Proof. We will only prove the statement for compact operators, the proof being verbatim the same in the case of weakly compact operators. Notice that

\[ \left\{ \dfrac{\delta(f(x)) - \delta(f(y))}{{\rm d}(x,y)} \; | \; x \neq y \in M \right\} = \widehat{f}(\mathcal{M}), \]

where $\mathcal {M} = \left \{{\rm d}(x,y)^{-1}(\delta (x) - \delta (y)) \; | \; x \neq y \in M \right \}$. Since $\mathcal {M} \subset B_{\mathcal {F}(M)}$, if $\widehat {f}$ is compact then $\widehat {f}(\mathcal {M})$ must be relatively compact. Conversely, it follows from the Hahn–Banach separation theorem that $B_{\mathcal {F}(M)} = \overline {\operatorname {conv}} \mathcal {M}$, the closure being taken for the norm topology. Now observe that

\[ \widehat{f}(B_{\mathcal{F}(M)}) \subset \widehat{f}(\overline{\operatorname{conv}}\mathcal{M}) \subset \overline{\operatorname{conv}} (\widehat{f}(\mathcal{M})) \subset \overline{\operatorname{conv}} \left(\overline{\widehat{f}(\mathcal{M})}\right). \]

So, if $\widehat {f}(\mathcal {M})$ is relatively compact, then $\overline {\operatorname {conv}} (\overline {\widehat {f}(\mathcal {M})})$ is compact (see e.g., [Reference Aliprantis and Border8, theorem 5.35] for norm compactness and [Reference Aliprantis and Burkinshaw9, theorem 10.15] for weak compactness), and therefore $\widehat {f}(B_{\mathcal {F}(M)})$ is relatively compact.

Proof. Let us write $\gamma = \sum _{i=1}^{l} a_i \delta (p_i)$ where $1\leq l \leq k$, $a_i \neq 0$ and $p_1, \ldots, p_l$ are pairwise distinct elements of $M \setminus \{0_M\}$. Aiming for a contradiction, assume that there exists $1\leq j \leq l$ such that none of the sequences $(x_{i}(n))_n$, $1\leq i \leq k$, has a subsequence converging to $p_j$. Then, there exists $\varepsilon > 0$ and a strictly increasing sequence $(n_m)_m \subset \mathbb {N}$ such that, for every $m$ and every $1\leq i\leq k$, ${\rm d}(x_{i}(n_m), p_j) \geq \varepsilon$. Hence, by lemma 1.2 we can find $h \in \operatorname {Lip}_0(M)$ such that $h(p_j) = 1, h(p_i) = 0$ if $i\neq j$ and $h = 0$ outside of $B(p_j, \varepsilon /2)$. Now, simply notice that since $\gamma _{n_m} \to \gamma$ in the weak topology we have

\[ 0 = \langle h , \gamma_{n_m} \rangle \to \langle h , \gamma \rangle = a_j, \]

which is a contradiction.

Proof. Notice that $(m_n)_n \subset \mathcal {FS}_2(N)$ which is weakly closed so $\gamma = a\delta (p)+b\delta (q) \in \mathcal {FS}_2(N)$ where either $p\neq q$ or $p=q=0$. Let us prove $(1)$. If $\gamma \neq 0$ then we can assume that $a\neq 0$, $p \neq 0_N$ and, according to lemma 2.2, that $(f(x_n))_n$ or $(f(y_n))_n$ has a subsequence converging to $p$. Since $d(x_n,y_n) \to 0$, both subsequences converge, that is, there exists an increasing sequence $(n_k)_k \subset \mathbb {N}$ such that $(f(x_{n_k}))_k$ and $(f(y_{n_k}))_k$ are converging to $p$. The same lemma ensures that $b=0$ so that $m_n \to a\delta (p)$. Now, let $h \in \operatorname {Lip}_0(N)$ be such that $h$ takes the value 1 on a ball around $p$. Then, for $k$ large enough, $\langle h , m_{n_k} \rangle = 0$ while the limit over $k$ of this term is $\langle h , a \delta (p) \rangle = a$, therefore $a=0$. This is a contradiction so we must have $\gamma = 0$.

We now prove $(2)$. We aim for a contradiction. If $\gamma \neq 0$, we may assume that $a\neq 0$ and $p\neq 0_N$ and by lemma 2.2, up to extracting a subsequence, that $(f(x_n))_n$ converges to $p$. Since $(f(x_n))_n$ converges and ${\rm d}(x_n,y_n) \to + \infty$, we have

\[ \|\cdot\| - \lim\limits_{n \to \infty} \dfrac{\delta(f(x_n))}{d(x_n,y_n)} = 0. \]

Therefore $({\rm d}(x_n,y_n)^{-1}\delta (f(y_n)))_n \subset \mathcal {FS}_1(N)$ must converge to an element $\gamma '=c\delta (r)$. We then distinguish two cases :

• • If for some subsequence, $(f(y_{n_k}))_k$ is bounded, then $m_{n_k} \to 0$ and this is a contradiction.

