Proof We prove the direction (i) $\Longrightarrow $ (ii) first. Let $\varepsilon \in \{-1,1\}$ and $a \in \mathbb {R}$ be fixed, let $\mu \in \mathcal {W}_1(\mathbb {R}^2,d_{\mathrm {{m}}})$ such that

$$ \begin{align*}\mathrm{supp}(\mu) \subset L=L_{\varepsilon, a}=\left\{ (x_1, x_2) \in \mathbb{R}^2 \, \middle| \, x_2= \varepsilon x_1 +a \right\}. \end{align*} $$

Let us construct the following map, which we will call the allocation of directions in the sequel:

(3.3) $$ \begin{align} e: \mathbb{R}^2 \rightarrow \mathbb{R}^2; \, (y_1,y_2) \mapsto e((y_1,y_2)):= \begin{cases} (-\varepsilon, 1) & \text{if } y_2 \geq \varepsilon y_1+a, \\ (\varepsilon, -1) & \text{if } y_2 < \varepsilon y_1+a. \end{cases} \end{align} $$

See Figure 5 for an illustration the map given above.

Figure 5:

The allocation of directions in $\mathbb {R}^2$ according to (3.3).

The above allocation of directions has the crucial property that for all ${x=(x_1,x_2) \in L,}$ for all $y=(y_1,y_2) \in \mathbb {R}^2,$ and for all $t \geq 0$ we have

(3.4) $$ \begin{align} d_{\mathrm{{m}}}(x,y+t e(y))=d_{\mathrm{{m}}}(x,y)+d_{\mathrm{{m}}}(y, y+t e(y))=d_{\mathrm{{m}}}(x,y)+t. \end{align} $$

Let us justify (3.4) only in the sub-case $\varepsilon =1$ and $y_2 \geq \varepsilon y_1 +a$ as the other three sub-cases are very similar. We know that $x_2=x_1+a$ and $y_2 \geq y_1+a$ which implies that $y_2-x_2 \geq y_1 - x_1,$ or equivalently, $x_1-y_1 \geq x_2 - y_2.$ Therefore,

(3.5) $$ \begin{align} \begin{aligned} d_{\mathrm{{m}}}(x,y)&=d_{\mathrm{{m}}}((x_1,x_2),(y_1,y_2)) \\ &=\max\{x_1-y_1, y_1-x_1,x_2-y_2,y_2-x_2\}\\ &=\max\{x_1-y_1,y_2-x_2\}. \end{aligned} \end{align} $$

Moreover,

(3.6) $$ \begin{align} \begin{aligned} d_{\mathrm{{m}}}(x, y+t e(y)) &=d_{\mathrm{{m}}}((x_1,x_2), (y_1,y_2)+t(-1,1)) \\ &=d_{\mathrm{{m}}}((x_1,x_2), (y_1-t,y_2+t)) \\ &=\max\{x_1-y_1+t, y_1-x_1-t, y_2-x_2+t, x_2-y_2-t\} \\ &=\max\{x_1-y_1+t, y_2-x_2+t\} \\ &=\max\{x_1-y_1, y_2-x_2\}+t. \end{aligned} \end{align} $$

That is, (3.5) and (3.6) show that $d_{\mathrm {{m}}}(x, y+t e(y))=d_{\mathrm {{m}}}(x,y)+t$ indeed, and it is clear by the definition (3.3) that $d_{\mathrm {{m}}}(y, y+ t e(y))=t$ for every nonnegative $t.$

It is a straightforward consequence of the definition of $e(y)$ —see eq. (3.3)—that for every $t \geq 0$ the map $y \mapsto y+t e(y)$ is an injection of $\mathbb {R}^2$ and hence invertible on its range.

Let $\nu \in \mathcal {W}_1(\mathbb {R}^2,d_{\mathrm {{m}}})$ and let $t_0:=d_{W_1}(\mu , \nu ).$ Let us define

(3.7) $$ \begin{align} \eta_t:=\left(y \mapsto y + t e(y) \right)_{\#} \nu \end{align} $$

for $t \geq 0.$ As the map $y \mapsto y+t e(y)$ is invertible, the couplings of $\mu $ and $\nu $ are in a one-by-one correspondence with the couplings of $\mu $ and $\eta _t$ (for every $t\geq 0$ ), and this correspondence is given by

(3.8) $$ \begin{align} \pi_{(\mu, \eta_{t})}=\left(\mathrm{id}_{\mathbb{R}^2} \times \left(y \mapsto y+ t e(y) \right)\right)_{\#}\pi_{(\mu, \nu)} \qquad (\pi_{(\mu, \nu)} \in C(\mu,\nu), \, \pi_{(\mu, \eta_{t})} \in C(\mu, \eta_t)). \end{align} $$

Therefore,

(3.9) $$ \begin{align} \begin{aligned} d_{W_1}(\mu,\eta_t)&=\inf \left\{ \iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,z) ~\mathrm{d} \pi_{(\mu, \eta_{t})} (x,z) \, \middle| \, \pi_{(\mu, \eta_{t})} \in C(\mu, \eta_t) \right\} \\ &=\inf \left\{ \iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,y+t e(y)) ~\mathrm{d} \pi_{(\mu, \nu)} (x,y) \, \middle| \, \pi_{(\mu, \nu)} \in C(\mu, \nu) \right\} \\ &=\inf \left\{ \iint_{\mathbb{R}^2 \times \mathbb{R}^2} (d_{\mathrm{{m}}}(x,y)+t) ~\mathrm{d} \pi_{(\mu, \nu)} (x,y) \, \middle| \, \pi_{(\mu, \nu)} \in C(\mu, \nu) \right\}\\ &=\inf \left\{ \iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,y) ~\mathrm{d} \pi_{(\mu, \nu)} (x,y) \, \middle| \, \pi_{(\mu, \nu)} \in C(\mu, \nu) \right\}+t \\ &=d_{W_1}(\mu, \nu)+t=t_0+t \end{aligned} \end{align} $$

for every $t \geq 0.$

Note, that in the above computation, we heavily relied on the identity (3.4). The reversed triangle inequality implies that

$$ \begin{align*}d_{W_1}(\nu, \eta_t) \geq \left| d_{W_1}(\mu, \eta_t)- d_{W_1}(\mu, \nu)\right|=(t_0+t)-t_0=t. \end{align*} $$

On the other hand, the cost of the coupling $\left (y \mapsto (y, y+ t e(y))\right )_{\#} \nu \in C(\nu , \eta _t)$ is simply $t,$ and hence $d_{W_1}(\nu , \eta _t)=t.$ Therefore, with the particular choice $t:=t_0=d_{W_1}(\mu ,\nu )$ the triple $(\mu , \nu , \eta _{t_0})$ satisfies the requirement

(3.10) $$ \begin{align} d_{W_1}(\mu, \nu)=d_{W_1}(\nu, \eta_{t_0})=\frac{1}{2}d_{W_1}(\mu,\eta_{t_0}) \end{align} $$

as every expression in (3.10) is equal to $t_0.$

We turn to the proof of the direction (ii) $\Longrightarrow $ (i). The assumption (ii) implies in particular that for every $y \in \mathbb {R}^2$ there exists an $\eta \in \mathcal {W}_1(\mathbb {R}^2,d_{\mathrm {{m}}})$ such that

(3.11) $$ \begin{align} d_{W_1}(\mu, \delta_y)=d_{W_1}(\delta_y, \eta)=\frac{1}{2}d_{W_1}(\mu, \eta). \end{align} $$

Note that

$$ \begin{align*}d_{W_1}(\mu, \delta_y)=\int\limits_{\mathbb{R}^2} d_{\mathrm{{m}}}(x,y) ~\mathrm{d} \mu(x) \text{ and } d_{W_1}(\delta_y,\eta)=\int\limits_{\mathbb{R}^2} d_{\mathrm{{m}}}(y,z) ~\mathrm{d} \eta(z). \end{align*} $$

Moreover, let $\pi _{(\mu ,\eta )}^{*}$ denote an optimal coupling of $\mu $ and $\eta ,$ and let us note that we have the following chain of inequalities:

(3.12) $$ \begin{align} \begin{aligned} d_{W_1}(\mu, \eta)&=\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,z) ~\mathrm{d} \pi_{(\mu,\eta)}^{*}(x,z) \\ &\leq \iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,z) ~\mathrm{d} (\mu \otimes \eta)(x,z) \\ &\leq \iint_{\mathbb{R}^2 \times \mathbb{R}^2} (d_{\mathrm{{m}}}(x,y)+d_{\mathrm{{m}}}(y,z)) ~\mathrm{d} \mu(x) ~\mathrm{d} \eta(z)\\ &=d_{W_1}(\mu, \delta_y)+ d_{W_1}(\delta_y,\eta). \end{aligned} \end{align} $$

Therefore, (3.11) implies that both inequalities of (3.12) are saturated. The saturation of the first inequality means that $\mu \otimes \eta $ is an optimal coupling of $\mu $ and $\nu $ with respect to the transport cost $c(x,y)=d_{\mathrm {{m}}}(x,y),$ while the saturation of the second inequality means that

(3.13) $$ \begin{align} d_{\mathrm{{m}}}(x,z)=d_{\mathrm{{m}}}(x,y)+d_{\mathrm{{m}}}(y,z) \text{ for } \mu \otimes \eta \text{-almost every } (x,z) \in \mathbb{R}^2 \times \mathbb{R}^2. \end{align} $$

In order to get a contradiction, assume that $\mu $ is not supported on a diagonal line, and let x and $x'$ be points of the support of $\mu $ that do not lie on a common diagonal line. Now let us choose y to be $y:=\frac {1}{2}(x+x').$ With this choice we get

(3.14) $$ \begin{align} d_{\mathrm{{m}}}(x,z) & < d_{\mathrm{{m}}}(x,y)+d_{\mathrm{{m}}}(y,z) \text{ or } d_{\mathrm{{m}}}(x',z) \nonumber \\ & < d_{\mathrm{{m}}}(x',y)+d_{\mathrm{{m}}}(y,z) \text{ for all } z \in \mathbb{R}^2 \setminus\{y\}. \end{align} $$

Indeed, it is easy to check—see also Figure 6—that if both triangle inequalities in (3.14) are saturated, then $z=y$ by necessity.

Figure 6:

Illustration for eq. (3.14).

Consequently, (3.13) forces $\eta $ to be $\eta =\delta _y.$ But then $d_{W_1}(\delta _y,\eta )=0,$ which contradicts to (3.11), because $d_{W_1}(\mu , \delta _y)>0$ as $\mu $ is not diagonally supported and hence not a Dirac. This contradiction completes the proof of the implication (ii) $\Longrightarrow $ (i).

Proof Let us start with the proof of the direction (i) $\Longrightarrow $ (ii). Assume that $\mu _1$ and $\mu _2$ are supported on the diagonal line $L=\left \{(x_1, x_2) \in \mathbb {R}^2 \, \middle | \, x_2= \varepsilon x_1 +a\right \}.$ Let us recall the allocation of directions (3.3) and its crucial property (3.4). Let $\eta $ be defined by

(3.16) $$ \begin{align} \eta:=\left(y \mapsto y + e(y) \right)_{\#} \nu. \end{align} $$

Note that (3.16) is a special case of (3.7) with $t=1,$ and hence $d_{W_1}(\mu _i, \eta )=d_{W_1}(\mu _i, \nu )+1$ for $i=1,2.$

Similarly as in the previous proposition, the reverse triangle inequality ensures that $d_{W_1}(\nu , \eta ) \geq \left |d_{W_1}(\mu _i,\eta )-d_{W_1}(\mu _i,\nu )\right |=1,$ and the transport map $(y \mapsto (y,y+e(y)))_{\#}\nu $ between $\nu $ and $\eta $ shows that $d_{W_1}(\nu , \eta )=1$ which completes the proof of this direction.

To prove the direction (ii) $\Longrightarrow $ (i), note that by the previous statement, both of the measures $\mu _1$ and $\mu _2$ are supported on some diagonal line. Assume by contradiction that $\mu _1$ and $\mu _2$ are not supported on the same diagonal line, and hence in particular there exist points $x_1 \in \mathrm {supp}(\mu _1)$ and $x_2 \in \mathrm {supp}(\mu _2)$ that do not lie on a common diagonal line. As in the proof of Proposition 3.1 let us choose $\nu :=\delta _y$ where ${y=\frac {1}{2}(x_1+x_2).}$ With this choice, (3.15) implies that

$$ \begin{align*}d_{\mathrm{{m}}}(x,z) & =d_{\mathrm{{m}}}(x,y) \\ &\quad +d_{\mathrm{{m}}}(y,z) \text{ for } \mu_i \otimes \eta \text{-almost every } (x,z) \in \mathbb{R}^2 \times \mathbb{R}^2 \qquad (i=1,2). \end{align*} $$

In particular, $d_{\mathrm {{m}}}(x_i,z)=d_{\mathrm {{m}}}(x_i,y)+d_{\mathrm {{m}}}(y,z)$ for $i=1,2$ which implies $z=y$ for $\eta $ -almost every $z,$ and hence forces $\eta $ to be $\delta _y.$ However, this contradicts the requirement $d_{W_1}(\nu ,\eta )=1,$ so we got the desired contradiction.

Proof Let $\Phi : \mathcal {W}_1(\mathbb {R}^2,d_{\mathrm {{m}}}) \to \mathcal {W}_1(\mathbb {R}^2,d_{\mathrm {{m}}})$ be an isometry, and let $\mu ,\mu ' \in \mathcal {W}_1(L_{+})$ be two measures, $\mu \neq \mu '$ . According to Proposition 3.2, their images $\Phi (\mu )$ and $\Phi (\mu ')$ are supported on a diagonal line $L_{\varepsilon ,a}$ for a suitable $\varepsilon \in \{-1,1\}$ and $a\in \mathbb {R}$ . Since for every $\varepsilon \in \{-1,1\}$ and $a\in \mathbb {R}$ there is an isometry $T: \mathbb {R}^2 \to \mathbb {R}^2 $ that maps $L_{\varepsilon , a} $ onto $L_+$ , we can assume that $\mathrm {supp}\big (\Phi (\mu )\big )\subseteq L_+$ and $\mathrm {supp}\big (\Phi (\mu ')\big )\subseteq L_+$ . In fact, for every $\nu $ with $\mathrm {supp}(\nu )\subseteq L_+$ we conclude that $\mathrm {supp}\big (\Phi (\nu )\big )\subseteq L_+$ . Indeed, let us repeat the above argument for $\mu $ and $\nu $ . Since they are both supported on $L_+$ their images are supported on the same diagonal line. We already know that $\mathrm {supp}\big (\Phi (\mu )\big )\subseteq L_+$ , and therefore if $\Phi (\mu )$ is not a Dirac mass, then Proposition 3.2 guarantees that $\mathrm {supp}\big (\Phi (\nu )\big )\subseteq L_+$ . If $\Phi (\mu )$ is a Dirac mass, say $\Phi (\mu )=\delta _{(x,x)}$ then we have to exclude the possibility of $\mathrm {supp}\big (\Phi (\nu )\big )\subseteq L_{-1,2x}$ . To this aim, consider $\nu $ and $\mu '$ , and again, apply Proposition 3.2 to conclude that $\mathrm {supp}\big (\Phi (\mu ')\big )\subseteq L_{-1,2x}$ . But this is a contradiction, as we already know that $\mathrm {supp}\big (\Phi (\mu ')\big )\subseteq L_+$ and therefore $\mathrm {supp}\big (\Phi (\mu ')\big )=\{(x,x)\}$ , or equivalently $\Phi (\mu ')=\delta _{(x,x)}=\Phi (\mu )$ .

