Proof. By [Reference Godefroy and Saint-Raymond19, Theorem 4.1], there are continuous functions $(f_{n})_{n\in \mathbb {N}}$ on $SB(X)$ with values in X such that for each $F\in SB(X)$, we have $\overline {\{f_{n}(F):\; n\in \mathbb {N}\}} = F$. Consider the mapping $\Phi $ given by $\Phi (F)(\sum _{n=1}^{k} a_{n} e_{n}):=\|\sum _{n=1}^{k} a_{n}f_{n}(F)\|_{X}$ for every $F\in SB(X)$ and $a_{1},\ldots ,a_{k}\in \mathbb {Q}$. Then it is easy to see that $\Phi $ is the mapping we need.

Proof. Let $j:X\to Y$ be an isometry (not necessarily surjective). Then the mapping f given by $f(F):= j(F)$, $F\in SB(X)$, does the job, because $f^{-1}(E^{+}(U)) = E^{+}(j^{-1}(U))$ for every open set $U\subseteq Y$.

Proof. For each $\mu \in \mathcal {P}_{\infty }$, let us inductively define natural numbers $(n_{k}(\mu ))_{k\in \mathbb {N}}$ by

$$ \begin{align*}\begin{split} n_{1}(\mu) & :=\min\{n\in\mathbb{N}:\; \mu(e_{n})\neq 0\},\\ n_{k+1}(\mu) & :=\min\{n\in\mathbb{N}:\; e_{n_{1}(\mu)},\ldots,e_{n_{k}(\mu)}, e_{n}\text{ are linearly independent}\}. \end{split} \end{align*} $$

Consider the mapping $\Phi $ given by $\Phi (\mu )(\sum _{i=1}^{k} a_{n} e_{n}):=\mu (\sum _{i=1}^{k} a_{i}e_{n_{i}(\mu )})$ for every $\mu \in \mathcal {P}_{\infty }$ and $a_{1},\ldots ,a_{k}\in \mathbb {Q}$. It is easy to see that $\Phi (\mu )\in \mathcal {B}$ and that $X_{\mu }$ is isometric to $X_{\Phi (\mu )}$ for each $\mu \in \mathcal {P}_{\infty }$.

For all natural numbers $N_{1}<\ldots <N_{k}$, the set $\{\mu \in \mathcal {P}_{\infty }:\; n_{1}(\mu )=N_{1},\ldots ,n_{k}(\mu )=N_{k}\}$ is a $\boldsymbol{\Delta}_{2}^{0}$ set in $\mathcal {P}_{\infty }$. Indeed, we may prove it by induction on k because for each $k\in \mathbb {N}$ and each $\mu \in \mathcal {P}_{\infty }$, we have that $n_{1}(\mu )=N_{1},\ldots ,n_{k+1}(\mu )=N_{k+1}$ iff

$$ \begin{align*}\begin{split} n_{1}(\mu) & =N_{1}, \ldots, n_{k}(\mu)=N_{k}\quad\&\\ &\forall n=N_{k}+1,\ldots,N_{k+1}-1: e_{N_{1}},\ldots,e_{N_{k}},e_{n}\text{ are linearly dependent}\\ & \&\quad e_{N_{1}},\ldots,e_{N_{k+1}}\text{ are linearly independent}, \end{split}\end{align*} $$

which is an intersection of a $\boldsymbol{\Delta}_{2}^{0}$-condition (by the inductive assumption) with a closed and an open condition (by Lemma 2.4).

Let us pick $v = \sum _{i=1}^{k} a_{n} e_{n}\in V$ and an open interval I. Then

$$ \begin{align*} \Phi^{-1}(U[v,I]) = \{\mu\in\mathcal{P}_{\infty}:\; \mu(\sum_{i=1}^{k} a_{i}e_{n_{i}(\mu)})\in I\}, \end{align*} $$

which is a $\boldsymbol {\Delta }_{2}^{0}$ set in $\mathcal {P}_{\infty }$. Indeed, on one hand we have $\mu \in \Phi ^{-1}(U[v,I])$ iff there are natural numbers $N_{1}<N_{2}<\ldots <N_{k}$ such that $n_{1}(\mu )=N_{1},\ldots ,n_{k}(\mu )=N_{k}$ and $\mu (\sum _{i=1}^{k} a_{i} e_{N_{i}})\in I$, which witnesses that $\Phi ^{-1}(U[v,I])\in \boldsymbol {\Sigma }_{2}^{0}(\mathcal {P}_{\infty })$ as it is a countable union of $\boldsymbol{\Delta}_{2}^{0}$ sets. On the other hand, we have that $\mu \in \Phi ^{-1}(U[v,I])$ iff for each $l\in \mathbb {N}$, we have that either $n_{k}(\mu )> l$ or there are natural numbers $N_{1}<N_{2}<\ldots <N_{k}\leq l$ such that $n_{1}(\mu )=N_{1},\ldots ,n_{k}(\mu )=N_{k}$ and $\mu (\sum _{i=1}^{k} a_{i} e_{N_{i}})\in I$, which witnesses that $\Phi ^{-1}(U[v,I])\in \boldsymbol {\Pi }_{2}^{0}(\mathcal {P}_{\infty })$ as it is a countable intersection of $\boldsymbol{\Delta}_{2}^{0}$ sets.

This proves the ‘Moreover’ part, from which it easily follows that $\Phi $ is $\boldsymbol {\Sigma }_{2}^{0}$-measurable.

Proof. Consider the mapping $\Phi :\mathcal {B}\to (SB(X),\tau )$ defined as

$$ \begin{align*}\Phi(\nu):=\overline{\operatorname{span}}\{\chi_{k}(\nu):\; k\in\mathbb{N}\},\;\;\; \nu\in\mathcal{B}.\end{align*} $$

We have $X_{\nu }\equiv \Phi (\nu )$. For every open set $U\subseteq X$, using the $\boldsymbol {\Sigma }_{n}^{0}$-measurability of $\chi _{k}$’s, it is easy to see that $\Phi ^{-1}(E^{+}(U))$ is a $\boldsymbol {\Sigma }_{n}^{0}$-set in $\mathcal {B}$. Thus, by Lemma 3.5, the mapping $\Phi $ is $\boldsymbol {\Sigma }_{n+1}^{0}$-measurable.

Sketch of the proof.

By Remark 3.7, it suffices to find, for every $d\in \mathbb {N}\cup \{\infty \}$, a $\boldsymbol {\Sigma }_{3}^{0}$-measurable reduction from $\mathcal {B}_{d}$ to $SB(X)$, where $\mathcal {B}_{\infty } = \mathcal {B}$. This is done for every $d\in \mathbb {N}\cup \{\infty \}$ in a similar way. Let us concentrate further only on the case of $d=\infty $; the other cases are similar. From the proof of [Reference Kurka24, Lemma 2.4], it follows that there are Borel measurable mappings $\chi _{k} : \mathcal {B} \to X$, $k\in \mathbb {N}$, such that $X_{\mu }\equiv \overline {\operatorname {span}} \{\chi _{k}(\mu ):\; k\in \mathbb {N}\}$ for every $\mu \in \mathcal {B}$. A careful inspection of the proof actually shows that the mappings $\chi _{k}$ are $\boldsymbol {\Sigma }_{2}^{0}$-measurable (since this part is improved in the next subsection [see Proposition 3.12], we do not give any more details here). Thus, an application of Lemma 3.8 finishes the proof.

Proof of Theorem 3.11.

Follows immediately from Lemma 3.8 and Proposition 3.12.

Proof. This is similar to the proof of Theorem 3.10; the only modification is that we use Proposition 3.12 and Remark 3.13 instead of the reference to the proof of [Reference Kurka24, Lemma 2.4].

Proof of Proposition 3.15.

