Proof We give a few references and sketches for the proof.

(1) The inclusion in (2.5) follows by [Reference Fitzpatrick19, Theorem 2.4] and shows in particular that ${\boldsymbol {A}}_f$ is monotone since $\mathbf {T}_f$ is monotone by [Reference Fitzpatrick19, Proposition 2.2]. Clearly, $f^{\top } \le \mathsf {c}_{{\boldsymbol {A}}_f}$ so that, passing to the conjugates, the reverse inequality follows.

(2) The result in (2.6) is [Reference Penot26, Proposition 4(c)], while (2.7) is [Reference Penot26, Proposition 4(f)].

(3) This follows by (1) and (2).

(4) The first implication follows by [Reference Penot26, Theorem 5]. The second implication follows by [Reference Penot26, Theorem 6] choosing $g:=f_{\boldsymbol {A}}$ and [Reference Burachik and Svaiter14, Theorem 3.1].

(5) The first part of the sentence is [Reference Penot26, Theorem 6], while the equality ${\boldsymbol {A}}_f={\boldsymbol {A}}_{(f^{*})^{\top }}$ can be found, e.g., in [Reference Burachik and Svaiter14, Theorem 3.1].

(6) This is contained in the statement and in the proof of [Reference Bauschke and Wang8, Fact 5.6].

Proof (1) It is clear that f is convex and lower semicontinuous and $f \ge \mathsf {f}_{\boldsymbol {A}}$ . On the other hand, $f^{\top }\le \mathsf {c}_{\boldsymbol {A}}$ by (2.6) and since $\mathrm {I}_C=0$ on ${\boldsymbol {A}}$ . It follows that $f^{\top }\le \mathsf {c}_{\boldsymbol {A}}^{**}=\mathsf {p}_{\boldsymbol {A}}.$

(2) It is clear from the definition of f that $f(x,v)=+\infty $ if $x \notin D$ . Let us compute the conjugate $f^{*}$ of f: by the Fenchel–Rockafellar duality theorem (see, e.g., [Reference Rockafellar29, Theorem 16.4]), we have that

$$\begin{align*}f^{*}=(\mathsf{f}_{\boldsymbol{A}}+\mathrm{I}_C)^{*} = \operatorname{\mathrm{cl}}(\mathsf{f}_{\boldsymbol{A}}^{*}\mathbin\Box \mathrm{I}^{*}_C)= \operatorname{\mathrm{cl}}(\mathsf{p}_{\boldsymbol{A}}\mathbin\Box \mathrm{I}^{*}_C ),\end{align*}$$

where $\operatorname {\mathrm {cl}}$ denotes the lower semicontinuous envelope of a given function (i.e., $\operatorname {\mathrm {cl}}(h)$ is the largest lower semicontinuous function staying below h) and $\mathbin \Box $ denotes the inf-convolution (or epigraphical sum) of two functions: if $k,j:\mathcal {Z}^{*} \to (-\infty , + \infty ]$ , then $k\mathbin \Box j$ is defined as

$$\begin{align*}(k\mathbin\Box j)(z):=\inf_{z_1,z_{2}\in \mathcal{Z}^{*}, z_1+z_{2}=z} k(z_1)+j(z_{2}), \quad z\in \mathcal{Z}^{*}.\end{align*}$$

Since $\mathrm {I}_C(x,v)=\mathrm {I}_D(x)$ , we can easily compute its dual for every $z=(v,x)\in \mathcal {X}^{*} \times \mathcal {X}$

$$ \begin{align*} \mathrm{I}_C^{*}(v,x)&= \sup_{(x_{0},v_{0}) \in \mathcal{X}^{*} \times \mathcal{X}} \langle v,x_{0} \rangle + \langle v_{0},x\rangle -\mathrm{I}_D(x_{0}) = \sup_{x_{0}\in D, v_{0} \in \mathcal{X}^{*}} \langle v_{0},x \rangle + \langle v,x_{0}\rangle\\ &= \mathrm{I}_{0}(x)+\sigma_D(v), \end{align*} $$

where $\mathrm {I}_{0}$ is the indicator function of the singleton $\{0\}$ and $\sigma _D$ is the support function of D defined by

$$\begin{align*}\sigma_D(v):= \sup_{x_{0} \in D} \langle v,x_{0} \rangle, \quad v \in \mathcal{X}^{*}.\end{align*}$$

We thus have

$$ \begin{align*} (\mathsf{p}_{\boldsymbol{A}} \mathbin\Box \mathrm{I}^{*}_C)(v,x) &= \inf_{x_1+x_{2}=x,v_1+v_{2}=v} \mathsf{p}_{\boldsymbol{A}}(v_1,x_1)+\mathrm{I}_{0}(x_{2}) +\sigma_D(v_{2}) \\ &= \inf_{v_1+v_{2}=v} \mathsf{p}_{\boldsymbol{A}}(v_1,x)+\sigma_D(v_{2}). \end{align*} $$

Since $\mathsf {p}_{\boldsymbol {A}} =\mathsf {c}_{\boldsymbol {A}}^{**}$ and $\mathsf {c}_{{\boldsymbol {A}}}(v,x)=+\infty $ , if $x\not \in D$ , we deduce that $\mathsf {p}_{\boldsymbol {A}}(v,x)=+\infty $ if $x\not \in D$ , $\mathrm {D}(\mathsf {p}_{\boldsymbol {A}} \mathbin \Box \mathrm {I}^{*}_C)\subset \mathfrak {s}(C)= \mathcal {X}^{*}\times D$ , and therefore that $\mathrm {D}(f^{*})\subset \operatorname {\mathrm {cl}}( \mathcal {X}^{*}\times D)=\mathcal {X}^{*} \times D$ .

(3) By the first claim and Theorem 2.2 with f as in (2.16), we obtain that $\tilde {{\boldsymbol {A}}}$ is a maximal monotone extension of ${\boldsymbol {A}}$ . We only need to check that $\mathrm {D}(\tilde {{\boldsymbol {A}}}) \subset D=\overline {\operatorname {co}}\left (\mathrm {D}({\boldsymbol {A}})\right )$ . Since it is clear from the definition of contact set that $\mathrm {D}(\tilde {\boldsymbol {A}})\subset \pi ^{\mathcal {X}}(\mathrm {D}(R_f))$ , it is sufficient to check that $\mathrm {D}(R_f)\subset C$ .