• • If for some subsequence, ${\rm d}(f(y_{n_k}),0) \to +\infty$, then $(f(y_{n_k}))_k$ is eventually far from $r$. Similarly as in $(1)$, one can show that $c$ must be equal to 0 by using a Lipschitz map taking the value 1 at $r$ and 0 outside of a ball centred at $r$. So $m_{n_k} \to 0$, yet another contradiction.

Let us finish with the proof of $(3)$. As above, since $\gamma \neq 0$ we can assume that $a\neq 0$, $p\neq 0_N$ and $(f(x_n))_n$ converges to $p$. We only need to show that $(f(y_n))_n$ has a convergent subsequence. If $b\neq 0$ and $q$ is not equal to $0_N$ or $p$, then lemma 2.2 ensures that $(f(y_n))_n$ has a subsequence converging to $q$. So assume that $b=0$ or $q=0_N$, that is, $m_n \to a\delta (p)$. Up to extracting another subsequence, we may assume that $d(x_n,y_n)$ converges to $\rho \in (0 , +\infty ]$. If $\rho = + \infty$ then $\gamma = 0$ by $(2)$, so we actually have that $\rho \in (0 , +\infty )$. Therefore

\[ \delta(f(x_n)) - \delta(f(y_n)) \to a' \delta(p) \ \ \text{weakly} \]

where $a'=a\rho$. Since $\delta (f(x_n)) \to \delta (p)$, we have

\[ \delta(f(y_n)) \to a'' \delta(p) \ \ \text{weakly} \]

where $a''=1-a'$. If $a'' \neq 0$ then by lemma 2.2, $f(y_n)$ has a subsequence converging to $p$, and if $a''=0$ then $f(y_n) \to 0_N$.

Proof. Assume that $f$ satisfies $(P_3)$. Let $(x_n)_n \subset M$ be such that ${\rm d}(x_n,0) \to +\infty$. We will show that there exists a subsequence $(x_{n_k})_k$ such that

\[ \dfrac{{\rm d}(f(x_{n_k}),0)}{{\rm d}(x_{n_k},0)} \mathop{\longrightarrow}\limits_{k\to +\infty} 0. \]

In view of applying Property $(P_3)$, we first construct by induction an increasing sequence $(n_k)_k \subset \mathbb {N}$ and a sequence $(y_{n_k})_k \subset M$ such that for every $k \in \mathbb {N}$

1. (i) ${\rm d}(y_{n_k},0) \geq k$;

2. (ii) ${\rm d}(x_{n_k},y_{n_k}) \geq k$;

3. (iii) $\dfrac {{\rm d}(x_{n_k}, y_{n_k})}{{\rm d}(x_{n_k}, y_{n_k}) - {\rm d}(y_{n_k}, 0)} \leq 2$;

4. (iv) $\dfrac {{\rm d}(f(y_{n_k}), 0)}{{\rm d}(x_{n_k}, y_{n_k}) - {\rm d}(y_{n_k}, 0)} \leq \dfrac {1}{k}$.

We proceed by induction and start with the base case $k=1$. Since $M$ is unbounded, there is an element $y\in M$ such that $d(y,0) \geq 1$. We fix such $y$. The inequality

\[ {\rm d}(x_n,y) \geq {\rm d}(x_n,0)-{\rm d}(y,0) \]

yields ${\rm d}(x_n,y) \mathop {\longrightarrow }\limits _{n\to +\infty } +\infty$ so that

\[ \dfrac{{\rm d}(x_n, y)}{{\rm d}(x_n, y) - {\rm d}(y, 0)} \mathop{\longrightarrow}\limits_{n\to +\infty} 1 \ \ \ \text{and} \ \ \ \dfrac{{\rm d}(f(y), 0)}{{\rm d}(x_n, y) - {\rm d}(y, 0)} \mathop{\longrightarrow}\limits_{n\to +\infty} 0. \]

Hence, we can find $n_1 \in \mathbb {N}$ large enough so that

\[ {\rm d}(x_{n_1},y) \geq 1, \ \dfrac{{\rm d}(x_{n_1}, y)}{{\rm d}(x_{n_1}, y) - {\rm d}(y, 0)} \leq 2 \ \ \ \text{and} \ \ \ \dfrac{{\rm d}(f(y), 0)}{{\rm d}(x_{n_1}, y) - {\rm d}(y, 0)} \leq 1. \]