(1) $ \Rightarrow $ (2): Given such $ U $ and $ \chi _{k} : \mathcal {A} \to U, k \in \mathbb {N} $, we put

$$ \begin{align*} \alpha_{k}(\nu, \mu) = \Vert \chi_{k}(\nu) - \chi_{k}(\mu) \Vert, \quad \nu, \mu \in \mathcal{A}, k \in \mathbb{N}. \end{align*} $$

Denote by $ I_{\mu } : (c_{00}, \mu ) \to U $ the isometry given by $ e_{k} \mapsto \chi _{k}(\mu ) $. Let $ \nu \in \mathcal {A} $ and $ z^{*} \in (c_{00})^{\#} $ satisfying $ |z^{*}(x)| \leq \nu (x) $ be given. For $ x, y \in c_{00} $ with $ I_{\nu }x = I_{\nu }y $, we have $ |z^{*}(y - x)| \leq \nu (y - x) = \Vert I_{\nu }(y - x) \Vert = 0 $, and so $ z^{*}(x) = z^{*}(y) $. Thus, the formula

$$ \begin{align*} u^{*}(I_{\nu}x) = z^{*}(x), \quad x \in c_{00}, \end{align*} $$

defines a functional on $ I_{\nu }(c_{00}) $ such that $ |u^{*}(I_{\nu }x)| = |z^{*}(x)| \leq \nu (x) = \Vert I_{\nu }x \Vert $. By the Hahn-Banach theorem, we can extend $ u^{*} $ to the whole $ U $ in the way that

$$ \begin{align*} |u^{*}(u)| \leq \Vert u \Vert, \quad u \in U. \end{align*} $$

For every $ \mu \in \mathcal {A} $, let us put

$$ \begin{align*} \Gamma(\mu)(x) = u^{*}(I_{\mu}x), \quad x \in c_{00}. \end{align*} $$

We obtain $ |\Gamma (\mu )(x)| = |u^{*}(I_{\mu }x)| \leq \Vert I_{\mu }x \Vert = \mu (x) $ and $ |\Gamma (\mu )(e_{k}) - \Gamma (\lambda )(e_{k})| = |u^{*}(I_{\mu }e_{k}) - u^{*}(I_{\lambda }e_{k})| = |u^{*}(\chi _{k}(\mu ) - \chi _{k}(\lambda ))| \leq \Vert \chi _{k}(\mu ) - \chi _{k}(\lambda ) \Vert = \alpha _{k}(\mu , \lambda ) $ for $ \mu , \lambda \in \mathcal {A} $ and $ k \in \mathbb {N} $.

(2) $ \Rightarrow $ (1): Given such $ \alpha _{k} : \mathcal {A}^{2} \to [0, \infty ), k \in \mathbb {N} $, we define a subset of $ c_{00}(\mathcal {A} \times \mathbb {N}) $ by

$$ \begin{align*} \Omega = \mathrm{co} \bigg( \bigcup_{\mu} \Big\{ \sum_{k} a_{k} e_{\mu, k} : \mu \Big( \sum_{k} a_{k} e_{k} \Big) \leq 1 \Big\} \cup \bigcup_{\mu, \lambda, k} \big\{ c \cdot (e_{\mu, k} - e_{\lambda, k}) : |c| \cdot \alpha_{k}(\mu, \lambda) \leq 1 \big\} \bigg), \end{align*} $$

and denote the corresponding Minkowski functional by $ \varrho $. Let $ U $ be the completion of the quotient space $ X/N $, where $ X = (c_{00}(\mathcal {A} \times \mathbb {N}), \varrho ) $ and $ N = \{ x \in c_{00}(\mathcal {A} \times \mathbb {N}) : \rho (x) = 0 \} $. In what follows, we identify every $ x \in c_{00}(\mathcal {A} \times \mathbb {N}) $ with its equivalence class $ [x]_{N} \in U $. Let us define

$$ \begin{align*} \chi_{k} : \mathcal{A} \to U, \quad \nu \mapsto e_{\nu, k}. \end{align*} $$

As $ c \cdot (e_{\mu , k} - e_{\lambda , k}) \in \Omega $ whenever $ |c| \cdot \alpha _{k}(\mu , \lambda ) \leq 1 $, we obtain $ \varrho (e_{\mu , k} - e_{\lambda , k}) \leq \alpha _{k}(\mu , \lambda ) $: that is, $ \varrho (\chi _{k}(\mu ) - \chi _{k}(\lambda )) \leq \alpha _{k}(\mu , \lambda ) $. For a fixed $ \mu $, we have $ \alpha _{k}(\mu , \lambda ) \to \alpha _{k}(\mu , \mu ) = 0 $ as $ \lambda \to \mu $, and consequently $ \varrho (\chi _{k}(\mu ) - \chi _{k}(\lambda )) \to 0 $ as $ \lambda \to \mu $. Therefore, $ \chi _{k} $ is continuous on $ \mathcal {A} $. It follows that the image of $ \chi _{k} $ is separable. As these images contain all basic vectors $ e_{\nu , k} $, the space $ U $ is separable.

We need to show that

$$ \begin{align*} \nu(x) = \varrho(\overline{x}) \end{align*} $$

for fixed $ \nu \in \mathcal {A} $, $ x = \sum _{k \in \mathbb {N}} a_{k} e_{k} \in c_{00} $ and its image $ \overline {x} = \sum _{k \in \mathbb {N}} a_{k} e_{\nu , k} $. The inequality $ \nu (x) \geq \varrho (\overline {x}) $ follows immediately from the definition of $ \Omega $ (for any $ c \geq \nu (x) $ with $ c> 0 $, we have $ \nu (\frac {1}{c}x) \leq 1 $, and so $ \frac {1}{c}\overline {x} \in \Omega $, hence $ \varrho (\frac {1}{c}\overline {x}) \leq 1 $ and $ \varrho (\overline {x}) \leq c $). Let us show the opposite inequality $ \nu (x) \leq \varrho (\overline {x}) $. Using the Hahn-Banach theorem, we can pick $ z^{*} \in (c_{00})^{\#} $ satisfying $ z^{*}(x) = \nu (x) $ and $ |z^{*}(y)| \leq \nu (y) $ for every $ y \in c_{00} $. Let $ \Gamma : \mathcal {A} \to (c_{00})^{\#} $ be the mapping provided for $ \nu $ and $ z^{*} $, and let $ u^{*} \in (c_{00}(\mathcal {A} \times \mathbb {N}))^{\#} $ be given by

$$ \begin{align*} u^{*}(e_{\mu, k}) = \Gamma(\mu)(e_{k}), \quad \mu \in \mathcal{A}, k \in \mathbb{N}. \end{align*} $$

Then $ u^{*}(\overline {x}) = u^{*}(\sum _{k} a_{k} e_{\nu , k}) = \sum _{k} a_{k} u^{*}(e_{\nu , k}) = \sum _{k} a_{k} \Gamma (\nu )(e_{k}) = \Gamma (\nu )(\sum _{k} a_{k} e_{k}) = z^{*}(x) = \nu (x) $. It is sufficient to show that $ u^{*} \leq 1 $ on $ \Omega $ (equivalently $ |u^{*}(y)| \leq \varrho (y) $ for every $ y \in c_{00}(\mathcal {A} \times \mathbb {N}) $), since it follows that $ \nu (x) = u^{*}(\overline {x}) \leq \varrho (\overline {x}) $.

To show that $ u^{*} \leq 1 $ on $ \Omega $, we need to check that

$$ \begin{align*} \mu \Big( \sum_{k} b_{k} e_{k} \Big) \leq 1 \quad \Rightarrow \quad u^{*} \Big( \sum_{k} b_{k} e_{\mu, k} \Big) \leq 1 \end{align*} $$

and

$$ \begin{align*} |c| \cdot \alpha_{k}(\mu, \lambda) \leq 1 \quad \Rightarrow \quad u^{*} \big( c \cdot (e_{\mu, k} - e_{\lambda, k}) \big) \leq 1. \end{align*} $$

Concerning the first implication, we compute $ u^{*}(\sum _{k} b_{k} e_{\mu , k}) = \sum _{k} b_{k} u^{*}(e_{\mu , k}) = \sum _{k} b_{k} \Gamma (\mu )(e_{k}) = \Gamma (\mu )(\sum _{k} b_{k} e_{k}) \leq \mu (\sum _{k} b_{k} e_{k}) \leq 1 $. Concerning the second implication, we compute $ u^{*}(c \cdot (e_{\mu , k} - e_{\lambda , k})) = c u^{*}(e_{\mu , k}) - c u^{*}(e_{\lambda , k}) = c \Gamma (\mu )(e_{k}) - c \Gamma (\lambda )(e_{k}) \leq |c| \alpha _{k}(\mu , \lambda ) \leq 1 $.

(2) $ \Rightarrow $ (2’): The choice $ \beta _{k} = \alpha _{k} $ works. Indeed, if $ \Gamma $ is provided by (2), we can take $ \eta \cdot \Gamma $.