We obtain in this way, that $\Phi $ restricted to $\mathcal {W}_1(L_{+})$ is a (bijective) isometry of $\mathcal {W}_1(L_{+}),$ which is isomorphic to $W_1(\mathbb {R}, |\cdot |)$ which is known to be isometrically rigid — see [Reference Gy, Gehér and Titkos10]. Therefore measures in $\mathcal {W}_1(L_{+})$ are left invariant by $\Phi .$

Finally we need to show that measures in $\mathcal {W}_1(L_{-})$ are also left invariant by $\Phi .$ To see this note that $(0,0) \in L_{+} \cap L_{-}$ and considering the measure $\mu = \delta _{(0,0)}$ together with another measure $\nu $ supported on $L_{-}$ we conclude by applying Proposition 3.2 that both $\Phi (\mu )$ and $\Phi (\nu )$ are supported on the same diagonal line. Since we know already that $\Phi (\mu ) = \mu = \delta _{(0,0)}$ we can conclude that the support of $\Phi (\nu )$ is in $L_+$ or it is in $L_-$ . Since the first option cannot hold as the pre-images of measures supported on $L_+$ are supported on $L_+$ by the previous paragraph, we are left with the second one. This shows that if $ \nu \in \mathcal {W}_1(L_{-})$ then so is $\Phi (\nu )$ . By possibly applying another isometry of $\mathbb {R}^2$ we obtain that $\Phi $ fixes the elements of $ \nu \in \mathcal {W}_1(L_{-})$ as well.

Proof Let us prove the direction (i’) $\Longrightarrow $ (ii’) first. Let $x \in \mathbb {R},$ let $\mu =\delta _x,$ and let us define the following dilation with center x on $\mathbb {R}^2$ :

(3.18) $$ \begin{align} D_x: \mathbb{R}^2 \to \mathbb{R}^2; \, y \mapsto D_x(y):=x+2(y-x). \end{align} $$

Now, for any $\nu \in \mathcal {W}_p(\mathbb {R}^2,d_{\mathrm {{m}}}),$ let us define the corresponding $\eta _{\nu }$ by

(3.19) $$ \begin{align} \eta_{\nu}:=\left(D_x\right)_{\#}\nu. \end{align} $$

It is clear that $d_{W_p}(\mu , \eta _{\nu })=2 d_{W_p}(\mu , \nu ).$ Indeed,

$$ \begin{align*} \begin{aligned} d_{W_p}(\delta_x, \eta_{\nu})&=\left(\int\limits_{\mathbb{R}^2}d_{\mathrm{{m}}}^p(x,z)~\mathrm{d}(D_x)_{\#}\nu (z)\right)^{\frac{1}{p}} \\ &=\left(\int\limits_{\mathbb{R}^2}d_{\mathrm{{m}}}^p(x,x+2(y-x))~\mathrm{d}\nu (y)\right)^{\frac{1}{p}} \\ &=\left(\int\limits_{\mathbb{R}^2}2^p d_{\mathrm{{m}}}^p(x,y))~\mathrm{d}\nu (y)\right)^{\frac{1}{p}} \\ &=2 d_{W_p}(\delta_x, \nu). \end{aligned} \end{align*} $$

Moreover, by the reversed triangle inequality, $d_{W_p}(\nu , \eta _{\nu }) \geq d_{W_p}(\mu , \eta _{\nu })-d_{W_p}(\mu , \nu )=d_{W_p}(\mu , \nu ),$ while the obvious coupling of $\nu $ and $\eta _{\nu }$ given by the dilation $D_x$ guarantees that $d_{W_p}(\nu , \eta _{\nu }) \leq d_{W_p}(\mu , \nu ),$ and hence the direction (i’) $\Longrightarrow $ (ii’) is proved.

To prove the other direction (ii’) $\Longrightarrow $ (i’), let $\nu $ be a Dirac mass, $\nu =\delta _y,$ and let $\eta $ be as in condition (ii’). Then

$$ \begin{align*} \begin{aligned} &d_{W_p}(\mu, \eta)=\left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}^p(x,z) ~\mathrm{d} \pi_{(\mu,\eta)}^{*}(x,z)\right)^{\frac{1}{p}} \\ &\leq \left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}^p(x,z) ~\mathrm{d} (\mu \otimes \eta)(x,z)\right)^{\frac{1}{p}} \\ &\leq \left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} (d_{\mathrm{{m}}}(x,y)+d_{\mathrm{{m}}}(y,z))^p ~\mathrm{d} (\mu \otimes \eta)(x,z)\right)^{\frac{1}{p}} \\ &\leq \left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,y)^p ~\mathrm{d} (\mu \otimes \eta)(x,z)\right)^{\frac{1}{p}}+ \left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}^p(y,z) ~\mathrm{d} (\mu \otimes \eta)(x,z)\right)^{\frac{1}{p}}\\ &= \left(\int\limits_{\mathbb{R}^2} d_{\mathrm{{m}}}(x,y)^p ~\mathrm{d} \mu(x)\right)^{\frac{1}{p}}+ \left(\int\limits_{\mathbb{R}^2} d_{\mathrm{{m}}}^p(y,z) ~\mathrm{d} \eta(z)\right)^{\frac{1}{p}} \\ &=\left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}(x,z)^p ~\mathrm{d} (\mu \otimes \delta_y)(x,z)\right)^{\frac{1}{p}}+ \left(\iint_{\mathbb{R}^2 \times \mathbb{R}^2} d_{\mathrm{{m}}}^p(x,z) ~\mathrm{d} (\delta_y \otimes \eta)(x,z)\right)^{\frac{1}{p}} \\ &=d_{W_p}(\mu,\nu)+d_{W_p}(\nu,\eta). \end{aligned} \end{align*} $$

Since we assumed that $d_{W_p}(\mu , \nu )=d_{W_p}(\nu , \eta )=\frac {1}{2}d_{W_p}(\mu , \eta )$ , all the inequalities in the above chain are saturated. In particular, by the saturation of the $L_p$ -Minkowski inequality for $p>1$ (strictly convex norm), we get that there is a nonnegative constant $\alpha \geq 0$ such that

(3.20) $$ \begin{align} d_{\mathrm{{m}}}(y,z)=\alpha d_{\mathrm{{m}}}(x,y) \qquad \text{ for } \mu\text{-a.e. } x \in \mathbb{R}^2 \text{ and for } \eta\text{-a.e. } z \in \mathbb{R}^2. \end{align} $$

If $\mu $ is not a Dirac mass, then let $x_1$ and $x_2$ be two different points in its support, and let $y:=\frac 23 x_1 +\frac 13 x_2.$ Then the left-hand side of (3.20) is independent of $x,$ while the right-hand side is not—a contradiction.

Proof To prove the statement let us note that Corollary 7.22 in Villani’s book [Reference Villani17] says that if $p>1$ , and the underlying metric space is a complete, separable, and locally compact length space, then constant-speed geodesics connecting two measures are all displacement interpolations, i.e., geodesics are always constructed from optimal transport plans.

Therefore (see [Reference Villani17, Corollary 7.23]), if we want to guarantee that there is only one geodesic between two measures $\mu ,\nu $ , we need two properties:

• • we need a unique optimal transport plan $\widetilde {\pi }$ ,

• • and for $\widetilde {\pi }$ -almost every $(x,y)$ , x and y must be joined by a unique geodesic.

Let us note that the first property is automatically satisfied since one of the masses that we consider is a Dirac mass. Furthermore, note that in $(\mathbb {R}^2,d_{\mathrm {{m}}})$ , the second property means exactly that x and y are diagonally aligned, i.e., both points lie on the same diagonal line. For a fixed $x\in \mathbb {R}^2$ let us denote by $D_x$ the set of those points that are diagonally aligned with x. Of course, $D_x$ is the union of the two diagonal lines $L_++x$ and $L_{-}+x$ passing through x concluding the proof.

