Exponential contraction of projection away from a quasiconvex set

We first justify projections to any specified geodesic segment $H=\gamma ([0,\mathrm{\ell }])$. The function $f(t)=d(x,\gamma (t))$ is $1$-Lipschitz. Let $m=\inf f([0,\mathrm{\ell }])$. Bisect the interval, keeping the left half if its infimum is $m$, and otherwise the right half, whose infimum must be $m$ since the original interval is their union. Iteration gives nested closed intervals of length $\mathrm{\ell }\mathrm{/}{2}^n$, each with infimum $m$. Their left endpoints have a supremum $r$ belonging to all the intervals, by completeness. Lipschitz control gives $0\le f(r)-m\le \mathrm{\ell }\mathrm{/}{2}^n$, hence $f(r)=m$. This includes $\mathrm{\ell }=0$. Moreover, if $p$ is a closest point of $x$ in $H$ and $x^{\mathrm{\prime }}$ lies on $[p,x]$, then $p$ is still closest to $x^{\mathrm{\prime }}$: for $z\in H$, $d(x^{\mathrm{\prime }},z)\ge d(x,z)-d(x,x^{\mathrm{\prime }})\ge d(x,p)-d(x,x^{\mathrm{\prime }})=d(x^{\mathrm{\prime }},p)$.

For a closest point $p$ of $x$ on a segment $H$ and $z\in H$, put $t=(x\mathrm{\mid }z{)}_p$ and choose $u\in [p,z]$ at radius $t$. 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 $d(x,u)\le d(x,p)-t+2\kappa$. Minimality of $p$ forces $t\le 2\kappa$, or equivalently $d(x,p)+d(p,z)\le d(x,z)+4\kappa$. For projections $p,r$ of $x,y$ to the same segment, put $A=d(x,p)$, $B=d(y,r)$ and $s=d(p,r)$. The just-proved estimate gives $d(x,r)+d(y,p)\ge A+B+2s-8\kappa$. By the four-point inequality this sum is at most $\max \{A+B,d(x,y)+s\}+2\kappa$. If $s>5\kappa$, the first entry of the maximum cannot suffice, and consequently $s\le d(x,y)-A-B+10\kappa$. Thus always $s\le \max \{5\kappa ,d(x,y)-A-B+10\kappa \}$.

Here is the halving estimate. Suppose projections $p,r$ of $x,y$ lie on a segment $H$, $d(x,y)\le 10\rho +c$, $c\ge 0$, and $A,B\ge M+5\rho +c\mathrm{/}2$, with $M\ge 15\rho \mathrm{/}2$. Step 1.2 gives $s=d(p,r)\le 5\kappa \le 5\rho$. Let $x^{\mathrm{\prime }},y^{\mathrm{\prime }}$ be the radius-$M$ points on $[p,x],[r,y]$. Step 1.1 preserves their projections. At basepoint $p$, all three products $(x^{\mathrm{\prime }}\mathrm{\mid }x{)}_p,(x\mathrm{\mid }y{)}_p,(y\mathrm{\mid }{y}^{\mathrm{\prime }}{)}_p$ are at least $M$: the first is exactly $M$; the second follows from $d(p,y)\ge B$ and $A+B-d(x,y)\ge 2M$; the third follows from $d(p,y^{\mathrm{\prime }})\ge M$, $d(p,y)\ge B$ and $d(y,y^{\mathrm{\prime }})=B-M$. Two product inequalities give $(x^{\mathrm{\prime }}\mathrm{\mid }{y}^{\mathrm{\prime }}{)}_p\ge M-2\rho$. The same argument at $r$ gives the analogous bound. Writing $v=d(x^{\mathrm{\prime }},y^{\mathrm{\prime }})$, we obtain $d(p,y^{\mathrm{\prime }})\ge M-4\rho +v$ and $d(r,x^{\mathrm{\prime }})\ge M-4\rho +v$. Four-point control now yields $2M-10\rho +2v\le \max \{s+v,2M\}$. Since $s\le 5\rho \le 2M-10\rho$, if the first entry is maximal this forces $v=0$; otherwise it gives $v\le 5\rho$. In either case $d(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\le 5\rho$.

