Asymptotic Cones and the Sublinear Triangle Criterion: Examples

If $(x_n)$ is admissible, $a_n={\lambda }_n(x_n-e_n)$ is bounded in absolute value, so $T([x_n])={\lim }_{\omega }{a}_n$ exists. For two admissible sequences, F2 gives $\mathrm{\mid }T([x_n])-T([y_n])\mathrm{\mid }={\lim }_{\omega }\mathrm{\mid }{\lambda }_n(x_n-y_n)\mathrm{\mid }=d_{\omega }([x_n],[y_n])\mathrm{.}$ Thus $T$ is constant on equivalent representatives, and equal $T$ values imply zero ultradistance and the same cone point. It is an isometric injection.

In a real tree every chosen triangle is a tripod by F4, and its branchpoint belongs to all three sides. Its minsize is therefore zero, giving the sublinear function $m_X(P)=0$. F3 makes the cone geodesic, and F5 applies under AC: each of its geodesic segments is unique and is the limit of any prescribed representative segments. This includes arbitrary moving basepoints and arbitrary positive scales tending to zero. The next branchpoint calculation establishes the additional tripod conclusion.

For $t\in \mathbb{R}$ set $x_n=e_n+t\mathrm{/}{\lambda }_n$. Then ${\lambda }_n\mathrm{\mid }{x}_n-e_n\mathrm{\mid }=\mathrm{\mid }t\mathrm{\mid }$, so this is admissible and $T([x_n])=t$. Conversely, for an admissible $(y_n)$ with $t=T([y_n])$, the distance to this inverse representative is the ultralimit of $\mathrm{\mid }{\lambda }_n(y_n-e_n)-t\mathrm{\mid }$, which is zero. Both inverse identities hold. For the concrete choice $e_n=(n+1{)}^2$, ${\lambda }_n=1\mathrm{/}(n+1)$, the sequence $x_n=(n+1{)}^2+3(n+1)$ maps to $3$, and $y_n=(n+1{)}^2-2(n+1)$ maps to $-2$; their rescaled distance is $5$ for every $n\in \mathbb{N}$, including $n=0$.

More explicitly, for three representative vertices $x_n,y_n,z_n$ choose their tree branchpoint $b_n$. It satisfies $d(b_n,x_n)\le d(x_n,y_n)$ and hence ${\lambda }_{n}d(b_n,e_n)\le {\lambda }_{n}d(x_n,e_n)+{\lambda }_{n}d(x_n,y_n)$, a bounded sequence. Thus $[b_n]$ exists. It is on each limit side, since $b_n$ belongs to each original side, and F3 parameterizes those sides by their rescaled distances from an endpoint. By step 1.2 these limit sides are the actual chosen cone sides. Write $b=[b_n]$. Each side splits at $b$ into its two endpoint legs by distance additivity. Two distinct legs toward vertices $x,y$ cannot share $q\mathrm{\ne }b$, since then $d(x,y)\le d(x,q)+d(q,y)=d(x,b)+d(b,y)-2d(b,q)<d(x,y)$. Thus they form a tripod, with zero legs permitted. Every chosen cone triangle is therefore a tripod, and F4 makes the cone a real tree. For a concrete change of isometry type, take the real tree $[0,1]$, basepoint $0$ and scales $1\mathrm{/}(n+1)$. Every representative has rescaled distance at most $1\mathrm{/}(n+1)$, so its cone is a single point, whereas the original interval has two points at distance one.

Parameterize the two axis sides by ${\gamma }_1(u)=(u,0)$ and ${\gamma }_2(u)=(0,u)$ for $0\le u\le s$. Their pairwise parameter distances are $\sqrt{(u-v{)}^2}=\mathrm{\mid }u-v\mathrm{\mid }$. Parameterize the third side by ${\gamma }_3(u)=(u\mathrm{/}\sqrt{2},s-u\mathrm{/}\sqrt{2})$ for $0\le u\le s\sqrt{2}$. Its squared distance between parameters $u,v$ is $2((u-v)\mathrm{/}\sqrt{2}{)}^2=(u-v{)}^2$, so it too has distance $\mathrm{\mid }u-v\mathrm{\mid }$. These are isometric segments with exactly the displayed endpoints. Their lengths sum to $(2+\sqrt{2})s$.