Proof Let $\Phi $ be an isometry. Since Proposition 3.4 is a metric characterization of Dirac masses, we know that $\Phi $ maps the set of Dirac masses onto itself. That is, there exists a bijection $T:\mathbb {R}^2\to \mathbb {R}^2$ such that $\Phi (\delta _x)=\delta _{T(x)}$ . In fact, T is an isometry, as $d_{W_p}(\delta _x,\delta _y)=d_{\mathrm {{m}}}(x,y)$ for all $x,y\in \mathbb {R}^2$ . Without loss of generality, we can assume that $T(x)=x$ , and thus $\Phi (\delta _x)=\delta _x$ for all $x\in \mathbb {R}^2$ . Next, consider the diagonal line $L_+$ . (The case of $L_-$ is completely analogous.) Fix an arbitrary $x\in L_+$ , and observe that according to Proposition 3.5, for any $\mu $ such that $\mathrm {supp}(\mu )\subseteq L$ , there is only one geodesic connecting $\delta _x$ and $\mu $ . Therefore, there must be only one geodesic between $\Phi (\delta _x)=\delta _x$ and $\Phi (\mu )$ . Again, according to Proposition 3.5 this means that $\mathrm {supp}\big (\Phi (\mu )\big )\subseteq D_x$ . Now choose a $y\in L_+$ ( $y\neq x$ ) and repeat the argument. The conclusion is that $\mathrm {supp}\big (\Phi (\mu )\big )\subseteq D_y$ . But $D_x\cap D_y=L_+$ , and therefore $\mathrm {supp}\big (\Phi (\mu )\big )\subseteq L_+$ . Now we know that $\Phi $ sends measures supported on $L_+$ into measures supported on $L_+$ . Since $L_+$ endowed with the restriction of $d_{\mathrm {{m}}}:\mathbb {R}^2\times \mathbb {R}^2\to \mathbb {R}_+$ onto $L_+\times L_+$ is nothing else but $(\mathbb {R},|\cdot |)$ , the set

$$ \begin{align*}\mathcal{W}_p(L_+,d_{\mathrm{{m}}}):=\{\mu\in\mathcal{W}_p(\mathbb{R}^2,d_{\mathrm{{m}}})\,|\,\mathrm{supp}\mu\subseteq L_+\}\end{align*} $$

endowed with the Wasserstein distance is isometrically isomorphic to the Wasserstein space $\mathcal {W}_p(\mathbb {R},d_{|\cdot |})$ investigated in [Reference Gy, Gehér and Titkos10, Reference Kloeckner13], which is isometrically rigid if $p\neq 2$ . Since we assumed that $\Phi (\delta _x)=\delta _x$ , isometric rigidity forces the restriction $\Phi |_{\mathcal {W}_p(L_+,d_{\mathrm {{m}}})}$ to be the identity, i.e., $\Phi (\mu )=\mu $ for all $\mu $ supported on $L_+$ . The same argument with $L_-$ completes the proof in the $p\neq 2$ case.

If $p=2$ we need to use a more involved argument, since $\mathcal {W}_2(\mathbb {R},d_{|\cdot |})$ is not isometrically rigid. In fact, if $\Psi :\mathcal {W}_2(\mathbb {R},d_{|\cdot |})\to \mathcal {W}_2(\mathbb {R},d_{|\cdot |})$ is an isometry, then $\Psi (\delta _x)=\delta _x$ for all $x\in \mathbb {R}$ itself does not imply $\Psi (\mu )=\mu $ for all $\mu \in \mathcal {W}_2(\mathbb {R},d_{|\cdot |})$ , as $\mathcal {W}_2(\mathbb {R},d_{|\cdot |})$ admits exotic isometries and a non-trivial shape-preserving isometry.

Therefore, even if we know that $\Phi (\delta _{(x,x)})=\delta _{(x,x)}$ for all $(x,x)\in L_+$ we cannot guarantee yet that $\Phi (\mu )=\mu $ for all $\mu $ with $\mathrm {supp}(\mu )\subseteq L_+$ . We have to rule out that the restriction of $\Phi $ onto $\mathcal {W}_p(L_+,d_{\mathrm {{m}}})$ does not act like a non-trivial isometry. This boils down to investigating the action on measures whose support consists of two points of $L_+$ as follows: using the isometric identification $t\mapsto (t,t)$ between the real line and $L_+$ , we will use Kloeckner’s result which tells us how the image of a $\mu =\alpha \delta _{(x,x)}+(1-\alpha )\delta _{(y,y)}$ would look like if $\Phi $ would act on $L_+$ like non-trivial isometry. Then we will choose a special $\mu $ and a special Dirac measure $\delta _{(u,v)}$ (not supported on $L_+$ ) with the property that $d_{W_2}(\mu ,\delta _{(u,v)})\neq d_{W_2}(\Phi (\mu ),\delta _{(u,v)})=d_{W_2}(\Phi (\mu ),\Phi (\delta _{(u,v)}))$ , a contradiction.

Let us introduce some notations. For the diagonal line $L_+$ the set of measures supported on two points of $L_+$ will be denoted by $\Delta _2$

(3.21) $$ \begin{align} \Delta_2=\{\alpha\delta_{(x,x)}+(1-\alpha)\delta_{(y,y)}\,|\, \alpha\in(0,1),\, x,y\in \mathbb{R}\}. \end{align} $$

Following the notations in Kloeckner’s paper [Reference Kloeckner13], elements of $\Delta _2$ will be parametrized by three parameters $m\in \mathbb {R}$ , $\sigma \geq 0$ , and $r\in \mathbb {R}$ as follows:

(3.22) $$ \begin{align} \mu(m,\sigma,r)=\frac{e^{-r}}{e^r+e^{-r}}\delta_{(m-\sigma e^r,m-\sigma e^r)}+\frac{e^{r}}{e^r+e^{-r}}\delta_{(m+\sigma e^{-r},m+\sigma e^{-r})}. \end{align} $$

According to [Reference Kloeckner13, Lemma 5.2], if an isometry $\Phi $ fixes all Dirac masses, then its action on $\Delta _2$ is

$$ \begin{align*}\Phi\big(\mu(m,\sigma,r)\big):=\mu(m,\sigma,\varphi(r)),\end{align*} $$

where $\varphi :\mathbb {R}\to \mathbb {R}$ is an isometry. In other words, $\Phi $ is equal to the shape-preserving isometry

(3.23) $$ \begin{align} \Phi^*:\mathcal{W}_2(L_+)\to\mathcal{W}_2(L_+),\quad\Phi^*\big(\mu(m,\sigma,r)\big):=\mu(m,\sigma,-r), \end{align} $$

or $\Phi $ is equal to an exotic isometry

(3.24) $$ \begin{align} \Phi_t:\mathcal{W}_2(L_+)\to\mathcal{W}_2(L_+),\quad\Phi^t\big(\mu(m,\sigma,r)\big):=\mu(m,\sigma,r+t) \end{align} $$

for some $t\neq 0$ , or $\Phi $ is the composition $\Phi _t\circ \Phi ^*$ for some $t\neq 0$ . Note, that if $t=0$ , then $\Phi _t$ is the identity, so $\Phi _0$ is not an exotic isometry.

To handle the case $\Phi ^*$ , choose $\mu =\mu (0,1,\ln 2)=\frac 15\delta _{(-2,-2)}+\frac 45\delta _{(\frac 12,\frac 12)}$ . Then

$$ \begin{align*}\Phi^*(\mu)=\mu(0,1,-\ln2)=\frac45\delta_{(-\frac12,-\frac12)}+\frac15\delta_{(2,2)}.\end{align*} $$

Calculating the Wasserstein distance of $\mu $ and $\Phi (\mu )$ from $\delta _{(2,0)}$ , we obtain that

$$ \begin{align*}d_{W_2}\big(\delta_{(2,0)},\mu(0,1,\ln2)\big)=\sqrt{5}\quad\mbox{and}\quad d_{W_2}\big(\delta_{(2,0)},\mu(0,1,-\ln2)\big)=\sqrt{5+\frac45}.\end{align*} $$

So if $\Phi $ is an isometry, it cannot act on $L_+$ like $\Phi ^*$ .