(2’) $ \Rightarrow $ (2): For every $ n \in \mathbb {N} $, let $ \beta _{k}^{n} : \mathcal {A}^{2} \to [0, \infty ), k \in \mathbb {N}, $ be provided by (2’) for $ \eta = (1 - 2^{-n}) $. We can assume that each $ \beta _{k}^{n} $ is a pseudometric. Indeed, instead of $ \beta _{k}^{n} $, we can take the maximal minorizing pseudometric $ \widetilde {\beta }^{n}_{k} $ (in such a case, $ \widetilde {\beta }^{n}_{k} $ is continuous, and since the function $ (\mu , \lambda ) \mapsto |\Gamma (\mu )(e_{k}) - \Gamma (\lambda )(e_{k})| $ is a pseudometric, if it minorizes $ \beta _{k}^{n} $, then it minorizes $ \widetilde {\beta }^{n}_{k} $ as well). Moreover, we can assume that $ \beta _{1}^{1} $ is a metric (it is possible to add a compatible metric on $ \mathcal {A} $ to $ \beta _{1}^{1} $).

Let us define

$$ \begin{align*} \alpha(\mu, \lambda) = \max_{n, k} \min \{ \beta_{k}^{n}(\mu, \lambda), 2^{- \max\{n, k\}} \}, \quad \mu, \lambda \in \mathcal{A}. \end{align*} $$

It is easy to check that $ \alpha $ is continuous. Due to our additional assumptions, $ \alpha $ is a metric on $ \mathcal {A} $. We want to show that there are some constants $ c_{k} $ such that the choice $ \alpha _{k} = c_{k} \cdot \alpha $ works.

Let $ \nu \in \mathcal {A} $ and $ z^{*} \in (c_{00})^{\#} $ satisfy $ |z^{*}(x)| \leq \nu (x) $ for every $ x \in c_{00} $. For every $ n \in \mathbb {N} $, there is a mapping $ \Gamma ^{n} : \mathcal {A} \to (c_{00})^{\#} $ such that

$$ \begin{align*} \Gamma^{n}(\nu) = (1 - 2^{-n}) \cdot z^{*}, \end{align*} $$

$$ \begin{align*} |\Gamma^{n}(\mu)(x)| \leq \mu(x), \quad \mu \in \mathcal{A}, \; x \in c_{00}, \end{align*} $$

and, if we denote

$$ \begin{align*} \gamma_{k}^{n}(\mu) = \Gamma^{n}(\mu)(e_{k}), \end{align*} $$

then

$$ \begin{align*} |\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\lambda)| \leq \beta_{k}^{n}(\mu, \lambda) \end{align*} $$

for every $ \mu , \lambda \in \mathcal {A} $ and every $ k $. Let us note that

$$ \begin{align*} |\gamma_{k}^{n}(\mu)| \leq 1, \end{align*} $$

as $ |\gamma _{k}^{n}(\mu )| = |\Gamma ^{n}(\mu )(e_{k})| \leq \mu (e_{k}) \leq 1 $ by the assumption on $ \mathcal {A} $.

Now, we define the desired mapping $ \Gamma $. For practical purposes, we first define

$$ \begin{align*} \gamma_{k}^{n}(\mu) = 0 \quad \text{for } n \in \mathbb{Z}, n \leq 0. \end{align*} $$

For every $ n \in \mathbb {Z} $, let $ f_{n} $ denote the piecewise linear function supported by $ [2^{-n-3}, 2^{-n-1}] $, which is linear on $ [2^{-n-3}, 2^{-n-2}] $ and $ [2^{-n-2}, 2^{-n-1}] $, and for which $ f_{n}(2^{-n-2}) = 1 $. In this way, we have $ \sum _{n \in \mathbb {Z}} f_{n} = 1 $ on $ (0, \infty ) $. We define

$$ \begin{align*} \gamma_{k}(\nu) = z^{*}(e_{k}) \end{align*} $$

and

$$ \begin{align*} \gamma_{k}(\mu) = \sum_{n \in \mathbb{Z}} f_{n}(\alpha(\mu, \nu)) \gamma_{k}^{n}(\mu), \quad \mu \neq \nu. \end{align*} $$

Finally, we put $ \Gamma (\mu )(e_{k}) = \gamma _{k}(\mu ) $, so $ \Gamma (\nu ) = z^{*} $ and $ \Gamma (\mu ) = \sum _{n \in \mathbb {N}} f_{n}(\alpha (\mu , \nu )) \Gamma ^{n}(\mu ) $ for $ \mu \neq \nu $. In both cases $ \mu = \nu $ and $ \mu \neq \nu $, it follows that $ |\Gamma (\mu )(x)| \leq \mu (x) $ for every $ x \in c_{00} $. It remains to prove the inequality

$$ \begin{align*} |\gamma_{k}(\mu) - \gamma_{k}(\lambda)| \leq c_{k} \cdot \alpha(\mu, \lambda) \end{align*} $$

for some suitable constants $ c_{k} $.

Let us show that the implication

(3.1)$$ \begin{align} \alpha(\mu, \lambda) < 2^{-n} \quad \Rightarrow \quad |\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\lambda)| \leq 2^{k+1} \alpha(\mu, \lambda) \end{align} $$

holds. Clearly, we can suppose that $ n \geq 1 $. If $ \alpha (\mu , \lambda ) \geq 2^{-k} $, then $ 2^{k+1} \alpha (\mu , \lambda ) \geq 2 \geq |\gamma _{k}^{n}(\mu ) - \gamma _{k}^{n}(\lambda )| $. So, let us assume that $ \alpha (\mu , \lambda ) < 2^{-k} $. Since $ \min \{ \beta _{k}^{n}(\mu , \lambda ), 2^{- \max \{n, k\}} \} \leq \alpha (\mu , \lambda ) < 2^{- \max \{n, k\}} $, we have $ \min \{ \beta _{k}^{n}(\mu , \lambda ), 2^{- \max \{n, k\}} \} = \beta _{k}^{n}(\mu , \lambda ) $, and so $ |\gamma _{k}^{n}(\mu ) - \gamma _{k}^{n}(\lambda )| \leq \beta _{k}^{n}(\mu , \lambda ) = \min \{ \beta _{k}^{n}(\mu , \lambda ), 2^{- \max \{n, k\}} \} \leq \alpha (\mu , \lambda ) \leq 2^{k+1} \alpha (\mu , \lambda ) $.

Next, we show that

(3.2)$$ \begin{align} 2^{-n-4} \leq \alpha(\mu, \nu) < 2^{-n} \quad \Rightarrow \quad |\gamma_{k}^{n}(\mu) - \gamma_{k}(\nu)| \leq (2^{k+1} + 16) \alpha(\mu, \nu). \end{align} $$

If $ n \geq 1 $, then $ \gamma _{k}(\nu ) - \gamma _{k}^{n}(\nu ) = z^{*}(e_{k}) - (1 - 2^{-n}) z^{*}(e_{k}) = 2^{-n} z^{*}(e_{k}) $. If $ n \leq 0 $, then $ \gamma _{k}(\nu ) - \gamma _{k}^{n}(\nu ) = z^{*}(e_{k}) $. In both cases, $ |\gamma _{k}(\nu ) - \gamma _{k}^{n}(\nu )| \leq 2^{-n} |z^{*}(e_{k})| \leq 2^{-n} \nu (e_{k}) \leq 2^{-n} \leq 2^{4} \alpha (\mu , \nu ) $. Using formula (3.1), we can compute

$$ \begin{align*} |\gamma_{k}^{n}(\mu) - \gamma_{k}(\nu)| \leq |\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\nu)| + |\gamma_{k}^{n}(\nu) - \gamma_{k}(\nu)| \leq (2^{k+1} + 2^{4}) \alpha(\mu, \nu). \end{align*} $$

Further, it follows from formula (3.2) that

(3.3)$$ \begin{align} |\gamma_{k}(\mu) - \gamma_{k}(\nu)| \leq (2^{k+1} + 16) \alpha(\mu, \nu). \end{align} $$

Indeed, since $ f_{n} $ is supported by $ [2^{-n-3}, 2^{-n-1}] $, we always have

$$ \begin{align*} f_{n}(\alpha(\mu, \nu)) |\gamma_{k}^{n}(\mu) - \gamma_{k}(\nu)| \leq f_{n}(\alpha(\mu, \nu)) (2^{k+1} + 16) \alpha(\mu, \nu), \end{align*} $$

and it is sufficient to use that $ \gamma _{k}(\mu ) - \gamma _{k}(\nu ) = \sum _{n \in \mathbb {Z}} f_{n}(\alpha (\mu , \nu )) (\gamma _{k}^{n}(\mu ) - \gamma _{k}(\nu )) $ for $ \mu \neq \nu $.