Let $(x,v) \in \mathrm {D}(R_f)$ , then by (2.14) we can find $x_1,x_{2} \in \mathcal {X}$ and $v_1,v_{2} \in \mathcal {X}^{*}$ such that $(x,v)=\frac {1}{2}(x_1+x_{2},v_1+v_{2})$ and $(x_1,v_1)\in \mathrm {D}(f)$ , $(v_{2},x_{2})\in \mathrm {D} (f^{*})$ : in particular, $x_1,x_{2}$ belong to the convex set D by (2) and therefore $x\in D$ as well.

Proof (1) We simply have, for every $\mathsf {U}\in \mathsf {G}$ ,

$$ \begin{align*} f^{*}(\mathsf{U}^{\top} z^{*}) &= \sup_{z\in \mathcal{Z}}\left\{ \langle \mathsf{U}^{\top} z^{*},z\rangle -f(z)\right\}= \sup_{z\in \mathcal{Z}}\left\{ \langle z^{*},\mathsf{U}^{-1} z\rangle -f(z)\right\} \\& = \sup_{z\in \mathcal{Z}}\left\{ \langle z^{*},\mathsf{U}^{-1} z\rangle -f(\mathsf{U}^{-1} z)\right\}= \sup_{\tilde z\in \mathcal{Z}}\left\{ \langle z^{*},\tilde z\rangle -f(\tilde z)\right\} =f^{*}(z^{*}), \end{align*} $$

where we applied (2.20), the fact that $\mathsf {U}^{-1}\in \mathsf {G}$ , the $\mathsf {G}$ -invariance of f, and the fact that $\{\tilde z=\mathsf {U}^{-1}z:z\in \mathcal {Z}\}=\mathcal {Z}.$

(2) One immediately sees that $\mathsf {c}_{\boldsymbol {A}}$ is $\mathsf {G}$ invariant thanks to the $\mathsf {G}$ -invariance of ${\boldsymbol {A}}$ and (2.18). The invariance of $\mathsf {f}_{\boldsymbol {A}}$ and of $\mathsf {p}_{\boldsymbol {A}}$ then follows by the previous claim.

(3) If $z=(x,v)\in {\boldsymbol {A}}_f$ , we know that $f(z)=\mathsf {c}^{\top }(z)$ . Since f is $\mathsf {G}$ -invariant, (2.18) yields, for every $\mathsf {U}\in \mathsf {G}$ ,

$$ \begin{align*} f(\mathsf{U}z)= f(z)=\mathsf{c}^{\top}(z)= \mathsf{c}^{\top}(\mathsf{U}z) \end{align*} $$

so that $\mathsf {U}z\in {\boldsymbol {A}}_f.$

(4) We first observe that the function

$$ \begin{align*} P(z_1,z_{2}):= \frac 12 f(z_1)+ \frac 12 f^{*}(\mathfrak{s}(z_{2}))+ \frac 14 \psi(z_1-z_{2}) \end{align*} $$

satisfies

(2.21) $$ \begin{align} P(\mathsf{U}z_1,\mathsf{U}z_{2})= P(z_1,z_{2}), \end{align} $$

thanks to the invariance of f, the invariance of $f^{*}$ from claim (1), and the invariance property of $\psi $ stated in (2.18).

Since $\mathsf {U}$ is a linear isomorphism, we also have

$$ \begin{align*} z=\frac 12 z_1+\frac 12 z_{2} \quad\Leftrightarrow\quad \mathsf{U}z=\frac 12 \mathsf{U}z_1+\frac 12 \mathsf{U}z_{2}. \end{align*} $$

Combining the above identities, we immediately get $R_f(\mathsf {U}z)= R_f(z).$

Proof Let us first observe that $C':=\mathrm {D}({\boldsymbol {A}})\times \mathcal {X}^{*}$ is $\mathsf {G}$ -invariant, thanks to the invariance of ${\boldsymbol {A}}.$ Since every element of $\mathsf {G}$ is linear, also $\operatorname {co}(C')$ is $\mathsf {G}$ -invariant. Eventually, since every element of $\mathsf {G}$ is continuous, $C:=\operatorname {\mathrm {cl}}(\operatorname {co}(C'))$ is $\mathsf {G}$ -invariant as well.

By claim (2) of Proposition 2.4, we deduce that the function f given by (2.16) is $\mathsf {G}$ -invariant. The invariance of $\tilde {\boldsymbol {A}}$ then follows by applying Claims (4) and (3) of Proposition 2.4, recalling that $R_f\ge \mathsf {c}^{\top }$ by Theorem 2.1(6). We conclude by Theorem 2.3.

Proof We can choose the self-dual kernel

$$ \begin{align*} \psi(x,v):=\frac 12 |x|^{2}+\frac 12 |v|^{2}, \end{align*} $$

and we immediately get that $\mathsf {G}$ satisfies (2.18). The statement is then an immediate application of Theorem 2.5 to the $\mathsf {G}$ -invariant monotone operator ${\boldsymbol {A}}:=-\boldsymbol {B}^\lambda .$

Proof The identities ${\boldsymbol {J}_{\tau }} (Ux)=U({\boldsymbol {J}_{\tau }} x)$ and $\boldsymbol {B}^\circ (Ux) = U(\boldsymbol {B}^\circ x)$ come from the $\mathsf {G}$ -invariance of $\boldsymbol {B}$ and the uniqueness property of the resolvent operator. The exponential formula (cf. (2.28))

$$ \begin{align*} \boldsymbol{S}_t x=\lim_{n\to\infty} ({\boldsymbol{J}_{t/n}})^n(x) \end{align*} $$

yields the $\mathsf {G}$ -invariance of $\boldsymbol {S}_t$ .