To handle the case $\Phi _t$ for $t>0$ , choose $\mu =\mu (0,1,0)=\frac 12\delta _{(-1,-1)}+\frac 12\delta _{(1,1)}$ . Then $\Phi (\mu (0,1,0))=\mu (0,1,t)$ . Now fix the Dirac measure $\delta _{(-1,0)}$ and calculate $d_{W_2}(\delta _{(-1,0)},\mu (0,1,0)$ and $d_{W_2}(\delta _{(-1,0)},\mu (0,1,t)$ . Again, if $\Phi $ would act like $\Phi _t$ on $L_+$ , we should get the same result, since $\Phi $ is an isometry. The calculation shows that $d_{W_2}\big (\delta _{(-1,0)},\mu (0,1,0)\big )=\sqrt {\frac 52}$ and

$$ \begin{align*}d_{W_2}\big(\delta_{(-1,0)},\mu(0,1,t)\big)=\sqrt{2+\frac{2-e^{-t}}{e^t+e^{-t}}}.\end{align*} $$

These two numbers are equal if and only if $t=0$ or $t=\ln 3$ . We assumed that $t>0$ , so one single Dirac $\delta _{(-1,0)}$ excluded all $\Phi _t$ except $t=\ln 3$ . Choosing a different Dirac measure, say $\delta _{(-\frac 12,0)}$ , a simple calculation shows that

$$ \begin{align*}\sqrt{\frac{13}{8}}=d_{W_2}\big(\delta_{(-\frac12,0)},\mu(0,1,0)\big)=d_{W_2}\big(\delta_{(-\frac12,0)},\mu(0,1,\ln3)\big)=\sqrt{\frac{12}{8}+\frac{1}{40}},\end{align*} $$

a contradiction. Similar calculations for $\Phi _t$ with negative t and for $\Phi ^*\circ \Phi _t$ show that the only case when we don’t get a contradiction is when $\Phi $ acts as $\Phi _0$ , which is the identity.

Proof Let us note that if $x,y \in Q$ are any two points then $d_{\mathrm {{m}}}(x,y) \leq 2 $ with equality if and only if x and y lie on two opposite sides of Q.

This implies that if $\mu , \nu \in \mathcal {W}_p(Q,d_{\mathrm {{m}}})$ , then

(3.25) $$ \begin{align} d_{W_p}(\mu, \nu) \leq 2, \end{align} $$

with equality if and only if the supports $\mathrm {supp}(\mu )$ and $\mathrm {supp}(\nu ) $ are contained in two opposite sides of Q. To see this, note that inequality (3.25) follows immediately from the definition of the Wasserstein metric $d_{W_p}$ and the fact that $d_{\mathrm {{m}}}(x,y) \leq 2$ for all ${x,y\in Q}$ . Furthermore, if $\mathrm {supp}(\mu )$ and $\mathrm {supp}(\nu ) $ are contained in two opposite sides of Q, then $d_{\mathrm {{m}}}(x,y) = 2$ for all $x\in \mathrm {supp} (\mu )$ and $y\in \mathrm {supp}(\nu )$ .

Let $\pi _0 $ be an optimal coupling of $\mu $ and $\nu $ . Since $\mathrm {supp}(\pi _0) \subseteq \mathrm {supp} (\mu ) \times \mathrm {supp} (\nu )$ we have that $d_{\mathrm {{m}}}(x,y)= 2$ for any $(x,y) \in \mathrm {supp} (\pi _0)$ and therefore

$$ \begin{align*}d_{W_p}^p(\mu, \nu) = \int\limits_{Q\times Q} d_{\mathrm{{m}}}^p(x,y) ~\mathrm{d}\pi_0(x,y) = 2^p.\end{align*} $$

To show the converse, let us assume that $d_{W_p}^p(\mu , \nu ) = 2^p$ . Then for all couplings ${\pi \in C(\mu , \nu )}$ we have

$$ \begin{align*}2^p =d_{W_p}^p(\mu, \nu)= \int\limits_{Q\times Q} d_{\mathrm{{m}}}^p(x,y) ~\mathrm{d}\pi(x,y) \leq \int\limits_{Q\times Q} 2^p ~\mathrm{d}\pi(x,y) = 2^p,\end{align*} $$

thus $d_{\mathrm {{m}}}(x,y) =2 $ for $\pi $ almost all $(x,y)$ . This applies for $\pi = \mu \otimes \nu $ , and so $\mu $ and $\nu $ must be concentrated on the opposite sides of Q.

The statement of the lemma is now an immediate consequence of this claim. Indeed, assume that $\mathrm {supp}(\mu )$ and $\mathrm {supp}(\nu ) $ are contained in two opposite sides of Q. Then we have $d_{W_p}^p(\mu , \nu )=2^p$ . Since $\Phi : \mathcal {W}_p(Q,d_{\mathrm {{m}}}) \to \mathcal {W}_p(Q,d_{\mathrm {{m}}})$ is an isometry, we have $d_{W_p}^p(\Phi (\mu ), \Phi (\nu ))=2^p.$ But, then the supports of the two measures $\Phi (\mu )$ and $\Phi (\nu )$ must be contained in two opposite sides of Q.

Proof Let us denote by $S_1, S_3$ the two vertical and by $S_2, S_4$ the two horizontal sides of Q such that $S_1$ is the left vertical and $S_4$ is the top horizontal side.

We apply Lemma 3.7 to every pair of elements of $V,$ and we conclude that for every $j \in \{1,2,3,4\},$ the measure $\Phi (\delta _{x_j})$ is supported on a certain side $S_i$ of $Q.$

We claim that each $\Phi (\delta _{x_j})$ must be in fact supported on some vertex. To see this we argue by contradiction. Let us assume that for example $\Phi (\delta _{x_1})$ is supported on one of the sides, say $S_1$ but not on any of the vertices of $S_1$ .

It is clear by Lemma 3.7 that all other measures $\Phi (\delta _{x_i})$ for $i=2,3,4$ must be supported on the opposite side of $S_1$ , that is $S_3.$ But the mutual distance of any pair of these measures must be equal to $2$ which shows that any two of these three measures must be lying on opposite sides again, which is $S_2$ and $S_4$ . But there are only two possibilities for measures with support in $S_2\cap S_3$ and $S_4 \cap S_3$ , namely the two Dirac masses on the vertices of $S_3$ which gives a contradiction.

It is easy to see that there exists an isometry $\Psi : Q \to Q$ such that $\Phi \circ \Psi _{\#}$ fixes $\delta _{x_i}$ and $\delta _{x_{i+1}}$ for some $i \in \{1,2,3,4\},$ and therefore we can assume without loss of generality that $(\Phi \circ \Psi _{\#})(\delta _{x_i})=\delta _{x_i}$ for $i=1,2.$ This will imply that all measures supported on $S_4$ are fixed. We must show that $\Phi \circ \Psi _{\#} (\delta _{x_i}) = \delta _{x_i}$ for $i=3,4$ . Note, that the map

$$ \begin{align*}\Phi \circ \Psi_{\#}: \mathcal{W}_p(Q,d_{\mathrm{{m}}}) \to \mathcal{W}_p(Q,d_{\mathrm{{m}}})\end{align*} $$

is itself an isometry. Let us assume indirectly that $\Phi \circ \Psi _{\#} (\delta _{x_3}) = \delta _{x_4}$ (which implies that and $\Phi \circ \Psi _{\#} (\delta _{x_4}) = \delta _{x_3}$ ), and take a $\xi $ such that $\mathrm {supp}(\xi )\subseteq S_3$ , $\xi \notin \{\delta _{x_2},\delta _{x_3}\}$ . Since $d_{W_p}(\delta _{x_1},\xi )=2$ , we have

$$ \begin{align*}2=d_{W_p}(\delta_{x_1},\xi)=d_{W_p}(\Phi(\delta_{x_1}),\Phi(\xi))=d_{W_p}(\delta_{x_1},\Phi(\xi)), \end{align*} $$

which implies that $\mathrm {supp}(\Phi (\xi ))\subseteq S_3\cup S_4$ . Similarly, $d_{W_p}(\delta _{x_4},\xi )=2$ , and thus

$$ \begin{align*}2=d_{W_p}(\delta_{x_4},\xi)=d_{W_p}(\Phi(\delta_{x_4}),\Phi(\xi))=d_{W_p}(\delta_{x_3},\Phi(\xi)), \end{align*} $$

