A pointed metric space is a metric space $(M,d)$ together with a chosen basepoint $o$. For a Lipschitz map $f:M\to R$, write $Lip(f)$ for the least $C$ with $|f(x)-f(y)|\le C\,d(x,y)$ for all $x,y$, and let $Lip_0(M)$ be the Banach space of those $f$ with $f(o)=0$, normed by $Lip$. The Lipschitz-free space $F(M)$ is the closed linear span of the evaluation functionals $\delta_x(f)=f(x)$ inside the dual $Lip_0(M)^*$, so $\delta_o=0$. Two facts make $F(M)$ useful: the assignment $x\mapsto\delta_x$ is isometric, and every pointed $1$-Lipschitz map $u:M\to N$ induces a contraction $F(u)$ sending $\delta_x$ to $\delta_{u(x)}$; metric questions thereby become linear ones. Arens and Eells introduced these spaces, and Godefroy and Kalton developed their linear structure.
(Arens and Eells, Pacific J. Math. 6 (1956), 397-403.)
./arens-eells-1956-pjm-embedding-uniform.pdf
(Godefroy and Kalton, Studia Math. 159 (2003), 121-141.)
./godefroy-kalton-2003-studia-lipschitz-free-banach-spaces.pdf

The distance lemma. For all $x,y\in M$, $\|\delta_x-\delta_y\|_{F(M)}=d(x,y)$.
Proof. If $Lip(f)\le1$ then $|f(x)-f(y)|\le d(x,y)$, so $\|\delta_x-\delta_y\|\le d(x,y)$. Conversely $g(z)=d(z,y)-d(y,o)$ has $Lip(g)\le1$ and $g(x)-g(y)=d(x,y)$, giving the reverse inequality. QED

The linearization lemma. A pointed $1$-Lipschitz $u:(M,d,o)\to(N,\rho,o')$ extends from $\sum_x c_x\delta_x\mapsto\sum_x c_x\delta_{u(x)}$ to a unique contraction $F(u):F(M)\to F(N)$, with $F(\mathrm{id})=\mathrm{id}$ and $F(v\circ u)=F(v)F(u)$.
Proof. Take $\varphi\in Lip_0(N)$ with $Lip(\varphi)\le1$. The composition $\varphi\circ u$ lies in $Lip_0(M)$ with $Lip(\varphi\circ u)\le1$, so by duality
$$\Big\langle\textstyle\sum_x c_x\delta_{u(x)},\varphi\Big\rangle=\sum_x c_x\varphi(u(x))=\Big\langle\textstyle\sum_x c_x\delta_x,\ \varphi\circ u\Big\rangle\le\big\|\textstyle\sum_x c_x\delta_x\big\|.$$
Taking the supremum over such $\varphi$ gives $\|\sum_x c_x\delta_{u(x)}\|\le\|\sum_x c_x\delta_x\|$, so the map is well defined on finite combinations and extends to a contraction. Functoriality holds on evaluations, hence everywhere by continuity. QED

A bounded linear operator is finite rank when its image is finite dimensional. A Banach space $X$ has the approximation property (AP) if for every norm-compact $C\subseteq X$ and every $\varepsilon>0$ there is a finite-rank operator $T$ with $\sup_{u\in C}\|Tu-u\|<\varepsilon$, the norm of $T$ allowed to depend on $C$ and $\varepsilon$. For $\lambda\ge1$, $X$ has $\lambda$-BAP if such $T$ can always be taken with $\|T\|\le\lambda$; BAP means $\lambda$-BAP for some finite $\lambda$, and MAP means $1$-BAP. Thus AP says approximation is always possible, while BAP says one bound serves every compact set.

Kalton proved that over a uniformly discrete space the free space always has AP, and asked whether a uniform bound always exists; the question was restated by Godefroy and Ozawa and in Godefroy's survey. The answer is negative.
(Kalton, Collect. Math. 55 (2004), 171-217.)
./kalton-2004-collect-spaces-lipschitz-holder.pdf
(Godefroy and Ozawa, Proc. Amer. Math. Soc. 142 (2014), 1681-1687.)
./godefroy-ozawa-2014-1201.0847-free-banach-ap.pdf
(Godefroy, Comment. Math. 55 (2015), 89-118, doi:10.14708/cm.v55i2.1104.)

The main theorem. There is a countable, unbounded, nonproper pointed metric space $(M,d,o)$ with $d(x,y)\ge1$ for distinct $x,y$, such that $F(M)$ has AP but fails $\lambda$-BAP for every finite $\lambda\ge1$.

The space must be nonproper: Dalet showed that over a countable proper space $F(M)$ always has MAP. Uniform discreteness cannot be dropped either, since Smith produced countable discrete examples without a positive lower bound on all nonzero distances.
(Dalet, Mediterr. J. Math. 12 (2015), 973-986.)
./dalet-2015-1404.3939-free-spaces-proper.pdf
(Smith, Lipschitz-free spaces and Bossard's reduction argument, arXiv 2025.)
./smith-2026-2509.00722-lipschitz-free-bossard.pdf

The construction fixes a candidate bound $p$. A bounded space $M_p$ is built with a finite anchor set $A_p$ at which every finite-rank operator of norm at most $p$ moves some $\delta_a$, $a\in A_p$, by at least $1/2$. Such a block settles nothing alone, since a bounded uniformly discrete space has BAP; it is a trap at scale $p$. The blocks are then glued at a common basepoint into a wedge $M$. Each block is a $1$-Lipschitz retract of $M$, so $F(M_p)$ is an isometrically complemented subspace of $F(M)$; a $\lambda$-BAP of $F(M)$ would force a $\lambda$-BAP of $F(M_p)$ for every $p$, contradicting the trap once $p\ge\lambda$. Approximation on a compact set never needs more than finitely many blocks, each handled by its own BAP with a norm allowed to grow, which gives AP. The work lies in one block.