Proof Let us take a pair of elements $(x_i,v_i)=T(y_i,w_i)\in {\boldsymbol {A}}$ , $i=1,2$ ; we have $y_i=\frac 1{\sqrt 2} (x_i+v_i)$ , $w_i=\frac 1{\sqrt 2}(-x_i+v_i)$ so that

(2.31) $$ \begin{align} 2|y_1-y_{2}|^{2}= |x_1+v_1-(x_{2}+v_{2})|^{2} &= |x_1-x_{2}|^{2} + |v_1-v_{2}|^{2} +2\langle x_1-x_{2},v_1-v_{2}\rangle, \end{align} $$

(2.32) $$ \begin{align} 2|w_1-w_{2}|^{2} =|-x_1+v_1-(-x_{2}+v_{2})|^{2} &= |x_1-x_{2}|^{2} + |v_1-v_{2}|^{2} -2\langle x_1-x_{2},v_1-v_{2}\rangle, \end{align} $$

and then

(2.33) $$ \begin{align} |w_1-w_{2}|^{2} \le |y_1-y_{2}|^{2} \quad \Leftrightarrow \quad \langle x_1-x_{2},v_1-v_{2}\rangle \ge 0. \end{align} $$

This proves the first statement.

Concerning (i), it is sufficient to recall that the domain D of f coincides with the image of $h({\boldsymbol {A}})$ where $h(x,v):=\frac 1{\sqrt 2}(x+v)$ and we know that ${\boldsymbol {A}}$ is maximal monotone if and only if such an image coincides with $\mathcal {H}$ (cf. [Reference Brézis13, Proposition 2.2(ii)]).

The second property (ii) is an immediate consequence of the invertibility of T and of the linearity of the transformations of $\mathsf {G}$ so that for every $\mathsf {U}=(U,U)\in \mathsf {G}$ , we have $T\circ \mathsf {U}= \mathsf {U}\circ T$ .

Let us eventually consider claim (iii): we clearly have $\boldsymbol {F}'= T^{-1}({\boldsymbol {A}}')\supset T^{-1}({\boldsymbol {A}}) = \boldsymbol {F}$ and therefore $D'=\pi ^1(\boldsymbol {F}')\supset \pi ^1(\boldsymbol {F})=D$ . On the other hand, since both $\boldsymbol {F}'$ and $\boldsymbol {F}$ are the graph of a nonexpansive map, $\boldsymbol {F}'\cap (D\times \mathcal {H})= \boldsymbol {F}\cap (D\times \mathcal {H})$ and therefore the restriction of $f'$ to D coincides with f.

Proof Up to a rescaling, it is not restrictive to assume that $L=1$ so that f is nonexpansive.

Let $\boldsymbol {F}\subset \mathcal {H}\times \mathcal {H}$ be the graph of f, and let ${\boldsymbol {A}}:=T(\boldsymbol {F})$ . By Lemma 2.10, we know that ${\boldsymbol {A}}$ is a monotone $\mathsf {G}$ -invariant operator

We can now apply Theorem 2.7 to find a maximal monotone extension ${\boldsymbol {\hat {A}}}$ of ${\boldsymbol {A}}$ which is still $\mathsf {G}$ -invariant.

Setting $\boldsymbol {\hat F}:= T^{-1}({\boldsymbol {\hat {A}}})$ , we can eventually apply Lemma 2.10 to obtain that $\boldsymbol {\hat F}$ is the graph of a nonexpansive map $\hat f:\hat D\to \mathcal {H}.$ Moreover, Claim (i) shows that $\hat D=\mathcal {H}$ so that $\hat f$ is globally defined, Claim (ii) shows that $\hat f$ is $\mathsf {G}_{\mathcal {H}}$ -invariant, and Claim (iii) ensures that $\hat f$ is an extension of f.

Proof Let $([0,1), \mathcal {B}([0,1)), \lambda ^c, (\mathfrak {I}_N)_{N \in {\mathfrak {N}}})$ be the ${\mathfrak {N}}$ -refined standard Borel space of Example 3.2 with $c=\mathfrak {m}(\Omega )$ . Since $([0,1), \mathcal {B}([0,1)))$ is a standard Borel space endowed with the nonatomic, positive finite measure $\lambda ^c$ such that $\mathfrak {m}(\Omega ) = \lambda ^c([0,1))$ , by Theorem 3.3, we can find measurable maps $\varphi :[0,1) \to \Omega $ , $\psi : \Omega \to [0,1)$ and two subsets $\Omega _{0} \in {\mathcal {B}}$ , $U \in \mathcal {B}([0,1))$ such that $\mathfrak {m}(\Omega _{0})=\lambda ^c(U)=0$ , $\varphi \circ \psi = \boldsymbol {i}_{\Omega \setminus \Omega _{0}}$ , $\psi \circ \varphi = \boldsymbol {i}_{[0,1)\setminus U}$ , $\varphi _\sharp \lambda ^c=\mathfrak {m}$ and $\psi _\sharp \mathfrak {m}=\lambda ^c$ . We can thus define

$$\begin{align*}\Omega_{N,0} = \varphi(I_{N,0}\setminus U) \cup \Omega_{0}, \quad \Omega_{N,k} = \varphi(I_{N,k}\setminus U), \quad k \in I_N\setminus\{0\}, \, N \in {\mathfrak{N}}.\end{align*}$$

Setting $\mathfrak {P}_N := \{\Omega _{N,k}\}_{k \in I_N}$ for every $N \in {\mathfrak {N}}$ , it is easy to check that $(\mathfrak {P}_N)_{N \in {\mathfrak {N}}}$ is a ${\mathfrak {N}}$ -segmentation of $(\Omega , {\mathcal {B}}, \mathfrak {m})$ .

Proof Applying Theorem 3.3 to the standard Borel spaces $(\Omega _{\{k\}}, {\mathcal {B}}_{\{k\}})$ and $(\Omega ^{\prime }_{\{k\}}, {\mathcal {B}}^{\prime }_{\{k\}})$ endowed, respectively, with the nonatomic, positive finite measures $\mathfrak {m}_{\{k\}}$ and $\mathfrak {m}^{\prime }_{\{k\}}$ for every $k \in I_N$ , we obtain the existence of measurable functions $\varphi _{k}, \psi _{k}$ satisfying (3.3) for each couple $\Omega _{N,k}$ , $\Omega ^{\prime }_{N,k}$ . It is then enough to define

$$\begin{align*}\varphi(\omega) := \varphi_{k}(\omega) \quad \text{ if } \omega \in \Omega_{N,k}, \quad \psi(\omega') := \psi_{k}(\omega') \quad \text{ if } \omega' \in \Omega^{\prime}_{N,k}.\end{align*}$$

Notice that (3.4) is satisfied by construction.