For the original $Y$, choose its quasiconvexity segment $H=[p_a,p_b]$, and choose closest points $r_a,r_b$ of $q(a),q(b)$ on $H$ using step 1.1. For any $h\in H$ and any $e>0$, there exists $y\in Y$ with $d(h,y)<K+e$. Consequently $D\le d(q(t),y)\le d(q(t),h)+K+e$. Letting $e$ tend to zero and taking the infimum over $h$ shows $d(q(t),H)\ge D-K$. At $h=r_a$, minimality of $p_a$ in $Y$ similarly gives $d(q(a),p_a)\le d(q(a),r_a)+K$. Step 1.2 on $H$, with $x=q(a),p=r_a,z=p_a$, then gives $d(r_a,p_a)\le K+4\kappa \le K+4\rho$. The same estimate holds at $b$.

By finite induction, a chain $z_0,\dots ,z_{2^j}$ with projections $r_i$ to $H$, adjacent gaps at most $10\rho +c$, and projection distances at least $5\rho j+15\rho \mathrm{/}2+c\mathrm{/}2$, satisfies $d(r_0,r_{2^j})\le 5\kappa$. For $j=0$ this is step 1.2. For $j+1$, move each point along its projection segment to radius $M=5\rho j+15\rho \mathrm{/}2$. Step 2.1 makes adjacent new gaps at most $5\rho$. Discard odd-indexed points: adjacent retained gaps are at most $10\rho$, their projections are unchanged by step 1.1, and all their distances to $H$ equal $M$. The induction hypothesis with $c=0$ applies to this chain of $2^j$ gaps and proves the claim. This is an induction over finite chains, not a selection over all path parameters.

First consider a geodesic target $H$ with the whole image at distance at least $D_0\ge 15\rho \mathrm{/}2+\epsilon \mathrm{/}2$. Set $j=\lfloor (D_0-\epsilon \mathrm{/}2-15\rho \mathrm{/}2)\mathrm{/}(5\rho )\rfloor \ge 0$ and $T=\lambda (b-a)$. If $T\le 10\rho 2^j$, sample $q$ at $2^j+1$ equally spaced parameters and project to $H$, retaining the supplied endpoint projections. Step 1.1 supplies the finitely many other projections. Each adjacent gap is at most $10\rho +\epsilon$, and step 3.1 gives the bound $5\kappa$. Even when $a=b$, the two endpoint projections may be supplied differently; the finite chain argument still applies.

If $T>10\rho 2^j$, take the least integer $N$ with $T\le 10\rho 2^N$. Then $N>j$ and $2^N<T\mathrm{/}(5\rho )$, by minimality. Sample at $2^N+1$ equally spaced parameters, with finite projections as before. Each block of $2^j$ gaps satisfies step 3.1. Adding over the $2^{N-j}$ blocks gives $d(r_a,r_b)\le 5\kappa 2^{N-j}\le 5\rho 2^{N-j}\le T{2}^{-j}$.

The floor defining $j$ gives $j>(D_0-\epsilon \mathrm{/}2)\mathrm{/}(5\rho )-5\mathrm{/}2$. Using F3, $2^{-j}\le 4\sqrt{2}\exp (-(D_0-\epsilon \mathrm{/}2)\log 2\mathrm{/}(5\rho ))$. Here $\log 2>0$ because $\exp$ is increasing and $\exp (0)=1$; addition gives $\exp (-j\log 2)=2^{-j}$ and $\exp ((5\mathrm{/}2)\log 2)=4\sqrt{2}$ (the latter is positive and its square is $32$). Thus steps 4.1–4.2 give the asserted maximum bound for a segment target, without the $2K+8\rho$ term.

Apply step 5.1 with $D_0=D-K$, whose required lower bound is exactly the stated hypothesis. The triangle inequality through $r_a,r_b$ adds at most $2K+8\rho$ by step 2.2, yielding the displayed result. The proof uses finite choices and uniquely specified bisections only; it never assumes projections of arbitrary points onto $Y$. The auxiliary positive $\rho$ also permits $\kappa =0$ without division by zero.