The block theorem. For each integer $p\ge1$ there are a bounded countably infinite pointed metric space $(M_p,d_p,o)$ and a finite set $A_p\subseteq M_p$ containing $o$, with $d_p(x,y)\ge1$ for distinct points and $\mathrm{diam}\,M_p\le B_p=300p^2+150p+5$, such that every bounded finite-rank $T_0:F(M_p)\to F(M_p)$ with $\|T_0\|\le p$ satisfies
$$\max_{a\in A_p}\|T_0\delta_a-\delta_a\|\ge\tfrac12.$$

The block has three layers: sparse regular colored multigraphs whose small neighborhoods mimic a regular tree, a labeling of each vertex by capped distances to the color classes, and a metric that routes all vertices sharing a label through one anchor and links the anchors. The empirical rooted-neighborhood viewpoint is that of Benjamini and Schramm, and the independent-permutation model of random regular graphs is Friedman's.
(Benjamini and Schramm, Electron. J. Probab. 6 (2001), no. 23.)
./benjamini-schramm-2001-distributional-limits.pdf
(Friedman, A proof of Alon's second eigenvalue conjecture, arXiv 2004.)
./friedman-2004-cs0405020-second-eigenvalue.pdf

Multiple edges are distinct instances, and each instance has two oriented incidences $e,\bar e$ exchanged by reversal, even for a loop, so a loop contributes two to the degree. Write $e:x\to y$ for initial and terminal vertices and let $D(x)$ be the set of $\Delta$ outgoing incidences at $x$. A walk is non-backtracking if it never immediately reverses its preceding incidence; it may revisit vertices and retake loops. A rooted radius-$r$ ball is the induced multigraph on vertices within graph distance $r$ of the root, retaining colors, loops, and multiplicities. Only vertices are points of the metric spaces.

The colored-tree-limit lemma. For every even integer $\Delta\ge4$ and integer $k\ge1$ there is a deterministic sequence of nonempty finite $\Delta$-regular multigraphs $G_m$, colored by $\{1,\ldots,k\}$, such that for every fixed $r\ge0$ and every rooted colored radius-$r$ type $\tau$, the fraction of vertices of $G_m$ whose ball has type $\tau$ converges, as $m\to\infty$, to the probability of $\tau$ for the $\Delta$-regular tree with independent uniform vertex colors.
Proof. Fix $N$ and take $\Delta/2$ independent uniform permutations $\sigma_1,\ldots,\sigma_{\Delta/2}$ of an $N$-set. For each $a$ and $v$ create a distinct edge instance joining $v$ to $\sigma_a(v)$; each $a$ gives every vertex one incidence to its image and one to its preimage, so the multigraph is $\Delta$-regular, with fixed points giving loops and $2$-cycles giving parallel edges. Color vertices independently and uniformly in $\{1,\ldots,k\}$, independently of the permutations.
Fix $r$ and explore from a root by breadth-first search, revealing each relevant permutation image and preimage at vertices within distance $r$, and recording without expanding the vertices first met at distance $r+1$ so that every edge of the induced radius-$r$ ball is seen. The number of queries and revealed vertices is at most $C=C(\Delta,r)$, independent of $N$. Conditionally on the revealed data, each unrevealed part of a permutation is a uniform bijection between unused domain and range, so a fresh image or preimage is uniform among at least $N-C$ unused values, at most $C$ of which are discovered; its chance of hitting a known vertex is at most $C/(N-C)$. Before any collision each fresh query adds one vertex and one edge, and a loop or second parallel edge occurs exactly when a fresh query hits a known vertex. A union bound over the at most $C$ queries gives collision probability at most $C^2/(N-C)=O_{\Delta,r}(N^{-1})$. Off this event the radius-$r$ ball is exactly the regular-tree ball. For two distinct roots, run both explorations together with both roots initially discovered; a larger $C(\Delta,r)$ bounds the chance of any collision, and then the two balls are disjoint tree balls with independent colors.

Let $I_v$ indicate that the ball at $v$ has type $\tau$, and set $X_{\tau,N}=\frac1N\sum_v I_v$, $q_\tau=P(\text{tree ball has type }\tau)$. The one-root exploration gives $E X_{\tau,N}\to q_\tau$, and the two-root exploration gives $E X_{\tau,N}^2=\frac1N E I_1+(1-\frac1N)E(I_1I_2)\to q_\tau^2$, which also covers non-tree types where $q_\tau=0$. Hence $\mathrm{Var}(X_{\tau,N})\to0$ and each frequency converges in probability to $q_\tau$. At radius at most $m$ there are finitely many colored types, so finite union bounds show that for all large $N$, with positive probability every such frequency is within $1/m$ of its tree value. Choosing one such realization of size $N_m$, strictly increasing in $m$, makes every fixed radius-and-type frequency tend to $q_\tau$. QED