which implies that $\mathrm {supp}(\Phi (\xi ))\subseteq S_1\cup S_2$ . Combining $\mathrm {supp}(\Phi (\xi ))\subseteq S_1\cup S_2$ and $\mathrm {supp}(\Phi (\xi ))\subseteq S_3\cup S_4$ with $\xi \notin \{\delta _{x_2},\delta _{x_3}\}$ , we get that $\Phi (\xi )=\alpha ^{\ast }\delta _{x_2}+(1-\alpha ^{\ast }) \delta _{x_4}$ for some $\alpha ^{\ast }\in (0,1).$ (Recall that $\Phi (\delta _{x_2})=\delta _{x_2}$ , $\Phi (\delta _{x_3})=\delta _{x_4}$ , and $\Phi $ is injective.) If $p>1$ , then this is a contradiction. Indeed, choose $\xi :=\delta _y$ with $y\in S_3\setminus \{x_2,x_3\}$ , and observe that the triple $\delta _{x_2}$ , $\delta _y$ , and $\delta _{x_3}$ saturates the triangle inequality, but the triple $\Phi (\delta _{x_2})=\delta _{x_2}$ , $\Phi (\delta _y)=\alpha ^{\ast }\delta _{x_2}+(1-\alpha ^{\ast }) \delta _{x_4}$ and $\Phi (\delta _{x_3})=\delta _{x_4}$ does not, as $\sqrt [p]{2^p(1-\alpha ^{\ast })}+\sqrt [p]{2^p\alpha ^{\ast }}\neq 2$ . If $p=1$ , we need a different argument. The set $\mathcal {I}:=\{ \mu _{\alpha }:=\alpha \delta _{x_2}+(1-\alpha ) \delta _{x_4}\,|\,\alpha \in (0,1)\}$ is isometric to the set $\big ((0,1),2|\cdot |\big )$ , since ${d_{W_1}(\mu _{\alpha },\mu _{\beta })=|\alpha -\beta |}$ . Therefore for any three different elements $\mu _{\alpha _1}, \mu _{\alpha _3}$ , and $\mu _{\alpha _3}$ there exists a bijection $\sigma :\{1,2,3\}\to \{1,2,3\}$ such that $d_{W_1}(\mu _{\alpha _{\sigma (1)}},\mu _{\alpha _{\sigma (3)}})=(\mu _{\alpha _{\sigma (1)}},\mu _{\alpha _{\sigma (2)}})+(\mu _{\alpha _{\sigma (2)}},\mu _{\alpha _{\sigma (3)}})$ . Now choose a non-degenerate triangle $\xi _1,\xi _2,$ and $\xi _3$ supported on $S_3\setminus \{x_2,x_3\}$ in the sense that they do not saturate the triangle inequality in any order. The existence of such a triple is a contradiction, as an appropriate permutation of their image in $\mathcal {I}$ will saturate the triangle inequality.

Proof By Corollary 3.8, we can assume without loss of generality that $\Phi $ fixes the Dirac masses at the four corners of Q. It is enough to show that if $\mu \in \mathcal {W}_1(Q,d_{\mathrm {{m}}})$ supported on $L_+ $ , then $\Phi (\mu )=\mu $ as the case of $L_{-}$ is similar. We show first that if $\mathrm {supp}(\mu ) \subseteq L_+$ , then $\mathrm {supp}\big (\Phi (\mu )\big ) \subseteq L_+$ . This is based on the following observation: If $x \in Q $ then

(3.26) $$ \begin{align} d_{\mathrm{{m}}}((-1,-1), x ) + d_{\mathrm{{m}}}(x, (1,1))\geq d_{\mathrm{{m}}}((-1,-1), (1,1)) = 2 \end{align} $$

with equality if and only if $x \in L_+$ .

The above inequality follows simply by the triangle inequality applied for the metric $d_{\mathrm {{m}}}$ . The characterization of the equality case is slightly more tricky. It is based on the fact that the line segment $ L_+: t\to (t,t), t\in [-1,1] $ is the only geodesic with respect to the metric $d_{\mathrm {{m}}}$ connecting the endpoints $(-1,-1)$ and $(1,1)$ . This observation has the following consequence for measures: If $\mu \in \mathcal {W}_1(Q,d_{\mathrm {{m}}})$ , then

(3.27) $$ \begin{align} d_{W_1}(\delta_{(-1,-1)}, \mu) + d_{W_1}(\mu, \delta_{(1,1)}) \geq 2, \end{align} $$

with equality if and only if $\mathrm {supp}(\mu ) \subseteq L_+$ . To show inequality (3.27), we integrate inequality (3.26) with respect to $\mu $ . In this way, we obtain

(3.28) $$ \begin{align} \begin{aligned} & d_{W_1}(\delta_{(-1,-1)}, \mu) + d_{W_1}(\mu, \delta_{(1,1)})\\ &\quad = \int\limits_{Q}d_{\mathrm{{m}}}((-1,-1), x )~\mathrm{d}\mu(x) + \int\limits_{Q} d_{\mathrm{{m}}}(x, (1,1)) ~\mathrm{d}\mu(x) \\ &\quad \geq \int\limits_Q 2~\mathrm{d}\mu(x) = 2. \end{aligned} \end{align} $$

Let us assume that $\mathrm {supp}(\mu ) \subseteq L_+$ . Then equality holds true in (3.26) for every point $x\in \mathrm {supp} (\mu )$ and thus by integrating, we obtain that equality holds in (3.28) as well.

Conversely, let us assume, that

$$ \begin{align*}d_{W_1}( \delta_{(-1,-1)},\mu) + d_{W_1}(\mu, \delta_{(1,1)}) = 2\end{align*} $$

for some measure $\mu \in \mathcal {W}_1(Q,d_{\mathrm {{m}}})$ . We have to show that $\mathrm {supp}(\mu ) \subseteq L_+$ . We argue by contradiction: assume that the exists a point $x_0 \in \mathrm {supp}( \mu )$ that is not contained in $L_+$ . Then there exists a small radius $r>0$ and a small $\varepsilon>0$ with the property that $\delta = \mu (B(x_0, r)>0 $ and

$$ \begin{align*}d_{\mathrm{{m}}}((-1,-1), x) + d_{\mathrm{{m}}}(x,(1,1))>2 +\varepsilon , \ \textrm{for all} \ x \in B(x_0,r).\end{align*} $$

Using this relation, we obtain

$$ \begin{align*} &d_{W_1}( \delta_{(-1,-1)},\mu) + d_{W_1}(\mu, \delta_{(1,1)}) \\ &\quad = \int\limits_{\mathrm{supp}(\mu)}d_{\mathrm{{m}}}((-1,-1), x )\ ~\mathrm{d}\mu(x) + \int\limits_{\mathrm{supp}(\mu)} d_{\mathrm{{m}}}(x,(1,1)) \ ~\mathrm{d}\mu(x) \\ &\quad = \int\limits_{\mathrm{supp}(\mu)}[d_{\mathrm{{m}}}((-1,-1), x ) + d_{\mathrm{{m}}}(x, (1,1))] \ ~\mathrm{d}\mu(x) \\ &\quad = \int\limits_{\mathrm{supp}(\mu) \cap B(x_0,r)}[d_{\mathrm{{m}}}((-1,-1), x ) + d_{\mathrm{{m}}}(x,(1,1)) ]\ ~\mathrm{d}\mu(x) \\ & \qquad + \int\limits_{\mathrm{supp}(\mu) \setminus B(x_0,r)}[d_{\mathrm{{m}}}((-1,-1), x ) + d_{\mathrm{{m}}}(x,(1,1) ) ] \ ~\mathrm{d}\mu(x)\\ &\quad > (2+\varepsilon) \delta + 2(1-\delta)= 2+\varepsilon\delta >2, \end{align*} $$

which is a contradiction. Let us consider an isometry $\Phi : \mathcal {W}_1(Q,d_{\mathrm {{m}}}) \to \mathcal {W}_1(Q,d_{\mathrm {{m}}})$ that fixes the elements of V (i.e., the Dirac masses on the corners of Q). Then we have

$$ \begin{align*}d_{W_1}(\delta_{(-1,-1)},\Phi(\mu))+ d_{W_1}(\Phi(\mu),\delta_{(1,1)}) = d_{W_1}(\delta_{(-1,-1)},\mu))+ d_{W_1}(\mu,\delta_{(1,1)}).\end{align*} $$

Assuming that $\mathrm {supp}(\mu ) \subseteq L_+$ we obtain that by the above that

$$ \begin{align*}d_{W_1}(\delta_{(-1,-1)},\mu))+ d_{W_1}(\mu,\delta_{(1,1)})=2.\end{align*} $$

By the above equality we have, then

$$ \begin{align*}d_{W_1}(\delta_{(-1,-1)},\Phi(\mu))+ d_{W_1}(\Phi(\mu),\delta_{(1,1)})=2,\end{align*} $$

which implies in turn that $\mathrm {supp}\big (\Phi (\mu )\big ) \subseteq L_+$ .