Proof It is enough to apply Lemma 3.6 to the standard Borel spaces $(\Omega , {\mathcal {B}})$ and $(\Omega ', {\mathcal {B}}')$ = $(\Omega , {\mathcal {B}})$ endowed with the nonatomic, positive finite measures $\mathfrak {m}$ and $\mathfrak {m}'=\mathfrak {m}$ , respectively, with the N-partitions $\mathfrak {P}_N$ and $\mathfrak {P}^{\prime }_N=\{\Omega _{N,\sigma (k)}\}_{k \in I_N}$ , respectively.

Proof Applying Corollary 3.7 to the standard Borel space $(\Omega _{0} \cup \Omega _1,{\mathcal {B}}|_{\Omega _{0} \cup \Omega _1})$ endowed with the nonatomic, positive finite measure $ \mathfrak {m}|_{\Omega _{0} \cup \Omega _1}$ with the $2$ -Borel partition $\mathfrak {P}_{2}= \{\Omega _{k}\}_{k=0,1}$ and $\sigma $ sending $0$ to $1$ , we obtain the existence of a measure-preserving isomorphism $\tilde {g} \in \mathrm {S}(\Omega _{0} \cup \Omega _1,{\mathcal {B}}|_{\Omega _{0} \cup \Omega _1}, \mathfrak {m}|_{\Omega _{0} \cup \Omega _1})$ such that

$$\begin{align*}(\tilde{g}_{0})_\sharp \mathfrak{m} |_{\Omega_{0}} = \mathfrak{m} |_{\Omega_{1}}, \quad (\tilde{g}_1)_\sharp \mathfrak{m} |_{\Omega_{1}} = \mathfrak{m} |_{\Omega_{0}}, \end{align*}$$

where $\tilde {g}_i$ is the restriction of $\tilde {g}$ to $\Omega _{k}$ , $k=0,1$ . It is then enough to define $g: \Omega \to \Omega $ as

$$\begin{align*}g(\omega) = \begin{cases} \tilde{g}(\omega), \quad &\text{ if } \omega \in \Omega_{0} \cup \Omega_1, \\ \omega, \quad &\text{ if } \omega \in \Omega \setminus (\Omega_{0} \cup \Omega_1). \end{cases}\\[-34pt] \end{align*}$$

Proof Since ${\mathfrak {N}}$ is unbounded and directed, for every finite subset $\mathfrak {M} \subset {\mathfrak {N}}$ , the quantity

is well defined. Let $(a_n)_n \subset \mathbb {N}$ be an enumeration of ${\mathfrak {N}}$ and consider the following sequence defined by induction

$$\begin{align*}b_{0}=a_{0}, \quad b_{n+1}= \text{succ} \left (\{a_{n+1},b_n\} \right ), \quad n \in \mathbb{N}.\end{align*}$$

Then for every $n \in \mathbb {N}$ and (3.6) holds for $(b_n)_n$ and any ${\mathfrak {N}}$ -refined standard Borel measure space $(\Omega , {\mathcal {B}}, \mathfrak {m}, (\mathfrak {P}_N)_{N \in {\mathfrak {N}}})$ .

Proof Let $(b_n)_n\subset {\mathfrak {N}}$ be a totally ordered cofinal sequence as in Lemma 3.10, and let $\tau $ be a Polish topology on $\Omega $ such that ${\mathcal {B}}$ coincides with the Borel sigma algebra generated by $\tau $ . By [Reference Bogachev10, Proposition 6.5.4], there exists a countable family $\mathcal {F}$ of $\tau $ -continuous functions $f: \Omega \to [0,1]$ separating the points of $\Omega $ , meaning that for every $\omega ', \omega " \in \Omega $ , $\omega ' \ne \omega "$ , there exists $f \in \mathcal {F}$ such that $f(\omega ') \ne f(\omega ")$ . Since $\mathcal {F} \subset L^{2}(\Omega , {\mathcal {B}}, \mathfrak {m}; \mathbb {R})$ , by Theorem 3.9 with $\mathcal {F}_n:={\mathcal {B}}_{b_n}$ , for every $f \in \mathcal {F}$ , there exists an $\mathfrak {m}$ -negligible set $\Omega _f$ such that

$$\begin{align*}\lim_{n \to + \infty} {\mathbb{E}}_{\mathfrak{m}} \left [ f \mid\sigma \left ( \mathfrak{P}_{b_n} \right ) \right ](\omega) = f(\omega) \quad \forall \omega \in \Omega \setminus \Omega_f.\end{align*}$$

Let $\Omega _{0} := \cup _{f \in \mathcal {F}} \Omega _f$ , and let $\omega ', \omega " \in \Omega \setminus \Omega _{0}$ , $\omega ' \ne \omega "$ . We can find $f \in \mathcal {F}$ such that $f(\omega ') \ne f(\omega ")$ . Thus, there exists $M=b_m \in {\mathfrak {N}}$ such that

$$\begin{align*}{\mathbb{E}}_{\mathfrak{m}} \left [ f \mid\sigma \left ( \mathfrak{P}_{M} \right ) \right ](\omega') \ne {\mathbb{E}}_{\mathfrak{m}} \left [ f \mid\sigma \left ( \mathfrak{P}_{M} \right ) \right ](\omega").\end{align*}$$

Since ${\mathbb {E}}_{\mathfrak {m}} \left [ f \mid \sigma \left ( \mathfrak {P}_{M} \right ) \right ]$ is constant on every $\Omega _{M,k}$ , $k \in I_{M}$ , we conclude that the points $\omega '$ and $\omega "$ belong to different elements of $\mathfrak {P}_{M}$ , and therefore they also belong to different elements of $\mathfrak {P}_{N}$ for every $N\in {\mathfrak {N}}$ multiple of M.