Now, we are going to investigate the value $ |\gamma _{k}(\mu ) - \gamma _{k}(\lambda )| $. First, we have

(3.4)$$ \begin{align} \alpha(\lambda, \nu) \geq 2 \alpha(\mu, \nu) \quad \Rightarrow \quad |\gamma_{k}(\mu) - \gamma_{k}(\lambda)| \leq 3 \cdot (2^{k+1} + 16) \alpha(\mu, \lambda). \end{align} $$

Indeed, as $ \alpha (\mu , \lambda ) \geq \alpha (\lambda , \nu ) - \alpha (\mu , \nu ) \geq 2 \alpha (\mu , \nu ) - \alpha (\mu , \nu ) = \alpha (\mu , \nu ) $, we can apply inequality (3.3) and write

$$ \begin{align*} |\gamma_{k}(\mu) - \gamma_{k}(\lambda)| & \leq |\gamma_{k}(\mu) - \gamma_{k}(\nu)| + |\gamma_{k}(\lambda) - \gamma_{k}(\nu)| \\ & \leq (2^{k+1} + 16) (\alpha(\mu, \nu) + \alpha(\lambda, \nu)) \\ & = (2^{k+1} + 16) (\alpha(\lambda, \nu) - \alpha(\mu, \nu) + 2 \alpha(\mu, \nu)) \\ & \leq (2^{k+1} + 16)(1 + 2) \alpha(\mu, \lambda). \end{align*} $$

Now, we prove the last but most challenging implication:

(3.5)$$ \begin{align} \alpha(\mu, \nu) \leq \alpha(\lambda, \nu) < 2 \alpha(\mu, \nu) \quad \Rightarrow \quad |\gamma_{k}(\mu) - \gamma_{k}(\lambda)| \leq [12 (2^{k+1} + 16) + 2^{k+1}] \alpha(\mu, \lambda). \end{align} $$

Let us compute

$$ \begin{align*} & \gamma_{k}(\mu) - \gamma_{k}(\lambda) = \sum_{n \in \mathbb{Z}} \big[ f_{n}(\alpha(\mu, \nu)) \gamma_{k}^{n}(\mu) - f_{n}(\alpha(\lambda, \nu)) \gamma_{k}^{n}(\lambda) \big] \\ & = \sum_{n \in \mathbb{Z}} \big[ f_{n}(\alpha(\mu, \nu)) \gamma_{k}^{n}(\mu) - f_{n}(\alpha(\lambda, \nu)) \gamma_{k}^{n}(\mu) + f_{n}(\alpha(\lambda, \nu)) \gamma_{k}^{n}(\mu) - f_{n}(\alpha(\lambda, \nu)) \gamma_{k}^{n}(\lambda) \big] \\ & = \sum_{n \in \mathbb{Z}} [f_{n}(\alpha(\mu, \nu)) - f_{n}(\alpha(\lambda, \nu))] \gamma_{k}^{n}(\mu) + \sum_{n \in \mathbb{Z}} f_{n}(\alpha(\lambda, \nu)) [\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\lambda)] \\ & = \sum_{n \in \mathbb{Z}} [f_{n}(\alpha(\mu, \nu)) - f_{n}(\alpha(\lambda, \nu))] (\gamma_{k}^{n}(\mu) - \gamma_{k}(\nu)) + \sum_{n \in \mathbb{Z}} f_{n}(\alpha(\lambda, \nu)) [\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\lambda)]. \end{align*} $$

Hence, $ |\gamma _{k}(\mu ) - \gamma _{k}(\lambda )| $ is less than or equal to

$$ \begin{align*} \sum_{n \in \mathbb{Z}} |f_{n}(\alpha(\mu, \nu)) - f_{n}(\alpha(\lambda, \nu))| |\gamma_{k}^{n}(\mu) - \gamma_{k}(\nu)| + \sum_{n \in \mathbb{Z}} f_{n}(\alpha(\lambda, \nu)) |\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\lambda)|. \end{align*} $$

Let us notice that

• • $ f_{n}(\alpha (\mu , \nu )) \neq 0 $ iff $ 2^{-n-3} < \alpha (\mu , \nu ) < 2^{-n-1} $,

• • $ f_{n}(\alpha (\lambda , \nu )) \neq 0 $ iff $ 2^{-n-3} < \alpha (\lambda , \nu ) < 2^{-n-1} $, and $ 2^{-n-4} < \alpha (\mu , \nu ) < 2^{-n-1} $ in this case,

• • the function $ f_{n} $ is Lipschitz with the constant $ 2^{n+3} $.

So, if $ f_{n}(\alpha (\mu , \nu )) \neq 0 $ or $ f_{n}(\alpha (\lambda , \nu )) \neq 0 $, then $ 2^{-n-4} < \alpha (\mu , \nu ) < 2^{-n-1} $, and formula (3.2) can be applied. We obtain for the first sum that

$$ \begin{align*} \sum_{n \in \mathbb{Z}} |f_{n}(\alpha(\mu, \nu)) & - f_{n}(\alpha(\lambda, \nu))| |\gamma_{k}^{n}(\mu) - \gamma_{k}(\nu)| \\ & \leq \sum_{2^{-n-4} < \alpha(\mu, \nu) < 2^{-n-1}} 2^{n+3} |\alpha(\mu, \nu) - \alpha(\lambda, \nu)| (2^{k+1} + 16) \alpha(\mu, \nu) \\ & \leq \sum_{2^{-n-4} < \alpha(\mu, \nu) < 2^{-n-1}} 2^{n+3} \alpha(\mu, \lambda) (2^{k+1} + 16) 2^{-n-1} \\ & \leq 3 \cdot 2^{2} (2^{k+1} + 16) \alpha(\mu, \lambda). \end{align*} $$

Concerning the second sum, we notice that if $ f_{n}(\alpha (\lambda , \nu )) \neq 0 $, then $ \alpha (\lambda , \nu ) < 2^{-n-1} $, and so $ \alpha (\mu , \lambda ) \leq \alpha (\mu , \nu ) + \alpha (\lambda , \nu ) \leq 2\alpha (\lambda , \nu ) < 2^{-n} $. Applying formula (3.1), we obtain

$$ \begin{align*} \sum_{n \in \mathbb{Z}} f_{n}(\alpha(\lambda, \nu)) |\gamma_{k}^{n}(\mu) - \gamma_{k}^{n}(\lambda)| \leq \sum_{n \in \mathbb{Z}} f_{n}(\alpha(\lambda, \nu)) 2^{k+1} \alpha(\mu, \lambda) = 2^{k+1} \alpha(\mu, \lambda), \end{align*} $$

and formula (3.5) follows.

Finally, we finish the proof with the observation that formulas (3.4) and (3.5) provide

$$ \begin{align*} |\gamma_{k}(\mu) - \gamma_{k}(\lambda)| \leq [12 (2^{k+1} + 16) + 2^{k+1}] \alpha(\mu, \lambda). \end{align*} $$

We can suppose that $ \alpha (\mu , \nu ) \leq \alpha (\lambda , \nu ) $. If $ \alpha (\lambda , \nu ) \geq 2 \alpha (\mu , \nu ) $, we use formula (3.4), and if $ \alpha (\lambda , \nu ) < 2 \alpha (\mu , \nu ) $, we use formula (3.5).

Proof. Let $ \eta \in [0, 1) $ be given. We fix numbers $ \kappa _{k} < 1 $ such that $ \eta \leq \kappa _{1} < \kappa _{2} < \kappa _{3} < \dots $. For every $ \mu , \lambda \in \mathcal {B}_{(1)} $, we define recursively

$$ \begin{align*} \beta_{1}(\mu, \lambda) = 0 \end{align*} $$

and

$$ \begin{align*} \beta_{k+1}(\mu, \lambda) = \sup \bigg\{ & \Big| \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \lambda \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) \Big| + \sum_{i=1}^{k} |a_{i}| \beta_{i}(\mu, \lambda) : \\ & a_{1}, \dots, a_{k} \in \mathbb{R}, \; \min \Big\{ \mu \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big), \lambda \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big) \Big\} < \frac{2\kappa_{k+1}}{\kappa_{k+1} - \kappa_{k}} \bigg\}. \end{align*} $$

Clearly, $ \beta _{k}(\nu , \nu ) = 0 $ for every $ \nu \in \mathcal {B}_{(1)} $. Let us sketch a proof of continuity of the functions $ \beta _{k} $. The function $ \beta _{1} = 0 $ is obviously continuous. Assuming that $ \beta _{i} $ is continuous for every $ i \leq k $, we consider for $ \delta> 0 $ the set

$$ \begin{align*} \mathcal{U}_{\delta}^{k+1}(\mu) = \bigg\{ \mu^{\prime} \in \mathcal{B}_{(1)} : \Big( \forall x \in \mathrm{span} \{ e_{1}, \dots, e_{k+1} \} \setminus \{ 0 \} : (1 + \delta)^{-1} < \frac{\mu^{\prime}(x)}{\mu(x)} < 1 + \delta \Big) \bigg\}. \end{align*} $$

Using Lemma 2.2, it is easy to see that $ \mathcal {U}_{\delta }^{k+1}(\mu ) $ is an open neighborhood of $ \mu $ in $ \mathcal {B}_{(1)} $. Given $ \varepsilon> 0 $, we can find $ \delta> 0 $ such that $ |\beta _{k+1}(\mu ^{\prime }, \lambda ^{\prime }) - \beta _{k+1}(\mu , \lambda )| < \varepsilon $ for every $ (\mu ^{\prime }, \lambda ^{\prime }) \in \mathcal {U}_{\delta }^{k+1}(\mu ) \times \mathcal {U}_{\delta }^{k+1}(\lambda ) $. The details are left to the reader.