The scalar tests take values in $[-L,L]$, so a Lipschitz inequality holds automatically between opposite values at points at least $2L$ apart. The attachment length is taken above $2L$ so no short path crosses a thread, and the label slope $L/K$ is made tiny so that comparing across a thread, whose norm error is at most $2pR+1$, costs less than $L$. Set
$$L=10p,\ \ R=3L=30p=:R_p,\ \ K=10p(R+1),\ \ D=K+2L+5,\ \ B=4R+D,\ \ V=1+4pB.$$
Substitution gives $B=300p^2+150p+5=B_p$, with $R>2L$ and
$$\frac{2pR+1}{K}=\frac{60p^2+1}{300p^2+10p}\le\frac15,\qquad K+2L+1=D-4<D.$$
The last inequality leaves room, below the cap $D$, for a label gap below $K$, one source-edge step, and one support step of length at most $2L$. Total variation is bounded by $V$, and a large degree dilutes the loss from at most one exceptional direction per support point. Choose an even integer $\Delta\ge4$ with $V/\Delta<1/20$, and set
$$Q=\Delta^{2L+1},\qquad H=\Delta^{D+1}.$$
The geometry gives at most $Q$ nearby successors for a support point and at most $H$ label coordinates below the cap, bounding admissible source colors, while a walk has $\Delta$ choices per step. A larger degree helps the averaging but multiplies the walks in the matching count, so after fixing $\Delta$ choose an integer $k>2\Delta QH$, making $\Delta QH/k<1/2$. Fix a deterministic sequence $(G_m)_{m\ge1}$ from the colored-tree-limit lemma for this $\Delta$ and $k$; all constants and the whole sequence are fixed before any operator appears.

Take disjoint copies of the graphs $G_m$ and call their vertices source vertices. For a source vertex $x\in G_m$ define $h(x)\in\{0,\ldots,D\}^k$ by
$$h_i(x)=\min\big\{D,\ \mathrm{dist}_{G_m}(x,\{\text{vertices of color }i\})\big\},\qquad 1\le i\le k,$$
the graph distance being $+\infty$, hence the coordinate $D$, if the component of $x$ has no vertex of color $i$. This uses unit graph distances, before the metric below. For each label $v$ realized by some source vertex, add one new vertex $a_v$, called an anchor, with $h(a_v)=v$; set $A_p=\{a_v: v \text{ realized}\}$ and $a(x)=a_{h(x)}$. The edges are length $R$ from each source vertex $x$ to its anchor $a(x)$, length $2R+\|v-w\|_\infty$ between distinct anchors $a_v,a_w$, and length $1$ on each source edge. Let $M_p$ be the set of all source vertices and anchors, and let $d_p(u,v)$ be the infimum of lengths of finite paths from $u$ to $v$, the empty path giving $d_p(u,u)=0$. Choose an anchor as basepoint $o$, which exists because each $G_m$ is nonempty. Each source vertex hangs by a length-$R$ thread from the anchor bearing its label, and anchors are joined by edges whose length dominates the sup-distance of their labels; this bounds the space while keeping small balls inside a single source graph.

The geometry lemma. The preceding construction satisfies:
(a) $(M_p,d_p,o)$ is a countably infinite pointed metric space with separation at least one and diameter at most $B=B_p$; $A_p$ is finite and contains $o$; each source edge has endpoints at $d_p$-distance at most one.
(b) The label $h$ is $1$-Lipschitz for the sup norm; a source vertex $x$ of color $i$ has $h_i(x)=0$; and $h(a(x))=h(x)$.
(c) For every source vertex $x$ and anchor $a$, $d_p(x,a)\ge R$ and $d_p(x,a(x))=R$; in particular $d_p(x,o)\ge R_p=30p$.
(d) Every closed ball of radius $2L$ has at most $Q$ points, and at most $H$ coordinates of any $h(t)$ are strictly below $D$.
Proof. (a) There are at most $(D+1)^k$ labels, so $A_p$ is finite. The disjoint union of the nonempty finite graphs is countable, and adding finitely many anchors keeps it countable. The added edges connect all vertices. Every edge length is a positive integer: source and thread lengths are integers, and the sup-distance of integer labels is an integer. For distinct endpoints the lengths of connecting paths form a nonempty set of positive integers, which has a least element, so $d_p$ is attained and is a metric; distinct points are at distance at least one because every edge has length at least one, and a source edge gives distance at most one between its endpoints. The sup-distance of two labels is at most $D$, so two source vertices with different anchors join through the two threads and the anchor edge in length at most $R+(2R+D)+R=4R+D=B$, and with a common anchor in length $2R$; a source vertex and an anchor lie within $3R+D$, two anchors within $2R+D$. Hence the diameter is at most $B$.
(c) A path from a source vertex to an anchor must first enter the anchor set, and the only edges from source vertices to anchors are threads of length $R$; hence $d_p(x,a)\ge R$. The thread realizes length $R$, and $o$ is an anchor, so $d_p(x,o)\ge R$.
(b) Across a source edge the distances to a fixed color class differ by at most one whenever the color is present in the component, since the edge is appended to a realizing path, and capping at $D$ preserves this; if the color is absent both coordinates equal $D$. So adjacent source labels differ by at most one in sup norm. Along a thread the labels are equal, and along an anchor edge the length dominates the sup-distance of the labels. Summing along a shortest path gives $\|h(u)-h(v)\|_\infty\le d_p(u,v)$. The own-color coordinate is zero by definition, and anchor labels are copied.
(d) Every edge other than a source edge has length at least $R=3L>2L$, so a shortest path of length at most $2L$ uses only source edges. The radius-$2L$ ball about an anchor is a singleton, and that about a source vertex lies in its graph ball of radius $2L$ inside its own $G_m$. In a $\Delta$-regular multigraph every vertex within graph distance $s$ is reached by some walk of length at most $s$, and there are at most $\Delta^t$ walks of length $t$ from a fixed vertex, so the ball has at most $\sum_{t=0}^{2L}\Delta^t\le\Delta^{2L+1}=Q$ points. If $h_i(x)<D$ then $x$ is within graph distance $D-1$ of a color-$i$ vertex; witnesses for distinct colors are distinct, so at most $\sum_{t=0}^{D-1}\Delta^t\le\Delta^{D+1}=H$ coordinates are below $D$, at source vertices and by copying at anchors. QED

An operator that nearly fixes the anchors yields, at each source vertex, a signed measure of bounded total variation with a large integral against a bump set by the vertex's label. Finite rank does not mean finite support, so a small perturbation first buys a common finite support.