Proof By Lemma 3.10, it is enough to prove the statement in case ${\mathfrak {N}} = (b_n)_n$ , where $(b_n)_n \subset \mathbb {N}$ is strictly $\prec $ -increasing sequence and $(\Omega ', {\mathcal {B}}', \mathfrak {m}', (\mathfrak {P}^{\prime }_N)_{N \in {\mathfrak {N}}})$ is $([0,1), \mathcal {B}([0,1)), \lambda ^c, (\mathfrak {I}_N)_{N \in {\mathfrak {N}}})$ as in Example 3.2 with $c= \mathfrak {m}(\Omega )$ . By Lemma 3.6, we can find for every $n \in \mathbb {N}$ two measurable maps $\varphi _n: \Omega \to [0,1)$ and $\psi _n:[0,1) \to \Omega $ satisfying the thesis of Lemma 3.6 for the standard Borel spaces $(\Omega , {\mathcal {B}})$ and $([0,1), \mathcal {B}([0,1))$ endowed with nonatomic, positive, and finite measures $\mathfrak {m}$ and $\lambda ^c$ , respectively, and the $\mathfrak {m}-\lambda ^c$ compatible $b_n$ -partitions of $(\Omega , {\mathcal {B}})$ and $([0,1), \mathcal {B}([0,1)))$ given by $\mathfrak {P}_{b_n}$ and $\mathfrak {I}_{b_n}$ , where we recall from Example 3.2 that $\mathfrak {I}_{b_n}= (I_{b_n,k})_{k \in I_{b_n}}$ with $I_{b_n,k}=[k/b_n, (k+1)/b_n)$ . Since $\sum _n b_n^{-1} < + \infty $ , for every $\omega \in \Omega $ , the sequence $(\varphi _n(\omega ))_n \subset [0,1)$ is Cauchy, hence converges. We thus have the existence of a measurable map $\varphi : \Omega \to [0,1)$ such that

$$\begin{align*}\varphi(\omega) = \lim_n \varphi_n (\omega) \quad \forall \omega \in \Omega.\end{align*}$$

If $n \in \mathbb {N}$ , $k \in I_{b_n}$ , and $ \xi \in \mathrm {C}_b(I_{b_n,k})$ , then

$$ \begin{align*} \int_{I_{b_n, k}} \xi \,\mathrm{d} \varphi_\sharp \mathfrak{m} &= \int_{\Omega_{b_n,k}} \xi(\varphi(\omega)) \,\mathrm{d} \mathfrak{m}(\omega) = \lim_m \int_{\Omega_{b_n,k}} \xi(\varphi_m(\omega)) \,\mathrm{d} \mathfrak{m}(\omega) \\ & = \lim_m \int_{I_{b_n,k}} \xi \,\mathrm{d} \lambda^c = \int_{I_{b_n,k}} \xi\,\mathrm{d} \lambda^c, \end{align*} $$

since for m sufficiently large $(\varphi _m)_\sharp \mathfrak {m}|_{\Omega _{b_n,k}}=\lambda ^c|_{I_{b_n,k}}$ by Lemma 3.6. This shows that $\varphi _\sharp \mathfrak {m} |_{\Omega _{b_n,k}} = \lambda ^c|_{I_{b_n,k}}$ for every $k \in I_{b_n}$ and every $n \in \mathbb {N}$ . To conclude, it is enough to show that $\varphi $ is $\mathfrak {m}$ -essentially injective. Let $\Omega _{0} \subset \Omega $ be the $\mathfrak {m}$ -negligible subset of $\Omega $ given by Lemma 3.11, and let $\Omega _1:= \varphi ^{-1}(J)$ , where

$$\begin{align*}J := \left \{ k/b_n \mid k \in I_{b_n}, \, n \in \mathbb{N} \right \} \subset [0,1).\end{align*}$$

Since $\lambda ^c(J)=0$ , then $\mathfrak {m}(\Omega _1)=0$ ; let $\omega ', \omega " \in \Omega \setminus (\Omega _{0} \cup \Omega _1)$ . Then there exists $M \in \mathbb {N}$ such that $\omega '$ and $\omega "$ belong to different elements of $\mathfrak {P}_{b_n}$ for every $n \ge M$ . By (3.4) and Lemma 3.11, we can find $k',k" \in I_{b_M}$ with $k \ne k'$ such that $\varphi _n(\omega ') \in I_{b_{M}, k'}$ and $\varphi _n(\omega ") \in I_{b_{M}, k"}$ for every $n \ge M$ . Thus, $\varphi (\omega ') \in \overline {I_{b_{M}, k'}}$ and $\varphi (\omega ') \in \overline {I_{b_{M}, k"}}$ ; however, since

$$\begin{align*}\overline{I_{b_{M}, k'}} \cap \overline{I_{b_{M}, k"}} \subset J,\end{align*}$$

it must be that $\varphi (\omega ') \ne \varphi (\omega ").$

Proof By Lemma 3.10, it is not restrictive to assume that ${\mathfrak {N}}=(b_n)_n$ for a totally ordered strictly increasing sequence $(b_n)_n$ , $n\in \mathbb {N}$ . We divide the proof in several steps.

(1) Let $([0,1), \mathcal {B}([0,1)), \lambda ^1, (\mathfrak {I}_N)_{N \in {\mathfrak {N}}})$ be the ${\mathfrak {N}}$ -refined standard Borel probability space of Example 3.2 with $c=1$ . Then, for every $\boldsymbol {\gamma } \in \Gamma (\lambda ^1, \lambda ^1)$ , there exist a strictly increasing sequence $(N_n)_n \subset \mathbb {N}$ and maps $g_n \in \mathrm {S}([0,1), \mathcal {B}([0,1)), \lambda ^1; \sigma (\mathfrak {I}_{b_{N_n}}))$ such that

$$\begin{align*}(\boldsymbol{i}_{[0,1)}, g_n)_\sharp \lambda^1 \to \boldsymbol{\gamma}\ \textit{in}\ \mathcal{P}([0,1) \times [0,1)).\end{align*}$$

Let $\bar {\mathcal {L}}$ be the one-dimensional Lebesgue measure restricted to $[0,1]$ , and let $\boldsymbol {\gamma } \in \Gamma (\lambda ^1, \lambda ^1)$ . Let $\boldsymbol {\mu } \in \mathcal {P}([0,1] \times [0,1]) $ be an extension of $\boldsymbol {\gamma }$ to $[0,1] \times [0,1]$ such that $\boldsymbol {\mu } \in \Gamma (\bar {\mathcal {L}}, \bar {\mathcal {L}})$ . In [Reference Brenier and Gangbo12, Theorem 1.1], it is proven that it is possible to find a strictly increasing sequence $(N_n)_n \subset \mathbb {N}$ and maps $(f_n)_n \subset \mathrm {S}([0,1], \mathcal {B}([0,1]), \bar {\mathcal {L}})$ such that for every $n \in \mathbb {N}$ , there exists $\sigma _n \in {\mathrm {Sym}(I_{2^{N_n}})}$ such that