Let us prove that the functions $ \beta _{k} $ work. Given $ \nu \in \mathcal {B}_{(1)} $ and $ z^{*} \in (c_{00})^{\#} $ satisfying $ |z^{*}(x)| \leq \nu (x) $ for every $ x \in c_{00} $, we define first

$$ \begin{align*} \gamma_{1}(\mu) = \eta \cdot z^{*}(e_{1}), \quad \mu \in \mathcal{B}_{(1)}. \end{align*} $$

Recursively, we define for every $ k \in \mathbb {N} $ functions

$$ \begin{align*} u_{k+1}(\mu) = \sup_{a_{1}, \dots, a_{k}} \Big[ - \kappa_{k+1} \mu \Big( - e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) + \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu) \Big], \end{align*} $$

$$ \begin{align*} v_{k+1}(\mu) = \inf_{a_{1}, \dots, a_{k}} \Big[ \kappa_{k+1} \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu) \Big] \end{align*} $$

and

$$ \begin{align*} \gamma_{k+1}(\mu) = p_{k+1} u_{k+1}(\mu) + q_{k+1} v_{k+1}(\mu), \end{align*} $$

where numbers $ p_{k+1} \geq 0, q_{k+1} \geq 0 $ with $ p_{k+1} + q_{k+1} = 1 $ are chosen in the way that

(3.6)$$ \begin{align} \gamma_{k+1}(\nu) = \eta \cdot z^{*}(e_{k+1}). \end{align} $$

Let us check that it is possible to choose such numbers. Note first that $ \gamma _{1}(\nu ) = \eta \cdot z^{*}(e_{1}) $. Assuming that the functions $ \gamma _{i} $ are already defined and satisfy $ \gamma _{i}(\nu ) = \eta \cdot z^{*}(e_{i}) $ for $ i \leq k $, we notice that, for every $ a_{1}, \dots , a_{k} \in \mathbb {R} $,

$$ \begin{align*} \pm \eta \cdot z^{*}(e_{k+1}) + \sum_{i=1}^{k} a_{i} \gamma_{i}(\nu) = \eta \cdot z^{*} \Big( \pm e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) \leq \kappa_{k+1} \nu \Big( \pm e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big), \end{align*} $$

and consequently

$$ \begin{align*} \eta \cdot z^{*}(e_{k+1}) \geq - \kappa_{k+1} \nu \Big( - e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) + \sum_{i=1}^{k} a_{i} \gamma_{i}(\nu), \end{align*} $$

$$ \begin{align*} \eta \cdot z^{*}(e_{k+1}) \leq \kappa_{k+1} \nu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\nu). \end{align*} $$

This gives

$$ \begin{align*} u_{k+1}(\nu) \leq \eta \cdot z^{*}(e_{k+1}) \leq v_{k+1}(\nu), \end{align*} $$

and it follows that suitable $ p_{k+1} $ and $ q_{k+1} $ do exist.

Let us prove that

(3.7)$$ \begin{align} \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu) \leq \kappa_{k} \mu \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big) \end{align} $$

for every $ \mu \in \mathcal {B}_{(1)} $, $ k \in \mathbb {N} $ and $ a_{1}, \dots , a_{k} \in \mathbb {R} $. For $ k = 1 $, we just write $ a_{1} \gamma _{1}(\mu ) = a_{1} \eta \cdot z^{*}(e_{1}) \leq |a_{1}| \kappa _{1} \nu (e_{1}) = |a_{1}| \kappa _{1} = |a_{1}| \kappa _{1} \mu (e_{1}) = \kappa _{1} \mu (a_{1} e_{1}) $. Assume that inequality (3.7) is valid for $ k $. We show first that

$$ \begin{align*} u_{k+1}(\mu) \leq \gamma_{k+1}(\mu) \leq v_{k+1}(\mu). \end{align*} $$

Clearly, it is sufficient to show just that $ u_{k+1}(\mu ) \leq v_{k+1}(\mu ) $. Given $ b_{1}, \dots , b_{k} $ and $ c_{1}, \dots , c_{k} $, we need to check that

$$ \begin{align*} - \kappa_{k+1} \mu \Big( - e_{k+1} + \sum_{i=1}^{k} b_{i} e_{i} \Big) + \sum_{i=1}^{k} b_{i} \gamma_{i}(\mu) \leq \kappa_{k+1} \mu \Big( e_{k+1} + \sum_{i=1}^{k} c_{i} e_{i} \Big) - \sum_{i=1}^{k} c_{i} \gamma_{i}(\mu). \end{align*} $$

But this is easy, as

$$ \begin{align*} \sum_{i=1}^{k} b_{i} \gamma_{i}(\mu) + \sum_{i=1}^{k} c_{i} \gamma_{i}(\mu) & = \sum_{i=1}^{k} (b_{i} + c_{i}) \gamma_{i}(\mu) \leq \kappa_{k} \mu \Big( \sum_{i=1}^{k} (b_{i} +c_{i}) e_{i} \Big) \\ & \leq \kappa_{k+1} \Big[ \mu \Big( - e_{k+1} + \sum_{i=1}^{k} b_{i} e_{i} \Big) + \mu \Big( e_{k+1} + \sum_{i=1}^{k} c_{i} e_{i} \Big) \Big]. \end{align*} $$

Now, let us verify inequality (3.7) for $ k + 1 $. We can suppose that $ a_{k+1} = \pm 1 $. For $ a_{k+1} = 1 $, it is enough to use

$$ \begin{align*} \gamma_{k+1}(\mu) \leq v_{k+1}(\mu) \leq \kappa_{k+1} \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu), \end{align*} $$

and for $ a_{k+1} = -1 $, it is enough to use

$$ \begin{align*} \gamma_{k+1}(\mu) \geq u_{k+1}(\mu) \geq - \kappa_{k+1} \mu \Big( - e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) + \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu). \end{align*} $$

Next, let us prove that

(3.8)$$ \begin{align} |\gamma_{k}(\mu) - \gamma_{k}(\lambda)| \leq \beta_{k}(\mu, \lambda) \end{align} $$

for every $ \mu , \lambda \in \mathcal {B}_{(1)} $ and $ k \in \mathbb {N} $. This is clear for $ k = 1 $, as $ \gamma _{1} $ is constant. Assume that inequality (3.8) is valid for $ i \leq k $. To prove it for $ k + 1 $, it is sufficient to show the inequalities

$$ \begin{align*} |u_{k+1}(\mu) - u_{k+1}(\lambda)| \leq \beta_{k+1}(\mu, \lambda) \quad \text{and} \quad |v_{k+1}(\mu) - v_{k+1}(\lambda)| \leq \beta_{k+1}(\mu, \lambda). \end{align*} $$

We consider only the function $ v_{k+1} $, since the inequality for $ u_{k+1} $ can be shown in the same way. Let us note first that, in the definition of $ v_{k+1}(\mu ) $, it is possible to take the infimum only over $ k $-tuples with

$$ \begin{align*} \mu \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big) < \frac{2\kappa_{k+1}}{\kappa_{k+1} - \kappa_{k}}. \end{align*} $$

