Proof. Choose $\delta >0$ so that $S$ is $\delta$-uniformly discrete, and $0<\epsilon '<\epsilon$ so that

(2.2)\begin{equation} 0<\frac{1}{1-4\delta^{{-}1}\epsilon'-\epsilon'}<1+\epsilon. \end{equation}

Let $\{z_1,\ldots,z_n\}$ be an $\epsilon '$-net in $S_{\mathbb {F}}$, and $\{z_1^*,\ldots,z_n^*\}\subset S_{{\mathbb {X}}^*}$ so that $z_j^*(z_j)=1$ for all $1\le j\le n$. Since $S$ is bounded, there exists $z_0^{**}\in {\mathbb {X}}^{**}$ a $w^*$-accumulation point of $\widehat {S}\subset {\mathbb {X}}^{**}$. Hence, there are $x\ne y\in S$, such that for $1\le k\le n$,

\begin{align*} |z_0^{**}(z_k^{*})-z_k^{*}(x)|\le \epsilon' \quad \text{and}\quad |z_0^{**}(z_k^{*})-z_k^{*}(y)|\le \epsilon'. \end{align*}

Fix $z\in S_{\mathbb {F}}$, and choose $1\le k\le n$ so that

\begin{align*} \|z_k-z\|\le \epsilon'. \end{align*}

Now pick $b \in \mathbb {K}$. If $|b|\le 2\delta ^{-1}$, then

\begin{align*} \|z+b(x - y)\|& \ge \|z_k+b(x - y)\|-\epsilon'\ge |z_{k}^{*}(z_k+b(x - y)|-\epsilon'\\ & \ge 1-|b||z_k^{*}(x - y)|-\epsilon'\\ & \ge 1-2\delta^{{-}1}|z_0^{**}(z_k)-z_k^{*}(x) + z_{k}^{*}(y) - z_0^{**}(z_k)|-\epsilon'\\ & \ge 1-4\delta^{{-}1}\epsilon'-\epsilon'. \end{align*}

Hence, by (2.2),

\begin{align*} \|z\|=1\le (1+\epsilon)\|z+b(x - y)\|. \end{align*}

On the other hand, if $|b|>2\delta ^{-1}$, then

\begin{align*} \|z+b(x - y)\|\ge |b|\|x - y\|-\|z\|\ge 1=\|z\|. \end{align*}

This completes the proof for $z\in S_{\mathbb {F}}$, and hence by scaling for all $z\in \mathbb {F}$.

Proof. The argument is very similar to that of [Reference Berasategui and Lassalle14, lemma 3.5]; we give a proof for the sake of completeness.

Fix ${\mathbb {Z}}\subset {\mathbb {X}}$ a separable subspace and $\epsilon >0$. Choose a sequence $(v_j)_{j\in \mathbb {N}}\subset {\mathbb {Z}}$, dense in ${\mathbb {Z}}$, and a sequence of positive scalars $(\epsilon _j)_{j\in \mathbb {N}}$ so that $\prod \limits _{j=1}^{\infty }(1+\epsilon _j)\le (1+\epsilon )$. Let $j_0:=1$. Applying lemma 2.5 to the set $\{u_j\}_{j>j_0}$, we can find $j_0< j_1< j_2$ so that for all $y\in [v_k, u_k: 1\le k\le j_0 ]$ and all $b\in \mathbb {K}$,

\begin{align*} \|y\|\le (1+\epsilon_1)\|y+b (u_{j_1}-u_{j_2}) \|. \end{align*}

Similarly, we can find $j_2< j_3< j_4$ so that for all $y\in [v_k, u_k: 1\le k \le j_2]$ and all $b\in \mathbb {K}$,

\begin{align*} \|y\|\le (1+\epsilon_2)\|y+b (u_{j_3}-u_{j_4}) \|. \end{align*}

By an inductive argument, we obtain a strictly increasing sequence of positive integers $\{j_n\}_{n\in \mathbb {N}}$ such that for all $y\in [v_k, u_k\colon 1\le k\le j_{2n-2}]$, $b\in \mathbb {K}$ and $n\in \mathbb {N}$,

\begin{align*} \|y\|\le (1+\epsilon_n)\|y+b (u_{j_{2n-1}}-u_{j_{2n}})\|. \end{align*}