As mentioned at the beginning of the proof, the same argument shows that if we have $\mathrm {supp}(\mu ) \subseteq L_-$ , then it follows that $\mathrm {supp}\big (\Phi (\mu )\big ) \subseteq L_-$ as well. Since $\mu _0=\delta _{(0,0)}$ has its support in $L_+ \cap L_-$ , and it is the unique measure with this property, we conclude, that $\Phi (\mu _0) = \mu _0$ . Now let us consider the Wasserstein space $\mathcal {W}_1(L_+,d_{\mathrm {{m}}})$ of measures supported on $L_+$ . By the above consideration, we have

$$ \begin{align*}\Phi:\mathcal{W}_1(L_+,d_{\mathrm{{m}}}) \to \mathcal{W}_1(L_+,d_{\mathrm{{m}}}),\end{align*} $$

moreover, we know that $\Phi (\mu _0) = \mu _0$ for the measure $\mu _0=\delta _{(0,0)}\in \mathcal {W}_1(L_+,d_{\mathrm {{m}}}) $ . We cannot apply the characterization of Wasserstein isometries on a line segment (see [Reference Gy, Gehér and Titkos10, Theorem 2.5]) which implies that $\Phi (\mu ) = \mu $ for all measures $\mu \in \mathcal {W}_1(L_+,d_{\mathrm {{m}}}) $ . In the same way we can also conclude, that $\Phi (\mu ) = \mu $ for all measures $\mu \in \mathcal {W}_1(L_-,d_{\mathrm {{m}}}) $ .

Proof Let $\Phi $ be an isometry of the Wasserstein space $\mathcal {W}_p(Q,d_{\mathrm {{m}}})$ . Since the Wasserstein space $\mathcal {W}_p([-1,1],d_{|\cdot |})$ is isometrically rigid, Corollary 3.8 and the remark after that, implies that we can assume without loss of generality that $\Phi $ leaves every measure supported either on the right vertical side $[(1,-1),(1,1)]$ or on the top horizontal side $[(-1,1),(1,1)]$ fixed. We are going to show that $\Phi $ leaves every measure supported on the line diagonal segment $[(-1,-1),(1,1)] \subset Q$ fixed. The case of the other diagonal is analogous.

In the first step, we consider measures supported on the upper half of the diagonal, i.e., on the line segment $[(0,0),(1,1)]$ . We prove first that these measures will be fixed. In the second step, we proceed to consider general measures supported on the full segment $[(-1,-1),(1,1)]$ .

Let $\mu $ be a measure supported on the upper half of the diagonal of Q (that is, on $[(0,0),(1,1)]$ ), and consider the projections $p_r: \, (t,t) \mapsto (1, 2t-1)$ and $p_u: (t,t) \mapsto (2t-1,1)$ that map $[(0,0),(1,1)]$ onto the right vertical side $[(1,-1),(1,1)]$ and the top horizontal side $[(-1,1),(1,1)],$ respectively (Figure 7). Let us introduce the notation $\mu _r:=(p_r)_{\#}(\mu )\quad \mbox {and}$ and $\mu _u:=(p_u)_{\#}(\mu )$ .

Figure 7:

Illustration for the definition of $p_u$ , $p_r$ , $\mu _u$ , and $\mu _r$ .

Our goal is to show that $\mu $ is the unique minimizer of the functional

(3.29) $$ \begin{align} \mathcal{W}_p(Q,d_{\mathrm{{m}}}) \ni \nu \mapsto d_{W_p}^p(\mu_r, \nu)+d_{W_p}^p(\nu, \mu_u). \end{align} $$

Once this is proved, we are done, as in this case the equation

(3.30) $$ \begin{align} \begin{aligned} d_{W_p}^p(\mu_r, \Phi(\mu))+d_{W_p}^p(\Phi(\mu), \mu_u)&=d_{W_p}^p(\Phi(\mu_r), \Phi(\mu))+d_{W_p}^p(\Phi(\mu), \Phi(\mu_u)) \\ &=d_{W_p}^p(\mu_r,\mu)+d_{W_p}^p(\mu, \mu_u) \\ &=\min_{\nu \in \mathcal{W}_p(Q,d_{\mathrm{{m}}})}(d_{W_p}^p(\mu_r, \nu)+d_{W_p}^p(\nu, \mu_u)) \end{aligned} \end{align} $$

forces $\Phi (\mu )$ to be $\mu .$

Let $\nu \in \mathcal {W}_p(Q,d_{\mathrm {{m}}})$ be arbitrary, let $\pi _{(\mu _r,\nu )}^*$ be an optimal transport plan between $\mu _r$ and $\nu $ with respect to the cost $c(x,y)=d_{\mathrm {{m}}}^p(x,y)$ , let $\pi _{(\nu ,\mu _u)}^*$ be an optimal transport plan between $\nu $ and $\mu _u$ with respect to the same cost, and let $\pi _{(\mu _r,\nu ,\mu _u)}^* \in \mathrm {Prob}(Q^3)$ be the gluing of them—see the “gluing lemma” [Reference Villani16, Lemma 7.6] for further details of this construction. Moreover, set $\pi _{(\mu _r,\mu _u)}^*:=\left ( \pi _{(\mu _r,\nu ,\mu _u)}^* \right )_{13}.$ Now

(3.31) $$ \begin{align} \begin{aligned} & d_{W_p}^p(\mu_r, \nu)+d_{W_p}^p(\nu, \mu_u) \\ & \quad =\iint_{Q^2}d_{\mathrm{{m}}}^p(x,y) ~\mathrm{d} \pi_{(\mu_r,\nu)}^*(x,y)+\iint_{Q^2}d_{\mathrm{{m}}}^p(y,z) ~\mathrm{d} \pi_{(\nu,\mu_u)}^*(y,z) \\ &\quad =\iiint_{Q^3} (d_{\mathrm{{m}}}^p(x,y)+d_{\mathrm{{m}}}^p(y,z)) ~\mathrm{d} \pi_{(\mu_r,\nu,\mu_u)}^*(x,y,z) \\ &\quad \geq \iiint_{Q^3} \min_{y \in Q}\{d_{\mathrm{{m}}}^p(x,y)+d_{\mathrm{{m}}}^p(y,z)\} ~\mathrm{d} \pi_{(\mu_r,\nu,\mu_u)}^*(x,y,z) \\ &\quad = \iint_{Q^2} \min_{y \in Q}\{d_{\mathrm{{m}}}^p(x,y)+d_{\mathrm{{m}}}^p(y,z)\} ~\mathrm{d} \pi_{(\mu_r,\mu_u)}^*(x,z). \end{aligned} \end{align} $$

Let us compute $\min _{y \in Q}\{d_{\mathrm {{m}}}^p(x,y)+d_{\mathrm {{m}}}^p(y,z)\}$ for any $(x,z) \in Q^2$ — the case $x=z$ gives a trivial zero. By the reversed triangle inequality, we have

$$ \begin{align*}d_{\mathrm{{m}}}^p(x,y) \geq |d_{\mathrm{{m}}}(x,z)-d_{\mathrm{{m}}}(y,z)|^p=d_{\mathrm{{m}}}^p(x,z)\left| 1 -\frac{d_{\mathrm{{m}}}(y,z)}{d_{\mathrm{{m}}}(x,z)}\right|^p. \end{align*} $$