Indeed, for $ a_{1}, \dots , a_{k} $ that do not satisfy this condition, using inequality(3.7), we obtain

$$ \begin{align*} \kappa_{k+1} \mu \Big( e_{k+1} + & \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu) \\ & \geq \kappa_{k+1} \mu \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big) - \kappa_{k+1} \mu(e_{k+1}) - \kappa_{k} \mu \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big) \\ & = (\kappa_{k+1} - \kappa_{k}) \mu \Big( \sum_{i=1}^{k} a_{i} e_{i} \Big) - \kappa_{k+1} \\ & \geq 2 \kappa_{k+1} - \kappa_{k+1} = \kappa_{k+1} = \kappa_{k+1} \mu \Big( e_{k+1} + \sum_{i=1}^{k} 0 \cdot e_{i} \Big) - \sum_{i=1}^{k} 0 \cdot \gamma_{i}(\mu). \end{align*} $$

Now, inequality (3.8) is provided by the following computation, in which every sup/inf is meant over $ k $-tuples with $ \mu \Big ( \sum _{i=1}^{k} a_{i} e_{i} \Big ) < \frac {2\kappa _{k+1}}{\kappa _{k+1} - \kappa _{k}} $ or $ \lambda \Big ( \sum _{i=1}^{k} a_{i} e_{i} \Big ) < \frac {2\kappa _{k+1}}{\kappa _{k+1} - \kappa _{k}} $:

$$ \begin{align*} & |v_{k+1}(\mu) - v_{k+1}(\lambda)| \\ & = \bigg| \inf \Big[ \kappa_{k+1} \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu) \Big] \\ & \hspace{4cm} - \inf \Big[ \kappa_{k+1} \lambda \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\lambda) \Big] \bigg| \\ & \leq \sup \bigg| \Big[ \kappa_{k+1} \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\mu) \Big] \\ & \hspace{4cm} - \Big[ \kappa_{k+1} \lambda \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \sum_{i=1}^{k} a_{i} \gamma_{i}(\lambda) \Big] \bigg| \\ & = \sup \bigg| \kappa_{k+1} \Big[ \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \lambda \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) \Big] - \sum_{i=1}^{k} a_{i} \big( \gamma_{i}(\mu) - \gamma_{i}(\lambda) \big) \bigg| \\ & \leq \sup \bigg[ \Big| \mu \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) - \lambda \Big( e_{k+1} + \sum_{i=1}^{k} a_{i} e_{i} \Big) \Big| + \sum_{i=1}^{k} |a_{i}| \beta_{i}(\mu, \lambda) \bigg] = \beta_{k+1}(\mu, \lambda). \end{align*} $$

Finally, as usual, we put $ \Gamma (\mu )(e_{k}) = \gamma _{k}(\mu ) $. The required properties of $ \Gamma $ follow now from inequalities (3.6), (3.7) and (3.8). Thus, the functions $ \beta _{k} $ work, and the proof of the lemma is completed.

Proof of Proposition 3.12.

Let $ X $ be an isometrically universal separable Banach space. By Lemma 3.18, the condition (2’) from Proposition 3.15 is valid for $ \mathcal {A} = \mathcal {B}_{(1)} $. Hence, the condition (1) from this proposition is valid for $ \mathcal {A} = \mathcal {B}_{(1)} $ as well. There are a separable Banach space $ U $ and continuous mappings $ \chi _{k} : \mathcal {B}_{(1)} \to U, k \in \mathbb {N}, $ such that

$$ \begin{align*} \Big\Vert \sum_{k=1}^{n} a_{k} \chi_{k}(\nu) \Big\Vert = \nu \Big( \sum_{k=1}^{n} a_{k} e_{k} \Big) \end{align*} $$

for every $ \sum _{k=1}^{n} a_{k} e_{k} \in c_{00} $ and every $ \nu \in \mathcal {B}_{(1)} $. Since $ X $ contains an isometric copy of $ U $, we can suppose that $ U \subseteq X $.

Let us consider the continuous mapping $ \Psi : \mathcal {B} \to \mathcal {B}_{(1)} $ given by

$$ \begin{align*} \Psi(\mu) \Big( \sum_{k=1}^{n} a_{k} e_{k} \Big) = \mu \Big( \sum_{k=1}^{n} \frac{a_{k}}{\mu(e_{k})} e_{k} \Big). \end{align*} $$

If we define

$$ \begin{align*} \widetilde{\chi}_{k}(\mu) = \mu(e_{k}) \cdot \chi_{k}(\Psi(\mu)), \quad \mu \in \mathcal{B}, \end{align*} $$

for each $ k \in \mathbb {N} $, then we get

$$ \begin{align*} \Big\Vert \sum_{k=1}^{n} b_{k} \widetilde{\chi}_{k}(\mu) \Big\Vert = \Big\Vert \sum_{k=1}^{n} b_{k} \mu(e_{k}) \chi_{k}(\Psi(\mu)) \Big\Vert = \Psi(\mu) \Big( \sum_{k=1}^{n} b_{k} \mu(e_{k}) e_{k} \Big) = \mu \Big( \sum_{k=1}^{n} b_{k} e_{k} \Big) \end{align*} $$

for every $ \sum _{k=1}^{n} b_{k} e_{k} \in c_{00} $ and every $ \mu \in \mathcal {B} $.

Question 1. Does there exist a continuous mapping $\Phi :\mathcal {P}_{\infty }\to \mathcal {B}$ such that for every $\mu \in \mathcal {P}_{\infty }$, we have $X_{\mu }\equiv X_{\Phi (\mu )}$?

Question 2. Let X be an isometrically universal separable Banach space, and let $\tau $ be an admissible topology on $SB(X)$. Does there exist a $\boldsymbol {\Sigma }_{2}^{0}$-measurable mapping $\Phi :\mathcal {P}_{\infty } \to (SB(X),\tau )$ such that for every $\mu \in \mathcal {P}_{\infty }$, we have $X_{\mu }\equiv \Phi (\mu )$?

Proof. 1: Pick $C>0$ such that $C\sum _{i=1}^{n}|\lambda _{i}|\leq \|\sum _{i=1}^{n} \lambda _{i} e_{i}\|$ for every $(\lambda _{i})_{i=1}^{n}\in \mathbb {R}^{n}$. Then for any $x = \sum _{i=1}^{n} \lambda _{i} e_{i}$, we have

$$ \begin{align*}\|Tx - x\|\leq \sum_{i=1}^{n} |\lambda_{i}| \|x_{i}-e_{i}\| < \frac{\varepsilon}{C} \|x\|.\end{align*} $$

Thus, $\|T-Id_{E}\|<\frac {\varepsilon }{C}$, $\|T\|\leq 1+\frac {\varepsilon }{C}$ and $\|Tx\|\geq (1-\frac {\varepsilon }{C})\|x\|= (1+\frac {\varepsilon }{C-\varepsilon })^{-1}\|x\|$. Thus, we may put $\phi _{2}^{\mathfrak {b}_{E}}(\varepsilon ):=\frac {\varepsilon }{C-\varepsilon }$ for $\varepsilon \in [0,C)$.

2: Let $\varepsilon>0$, $T:X\to Y$ and N be as in the assumptions. Then for every $y\in S_{Y}$, there is $x\in N$ with $\|x-\frac {T^{-1}(y)}{\|T^{-1}(y)\|}\|<\varepsilon $. Thus, we have

$$ \begin{align*}\begin{split} \|y - Tx\|&\leq \Big\|y - \frac{y}{\|T^{-1}(y)\|}\Big\| + \|T\|\cdot\Big\|x-\frac{T^{-1}(y)}{\|T^{-1}(y)\|}\Big\|\\ & < \Big|1 - \frac{1}{\|T^{-1}y\|}\Big| + (1+\varepsilon)\varepsilon\leq \varepsilon + 2\varepsilon = 3\varepsilon. \end{split}\\[-46pt]\end{align*} $$

Proof. It suffices to define such a norm $\nu $ on $\operatorname {span} A$ since then we can easily find some extension to the whole V. We assume that $0\notin A$, and moreover, we can assume that no two elements of A lie in the same one-dimensional subspace: that is, are scalar multiples of each other. Indeed, otherwise we would find a subset $A^{\prime }\subseteq A$ where no elements are scalar multiples of each other and every element of A is a scalar multiple, necessarily rational scalar multiple, of some element from $A^{\prime }$. Then proving the fact for $A^{\prime }$ for sufficiently small $\delta $ automatically proves it for A and $\varepsilon $.

We enumerate A as $\{a_{1},\ldots ,a_{n}\}$ and so that the first k elements $a_{1},\ldots ,a_{k}$, for some $k\leq n$, are linearly independent and form a basis of $\operatorname {span} A$.