(3.13) $$ \begin{align} f_n(x) = x - x_{N_n, k} + x_{N_n, \sigma_n(k)}, \quad x \in I_{2^{N_n}, k}, \quad k \in I_{2^{N_n}}, \end{align} $$

with $x_{m,j}$ being the center of $I_{2^m, j}$ , and satisfying

(3.14) $$ \begin{align} (\boldsymbol{i}_{[0,1]}, f_n)_\sharp \bar{\mathcal{L}} \to \boldsymbol{\mu}\ {in}\ \mathcal{P}([0,1]\times [0,1]). \end{align} $$

If we call $g_n$ the restriction of $f_n$ to $[0,1)$ , $n \in \mathbb {N}$ , we get that $g_n \in \mathrm {S}([0,1), \mathcal {B}([0,1)), \lambda ^1; \sigma (\mathfrak {I}_{b_{N_n}}))$ for every $n \in \mathbb {N}$ and

$$\begin{align*}(\boldsymbol{i}_{[0,1)}, g_n)_\sharp \lambda^1 \to \boldsymbol{\gamma} \ {in}\ \mathcal{P}([0,1)\times [0,1)). \end{align*}$$

This proves the first step only in case $b_n=2^n$ . However, it can be easily checked that the proof of [Reference Brenier and Gangbo12, Theorem 1.1] does not depend on the specific choice of the sequence $b_n$ , but it is enough that for every $n \in \mathbb {N}$ so that the length of the interval $[k/b_n, (k+1)/b_n]$ goes to $0$ faster than $2^{-n}$ as $n \to + \infty $ . This concludes the proof of the first claim.

(2) Let $([0,1), \mathcal {B}([0,1)), \lambda ^1, (\mathfrak {I}_N)_{N \in {\mathfrak {N}}})$ be the ${\mathfrak {N}}$ -refined standard Borel probability space of Example 3.2 with $c=1$ . Then, for every $\boldsymbol {\gamma } \in \Gamma (\lambda ^1, \lambda ^1)$ , there exist a strictly increasing sequence $(N_n)_n \subset \mathbb {N}$ and maps $g_n \in \mathrm {S}([0,1), \mathcal {B}([0,1)), \lambda ^1; \sigma (\mathfrak {I}_{b_{N_n}}))$ such that, for every separable Banach spaces $\mathsf {Z}, \mathsf {Z}'$ and every $Z \in L^0([0,1), \mathcal {B}([0,1)), \lambda ^1; \mathsf {Z})$ , $Z' \in L^0([0,1), \mathcal {B}([0,1)), \lambda ^1; \mathsf {Z}')$ , it holds

$$\begin{align*}(Z \otimes Z')_\sharp (\boldsymbol{i}_{[0,1)}, g_n)_\sharp \lambda^1 \to (Z\otimes Z')_\sharp \boldsymbol{\gamma} \ {in}\ \mathcal{P}(\mathsf{Z}\times \mathsf{Z}').\end{align*}$$

Let $\boldsymbol {\gamma } \in \Gamma (\lambda ^1, \lambda ^1)$ , and let $(g_n)_n$ be the sequence given by claim (1) for $\boldsymbol {\gamma }$ . Let $\mathsf {Z}$ and $\mathsf {Z}'$ be separable Banach spaces, and let $Z \in L^0([0,1), \mathcal {B}([0,1)), \lambda ^1; \mathsf {Z})$ , $Z' \in L^0([0,1), \mathcal {B}([0,1)), \lambda ^1; \mathsf {Z}')$ . Observe that for every $\varepsilon>0$ , there exists a compact set $K_\varepsilon \subset [0,1)$ such that the restrictions of Z and $Z'$ to $K_\varepsilon $ are continuous in $K_\varepsilon $ and $\lambda ^1([0,1)\setminus K_\varepsilon )<\varepsilon $ , so that, setting $\gamma _n:=(\boldsymbol {i}_{[0,1)}, g_n)_\sharp \lambda ^1$ , $n \in \mathbb {N}$ , we have that $\gamma _n([0,1)^{2}\setminus K_\varepsilon ^{2})\le 2\varepsilon $ for every $n\in \mathbb {N}$ . By [Reference Ambrosio, Gigli and Savaré3, Proposition 5.1.10] and claim (1), $(Z\otimes Z')_\sharp (\boldsymbol {i}_{[0,1)}, g_n)_\sharp \lambda ^1 \to (Z\otimes Z')_\sharp \boldsymbol {\gamma } \text { in } \mathcal {P}(\mathsf {Z}\times \mathsf {Z}')$ .

(3) Conclusion. Let $\boldsymbol {\gamma } \in \Gamma (\mathbb {P}, \mathbb {P})$ , and let $\varphi :\Omega \to [0,1)$ and $\psi : [0,1) \to \Omega $ be the maps given by Proposition 3.12 for the ${\mathfrak {N}}$ -refined standard Borel probability spaces $(\Omega , {\mathcal {B}}, \mathbb {P}, (\mathfrak P_N)_{N \in {\mathfrak {N}}})$ and $([0,1), \mathcal {B}([0,1)), \lambda ^1, (\mathfrak {I}_N)_{N \in {\mathfrak {N}}})$ , where the latter is as in Example 3.2 with $c=1$ . If we define $\boldsymbol {\gamma }':=(\varphi \otimes \varphi )_\sharp \boldsymbol {\gamma }$ , we have that $\boldsymbol {\gamma }' \in \Gamma (\lambda ^1, \lambda ^1)$ so that we can find a strictly increasing sequence $(N_n)_n \subset \mathbb {N}$ and maps $g^{\prime }_n \in \mathrm {S}([0,1), \mathcal {B}([0,1)), \lambda ^1; \sigma (\mathfrak {I}_{b_{N_n}}))$ as in step (2). Let us define

$$\begin{align*}g_n:= \psi \circ g^{\prime}_n \circ \varphi, \quad n \in \mathbb{N}.\end{align*}$$