The finite-support lemma. Suppose a bounded finite-rank $T_0:F(M_p)\to F(M_p)$ satisfies $\|T_0\|\le p$ and $\|T_0\delta_a-\delta_a\|<1/2$ for all $a\in A_p$. Then there are a finite set $S\subseteq M_p$ containing $o$ and a bounded finite-rank $T:F(M_p)\to F(M_p)$ with $T F(M_p)\subseteq\mathrm{span}\{\delta_j:j\in S\}$, $\|T\|\le2p$, and $\|T\delta_a-\delta_a\|<1$ for all $a\in A_p$.
Proof. If $T_0=0$ take $T=0$ and $S=\{o\}$. Otherwise choose a basis $v_1,\ldots,v_r$ of the range, with bounded coordinate functionals composed with $T_0$ giving bounded $\phi_i$ and $T_0u=\sum_i\phi_i(u)v_i$. Take $0<\eta<\min\{p,1/(2B)\}$ and finite evaluation combinations $w_i$ with $\sum_i\|\phi_i\|\,\|w_i-v_i\|<\eta$, possible by density. Then $Tu=\sum_i\phi_i(u)w_i$ has $\|T-T_0\|<\eta$; let $S$ be $\{o\}$ together with the supports of the $w_i$. Then $T$ has the stated support and $\|T\|\le p+\eta<2p$. Since $\|\delta_a\|\le B$, $\|T\delta_a-\delta_a\|\le\|T_0\delta_a-\delta_a\|+\eta B<1/2+\eta B<1$. QED

For a source vertex $x$ define on all of $M_p$
$$b_x(t)=L\Big(1-\frac{\|h(x)-h(t)\|_\infty}{K}\Big)_+,\qquad s_+=\max\{s,0\}.$$
By the geometry lemma the label $h$ is $1$-Lipschitz, so $b_x\in[0,L]$ has Lipschitz constant at most $L/K\le1$ and equals $L$ at the anchor $a(x)$. For a signed measure $\nu$ on a finite set $S$ write $|\nu|(E)=\sum_{j\in E}|\nu(\{j\})|$ and $\|\nu\|_{TV}=|\nu|(S)$.

The measures lemma. Let finite $S\subseteq M_p$ contain $o$ and let bounded $T$ satisfy $T F(M_p)\subseteq\mathrm{span}\{\delta_j:j\in S\}$, $\|T\|\le2p$, and $\|T\delta_a-\delta_a\|<1$ for $a\in A_p$. Then there are signed measures $(\mu_x)_{x\in M_p}$ on $S$, each of total mass one, with
$$\int g\,d\mu_x=(T\delta_x)(g-g(o))+g(o)$$
for every Lipschitz $g$, with $\|\mu_x\|_{TV}\le V=1+4pB$, and with $\int b_x\,d\mu_x\ge 3L/4$ for every source vertex $x$.
Proof. Write $T\delta_x=\sum_{j\in S\setminus\{o\}}c_x(j)\delta_j$; set $\mu_x(\{j\})=c_x(j)$ for $j\ne o$ and $\mu_x(\{o\})=1-\sum_{j\ne o}c_x(j)$. These masses sum to one and give the affine formula. For the variation, let $g$ equal the sign of $\mu_x(\{j\})$ on $S$ and $0$ off $S$, with $\mathrm{sign}(0)=0$. Its values lie in $[-1,1]$, so by separation $Lip(g)\le2$, and
$$\|\mu_x\|_{TV}=\int g\,d\mu_x=(T\delta_x)(g-g(o))+g(o)\le2\|T\delta_x\|+1\le1+4pB,$$
using $\|T\|\le2p$ and $\|\delta_x\|\le B$. For a source vertex $x$, the geometry lemma gives $d_p(x,a(x))=R$, so $\|T\delta_x-\delta_{a(x)}\|\le\|T(\delta_x-\delta_{a(x)})\|+\|T\delta_{a(x)}-\delta_{a(x)}\|\le2pR+1$. Since $b_x(a(x))=L$,
$$\int b_x\,d\mu_x-L=(T\delta_x-\delta_{a(x)})(b_x-b_x(o))\ge-\frac{L}{K}(2pR+1)\ge-\frac{L}{5},$$
because $(2pR+1)/K\le1/5$. Hence $\int b_x\,d\mu_x\ge4L/5\ge3L/4$. QED

Fix a finite set $S\subseteq M_p$ containing $o$; its points are states and their labels and distances stay fixed. This is the support from the finite-support lemma, and the graph is chosen only after $S$ is fixed. A graph vertex $x$ and a state $j\in S$ are compatible when
$$\|h(x)-h(j)\|_\infty\le K+2L+1=D-4.$$
A non-backtracking walk $x_0,e_1,x_1,\ldots,e_t,x_t$ in one graph $G_m$ is matched if there are states $j_0,\ldots,j_t\in S$ with $(x_s,j_s)$ compatible for every $s$ and $d_p(j_{s-1},j_s)\le2L$ for $1\le s\le t$. Both the states and the source vertices may repeat; a return along a different parallel edge is allowed, only the reverse of the preceding incidence is forbidden. The threshold allows for a label gap below $K$ where a bump is positive, one source-edge step of length at most one, and one state step of length at most $2L$; the inequality $K+2L+1<D$ also restricts the possible source colors.