We then set $y_{n_1}=y$. Assume now that $y_{n_1}, \ldots, y_{n_k} \in M$ are constructed with $n_1 < n_2 < \cdots < n_k$. We can find $y\in M$ such that ${\rm d}(y,0) \geq k+1$. We now proceed as above, and we find $n_{k+1} \in \mathbb {N}$ such that $n_k < n_{k+1}$ and

\[ {\rm d}(x_{n_{k+1}},y) \geq k+1, \ \dfrac{{\rm d}(x_{n_{k+1}}, y)}{{\rm d}(x_{n_{k+1}}, y) - {\rm d}(y, 0)} \leq 2 \quad \text{and}\quad \dfrac{{\rm d}(f(y), 0)}{{\rm d}(x_{n_{k+1}}, y) - {\rm d}(y, 0)} \leq \dfrac{1}{k+1}. \]

We can now set $y_{n_{k+1}} = y$ and by construction, the sequence $(y_{n_k})_k \subset M$ satisfies the desired properties.

In particular, ${\rm d}(y_{n_k},0) \to +\infty$, so we can apply $(P_3)$ to $(x_{n_k})_k$ and $(y_{n_k})_k$ and we keep denoting by $(x_{n_k})_k$ and $(y_{n_k})_k$ the subsequences that we obtain. Hence, we either have $f(x_{n_k}) \to p$ and $f(y_{n_k}) \to q$ for some $p,q \in N$ or $\frac {{\rm d}(f(x_{n_k}) , f(y_{n_k}))}{{\rm d}(x_{n_k},y_{n_k})} \to 0$. Note that if we are in the first case, then we also have

(2.1)\begin{equation} \dfrac{{\rm d}(f(x_{n_k}) , f(y_{n_k}))}{{\rm d}(x_{n_k},y_{n_k})} \to 0 \end{equation}

because ${\rm d}(x_{n_k},y_{n_k}) \to +\infty$, and that is the property we will need. Indeed, by the triangle inequality

\begin{align*} \dfrac{{\rm d}(f(x_{n_k}), 0)}{{\rm d}(x_{n_k},0)} & \leq \dfrac{{\rm d}(f(x_{n_k}), f(y_{n_k}))}{{\rm d}(x_{n_k},y_{n_k}) - {\rm d}(y_{n_k},0)} + \dfrac{{\rm d}(f(y_{n_k}), 0)}{{\rm d}(x_{n_k},y_{n_k}) - {\rm d}(y_{n_k},0)} \\ & = \dfrac{{\rm d}(f(x_{n_k}), f(y_{n_k}))}{{\rm d}(x_{n_k}, y_{n_k})} \dfrac{{\rm d}(x_{n_k},y_{n_k})}{{\rm d}(x_{n_k},y_{n_k}) - {\rm d}(y_{n_k},0)} + \dfrac{{\rm d}(f(y_{n_k}), 0)}{{\rm d}(x_{n_k},y_{n_k}) - {\rm d}(y_{n_k},0)} \end{align*}

and the right-hand side converges to $0$ by $(iii)$, $(iv)$ and (2.1).

Proof of theorem A. We first prove the ‘$\implies$’ direction.

We start with $\widehat {f}$ compact implies $(P_1)$. Let $S$ be a bounded subset of $M$ and let $(x_n)_n$ be a sequence in $S$. By assumption (and proposition 2.1), the sequence

\[ (m_n)_n := \left(\widehat{f}\left(\frac{\delta(x_n)}{{\rm d}(x_n,0)} \right)\right)_n = \left(\frac{\delta(f(x_n))}{{\rm d}(x_n,0)}\right)_n \]

has a convergent subsequence $(m_{n_k})_k$. Denote by $\gamma$ the limit of $(m_{n_k})_k$. If $\gamma = 0$, then

\[ {\rm d}(f(x_{n_k}), 0_N) =\| m_{n_k} \| {\rm d}(x_{n_k}, 0_M) \mathop{\longrightarrow}\limits_{k \to \infty} 0 \]

because $(x_{n_k})_k$ is bounded. In that case, $f(x_{n_k}) \to 0_N$ and we are done. Hence, it only remains to consider the case when $\gamma \neq 0$. By lemma 2.3, this can only happen if $d(x_{n_k},0_M)$ does not tend to $0$. But then, we can find a subsequence, still denoted by $(n_k)_k$ for convenience, such that ${\rm d}(x_{n_k},0_M) \geq \alpha > 0$ for every $k$. By the same lemma, we then must have a subsequence of $(f(x_{n_k}))_k$ which converges and this finishes the proof of $(P_1)$.