Then, up to change each $g_n$ on a $\mathbb {P}$ -negligible set of points, we can assume that $g_n \in \mathrm {S}(\Omega , {\mathcal {B}}, \mathbb {P}; {\mathcal {B}}_{b_{N_n}})$ . Let $\mathsf {Z}$ and $\mathsf {Z}'$ be separable Banach spaces, and let $Z \in L^0(\Omega , {\mathcal {B}}, \mathbb {P}; \mathsf {Z})$ , $Z' \in L^0(\Omega , {\mathcal {B}}, \mathbb {P}; \mathsf {Z}')$ . If we define $Z_{0}:=Z \circ \psi $ and $Z^{\prime }_{0}:=Z' \circ \psi $ , we get that $Z_{0} \in L^0([0,1), \mathcal {B}([0,1)), \lambda ^1); \mathsf {Z})$ , $Z_{0}' \in L^0([0,1), \mathcal {B}([0,1)), \lambda ^1); \mathsf {Z}')$ . By step (2), we thus get

$$\begin{align*}(Z_{0}\otimes Z_{0}')_\sharp (\boldsymbol{i}_{[0,1)}, g^{\prime}_n)_\sharp \lambda^1 \to (Z_{0}\otimes Z_{0}')_\sharp \boldsymbol{\gamma}' \text{ in } \mathcal{P}(\mathsf{Z}\times \mathsf{Z}'), \end{align*}$$

which is equivalent to (3.12).

Proof By Theorem 3.13 and Remark 3.14, we have the existence of a strictly increasing sequence $(N_n)_n \subset \mathbb {N}$ and maps $g_n \in \mathrm {S}(\Omega , {\mathcal {B}}, \mathbb {P}; {\mathcal {B}}_{{N_n}})$ such that, choosing the separable Hilbert space $\mathbb {R}$ , we get

$$\begin{align*}(\varphi_1\otimes \varphi_{2})_\sharp (\boldsymbol{i}_{\Omega}, g_n)_{\sharp}\mathbb{P} \to (\varphi_1\otimes \varphi_{2})_\sharp \boldsymbol{\gamma} \text{ in } \mathcal{P}_{2}(\mathbb{R}^{2}) \end{align*}$$

for every $\varphi _1, \varphi _{2} \in \mathrm {C}_b(\Omega , \tau ) \subset L^{2}(\Omega ,{\mathcal {B}}, \mathbb {P}; \mathbb {R})$ . Since the range of $\varphi _i$ is bounded and thus relatively compact, by the $\mathcal {P}(\mathbb {R}^{2})$ convergence, we get that

$$\begin{align*}\int_{\Omega \times \Omega} h(\varphi_1(\omega_1), \varphi_{2}(\omega_{2})) \,\mathrm{d} \boldsymbol{\gamma}_n(\omega_1, \omega_{2}) \to \int_{\Omega \times \Omega} h(\varphi_1(\omega_1), \varphi_{2}(\omega_{2})) \,\mathrm{d} \boldsymbol{\gamma}(\omega_1, \omega_{2})\end{align*}$$

for every continuous function $h: \mathbb {R}^{2} \to \mathbb {R}$ where $\boldsymbol {\gamma }_n = (\boldsymbol {i}_{\Omega }, g_n)_{\sharp}\mathbb {P}$ , $n \in \mathbb {N}$ . Choosing $h(x,y):= xy$ , we get that

(3.15) $$ \begin{align} \int_{\Omega \times \Omega} \varphi_1(\omega_1) \varphi_{2}(\omega_{2}) \,\mathrm{d} \boldsymbol{\gamma}_n(\omega_1, \omega_{2}) \to \int_{\Omega \times \Omega} \varphi_1(\omega_1) \varphi_{2}(\omega_{2}) \,\mathrm{d} \boldsymbol{\gamma}(\omega_1, \omega_{2}) \quad \forall \varphi_1, \varphi_{2} \in \mathrm{C}_b(\Omega, \tau). \end{align} $$

Let $\mathcal {A} \subset \mathrm {C}_b(\Omega , \tau )$ be a unital subalgebra whose induced initial topology on $\Omega $ coincides with $\tau $ (e.g., the subset of $\mathsf {d}$ -Lipschitz continuous and bounded functions for a complete distance $\mathsf {d}$ inducing $\tau $ ). It is easy to check that

$$\begin{align*}\mathcal{A} \otimes \mathcal{A} := \left \{ \sum_{i=1}^n \varphi_1^i \otimes \varphi_{2}^i \mid (\varphi_1^i)_{i=1}^n, (\varphi_{2}^i)_{i=1}^n \subset \mathcal{A}, \, n \in \mathbb{N} \right \} \subset \mathrm{C}_b(\Omega \times \Omega, \tau \otimes \tau) \end{align*}$$

is a unital subalgebra whose induced initial topology on $\Omega \times \Omega $ coincides with $\tau \otimes \tau $ . By (3.15), we thus have that

$$\begin{align*}\int_{\Omega \times \Omega} \varphi \,\mathrm{d} \boldsymbol{\gamma}_n \to \int_{\Omega \times \Omega} \varphi \,\mathrm{d} \boldsymbol{\gamma} \quad \forall \varphi \in \mathcal{A} \otimes \mathcal{A}.\end{align*}$$

We conclude by [Reference Savaré and Sodini31, Lemma 2.3].

Proof Let $\mu :=Z_{\sharp}\mathbb {P}$ and $\mu ':=Z^{\prime }_{\sharp}\mathbb {P}$ ; let us first observe that there exists $\boldsymbol {\gamma }\in \Gamma ({\mathbb {P}},{\mathbb {P}})$ such that $(Z\otimes Z')_\sharp \boldsymbol {\gamma }=\boldsymbol {\mu }$ . In fact, we can disintegrate $\mathbb {P}$ w.r.t. Z and $\mu $ (see, e.g., [Reference Ambrosio, Gigli and Savaré3, Theorem 5.3.1] for details on the disintegration theorem), obtaining a Borel family of measures $(\theta _z)_{z\in \mathsf {Z}}\subset \mathcal {P}(\Omega )$ such that $\theta _z$ is concentrated on $Z^{-1}(z)$ for $\mu $ -a.e. $z \in \mathsf {Z}$ and $\mathbb {P}=\int \theta _z \,\mathrm {d}\mu (z)$ ; similarly, we can also find $(\theta ^{\prime }_{z'})_{z'\in \mathsf {Z}'}\subset \mathcal {P}(\Omega )$ such that $\theta _{z'}^{\prime }$ is concentrated on $(Z')^{-1}(z')$ for $\mu '$ -a.e. $z' \in \mathsf {Z}'$ and $\mathbb {P} = \int \theta ^{\prime }_{z'}\,\mathrm {d}\mu '(z')$ . We can thus define $\boldsymbol {\gamma }:=\int (\theta _{z}\otimes \theta ^{\prime }_{z'})\,{\mathrm {d}} \boldsymbol {\mu }(z,z')\in {\mathcal {P}}({\Omega }\times {\Omega })$ , and it is immediate to check that $(Z\otimes Z')_\sharp \boldsymbol {\gamma }=\boldsymbol {\mu }$ since for every function $\varphi _1\in \mathrm {C}_b(\mathsf {Z})$ , $\varphi _{2}\in \mathrm {C}_b(\mathsf {Z'})$ ,