Each state has at most $Q$ possible successors, by the radius-$2L$ ball bound, while a graph walk meets many possible colors; this disparity bounds the fraction of roots from which a long matched walk starts.

The rare-walk lemma. For fixed $n\ge1$, let $\beta_m$ be the fraction of vertices of $G_m$ from which a matched walk of length $n$ starts. Then
$$\limsup_{m\to\infty}\beta_m\le|S|\frac{H}{k}\Big(\frac{\Delta QH}{k}\Big)^n.$$
Hence one chooses $n\ge2$ and then $m$ so that $\beta_m<1/(20V)$.
Proof. For $j\in S$ set $C_j=\{i:h_i(j)<D\}$, so $|C_j|\le H$ by the geometry lemma. If $x$ has color $i$ and is compatible with $j$, then $h_i(x)=0$ forces $h_i(j)\le K+2L+1=D-4<D$, so $i\in C_j$; this is only a necessary condition, avoiding any independence claim about full labels.
Consider the weaker event that a non-backtracking length-$n$ walk $x_0,\ldots,x_n$ from the root has states $j_0,\ldots,j_n$ with $d_p(j_{s-1},j_s)\le2L$ for $1\le s\le n$ and with the color of $x_s$ in $C_{j_s}$ for $0\le s\le n$. This depends only on the induced colored radius-$n$ ball, of which there are finitely many rooted types. With $S,n$ fixed, the sets $C_j$ and the transition relation are fixed, so the event is a fixed finite union of types, and the colored-tree-limit lemma makes its frequency converge to its probability on the independently colored tree; the support may depend on the whole graph sequence, but once fixed it only selects types, and no colors are conditioned on it. Along a uniform non-backtracking walk with $\Delta$ choices per step, there are at most $\Delta^n$ paths reaching $x_n$; a fixed choice of all states gives probability at most $(H/k)^{n+1}$ for the required color memberships, since tree vertices are distinct and overlaps of the $C_j$ do not matter. A union bound gives $\Delta^n|S|Q^n(H/k)^{n+1}=|S|(H/k)(\Delta QH/k)^n$. Every matched walk satisfies the weaker event, proving the display. Finally $k>2\Delta QH$ gives $\Delta QH/k<1/2$; choose $n\ge2$ with the right side below $1/(40V)$, then $m$ with $\beta_m<1/(20V)$. QED

Call a vertex bad if a matched walk of length $n$ starts there, else good. At a good vertex the absence of long matched walks picks out at most one direction per state. For a good vertex $x$, an incidence $e\in D(x)$, and a state $j\in S$, let $L_x(e,j)$ be the set of positive lengths of matched walks starting through $e$ with initial state $j$, and set $r_x(e,j)=\max(\{0\}\cup L_x(e,j))$. A matched walk of length at least $n$ has a matched prefix of length $n$, so $L_x(e,j)\subseteq\{1,\ldots,n-1\}$ and the maximum is defined; the value $0$ says no positive-length walk exists. Call $e$ exceptional at $(x,j)$ if its record is strictly larger than every other outgoing record, so there is at most one such incidence; at a bad vertex call all outgoing incidences exceptional at every state. For an incidence $e:x\to y$ put $E_e=\{j\in S: e \text{ exceptional at } (x,j)\}$.

The exceptional-direction lemma. Let $e:x\to y$ be an incidence and $j,j'\in S$. If $(x,j)$ and $(y,j')$ are compatible and $d_p(j,j')\le2L$, then $j\in E_e$ or $j'\in E_{\bar e}$.
Proof. If either endpoint is bad this is the definition. Suppose both are good and put
$$A_x=\max_{g\in D(x)\setminus\{e\}}r_x(g,j),\qquad A_y=\max_{f\in D(y)\setminus\{\bar e\}}r_y(f,j').$$
The maxima range over nonempty finite sets, since $\Delta\ge4$. If $A_y>0$, choose a walk realizing it and prepend $e$ with state transition $j$ to $j'$; the old first incidence is not $\bar e$, so the result is non-backtracking and matched. If $A_y=0$, the edge alone with states $j,j'$ is a matched walk. The same after reversing $e$ gives
$$r_x(e,j)\ge1+A_y,\qquad r_y(\bar e,j')\ge1+A_x.$$
If neither incidence were exceptional, their records would be at most $A_x$ and $A_y$, forcing $A_x\ge1+A_y\ge2+A_x$, a contradiction. QED

For an incidence $e:x\to y$ the bump $b_x$ is prescribed off $E_e$ and $-b_y$ off $E_{\bar e}$; bump bounds control pairs inside one set and far cross pairs, and the exceptional-direction lemma controls close cross pairs. The finite-data extension used is McShane's.
(McShane, Bull. Amer. Math. Soc. 40 (1934), 837-842.)
./mcshane-1934-bams-extension-of-range-of-functions.pdf