We now show that $\widehat {f}$ compact implies $(P_2)$. Let $(x_n)_n$, $(y_n)_n$ be two sequences in $M$ such that $d(x_n, y_n) \to 0$. By proposition 2.1, the sequence

\[ \left(\frac{\delta(f(x_n)) - \delta(f(y_n))}{{\rm d}(x_n,y_n)}\right)_n \]

has a convergent subsequence. However it follows immediately from lemma 2.3(i) that the limit is $0$.

It remains to prove that $\widehat {f}$ compact implies $(P_3)$. We already know that if $\widehat {f}$ is compact then $f$ satisfies $(P_2)$, which will be of use. Let $(x_n)_n, (y_n)_n$ be two sequences in $M$ such that $x_n \neq y_n$ for every $n$. Assume that $(x_n)_n$ and $(y_n)_n$ tend to infinity, that is, $\lim \limits _{n \to \infty } {\rm d}(x_n,0) = \lim \limits _{n \to \infty } {\rm d}(y_n,0) = \infty$. Again, we let

\[ m_n := \dfrac{\delta(f(x_n)) - \delta(f(y_n))}{{\rm d}(x_n,y_n)} \]

and $(m_n)_n$ has a convergent subsequence, which we keep denoting by $(m_n)_n$, for simplicity. Let $\gamma$ be the limit of $(m_n)_n$. Notice that

\[ \|m_n \| = \dfrac{{\rm d}(f(x_n) , f(y_n))}{{\rm d}(x_n,y_n)}. \]

We distinguish two cases: up to extracting a further subsequence, we will need to consider the cases when ${\rm d}(x_n,y_n)$ converges to $0$ and when there exists $\alpha >0$ such that ${\rm d}(x_n,y_n) \geq \alpha$ for every $n$. In the first case, we get by $(P_2)$ that $m_n \to 0$ so that $\|m_n\| \to 0$. In the second case, if $\gamma \neq 0$, we have by lemma 2.3(iii) that there exist $p,q\in N$ and an increasing sequence $(n_k)_k \subset \mathbb {N}$ such that $f(x_{n_k}) \to p$ and $f(y_{n_k}) \to q$. Finally if $\gamma = 0$ then again $\|m_n\| \to 0$. In all cases, $f$ satisfies $(P_3)$.

Let us now prove the ‘$\impliedby$’ direction. We keep using the notation

\[ (m_n)_n :=\left(\frac{\delta(f(x_n)) - \delta(f(y_n))}{d(x_n,y_n)}\right)_n \]

where $x_n \neq y_n \in M$ for every $n \in \mathbb {N}$. By proposition 2.1, we have to show that this sequence admits a convergent subsequence in $\mathcal {F}(N)$. Up to extracting a subsequence, we only have to distinguish three cases : when both $(x_n)_n$ and $(y_n)_n$ are bounded, when one of them is bounded while the other one goes to $+\infty$, and when both go to $+\infty$.

1. (i) If $(x_n)_n$ and $(y_n)_n$ are bounded, by $(P_1)$ there exists an increasing sequence $(n_k)_k \subset \mathbb {N}$ such that $(f(x_{n_k}))_k$ converges to a point $p \in N$ and $(f(y_{n_k}))_k$ converges to some $q\in N$. Since the sequence $({\rm d}(x_{n_k},y_{n_k}))_k$ is bounded, up to a further extraction, we may assume that it converges to some $\rho \geq 0$. Since $f$ is uniformly locally flat, if $\rho = 0$ then $(m_{n_k})_k$ converges to 0. If $\rho >0$, then it is readily seen that $(m_{n_k})_k$ converges to $\rho ^{-1}(\delta (p) - \delta (q))$.

2. (ii) If $(x_n)_n$ is bounded while $d(y_n,0) \to \infty$, thanks to $(P_1)$ there exists an increasing sequence $(n_k)_k \subset \mathbb {N}$ such that $(f(x_{n_k}))_k$ converges to a point $p \in N$. Therefore we may write for every $k \in \mathbb {N}$:

   \begin{align*} m_{n_k} & = \frac{\delta(f(x_{n_k})) - \delta(f(y_{n_k}))}{{\rm d}(x_{n_k},y_{n_k})}\\ & = \frac{\delta(f(x_{n_k})) - \delta(0)}{{\rm d}(x_{n_k},y_{n_k})} + \frac{\delta(0) - \delta(f(y_{n_k}))}{{\rm d}(0,y_{n_k})} \frac{{\rm d}(0,y_{n_k})}{{\rm d}(x_{n_k},y_{n_k})}. \end{align*}
   On the one hand,
   \[ \left\|\frac{\delta(f(x_{n_k})) - \delta(0)}{{\rm d}(x_{n_k},y_{n_k})}\right\| = \frac{{\rm d}(f(x_{n_k}) , 0)}{{\rm d}(x_{n_k},y_{n_k})} \mathop{\longrightarrow}\limits_{k \to \infty} 0. \]
   On the other hand, $f$ is radially flat thanks to lemma 2.4 so that
   \[ \left\|\frac{\delta(0) - \delta(f(y_{n_k}))}{{\rm d}(0,y_{n_k})}\right\| = \frac{{\rm d}(f(y_{n_k}) , 0)}{{\rm d}(y_{n_k},0)} \mathop{\longrightarrow}\limits_{k \to \infty} 0. \]
   Since the triangle inequality implies that $\lim \limits _{k \to \infty }\,{\rm d}(0 , y_{n_k})^{-1}\,{\rm d}(x_{n_k},y_{n_k}) = 1$, we obtain that $(m_{n_k})_k$ converges to 0.