Write an arbitrary side triple as $p=(u,0)$, $q=(0,v)$, $r=(t,s-t)$ with $u,v,t\in [0,s]$. Then $d_2(p,r)\ge s-t$ and $d_2(q,r)\ge t$, since their squared distances include respectively $(s-t{)}^2$ and $t^2$ and all other summands are nonnegative. Therefore its diameter is at least $\max (t,s-t)\ge s\mathrm{/}2$. This holds for every triple, so $\mu (s)\ge s\mathrm{/}2$.

Take $p=q=(0,0)$ and $r=(s,0)$, which belong respectively to the two axis sides and the third side. Their three pair distances are $0,s,s$, so the diameter is $s$ and $\mu (s)\le s$. In particular at $s=1$ the computed bounds are $1\mathrm{/}2\le \mu (1)\le 1$.

Multiplication of both coordinates by $s$ bijects all triples for scale one with all triples for scale $s$, with inverse division by $s$. The distance formula gives $d_2(sx,sy)=s{d}_2(x,y)$ because $s>0$. Hence the set of admissible diameters is exactly $s$ times the scale-one set. Multiplication by a positive scalar commutes with its infimum: all scaled values are at least $s\mu (1)$, and a value less than $\mu (1)+\epsilon \mathrm{/}s$ scales to less than $s\mu (1)+\epsilon$. Thus $\mu (s)=s\mu (1)$, with a positive coefficient by step 2.2.

Put $H={\sup }_n{\lambda }_n{P}_n<\mathrm{\infty }$. Given $\epsilon >0$, sublinearity supplies $P_0>0$ with $m_X(P)\le \epsilon P$ whenever $P\ge P_0$. For $P<P_0$, F1 gives $m_X(P)\le P_0$. Separating these cases for each $P_n$ yields the single bound $0\le {\lambda }_n{m}_X(P_n)\le \epsilon {\lambda }_n{P}_n+{\lambda }_n{P}_0\le \epsilon H+{\lambda }_n{P}_0\mathrm{.}$ There is no assumption that $P_n$ tends to infinity.

For each $n\in \mathbb{N}$ in ${\mathbb{R}}^2$, take vertices $(0,0),(n+1,0),(0,n+1)$. The axis-side parameterizations $(u,0)$ and $(0,u)$ for $0\le u\le n+1$ have distance $\mathrm{\mid }u-v\mathrm{\mid }$. The third parameterization $(u\mathrm{/}\sqrt{2},n+1-u\mathrm{/}\sqrt{2})$ for $0\le u\le (n+1)\sqrt{2}$ has squared distance $2((u-v)\mathrm{/}\sqrt{2}{)}^2=(u-v{)}^2$. Thus these really are geodesic triangles and their perimeters are $P_n=(2+\sqrt{2})(n+1)$.

To make the last expression less than any $\eta >0$, first take $\epsilon =\eta \mathrm{/}(2(H+1))$, obtain its $P_0$, and then take $n$ so large that ${\lambda }_n{P}_0<\eta \mathrm{/}2$. This proves ordinary convergence to zero, and also ultralimit zero for any free ultrafilter by F2. If instead $P_n\le M$ for a fixed finite $M$, the simpler bound $0\le {\lambda }_n{m}_X(P_n)\le {\lambda }_{n}M\to 0$ works without sublinearity.

For any triple $p=(u,0)$, $q=(0,v)$, $r=(t,n+1-t)$ on these sides, the diameter is at least $\max (n+1-t,t)\ge (n+1)\mathrm{/}2$: the two terms are lower bounds for $d(p,r)$ and $d(q,r)$ respectively. Conversely the triple $(0,0),(0,0),(n+1,0)$ has diameter $n+1$. At ${\lambda }_n=1\mathrm{/}(n+1)$ the rescaled minsize of these triangles therefore lies in $[1\mathrm{/}2,1]$, and their rescaled perimeter is the constant $2+\sqrt{2}$. They do not vanish, whereas any fixed triangle in the very same plane has both its perimeter and its minsize multiplied by $1\mathrm{/}(n+1)$ and tending to zero. This is the claimed witness that controlling only a fixed unscaled perimeter cutoff cannot establish sublinear behavior at the moving scale.