Consequently,

$$ \begin{align*}d_{\mathrm{{m}}}^p(x,y)+d_{\mathrm{{m}}}^p(y,z) \geq d_{\mathrm{{m}}}^p(x,z) \left( \left| 1 -\frac{d_{\mathrm{{m}}}(y,z)}{d_{\mathrm{{m}}}(x,z)}\right|^p + \left| \frac{d_{\mathrm{{m}}}(y,z)}{d_{\mathrm{{m}}}(x,z)}\right|^p\right). \end{align*} $$

It is crucial that $p>1$ and hence the map $t \mapsto |t|^p$ is strictly convex on $\mathbb {R}.$ Therefore, the function $\mathbb {R} \ni \lambda \mapsto |1-\lambda |^p+|\lambda |^p$ has a unique minimizer which is $\lambda _0=\frac {1}{2},$ and the minimum is $2^{1-p}$ — this can be justified by simple one-variable calculus. To sum up,

$$ \begin{align*}\min_{y \in Q}\{d_{\mathrm{{m}}}^p(x,y)+d_{\mathrm{{m}}}^p(y,z)\}=2^{1-p} d_{\mathrm{{m}}}^p(x,z), \end{align*} $$

and this minimum is achieved if and only if $d_{\mathrm {{m}}}(x,y)=d_{\mathrm {{m}}}(y,z)=\frac {1}{2}d_{\mathrm {{m}}}(x,z)$ —note that this does not imply that $y=\frac {1}{2}(x+z).$ So we can continue (3.31) as follows:

(3.32) $$ \begin{align} \begin{aligned} \iint_{Q^2} \min_{y \in Q}\{d_{\mathrm{{m}}}^p(x,y)+d_{\mathrm{{m}}}^p(y,z)\} ~\mathrm{d} \pi_{(\mu_r,\mu_u)}^*(x,z)&=2^{1-p} \iint_{Q^2} d_{\mathrm{{m}}}^p(x,z) ~\mathrm{d} \pi_{(\mu_r,\mu_u)}^*(x,z) \\ &\geq 2^{1-p} d_{W_p}^p(\mu_r,\mu_u). \end{aligned} \end{align} $$

The inequality in (3.32) is saturated if and only if $\pi _{(\mu _r,\mu _u)}^*$ is an optimal transport plan between $\mu _r$ and $\mu _u$ with respect to the cost $c(x,z)=d_{\mathrm {{m}}}^p(x,z).$ Observe, that $(p_r \times p_u)_{\#}(\mu )$ is the unique optimal transport plan between $\mu _r$ and $\mu _u$ for this cost. Indeed, by the definition of the max norm we get

(3.33) $$ \begin{align} (1-(2t-1))^p \leq d_{\mathrm{{m}}}^p((1, 2t-1),(2s-1,1)) \end{align} $$

and

(3.34) $$ \begin{align} (1-(2s-1))^p \leq d_{\mathrm{{m}}}^p((1, 2t-1),(2s-1,1)) \end{align} $$

for all $(s,t) \in [0,1] \times [0,1].$ By the construction of $\mu _r$ and $\mu _u,$ the couplings of these measures are in $1-1$ correspondence with the couplings of $\mu $ with itself—now we consider $\mu $ as a measure on $[0,1].$ This correspondence is described as follows: if $\pi \in C(\mu _r,\mu _u),$ then let us define $\tilde {\pi } \in C(\mu ,\mu )$ by $\tilde {\pi }:=(p_r^{-1} \times p_u^{-1})_{\#} (\pi ).$ Then

(3.35) $$ \begin{align} \begin{aligned} \iint_{Q^2}d_{\mathrm{{m}}}^p(x,z) ~\mathrm{d} \pi(x,z)&=\iint_{[0,1]^2} d_{\mathrm{{m}}}^p((1, 2t-1),(2s-1,1)) ~\mathrm{d} \tilde{\pi}(t,s). \\ &\geq \iint_{[0,1]^2} (1-(2t-1))^p ~\mathrm{d} \tilde{\pi}(t,s) \\ &=\int\limits_{[0,1]} (1-(2t-1))^p ~\mathrm{d} \mu(t) \end{aligned} \end{align} $$

and (3.35) is saturated if and only if $s \geq t$ for $\tilde {\pi }-$ a.e. $(s,t).$ Using (3.34) we get

$$ \begin{align*}\iint_{Q^2}d_{\mathrm{{m}}}^p(x,z) ~\mathrm{d} \pi(x,z) \geq \int\limits_{[0,1]} (1-(2s-1))^p ~\mathrm{d} \mu(s) \end{align*} $$

which is saturated if and only if $t \geq s$ for $\tilde {\pi }-$ a.e. $(s,t).$ Therefore, if $\pi $ is an optimal coupling of $\mu _r$ and $\mu _u,$ then $\tilde {\pi }$ is supported on the diagonal of $[0,1]^2$ which implies that $\tilde {\pi }=(id \times id)_{\#}(\mu )$ . This means that $\pi = (p_r \times p_u)_{\#}(\mu ).$ So $(p_r \times p_u)_{\#}(\mu )$ is the only optimal transport plan.

At this point we know by (3.31) and (3.32) that

(3.36) $$ \begin{align} d_{W_p}^p(\mu_r, \nu)+d_{W_p}^p(\nu, \mu_u) \geq 2^{1-p} d_{W_p}^p(\mu_r,\mu_u), \end{align} $$

and (3.36) is saturated if and only if $\pi _{(\mu _r,\mu _u)}^*=(p_r \times p_u)_{\#}(\mu )$ and $d_{\mathrm {{m}}}(x,y)=d_{\mathrm {{m}}}(y,z)=\frac {1}{2}d_{\mathrm {{m}}}(x,z)$ for $\pi _{(\mu _r, \nu ,\mu _u)}^*-$ a.e. $(x,y,z) \in Q^3.$ Therefore, equality in (3.36) implies that

$$ \begin{align*}\mathrm{supp}(\pi_{(\mu_r,\nu,\mu_u)}^*) \subset \left\{((1, 2t-1),y,(2t-1,1)) \, \middle| \, t \in [0,1], \, y \in Q\right\}. \end{align*} $$

However, the unique metric midpoint of $(1, 2t-1)$ and $(2t-1,1)$ is $(t,t).$ Therefore, $y=\frac {1}{2}(x+z)$ must hold for $\pi _{(\mu _r,\nu ,\mu _u)}^*$ -a.e. $(x,y,z) \in Q^3,$ which forces $\nu $ to be

$$ \begin{align*}\left((x,z) \mapsto \frac{1}{2}(x+z) \right)_{\#} \pi_{(\mu_r,\mu_u)}^* =\left((x,z) \mapsto \frac{1}{2}(x+z) \right)_{\#} (p_r \times p_u)_{\#}(\mu)= \mu. \end{align*} $$

Now let us consider $\mu $ , a probability measure supported on the main diagonal $[(-1,-1),(1,1)].$ The displacement interpolation given by

$$ \begin{align*}\mu_s:=\left((x,x) \mapsto (1-s)(1,1)+s(x,x)\right)_{\#}(\mu) \qquad (s \in [0,1]) \end{align*} $$

is the unique geodesic line segment between $\mu _0=\delta _{(1,1)}$ and $\mu _1=\mu .$ Note that $\mu _{\frac {1}{2}}$ is supported on the “upper half of the diagonal” $[(0,0),(1,1)]$ and hence preserved by $\Phi .$ Moreover, $\left (\mu _s\right )_{0\leq s\leq \frac {1}{2}}$ is the only geodesic line segment between $\delta _{(1,1)}$ and $\mu _{\frac {1}{2}},$ and the unique extension of this geodesic segment to the parameter domain $[0,1]$ is $\left (\mu _s\right )_{0\leq s\leq 1}.$ Therefore, the geodesic line segment $\left (\phi (\mu _s)\right )_{0\leq s\leq 1}$ containing $\Phi (\mu _0)=\Phi (\delta _{(1,1)})=\delta _{(1,1)}$ and $\Phi (\mu _{\frac {1}{2}})=\mu _{\frac {1}{2}}$ must coincide with $\left (\mu _s\right )_{0\leq s\leq 1},$ in particular, $\Phi (\mu )=\Phi (\mu _1)=\mu _1=\mu .$