3. (iii) If ${\rm d}(x_n,0) \to +\infty$ and ${\rm d}(y_n,0)\to +\infty$, then by $(P_3)$ there exists $(n_k)_k \subset \mathbb {N}$ such that, either $\|m_{n_k}\| \to 0$ or $f(x_{n_k}) \to p$ and $f(y_{n_k}) \to q$ for some $p,q \in N$. In the first case we are done since $(m_{n_k})_k$ converges to 0. In the second case, up to further extraction, we may assume that ${\rm d}(x_{n_k},y_{n_k}) \to \rho \in [0,+\infty ]$. Hence, $m_{n_k}$ converges to $0$ if $\rho = 0$ or $\rho = +\infty$ and converges to $\rho ^{-1}(\delta (p) - \delta (q))$ otherwise.

In all cases, the sequence $(m_n)_n$ admits a convergent subsequence.

Proof. It is readily seen that $(P_4)$ implies $(P_3)$.

Proof. If $f(M)$ is totally bounded then clearly $f$ satisfies $(P_1)$. Moreover if $(x_n)_n$ and $(y_n)_n$ are two sequences in $M$ going to infinity, then the sequences $(f(x_n))_n$ and $(f(y_n))_n$ have a common convergent subsequence and so $f$ readily satisfies $(P_3)$ in theorem A.

Proof. The implication (i) $\implies$ (ii) is obvious. Next, (ii) $\implies$ (i) follows from theorem C and proposition 2.1. Indeed, thanks to the Eberlein–Šmulian theorem (see [Reference Albiac and Kalton4, theorem 1.6.3] e.g.), a subset $S$ of a Banach space $X$ is (relatively) weakly compact if and only if it is (relatively) weakly sequentially compact. So, theorem C implies that a subset $S \subset \mathcal {FS}_k(M)$ is weakly compact if and only if it is compact in the norm topology. Now observe that the set appearing in proposition 2.1 is a subset of $\mathcal {FS}_2(M)$ so that compactness and weak compactness are indeed equivalent. To conclude, (i) $\iff$ (iii) follows from Schauder's theorem (see e.g. [Reference Megginson19, theorem 3.4.15]), (ii) $\iff$ (iv) follows from Gantmacher's theorem (see e.g. [Reference Megginson19, theorem 3.5.13]), and (ii) $\iff$ (v) follows from a classical result [Reference Megginson19, theorem 3.5.14] due to Gantmacher in the separable case and Nakamura in the general case. □

Proof. Arguing by contradiction, we assume that $(\gamma _n)_n$ does not converge to 0 in the norm topology. We can extract a subsequence, still denoted the same way for simplicity, such that:

\[ \exists \delta >0, \; \forall n \in \mathbb{N},\quad \|\gamma_n\| > \delta. \]

Furthermore, if for some $k$ we have ${\rm d}(\{0\}, \operatorname {supp}(\gamma _k)) < \dfrac {\varepsilon }{2}$, then, by triangle inequality and using the assumption, we have, for any $i\neq k$ ${\rm d}(\{0\}, \operatorname {supp}(\gamma _i)) > \dfrac {\varepsilon }{2}$. Hence, we can as well assume that ${\rm d}(\{0\}, \operatorname {supp}(\gamma _n)) > \dfrac {\varepsilon }{2}$ for every $n\in \mathbb {N}$. In what follows, denote $\alpha = \dfrac {\varepsilon }{2}$ and we let $R:=\operatorname {rad}(M)$ be the radius of $M$.