Then, for any positive integers $m\le l$, any $y\in [v_k, u_k\colon 1\le k\le j_{2m-2}]$ and any scalars $(a_n)_{m \le n\le l}$,

\begin{align*} \|y\|& \le \prod\limits_{n=m}^{l}(1+\epsilon_{n})\left\|y+\sum\limits_{n=m}^{l}a_n (u_{j_{2n-1}}-u_{j_{2n}})\right\|\\ & \le \prod\limits_{n=m}^{\infty}(1+\epsilon_{n})\left\|y+\sum\limits_{n=m}^{l}a_n (u_{j_{2n-1}}-u_{j_{2n}})\right\|. \end{align*}

In particular, $(u_{j_{2n-1}}-u_{j_{2n}})_{n\in \mathbb {N}}$ is basic with basis constant no greater than $\prod \limits _{n=1}^{\infty }(1+\epsilon _n)\le 1+\epsilon$, and, given $\mathbb {F} \subset [ v_j\colon 1\le j\le n]$ for some $n\in \mathbb {N}$, we can pick $r_{\mathbb {F},\xi }$ using the above computation. Now, standard density arguments allow us to obtain the result for any finite dimensional subspace of ${\mathbb {Z}}=\overline {[v_j \colon j\in \mathbb {N}]}$.

Proof. Fix ${\mathbb {Z}}\subset {\mathbb {X}}$ a separable subspace and $\epsilon >0$. An application of proposition 2.3 gives a separating subsequence $(\mathbf {x}_{i_l})_{l\in \mathbb {N}}$ for $({\mathbb {Z}},\mathbf {M},\epsilon )$. Thus, for any finite dimensional $\mathbb {F}\subset {\mathbb {Z}}$, every $x\in \mathbb {F}$ and every $y \in \overline {[\mathbf {x}_{i_l}: l > j_{\mathbb {F},\epsilon }]}$,

\begin{align*} \|x\|\le (\mathbf{M}+\epsilon)\|x+y\|. \end{align*}

Let ${\mathcal {B}}^*=(\mathbf {x}_i^*)_{i\in \mathbb {N}}$ be the dual basis of ${\mathcal {B}}$. Since $(\mathbf {x}^*_i)_{i\in \mathbb {N}}$ is bounded, $(\mathbf {x}_{i_l})_{l\in \mathbb {N}}$ is uniformly discrete, then by lemma 2.6 we obtain a further subsequence $(\mathbf {x}_{i_{l_k}})_{k\in \mathbb {N}}$ such that for any finite dimensional $\mathbb {F}\subset {\mathbb {Z}}$, every $x\in \mathbb {F}$ and every $z \in \overline {[\mathbf {x}_{i_{l_{2k-1}}}-\mathbf {x}_{i_{l_{2k}}}: k> r_{\mathbb {F},\epsilon }]}$,

\begin{align*} \|x\|\le (1+\epsilon)\|x+z\|. \end{align*}

Taking for each $\mathbb {F}$, $s_{\mathbb {F},\epsilon }:=1+\max \{r_{\mathbb {F},\epsilon }, j_{\mathbb {F},\epsilon }\}$, it is immediate from the above that $(\mathbf {x}_{i_{l_k}})_{k\in \mathbb {N}}$ has the desired properties.

Proof. The implications (i)$\Longrightarrow$ (iii) $\Longrightarrow$ (v) and (ii) $\Longrightarrow$ (iv) $\Longrightarrow$ (vi), as well as (i) $\Longrightarrow$ (ii), together with (iii) $\Longrightarrow$ (iv) and (v) $\Longrightarrow$ (vi) are immediate.

Let us prove (v) $\Longrightarrow$ (i): Fix $A\in {\mathcal {G}}(x,m,1)$. We may assume $x\not =P_A(x)$ (else, we take $y=x$), so $|{\rm supp}(x)|>m$. For each $l\in \mathbb {N}$ choose $z_l\in {\mathbb {X}}$ with $w({\rm supp}(z_l))\le w(A)$ so that

\begin{align*} \|x-z_l\|\le \left(1+\frac{1}{l}\right) \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w(A)}}\|x-z\|, \end{align*}

which is possible because $A\subsetneq {\rm supp}(x)$ so the infimum above is not zero.