The edge-test lemma. For every incidence $e:x\to y$ of $G$ there is a $1$-Lipschitz $f_e:M_p\to[-L,L]$ with $f_e(j)=b_x(j)$ for $j\in S\setminus E_e$ and $f_e(j')=-b_y(j')$ for $j'\in S\setminus E_{\bar e}$.
Proof. Within either set the Lipschitz inequality follows from the bump bound. Across the two sets the value difference at $j,j'$ is $b_x(j)+b_y(j')\le2L$, so the inequality holds whenever $d_p(j,j')\ge2L$.
Suppose $d_p(j,j')<2L$ and $b_x(j)>0$. Then $\|h(x)-h(j)\|_\infty<K$, and since $d_p(x,y)\le1$ and $h$ is $1$-Lipschitz,
$$\|h(y)-h(j')\|_\infty\le\|h(y)-h(x)\|_\infty+\|h(x)-h(j)\|_\infty+\|h(j)-h(j')\|_\infty<1+K+2L.$$
Thus $(x,j)$ and $(y,j')$ are compatible with state distance at most $2L$, so by the exceptional-direction lemma both prescriptions cannot be present. The same holds when $b_y(j')>0$. Hence whenever two prescribed states, one from each set, are less than $2L$ apart, both prescribed values vanish; taking $j=j'$ makes the data consistent on the overlap.
Let $P\subseteq S$ be the union and $v:P\to[-L,L]$ the resulting data, which is $1$-Lipschitz. If $P$ is empty take $f_e=0$; otherwise extend and clip:
$$F(z)=\min_{j\in P}\{v(j)+d_p(z,j)\},\qquad f_e(z)=\max\{-L,\min\{L,F(z)\}\}.$$
For $z\in P$ the Lipschitz inequalities give $F(z)=v(z)$; the triangle inequality gives $F(z)\le F(w)+d_p(z,w)$ and symmetrically, so $F$ is $1$-Lipschitz, and clipping preserves both the bound and the values. QED

Each test compares the measures at the ends of its incidence, losing variation only on exceptional states. At a good vertex each state is exceptional in at most one of the $\Delta$ directions, so averaging controls the loss by the degree and the bad fraction.

Proof of the block theorem. For the integer $p$, the geometry lemma supplies the countably infinite block, the separation and diameter bounds, and the finite anchor set $A_p$. Suppose a bounded finite-rank $T_0$ of norm at most $p$ has error less than $1/2$ on every anchor. Take $T,S$ from the finite-support lemma and $(\mu_x)$ from the measures lemma. With $S$ fixed, choose $n\ge2$ and then $G=G_m$ by the rare-walk lemma so that the bad fraction satisfies $\beta<1/(20V)$, then apply the exceptional-direction and edge-test lemmas.
For an incidence $e:x\to y$, since $f_e-b_x$ vanishes off $E_e$ and is bounded by $2L$ there, and $f_e+b_y$ vanishes off $E_{\bar e}$, the test prescription gives
$$\int f_e\,d(\mu_x-\mu_y)\ge\frac{3L}{2}-2L\big(|\mu_x|(E_e)+|\mu_y|(E_{\bar e})\big).$$
On the other hand, by the affine formula and $Lip(f_e)\le1$ with values in $[-L,L]$,
$$\int f_e\,d(\mu_x-\mu_y)=\big(T(\delta_x-\delta_y)\big)(f_e-f_e(o)),\qquad \Big|\int f_e\,d(\mu_x-\mu_y)\Big|\le2p\,d_p(x,y)\le2p,$$
because the difference of the two affine pairings is a single pairing of $\delta_x-\delta_y$, and the Lipschitz norm of $f_e-f_e(o)$ is at most one.
Choose a uniform vertex $x$ of $G$, then a uniform incidence $e\in D(x)$. At a good vertex each state lies in at most one exceptional set, so
$$\frac1\Delta\sum_{e\in D(x)}|\mu_x|(E_e)\le\frac1\Delta\sum_{j\in S}|\mu_x|(\{j\})\le\frac{V}{\Delta},$$
while at a bad vertex the average is at most $V$; hence
$$E\,|\mu_x|(E_e)\le\frac{V}{\Delta}+V\beta<\frac{1}{10}.$$
By $\Delta$-regularity this sampling is uniform over all oriented incidences, and reversal preserves it, so $E\,|\mu_y|(E_{\bar e})<1/10$ as well. Taking expectations in the lower bound and using the upper bound gives
$$2p>\frac{3L}{2}-\frac{4L}{10}=\frac{11L}{10}=11p,$$
a contradiction. The construction fixed $p$, all constants, the whole graph sequence, $M_p$, and $A_p$ before $T_0$; only after $T_0$ produced $S$ were $n$ and $m$ chosen, so the same finite anchor set defeats every operator in the class. QED

A bounded uniformly discrete space has a concrete coefficient model and a controlled BAP. The identification with $\ell_1$ appears in the work of Albiac, Ansorena, Cuth, and Doucha, and the coefficient bounds are stated by Cuth and Kalenda.
(Albiac, Ansorena, Cuth, Doucha, Israel J. Math. 240 (2020), 65-98.)
./albiac-ansorena-cuth-doucha-2020-1811.01265-lipschitz-free-p-spaces.pdf
(Cuth and Kalenda, Lipschitz-free spaces over uniformly discrete metric spaces are 3-Schur, arXiv 2026.)
./cuth-kalenda-2026-2604.01875-3-schur.pdf