We let $(E_n)_{n \geq 0}$ be the increasing sequence of subsets of $M$ given by

\[ E_n := \{0\} \cup \bigcup_{k=1}^{n} \operatorname{supp}(\gamma_k). \]

By assumption and the remark above, we have that

\[ ({\star}) \quad \forall n \in \mathbb{N}, \quad {\rm d}(\operatorname{supp} \gamma_{n+1} , E_n) > \alpha. \]

Let $n_1=1$ and choose a positive $h_1\in B_{\operatorname {Lip}_0(M)}$ such that $\left |{\left <{h_1,\gamma _{n_1}}\right >}\right |> \delta /2$. By induction, we will build an increasing sequence of integers $(n_k)_{k \in \mathbb {N}}$ as well as Lipschitz maps $(h_k)_{k \in \mathbb {N}} \subset 2R\alpha ^{-1}B_{\operatorname {Lip}_0(M)}$ with the following properties:

• • $|\left <{h_k,\gamma _{n_k}}\right >| >\delta /2$,

• • $\left |{\left <{h_i,\gamma _{n_j}}\right >}\right |<\frac {\delta }{4k}$ whenever $i\leq k$ and $j>k$,

• • $h_{k} \geq 0$, $h_{k}$ is zero on $[E_{n_{k-1}}]_{\alpha /2}$ and $(h_k)_k$ have mutually disjoint supports.

Assume that $n_k$ has been selected, as well as $h_1,\ldots,h_k$. Since $(\gamma _n)$ is weakly-null, we can pick $n_{k+1}>n_k$ such that

\[ \forall n\geq n_{k+1}, \; \forall i\leq k, \quad \left|{\left<{h_i,\gamma_n}\right>}\right|<\dfrac{\delta}{4k}. \]

Using the Hahn–Banach theorem we then pick $f_{k+1} \in \operatorname {Lip}_0(M)$ such that $\operatorname {Lip}(f_{k+1}) \leq 1$ and $\langle f_{k+1}, \gamma _{n_{k+1}} \rangle = \| \gamma _{n_{k+1}} \| > \delta$. Next, we let $g_{k+1} \in \operatorname {Lip}_0(E_{n_{k+1}})$ be the map defined by $g_{k+1} \equiv f_{k+1}$ on $\operatorname {supp}(\gamma _{n_{k+1}})$ and $g_{k+1} \equiv 0$ on $[E_{n_{k+1}-1}]_{\alpha /2}$. We deduce from $(\star )$ that $g_{k+1}$ is $2R\alpha ^{-1}$-Lipschitz, and moreover $\langle g_{k+1}, \gamma _{n_{k+1}} \rangle = \langle f_{k+1}, \gamma _{n_{k+1}} \rangle = \|\gamma _{n_{k+1}}\|> \delta$. We let $\tilde {g}_{k+1}$ be either the positive or the negative part of $g_{k+1}$. The choice is made so that $\left |{\left <{\tilde {g}_{k+1},\gamma _{n_{k+1}}}\right >}\right | > \delta /2$. We now define

\[ \forall x \in M, \quad h_{k+1}(x) = \max\left( \sup_{y\in S} (\tilde{g}_{k+1}(y)-2R\alpha^{{-}1}\,{\rm d}(x,y)) \; , \; 0 \right), \]

where $S = E_{n_{k+1}} \cup [E_{n_k}]_{\alpha / 2}$. Notice that $h_{k+1}$ is the smallest positive $2R \alpha ^{-1}$ -Lipschitz extension to the whole space $M$ of $\tilde {g}_{k+1}$ (see [Reference Weaver20, theorem 1.33]). In particular, $h_{k+1}$ is zero on $[E_{n_k}]_{\alpha /2}$ (and everywhere where $\tilde {g}_{k+1}$ was zero). Now let $l< k$ and assume that $h_{k}(x)\neq 0$. Then clearly $x \notin [E_{n_l}]_{\alpha /2}$. But also $h_l(x)=0$ by the choice of the constant of the extension above. So $(h_k)_k$ have mutually disjoint supports. This finishes the inductive construction.

To conclude, letting $h=\sup _k h_k$ we have $h=\sum _{k=1}^{\infty } h_k$ pointwise together with $\left\|{h}\right\|_L\leq 2R\alpha ^{-1}$. Therefore $h\in \operatorname {Lip}_0(M)$ and for every $k$ we have

\begin{align*} \left|{\left<{\gamma_{n_k},h}\right>}\right| = \left|{\left<{\gamma_{n_k},\sum_{i=1}^{k} h_i}\right>}\right| & \geq \left|{\left<{\gamma_{n_k},h_k}\right>}\right| - \sum_{i=1}^{k-1}\left|{\left<{\gamma_{n_k},h_i}\right>}\right| \\ & \geq \frac{\delta}{2} - \sum_{i=1}^{k-1}\frac{\delta}{4(k-1)} \geq \frac{\delta}{4} \end{align*}

contradicting the fact that $(\gamma _n)_n$ is weakly null.

