f(u),H)$, possible since $d>0$. Stage $k_0$ is impossible at $t=u$. Only finitely many stages and selections are therefore required, and a stopping case must have occurred. Step 11.1 proves the desired bound in this large-left-minimum case.

If instead $d_{+}\ge D+4C$ and $d_{-}\le d_{+}$, reverse the real parameter by replacing $f(t)$ with $f(-t)$ on $[-b,-a]$, and replace $u,z,v$ by $-v,-z,-u$. Lipschitz and quasi-geodesic constants, products and interval lengths are unchanged; the large-right minimum becomes the large-left minimum. Steps 7.1–12.1 then shorten to an interval containing $-z$ by at least $h_0$. The same outer induction hypothesis applies because it concerns all intervals containing the fixed point and uses only their length and products; equivalently undo the reversal on the resulting shorter interval before applying it. This yields the identical bound for the original interval. Together with steps 3.1 and 6.1, all cases of the induction step are proved. Every finite $b-a$ is at most $n{h}_0$ for some integer $n$, so the induction applies to $[a,b]$.

For any specified geodesic $G=[f(a),f(b)]$, step 1.1 and the resulting product bound give $d(f(z),G)\le {\lambda }^2(D+3L\mathrm{/}2+\rho +11C\mathrm{/}2)+B_{\ast }$. Substituting the constants in step 2.2 and combining exponentials gives $B_{\ast }={\lambda }^2\rho \frac{3200\exp (-(459\mathrm{/}50)\log 2)}{\log 2}<8{\lambda }^2\rho \mathrm{.}$ Indeed $4+(L+2\rho )\mathrm{/}D=48\mathrm{/}11$, $(1-\alpha )D\mathrm{/}(5\rho )=242\mathrm{/}25$, and the $\sqrt{2}$ factor changes the exponential exponent to $-459\mathrm{/}50$. Step 1.4 proves the strict numerical bound. Since $D+3L\mathrm{/}2+\rho =83\rho$, we obtain $d(f(z),G)\le {\lambda }^2(11C\mathrm{/}2+91\rho )$. This holds for every $\rho >\kappa$, so real order gives $d(f(z),G)\le D_0:={\lambda }^2(11C\mathrm{/}2+91\kappa )$ by letting $\rho$ decrease to $\kappa$. No limiting choice of projections is required, because the resulting inequality concerns the same fixed distance.

Fix $x\in G$ and split that specified segment into $G_{-}$ from $f(a)$ to $x$ and $G_{+}$ from $x$ to $f(b)$. The two functions $t\mapsto d(f(t),G_{\pm })$ are continuous, and their minimum is at most $D_0$ by step 14.1 and the union $G=G_{-}\cup G_{+}$. Their difference is nonpositive at $a$ and nonnegative at $b$, so step 1.4 gives a parameter $t$ where they are equal and thus both at most $D_0$. F4 gives closest points $x_{-}\in G_{-}$, $x_{+}\in G_{+}$ to $f(t)$, each within $D_0$. The subsegment $[x_{-},x_{+}]$ of $G$ contains $x$. Step 1.1 gives $d(x,f(t))\le \max \{d(x_{-},f(t)),d(x_{+},f(t))\}+\kappa \le D_0+\kappa$. Together with step 14.1 and $\lambda \ge 1$, both Hausdorff inclusions give $d_H(f([a,b]),G)\le {\lambda }^2(11C\mathrm{/}2+92\kappa )\mathrm{.}$ This includes constant $G$, a one-point interval and $\kappa =0$.

Return to the possibly discontinuous $q$. If its endpoint distance is at least $2\epsilon$, apply F3 to obtain the Lipschitz map $f$ with $C=4\epsilon$. Step 15.1 and the two inclusions in F3 give, by adding approximate point-to-set witnesses, $d_H(q([a,b]),G)\le 2\epsilon +{\lambda }^2(22\epsilon +92\kappa )\le {\lambda }^2(24\epsilon +92\kappa )\le 92{\lambda }^2(\epsilon +\kappa )\mathrm{.}$ For precision, each point-to-set bound is composed with witnesses whose errors are arbitrarily small, and then the errors tend to zero; no nearest point on $q([a,b])$ is asserted. If the endpoint distance is less than $2\epsilon$, F3 places every image point within $3{\lambda }^2\epsilon +\epsilon \le 4{\lambda }^2\epsilon$ of $q(a)\in G$, while every point of $G$ is within $2\epsilon$ of $q(a)\in q([a,b])$. The same desired bound follows directly. When $\epsilon =0$, F3 uses $f=q$, and the zero-$\kappa$ limit above remains valid. Finally $\kappa =3\delta$ yields the exact stated $92{\lambda }^2(\epsilon +3\delta )$. AC was spent in the explicitly supplied closest-point and radial-segment families in steps 3.1 and 7.1; interval minima, the finite stopping construction and all boundary cases were justified locally.