$$ \begin{align*} &\int \varphi_1(Z(\omega))\varphi_{2}(Z'(\omega'))\,\mathrm{d}\boldsymbol{\gamma}(\omega,\omega')\\ &= \int \bigg(\int \varphi_1(Z(\omega))\varphi_{2}(Z'(\omega'))\,\mathrm{d}\theta_{z}(\omega)\,\mathrm{d}\theta^{\prime}_{z'}(\omega') \bigg)\,\mathrm{d} \boldsymbol{\mu}(z,z') \\&= \int \int \varphi_1(Z(\omega))\,\mathrm{d}\theta_z(\omega) \int\varphi_{2}(Z'(\omega'))\,\mathrm{d}\theta^{\prime}_{z'}(\omega') \,\mathrm{d} \boldsymbol{\mu}(z,z') \\&= \int \Big( \varphi_1(z) \varphi_{2}(z')\Big) \,\mathrm{d} \boldsymbol{\mu}(z,z'). \end{align*} $$

Notice also that it is enough to verify that $(Z,Z')_{\sharp}\mathbb {P}$ and $\boldsymbol {\gamma }$ have the same integral for every function of the form $\varphi _1 \otimes \varphi _{2}$ as above to conclude that the two measures coincide (see, e.g., the proof of the above Corollary 3.15). We can then apply (3.12) and obtain (3.16).

Let us show the last part of the statement: we take $\mathsf {Z}=\mathsf {Z}':=\mathsf {X} \times \mathsf {Y}$ and, by (3.16) with $Z:=(X,Y)$ , $Z':=(X',Y')$ and $\boldsymbol {\mu }:=(\boldsymbol {i}_{\mathsf {X} \times \mathsf {Y}},\boldsymbol {i}_{\mathsf {X} \times \mathsf {Y}})_\sharp (X,Y)_{\sharp}\mathbb {P}$ , we have that

$$\begin{align*}(X,Y,X'\circ g_n,Y'\circ g_n)_{\sharp}\mathbb{P} \to \boldsymbol{\mu} \text{ in } \mathcal{P}((\mathsf{X} \times \mathsf{Y})^{2}). \end{align*}$$

We thus have that $(X,X'\circ g_n)_{\sharp}\mathbb {P} \to (\boldsymbol {i}_{\mathsf {X}},\boldsymbol {i}_{\mathsf {X}})_\sharp X_{\sharp}\mathbb {P}$ in $\mathcal {P}(\mathsf {X}^{2})$ . By Remark 3.14 with $\psi ((x,y),(x',y'))= |x|_{\mathsf {X}}^{p} + |x'|_{\mathsf {X}}^{p}$ , $x,x \in \mathsf {X}, y,y' \in \mathsf {Y}$ , we get, also using (3.11), that $(X,X'\circ g_n)_{\sharp}\mathbb {P} \to (\boldsymbol {i}_{\mathsf {X}},\boldsymbol {i}_{\mathsf {X}})_\sharp X_{\sharp}\mathbb {P}$ in $\mathcal {P}_p(\mathsf {X}^{2})$ . As a consequence (see, e.g., [Reference Ambrosio, Gigli and Savaré3, Proposition 7.1.5 and Lemma 5.1.7]), we get

$$ \begin{align*} \lim_{n\to\infty} \int |X-X'\circ g_n|^{p}\,\mathrm{d}\mathbb{P} &= \lim_{n\to\infty} \int |x-x'|^{p}\,\mathrm{d}\big((X,X'\circ g_n)_{\sharp}\mathbb{P})(x,x')\\ &= \int |x-x'|^{p}\,\mathrm{d}((\boldsymbol{i}_{\mathsf{X}},\boldsymbol{i}_{\mathsf{X}})_\sharp X_{\sharp}\mathbb{P})(x,x')\\ &=0. \end{align*} $$

The proof for Y and $Y'$ is identical.

Proof Let us consider the ${\mathfrak {N}}$ -refined standard Borel probability space $(\Omega , {\mathcal {B}}, \mathbb {P}, (\mathfrak {P}_N)_{N \in {\mathfrak {N}}})$ with ${\mathfrak {N}} =(2^k)_{k \in \mathbb {N}}$ ; let $\boldsymbol {\mu } \in \Gamma _o(\mu , \nu )$ and let $X'\in L^{p}(\Omega ,{\mathcal {B}},\mathbb {P}; \mathsf {X})$ be such that $X^{\prime }_{\sharp}\mathbb {P}=\nu $ . By Corollary 3.16 and Remark 3.17, there exist a strictly increasing sequence $(N_n)_n \subset {\mathfrak {N}}$ and maps $g_n \in \mathrm {S}(\Omega , {\mathcal {B}}, \mathbb {P}; {\mathcal {B}}_{{N_n}})$ such that

$$ \begin{align*} (X,X'\circ g_n)_{\sharp}\mathbb{P} \to \boldsymbol{\mu} \text{ in } \mathcal{P}_p(\mathsf{X}^{2}). \end{align*} $$

We have

$$ \begin{align*} \lim_{n\to\infty} \int |X-X'\circ g_n|^{p}\,\mathrm{d}\mathbb{P} &= \lim_{n\to\infty} \int |x-y|^{p}\,\mathrm{d}\big((X,X'\circ g_n)_{\sharp}\mathbb{P})(x,y)\\&=\int |x-y|^{p}\,\mathrm{d}\boldsymbol{\mu}(x,y) =W_p^{p}(\mu,\nu). \end{align*} $$

Thus, given $\varepsilon>0$ , it is always possible to find $n \in \mathbb {N}$ sufficiently large such that $Y:=X' \circ g_n$ satisfies the thesis.