The coefficient proposition. Let $(N,d,o)$ be a countable pointed metric space with $d(x,y)\ge1$ for distinct points and $\mathrm{diam}\,N\le B<\infty$. The map $J:\ell_1(N\setminus\{o\})\to F(N)$, $Jc=\sum_{x\ne o}c_x\delta_x$, is a linear isomorphism onto $F(N)$ with
$$\tfrac12\|c\|_1\le\|Jc\|\le B\|c\|_1.$$
If $N\ne\{o\}$ then $F(N)$ has $2B$-BAP; if $N=\{o\}$ both spaces are zero.
Proof. For finitely supported $c$, $\|\delta_x\|\le B$ gives the upper bound. For the lower bound let $f_c:N\to R$ equal $\mathrm{sign}(c_x)/2$ on the support off $o$ and $0$ elsewhere: it vanishes at $o$, its values differ by at most one, so by separation it is $1$-Lipschitz, and testing gives $\|Jc\|\ge(Jc)(f_c)=\tfrac12\sum_{x\ne o}|c_x|$.
The upper bound makes $Jc$ converge in $F(N)$ for every $c\in\ell_1$; both bounds pass to the extension by approximation with finitely supported families. So $J$ is injective with inverse of norm at most two on its range. The range is closed since $\|c_m-c_j\|_1\le2\|Jc_m-Jc_j\|$; it contains every $\delta_x$, whose span is dense, hence it is all of $F(N)$.
For $N\ne\{o\}$ (so $B\ge1$), choose increasing finite $E_m$ exhausting $N\setminus\{o\}$ with coordinate projections $q_m$. Then $S_m=Jq_mJ^{-1}$ has finite rank and norm at most $2B$, and $S_mu\to u$ for all $u$ because $q_mc\to c$ in $\ell_1$. Convergence is uniform on compact sets: given compact $C$ and $\varepsilon>0$, take a finite $\varepsilon/(3(2B+1))$-net; for large $m$ the error is below $\varepsilon/3$ on the net, and since $\|S_m-\mathrm{id}\|\le2B+1$ it is below $\varepsilon$ on $C$. If $N$ is finite, eventually $q_m=\mathrm{id}$. The case $N=\{o\}$ is trivial. QED

The renorming corollary. For each integer $p\ge1$, real $\ell_1$ carries an equivalent norm $\|\cdot\|_p$ with $\tfrac12\|c\|_1\le\|c\|_p\le B_p\|c\|_1$ such that $(\ell_1,\|\cdot\|_p)$ has $2B_p$-BAP but fails $p$-BAP; for $p=1$ it fails MAP.
Proof. By the block theorem, $M_p\setminus\{o\}$ is countably infinite; enumerate it and let $J_p$ be the coefficient map of the coefficient proposition with $B=B_p$. Set $\|c\|_p=\|J_pc\|_{F(M_p)}$, so $J_p$ is a surjective isometry $(\ell_1,\|\cdot\|_p)\to F(M_p)$, and the coefficient bounds give the stated norm bounds. Conjugation by $J_p$ preserves norms, finite rank, and approximation errors, so it transfers $2B_p$-BAP to the new norm. If the new norm had $p$-BAP, conjugation would give $p$-BAP on $F(M_p)$; applying it with accuracy $\tfrac12$ to $\{\delta_a:a\in A_p\}$ contradicts the block theorem. For $p=1$ this is failure of MAP. QED

Take disjoint copies of the spaces $M_p$ and identify only their basepoints:
$$M=\{o\}\cup\bigsqcup_{p\ge1}\big(M_p\setminus\{o\}\big).$$
Keep $d_p$ within each block $M_p$, and for nonbasepoint points in distinct blocks set
$$d(x,y)=d_p(x,o)+d_q(y,o)\qquad(x\in M_p\setminus\{o\},\ y\in M_q\setminus\{o\},\ p\ne q).$$

The wedge lemma. $M$ is a countable uniformly discrete pointed metric space with $d(x,y)\ge1$ for distinct points; it is unbounded and nonproper; and for each $p$ there are pointed $1$-Lipschitz maps $i_p:M_p\to M$ and $r_p:M\to M_p$ with $r_p\circ i_p=\mathrm{id}$.
Proof. Symmetry and positivity are clear. For the triangle inequality, if the three points lie in one block that block's inequality applies; if the third point $z$ lies in another block, the detour through $o$ has length $d_p(x,o)+2d_q(z,o)+d_p(y,o)\ge d_p(x,y)$, and a third point equal to $o$ gives equality. So $d$ is a metric. Distinct points are at distance at least one: within a block by the block theorem, and across blocks because each nonbasepoint is at distance at least one from $o$.
Were $M$ of finite diameter $C_0$, every $M_p$ would have diameter at most $C_0$, so the coefficient proposition would give $2C_0$-BAP on every $F(M_p)$. For an integer $p\ge\max\{1,2C_0\}$, applying that property with accuracy $\tfrac12$ to $\{\delta_a:a\in A_p\}$ would produce a finite-rank operator of norm at most $2C_0\le p$ with error below $\tfrac12$ at every anchor, contradicting the block theorem. Hence $M$ is unbounded. The closed ball about $o$ of radius $B_1$ contains the countably infinite $1$-separated set $M_1$; if it were compact, its cover by balls of radius $1/3$ would have a finite subcover, each member holding at most one point of $M_1$, which is impossible. So $M$ is nonproper.
Let $i_p$ be the inclusion and let $r_p:M\to M_p$ fix $M_p$ and send every other block to $o$. Both are pointed and $1$-Lipschitz: inclusion preserves distances, and for the retraction the only case needing an inequality has $x\in M_p$ and $y\in M_q\setminus\{o\}$ with $q\ne p$, where $d_p(r_px,r_py)=d_p(x,o)\le d(x,y)$. QED

By the linearization lemma the maps $i_p,r_p$ give contractions $I_p=F(i_p):F(M_p)\to F(M)$ and $P_p=F(r_p):F(M)\to F(M_p)$. Since $r_p\circ i_p=\mathrm{id}$, one has $P_pI_p=\mathrm{id}$ on $F(M_p)$, and then for every $v\in F(M_p)$,
$$\|v\|=\|P_pI_pv\|\le\|I_pv\|\le\|v\|,$$
so $I_p$ is an isometry onto an isometrically complemented subspace.