Suppose the claim is proved. Then for every $i\leq n$, we set $\nu ^{\prime }(a_{i})$ to be an arbitrary positive rational number in the interval $\left [\mu (a_{i}),\min \{K_{i},\mu (a_{i})+\varepsilon \}\right )$. From the assumption it is now clear that for all $i\leq n$, we have

$$ \begin{align*}\nu^{\prime}(a_{i})\leq \inf\{\sum_{j=1}^{n} |\alpha_{j}|\nu^{\prime}(a_{j})\colon a_{i}=\sum_{j=1}^{n} \alpha_{j} a_{j}\}.\end{align*} $$

We extend $\nu ^{\prime }$ to a norm $\nu $ on $\operatorname {span} A$ by the formula

$$ \begin{align*}\nu(v):=\inf\{\sum_{i=1}^{n} |\alpha_{i}| \nu^{\prime}(a_{i})\colon v=\sum_{i=1}^{n} \alpha_{i} a_{i}\},\end{align*} $$

for $v\in \operatorname {span} A$. From the previous assumption, it follows that $\nu (a_{i})=\nu ^{\prime }(a_{i})$ for all $i\leq n$. Moreover, $\nu $ is indeed a norm since $\nu (a_{i})>0$ for all $i\leq n$, and the infimum in the definition of $\nu $ is, by compactness, always attained.

It remains to prove the claim. Let $\|\cdot \|_{2}$ be the $\ell _{2}$ norm on $\operatorname {span} A$ with $a_{1},\ldots ,a_{k}$ the orthonormal basis. For each $m\in \mathbb {N}$, set $\mu _{m}:=\mu +\frac {\|\cdot \|_{2}}{m}$. Clearly $\mu _{m}\to \mu $, so it suffices to show that each $\mu _{m}$ satisfies the condition from the claim. Suppose that for some $m\in \mathbb {N}$ and $i\leq n$, we have

$$ \begin{align*}\mu_{m}(a_{i})=\inf\{\sum_{j\in J}\mu_{m}(\alpha_{j} a_{j})\colon i\notin J\subseteq \{1,\ldots,n\}, a_{i}=\sum_{j\in J} \alpha_{j} a_{j}\}.\end{align*} $$

By compactness, the infimum is attained: that is, there exists $(\alpha _{j})_{j\leq n}$, with $\alpha _{i}=0$, $a_{i}=\sum _{j\leq n} \alpha _{j} a_{j}$ and $\mu _{m}(a_{i})=\sum _{j\leq n} \mu _{m}(\alpha _{j} a_{j})$. Indeed, if the infimum is approximated by a sequence $(\alpha ^{l}_{1},\ldots ,\alpha ^{l}_{n})_{l\in \mathbb {N}}\subseteq \mathbb {R}^{n}$, then since each coordinate is bounded (because up to finitely many ls, we have $\sum _{j=1}^{n} \mu _{m}(\alpha _{j}^{l} a_{j})\leq 2 \mu _{m}(a_{i})$), we may pass to a convergent subsequence and attain the infimum at the limit. The $\ell _{2}$ norm $\|\cdot \|_{2}$ is strictly convex, so $\|a_{i}\|_{2}{<}\sum _{j\leq n} \|\alpha _{j} a_{j}\|_{2}$, while $\mu $ by triangle inequality satisfies $\mu (a_{i})\leq \sum _{j\leq n} \mu (\alpha _{j} a_{j})$. Since $\mu _{m}$ is the sum of $\mu $ and a positive multiple of the $\ell _{2}$ norm, we must have $\mu _{m}(a_{i})<\sum _{j\leq n} \mu _{m}(\alpha _{j} a_{j})$, a contradiction.

Notation 4.5. For a finite set $A\subseteq V$ and $P,P^{\prime }$ partial functions on V (i.e., functions whose domains are subsets of V) with $A\subseteq \operatorname {dom}(P),\operatorname {dom}(P^{\prime })$, we put $d_{A}(P,P^{\prime }):=\max _{a\in A} |P(a) - P^{\prime }(a)|$.

Proof. In order to prove the first implication, let $\mu \in \mathcal {P}_{\infty }$ be such that $X_{\mu }$ is isometric to the Gurariĭ space, and let $(n,n^{\prime },P,P^{\prime },g)\in T$ be such that $d_{\operatorname {dom}(P)}(P,\mu ) < \tfrac {1}{n}$ and $\mu $ restricted to $\operatorname {span} (\operatorname {dom}(P))\subseteq c_{00}$ is a norm. Consider the finite-dimensional space $A:=(\operatorname {span}(\operatorname {dom}(P)),\mu )$. Let $\nu \in \mathcal {B}$ be as in 3. Put $B = (\operatorname {span}(\operatorname {dom} P^{\prime }),\nu )$, and pick a basis $\mathfrak {b}\subseteq V$ of B. By 6, there exists $T_{g}:A\to B$, which is a $(1+\tfrac {1}{n^{\prime }})$-isomorphism and $T_{g}\supseteq g$. By [Reference Kubiś and Solecki22, Lemma 2.2], there is a $(1+\tfrac {1}{3n^{\prime }})$-isomorphism $S:B\to X_{\mu }$ such that $\|ST_{g}-Id_{A}\|<\tfrac {1}{n^{\prime }}$. By Lemma 4.31, we may for every $b\in \mathfrak {b}$ find $x_{b}\in V$ such that the linear mapping $Q:S(B)\to X_{\mu }$ given by $Q(S(b)) = x_{b}$, $b\in \mathfrak {b}$, is a $(1+\tfrac {1}{3n^{\prime }})$-isomorphism with $\|Q-Id\|<\tfrac {1}{3n^{\prime }}$. Consider $\Phi = QS|_{V\cap \operatorname {span}(\operatorname {dom} P^{\prime })}$. This is indeed a $\mathbb {Q}$-linear map and since $QS$ is a $(1+\tfrac {1}{3n^{\prime }})^{2}$-isomorphism and $(1+\tfrac {1}{3n^{\prime }})^{2}<1+\tfrac {1}{n^{\prime }}$, we have $|\mu (\Phi (x)) - \nu (x)|<\tfrac {1}{n^{\prime }}\nu (x)$ for $x\in \operatorname {dom}(P^{\prime })$. Moreover, for every $x\in \operatorname {dom}(P)$, we have

$$ \begin{align*}\begin{split} \mu(\Phi(gx) - x) & = \mu(QST_{g}x - x)\leq \mu(QST_{g}x - ST_{g}x) + \mu(ST_{g}x - x)\\ & < \tfrac{1}{3n^{\prime}}\|ST_{g}\|\mu(x) + \tfrac{1}{n^{\prime}}\mu(x) \leq \tfrac{2}{n^{\prime}}\mu(x). \end{split}\end{align*} $$

This shows that $\mu \in G(n,n^{\prime },P,P^{\prime },g)$.

In order to prove the second implication, let $\mu \in \mathcal {P}_{\infty }$ be such that $\mu \in G(n,n^{\prime },P,P^{\prime },g)$ whenever $(n,n^{\prime },P,P^{\prime },g)\in T$. In what follows for $x\in c_{00}$, we denote by $[x]{\in} X_{\mu }$ the equivalence class corresponding to x. Pick a finite-dimensional space $A\subseteq X_{\mu }$ and an isometry $G:A\to B$, where B is a finite-dimensional Banach space; we may without loss of generality assume $B \subseteq X_{\mu _{B}}$ for some $\mu _{B}\in \mathcal {B}$. Let $\mathfrak {b}_{A}:=\{a_{1},\ldots ,a_{j}\}$ be a normalized basis of A, and extend $G(\mathfrak {b}_{A}) = \{G(a_{1}),\ldots ,G(a_{j})\}$ to a normalized basis $\mathfrak {b}_{B} = \{b_{1},\ldots ,b_{k}\}$ of B. Fix $\eta> 0$. It suffices to find a $(1+\eta )$-isomorphism $\Psi :B\to X_{\mu }$ with $\|\Psi G - I_{A}\|\leq \eta $. Consider the functions $\phi _{1}$ and $\phi _{2}^{\mathfrak {b}_{A}}$ from Lemma 2.2 and Lemma 4.31. Pick $\delta \in (0,1)$ such that $\max \{\phi _{1}(t),\phi _{2}^{\mathfrak {b}_{A}}(t)\}<\eta $ whenever $t<\delta $ and $\varepsilon \in (0,\tfrac {1}{20})$ such that $\phi _{1}(5\varepsilon )<\tfrac {1}{20}$ and $\varepsilon + 72\max \{\varepsilon ,\phi _{1}(5\varepsilon )\}<\delta $.