Proof. Since $\mathcal {FS}_k(M)$ is weakly closed by [Reference Abbar, Coine and Petitjean1, lemma 2.10], if a sequence $(\gamma _n)_n \subset \mathcal {FS}_k(M)$ weakly converges to some $\gamma \in \mathcal {F}(M)$, then $\gamma \in \mathcal {FS}_k(M)$. Therefore, for every $n \in \mathbb {N}$, $\gamma - \gamma _n \in \mathcal {FS}_{2k}(M)$. Consequently, to prove the result it is enough to show that for every complete metric space $M$ and for every $k \in \mathbb {N}$, any weakly null sequence in $\mathcal {FS}_k(M)$ is actually norm null. Furthermore, thanks to [Reference Albiac, Ansorena, Cúth and Doucha2, theorem A], there exists a bounded metric space $B(M)$ such that $\mathcal {F}(M)$ is linearly isomorphic to $\mathcal {F}(B(M))$. The isomorphism $T : \mathcal {F}(M) \to \mathcal {F}(B(M))$ preserves finitely supported elements is the sense that $\gamma \in \mathcal {FS}_k(M)$ if and only if $T(\gamma ) \in \mathcal {FS}_k(B(M))$. So, without loss of generality, we may assume that $M$ is a bounded metric space. Finally let us point out that it is enough to prove that $(\gamma _n)_n$ has a norm-null subsequence. Indeed, assume that $(\gamma _n)_n$ does not converge to 0. Then there exists $\varepsilon >0$ and a subsequence $(\gamma _{\varphi _1(n)})_n$ such that $\|\gamma _{\varphi _1(n)}\|> \varepsilon >0$ for every $n$. Since $(\gamma _{\varphi _1(n)})_n$ is still weakly null, the proof below will show that we may extract a further subsequence $(\gamma _{\varphi _1 \circ \varphi _2 (n)})_n$ which is norm-null, but this contradicts the fact that $\|\gamma _{\varphi _1(n)}\|> \varepsilon >0$. We will proceed by induction on $k \in \mathbb {N}$. If $k=1$ and $(\gamma _n)_n \subset \mathcal {FS}_1(M)$ is a weakly null sequence, we can write $\gamma _n = a_n \delta (x_n)$ where $a_n \in \mathbb {K}$ and $x_n \in M$. Let us denote $f := d(\cdot,0) \in \operatorname {Lip}_0(M)$. Since $(\gamma _n)_n$ is weakly null, it is readily seen that

\[ \|\gamma_n\| = |a_n|\,{\rm d}(x_n,0) = |\langle f , \gamma_n \rangle | \mathop{\longrightarrow}\limits_{n \to \infty} 0. \]

Let us fix $k \in \mathbb {N}$. Assume we have shown that, for every $j \leq k$, every weakly null sequence in $\mathcal {FS}_j(M)$ is in fact norm null. Let us consider a weakly null sequence $(\gamma _n)_n \subset \mathcal {FS}_{k+1}(M)$. For every $n \in \mathbb {N}$, we will write

\[ \gamma_n = \sum_{i=1}^{k+1} a_i(n) \delta(x_i(n)), \]

where $a_i(n) \in \mathbb {K}$ and $x_i(n) \in M$ for every $1 \leq i \leq k+1$. We will distinguish two cases:

• • There exists $i \in \{1, \ldots, k+1\}$ such that $(x_i(n))_n$ has a convergent subsequence to some $x \in M$. For simplicity, we still denote the subsequence by $(x_i(n))_n$. Notice that $i \in \{1, \ldots, k+1\}$ as above might not be unique. So, up to a further extraction, we may assume that there exists $\varepsilon >0$ and $i_1, \ldots, i_j$ such that $(x_{i}(n))_n$ converges to $x$ for every $i \in I:=\{i_1, \ldots, i_j\}$, while $(x_{i}(n))_n \subset M \setminus B(x,\varepsilon )$ whenever $i \in \{1 , \ldots, k+1\} \setminus I$. If $j=k+1$, that is $I = \{1 , \ldots, k+1\}$, then the set $K:=\{x_i(n) \; | \; n \in \mathbb {N} \text { and } 1 \leq i \leq k+1 \} \cup \{x\} \cup \{0\}$ is a countable compact metric space such that $(\gamma _n)_n \subset \mathcal {F}(K)$. Thanks to [Reference Hájek, Lancien and Pernecká13, theorem 3.1] (see also [Reference Aliaga, Gartland, Petitjean and Procházka5]), $\mathcal {F}(K)$ has the Schur property so that $(\gamma _n)$ is actually norm null, which is what we wanted to prove. If $j< k+1$, we let $h$ be the map defined by $h(z)=1$ if $z \in B(x, \varepsilon /2)$ and $h(z) = 0$ if $z \in M \setminus B(x, \varepsilon )$. It is easy to prove that $h$ is Lipschitz on $B(x,\varepsilon /2) \cup M \setminus B(x, \varepsilon )$ and using McShane's extension theorem (see e.g. [Reference Weaver20, theorem 1.33 and corollary 1.34]), we can extend $h$ to the all $M$. Clearly, $\operatorname {supp}(h) \subset K:= B(x,\varepsilon )\cup \{0\}$. Now let $T_h$ be as in lemma 3.1 and $W_h : \mathcal {F}(M) \to \mathcal {F}(K)$ be its pre-adjoint operator. It is a routine check to see that if $\mu \in \mathcal {F}(B(x,\varepsilon /2) \cup \{0\} )$ then $W_h(\mu ) = \mu$. Furthermore, there exists $N_0 \in \mathbb {N}$ such that for every $n \geq N_0$ and every $i \in \{i_1, \ldots, i_j\}$, $(x_{i}(n))_n \subset B(x,\varepsilon /2)$. Thus, by construction, we have:

  \[ \forall n \geq N_0, \quad W_h\gamma_n = \mathop{\sum_{i=1}^{k+1}}\limits_{i \in I} a_{i}(n) \delta(x_{i}(n)). \]
  Since $W_h$ is continuous and since $(\gamma _n)_n$ is weakly null, the sequence $(W_h\gamma _n)_n \subset \mathcal {F}(K)$ is weakly null as well. As $j< k+1$, we may use the induction hypothesis to deduce that $(W_h\gamma _n)_n$ is norm null in $\mathcal {F}(K)$. Recall that $\mathcal {F}(K)$ is a closed subspace of $\mathcal {F}(M)$ so that $(W_h\gamma _n)_n$ can be seen as a norm null sequence in $\mathcal {F}(M)$, which in turn implies that the sequence $(\mu _n)_n$ given by
  \[ \mu_n := \mathop{\sum_{i=1}^{k+1}}\limits_{i \not\in I} a_i(n) \delta(x_i(n)) \]
  has to be weakly null. So we use once more our induction hypothesis to get that $(\mu _n)_n$ is norm null and finally
  \[ (\gamma_n)_n = (W_h\gamma_n)_n + (\mu_n)_n \]
  is norm convergent to 0 as the sum of two such sequences.

• • There is no $i \in \{1, \ldots, k+1\}$ such that $(x_i(n))_n$ has a convergent subsequence. Then each set $\{x_i(n) \; | \; n \in \mathbb {N} \}$, $1 \leq i \leq k+1$, is not totally bounded. Hence there exists $\varepsilon >0$ and an infinite subset $\mathbb {M}$ of $\mathbb {N}$ such that for every $i$ and every $n \neq m \in \mathbb {M}$: $d(x_i(n) , x_i(m))> \varepsilon$. We now claim that we can extract an infinite subset $\mathbb {M}_1$ of $\mathbb {M}$ such that for every $i\neq j$ and every $n \neq m \in \mathbb {M}_1$: ${\rm d}(x_i(n) , x_j(m))> \varepsilon /2$. Let us briefly sketch this extraction. We write $\mathbb {M}= \{n_1 , n_2 , \ldots \}$ and we let $m_1:=n_1$. Since the sequences $(x_i(n_\ell ))_\ell$, $1\leq i \leq k+1$, are $\varepsilon$-separated, by the triangle inequality they must ‘escape’ the balls $B(x_j(m_1), \varepsilon /2)$, $1\leq j \leq k+1$, eventually. In other words, there exists $m_2 \in \mathbb {M}$ such that $m_1< m_2$ and for every $n \in \mathbb {M}$ and $1 \leq i,j \leq k+1$, $n \geq m_2 \implies d(x_i(n) , x_j(m_1)) \geq \varepsilon /2$. By the same argument, there exists $m_3 \in \mathbb {M}$ such that $m_3 > m_2$ and for every $i,j \in \{1 , \ldots, k+1\}$ and every $n \in \mathbb {M}$, $n \geq m_3 \implies d(x_i(n) , x_j(m_1)) \geq \varepsilon /2$ and $d(x_i(n) , x_{j}(m_2)) \geq \varepsilon /2$. Continuing this construction by induction provides the required $\mathbb {M}_1 = \{m_1, m_2 , \ldots \}$. To conclude, notice that for every $n \neq m \in \mathbb {M}_1$, $d(\operatorname {supp}(\gamma _n) , \operatorname {supp}(\gamma _m)) > \varepsilon /2$. Since $(\gamma _n)_{n \in \mathbb {M}_1}$ is weakly null, we may apply lemma 3.2 to conclude that $(\gamma _n)_{n \in \mathbb {M}_1}$ is norm null.