Proof of the main theorem. Suppose $F(M)$ had $\lambda$-BAP for a finite $\lambda\ge1$. Choose an integer $p\ge\lambda$ and apply the property with accuracy $\tfrac12$ to the finite set $\{I_p\delta_a:a\in A_p\}$, getting $U$ with $\|U\|\le\lambda$. The compression $P_pUI_p$ has finite rank and norm at most $\lambda\le p$, and for every $a\in A_p$
$$\|P_pUI_p\delta_a-\delta_a\|\le\|UI_p\delta_a-I_p\delta_a\|<\tfrac12,$$
contradicting the block theorem. All blocks and anchors were fixed before $\lambda$ and $U$.
For AP, define $r_{\le N}:M\to M$ to keep the first $N$ blocks and send the rest to $o$; it is pointed and $1$-Lipschitz, so $R_N=F(r_{\le N})=\sum_{p\le N}I_pP_p$ is a contractive projection. On an evaluation $\delta_x$ the sum keeps $\delta_x$ if $x$ lies in the first $N$ blocks and sends it to $0$ otherwise, so $R_N$ fixes every finite evaluation combination once $N$ is large. Such combinations are dense and $\|\mathrm{id}-R_N\|\le2$, so $R_Nu\to u$ for every $u$; the finite-net argument of the coefficient proposition makes this uniform on compact sets.
Fix compact $C\subseteq F(M)$ and $\varepsilon>0$. Choose $N$ with $\sup_{u\in C}\|R_Nu-u\|<\varepsilon/2$. For $1\le p\le N$ the image $P_p(C)$ is compact, so by the coefficient proposition pick finite-rank $T_p:F(M_p)\to F(M_p)$ with $\|T_p\|\le2B_p$ and $\sup_{u\in C}\|T_pP_pu-P_pu\|<\varepsilon/(2N)$. Then $T=\sum_{p\le N}I_pT_pP_p$ has finite rank and bound $\sum_{p\le N}2B_p<\infty$, and for $u\in C$,
$$\|Tu-u\|\le\|R_Nu-u\|+\sum_{p\le N}\|T_pP_pu-P_pu\|<\varepsilon.$$
The combined bound may grow with $N$, which AP permits, so $F(M)$ has AP. With the uniform discreteness, unboundedness, and nonproperness of the wedge lemma, this proves the main theorem. QED

A nonlinear consequence follows. Kalton asked whether every separable Banach space is approximable: whether there is an equi-uniformly continuous sequence $\psi_n:X\to X$ with relatively compact ranges and $\psi_n(x)\to x$ pointwise. The space $F(M)$ above is not approximable, although it has AP. Its $1$-separated image $\{\delta_x\}$ extends to a maximal $1$-separated net $N_X\subseteq X$, so every point of $X$ lies within distance one of $N_X$. If $X$ were approximable, one Lipschitz bound $\Lambda\ge1$ would serve for maps $g:N_X\to N_X$ with relatively compact range, and separation makes such a range finite, with approximation within distance below one becoming exact fixing. The question and the net criterion are Kalton's.
(Kalton, Math. Ann. 354 (2012), 1247-1288.)
./kalton-2012-uniform-structure-of-banach-spaces.pdf
The nonlinear proposition. The space $X=F(M)$ of the main theorem is not approximable.
Proof. Fix a finite $F\subseteq M$ containing $o$ and choose $g$ fixing every $\delta_x$, $x\in F$. The restriction $x\mapsto g(\delta_x)$ is a pointed $\Lambda$-Lipschitz map $M\to X$, so by the linearization property of free spaces into a Banach space it extends to a linear $T_F:X\to X$ of norm at most $\Lambda$ with $T_F\delta_x=g(\delta_x)$ for all $x$; its range is finite, since the relatively compact range of $g$ lies in the $1$-separated set $N_X$ and is therefore finite. To approximate a compact set, approximate a finite net in it by finite evaluation combinations and put all their supports in $F$; then $T_F$ fixes those combinations, and $\|T_F-\mathrm{id}\|\le\Lambda+1$ controls the remaining errors. These $T_F$ would be finite-rank operators of uniformly bounded norm approximating on compacts, that is BAP, contradicting the main theorem. QED

The block rests on classical and modern ingredients. The finite metric-extension lower bounds that separate projections from general operators come from Brudnyi and Brudnyi, and the signed-versus-probability distinction in almost extension is Naor's; the comparison of nonbacktracking continuation lengths that underlies the parent-direction records appears in the work of Ben-David, Childs, Gilyen, Kretschmer, Podder, and Wang.
(Brudnyi and Brudnyi, Amer. J. Math. 129 (2007), 217-314.)
./brudnyi-brudnyi-2007-0404304-metric-linear-extensions.pdf
(Naor, Adv. Math. 384 (2021), 107761.)
./naor-2021-2009.11373-almost-extension.pdf
(Ben-David, Childs, Gilyen, Kretschmer, Podder, Wang, SIAM J. Comput. 53 (2024), FOCS20-368--415.)
./ben-david-childs-gilyen-kretschmer-podder-wang-2020-2006.12760-symmetries-graph-properties.pdf