By Lemma 4.4, there is $\nu \in \mathcal {B}$ having rational values on M with $d_{M}(\nu ,\mu _{B}\circ (T_{B})^{-1})<\tfrac {\varepsilon }{2}$. Put $P^{\prime } = \nu |_{M}$, consider the one-to-one map $g:N\to M$ given by $g:=T_{B}G(T_{A})^{-1}|_{N}$ and put $P = P^{\prime }\circ g$. Let $n\in \mathbb N$ be the integer part of $\tfrac {2}{3\varepsilon }$ and $n^{\prime }\in \mathbb N$ be the integer part of $\frac {1}{9\max \{\varepsilon ,\phi _{1}(5\varepsilon )\}}$. Easy computations show that $\tfrac {3}{2}\varepsilon \leq \frac {1}{n}<2\varepsilon $ and $9\max \{\varepsilon ,\phi _{1}(5\varepsilon )\}\leq \tfrac {1}{n^{\prime }} < 18\max \{\varepsilon ,\phi _{1}(5\varepsilon )\}$ (in the last inequality, we are using that $\max \{\varepsilon ,\phi _{1}(5\varepsilon )\}<\tfrac {1}{20}$).

Note that for every $x\in M$, we have

(4.1)$$ \begin{align} \max\{|\nu(T_{B}x)-1|,|\nu(x)-1|\}\leq \tfrac{\varepsilon}{2} + \max\{|\mu_{B}(x)-1|,|\mu_{B}((T_{B})^{-1}x)-1|\} < \varepsilon. \end{align} $$

Proof of Claim 2.

In order to see that $d_{N}(P,\mu )<\tfrac {1}{n}$, pick $x\in N$. Then

$$ \begin{align*}\begin{split} |P(x) - \mu(x)| \leq \tfrac{\varepsilon}{2} + |\mu_{B}(G(T_{A})^{-1}(x)) - \mu(x)| = \tfrac{\varepsilon}{2} + |\|(T_{A})^{-1}(x)\|_{X_{\mu}} - \|[x]\|_{X_{\mu}}| < \tfrac{3}{2}\varepsilon. \end{split}\end{align*} $$

In order to see that $(n,n^{\prime },P,P^{\prime },g)\in T$, let us verify condition 6. Let $\mu ^{\prime }\in \mathcal {P}$, $\nu ^{\prime }\in \mathcal B$ be such that $P^{\prime }\subseteq \nu ^{\prime }$, $d_{N}(P,\mu ^{\prime })<\tfrac {1}{n} < 2\varepsilon $ and $\mu ^{\prime }$ restricted to $\operatorname {span} N\subseteq c_{00}$ is a norm. Note that $|\mu ^{\prime }(x)-1|<5\varepsilon $ for every $x\in N$, and so, since N is $\varepsilon $-dense for the sphere of $T_{A}(A) = (\operatorname {span} N,\mu )$, the mapping $id:(\operatorname {span} N,\mu )\to (\operatorname {span} N,\mu ^{\prime })$ is a $(1+\phi _{1}(5\varepsilon ))$-isomorphism. Further, $|\nu ^{\prime }(x)-1|=|\nu (x)-1|<\varepsilon $ for every $x\in M$, and so the mapping $id:(\operatorname {span} M,\mu _{B})\to (\operatorname {span} M,\nu ^{\prime })$ is a $(1+\phi _{1}(5\varepsilon ))$-isomorphism as well. Finally, since $T_{B}G(T_{A})^{-1}$ is a $(1+\varepsilon )^{2}$-isomorphism between $(\operatorname {span} N,\mu )$ and $(\operatorname {span} g(N),\mu _{B})$ and

$$ \begin{align*} (1+\phi_{1}(5\varepsilon))^{2}(1+\varepsilon)^{2} \leq (1+3\phi_{1}(5\varepsilon))(1+3\varepsilon)\leq 1 + 9\max\{\varepsilon,\phi_{1}(5\varepsilon)\}\leq 1+\tfrac{1}{n^{\prime}}, \end{align*} $$

we have that $T_{g}:=id\circ T_{B}\circ G\circ (T_{A})^{-1}\circ id:(\operatorname {span} N,\mu ^{\prime })\to (\operatorname {span} M,\nu ^{\prime })$ is a $(1+\tfrac {1}{n^{\prime }})$-isomorphism.

Since $\mu \in G(n,n^{\prime },P,P^{\prime },g)$, there is a $\mathbb {Q}$-linear mapping $\Phi :V\cap (\operatorname {span} M,\nu )\to V$ such that $\mu (\Phi (gx)-x) < \tfrac {2}{n^{\prime }}\mu (x)$ for every $x\in N$ and $|\nu (x) - \mu (\Phi (x))|<\tfrac {1}{n^{\prime }}\nu (x)$ for every $x\in M$. It is easy to see that $\Phi $ extends to a bounded linear operator $\Phi ^{\prime }:(\operatorname {span} M,\nu )\to X_{\mu }$. Finally, consider $\Psi :=\Phi ^{\prime }\circ T_{B}:B\to X_{\mu }$.

For every $x\in M$, we have

$$ \begin{align*}|\mu(\Phi(x)) - 1| \leq |\mu(\Phi(x)) - \nu(x)| + |\nu(x) - 1| \stackrel{(4.1)}{\leq} \tfrac{1}{n^{\prime}}\nu(x) + \varepsilon \stackrel{(4.1)}{\leq} \tfrac{1}{n^{\prime}}(1+\varepsilon) + \varepsilon< \delta;\end{align*} $$

thus, $|\|\Psi (x)\|_{X_{\mu }} - 1| < \delta $ for every $x\in (T_{B})^{-1}(M)$ and so $\Psi $ is a $(1+\eta )$-isomorphism.

Further, we have

$$ \begin{align*}\begin{split} \|\Psi G(a_{i}) - a_{i}\|_{X_{\mu}} & \leq \|\Phi(g(T_{A} a_{i})) - [T_{A} a_{i}]\|_{X_{\mu}} + \|[T_{A} a_{i}] - a_{i}\|_{X_{\mu}}\\ & < \tfrac{2}{n^{\prime}}(1+\varepsilon)+\varepsilon\le\tfrac{4}{n^{\prime}}+\varepsilon<\delta; \end{split}\end{align*} $$

hence, by Lemma 4.31, we have $\|\Phi ^{\prime } T_{B} G - I_{A}\|\leq \phi _{2}^{\mathfrak {b}_{A}}(\tfrac {2}{n^{\prime }}(1+\varepsilon )+\varepsilon ) < \eta $.

Proof. By Proposition 4.6, we have for the countable set T defined before Proposition 4.6 that

$$ \begin{align*}{\langle \mathbb{G}\rangle_{\equiv}^{\mathcal{P}_{\infty}}} = \bigcap_{(n,n^{\prime},P,P^{\prime},g)\in T} G(n,n^{\prime},P,P^{\prime},g),\end{align*} $$

where $G(n,n^{\prime },P,P^{\prime },g)$ is the union of a closed and an open set in $\mathcal {P}_{\infty }$ (here we use the observation that the set $\{\mu \in \mathcal {P}_\infty \colon \mu \text { restricted to } \operatorname {span}(\operatorname {dom} P)\subseteq c_{00} \text { is a norm}\}$ is open due to Lemma 2.4); thus it is the countable intersection of $G_{\delta }$-sets.

Proof of Theorem 4.1.

Let us recall that $\mathcal {P}_{\infty }$ and $\mathcal {B}$ are $G_{\delta }$-sets in $\mathcal {P}$; see Corollary 2.5. Thus, since we have ${\langle \mathbb {G}\rangle _{\equiv }^{\mathcal {B}}} = {\langle \mathbb {G}\rangle _{\equiv }^{\mathcal {P}_{\infty }}}\cap \mathcal {B}$, it follows from Proposition 4.6 that ${\langle \mathbb {G}\rangle _{\equiv }^{\mathcal I}}$ is a $G_{\delta }$-set in any $\mathcal {I}\in \{\mathcal {P},\mathcal {P}_{\infty },\mathcal {B}\}$.

By Corollary 2.10, we also have that ${\langle \mathbb {G}\rangle _{\equiv }^{\mathcal I}}$ is dense in $\mathcal I$ for every $\mathcal {I}\in \{\mathcal {P},\mathcal {P}_{\infty },\mathcal {B}\}$.