For each $n \in \mathbb {N}$, set $x_{n}:=x+ \frac {1}{n}P_A(x)$. As ${\mathcal {G}}(x_n,m,1)=\{A\}$ for each $n\in \mathbb {N}$, for each $l\in \mathbb {N}$ there is $y_{n,l}$ with ${\rm supp}(y_{n,l})\subset A$ such that

\begin{align*} \|x_n-y_{n,l}\|\le \mathbf{C}\|x_n-z_l\|. \end{align*}

Given that $A\in \mathbb {N}^{<\infty }$, for fixed $l\in \mathbb {N}$ there is $y_l$ with ${\rm supp}(y_l)\subset A$ and a subsequence $(y_{n_{k,l},l})_{k\in \mathbb {N}}$ convergent to $y_l$. Letting $k\to \infty$ in

\begin{align*} \|x_{n_{k,l}}-y_{n_{k,l},l}\|\le \mathbf{C}\|x_{n_{k,l}}-z_l\|, \end{align*}

we obtain

\begin{align*} \|x-y_l\|\le \mathbf{C}\|x-z_l\|\le \mathbf{C} \left(1+\frac{1}{l}\right) \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w(A)}}\|x-z\|. \end{align*}

Reasoning as before and taking a subsequence if necessary, we may assume that $(y_l)_{l\in \mathbb {N}}$ is convergent to some $y$ with ${\rm supp}(y)\subset A$, so we complete the step letting $l\rightarrow \infty$.

The implication (vi) $\Longrightarrow$ (ii) is proven by the same argument as that given above to prove (v) $\Longrightarrow$ (i).

Finally, we show that (iv) $\Longrightarrow$ (iii). Fix $x\in {\mathbb {X}}$ and $m\in \mathbb {N}$. Suppose there is $z\in {\mathbb {X}}$ with $|{\rm supp}(z)|=\infty$ and $w({\rm supp}(z))\le w(\Lambda _m(x))$. Then by remark 3.3, given $\epsilon >0$ there is a finite set $B\subset {\rm supp}(z)$ such that

\begin{align*} \|z-P_B(z)\|\le \epsilon. \end{align*}

It follows that

\begin{align*} \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w\left(\Lambda_m(x)\right)}}\|x-z\|=\inf_{\substack{z\in {\mathbb{X}}\\|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w\left(\Lambda_m(x)\right)}}\|x-z\|, \end{align*}

so the proof is complete.

Proof. Suppose ${\mathcal {B}}$ is $\mathbf {C}$-$\mathbf {w}$-disjoint almost greedy, fix $x\in {\mathbb {X}}$, $A\in {\mathcal {G}}(x,1)$, and $B\subset \mathbb {N}$ with $w(B)\le w(A)$. If $B\cap A=\varnothing$ or $A=B$, there is nothing to prove. Else, since $A\setminus B\in {\mathcal {G}}(x-P_{A\cap B}(x),1)$ and $w(B\setminus A)\le w(A\setminus B)$, we have

\begin{align*} \|x-P_A(x)\|& =\|x-P_{A\cap B}(x)- P_{A\setminus B}(x)\|\le \mathbf{C} \|x-P_{A\cap B}(x)- P_{B\setminus A}(x)\|\\ & = \mathbf{C}\|x-P_B(x)\|. \end{align*}

Proof. Let

\begin{align*} B:=\bigcup_{A\in {\mathcal{G}}(x,m,t)}A\quad\text{and}\quad b:=\inf_{i\in B}|\mathbf{x}_i^*(x)|, \end{align*}

and pick any $j\in {\rm supp}(x)$. Since ${\mathcal {B}}^*$ is weak star null, there is $i_0\in \mathbb {N}$ such that

\begin{align*} |\mathbf{x}_i^*(x)|< t |\mathbf{x}_j^*(x)|,\quad \forall i\ge i_0. \end{align*}

Hence, $B\subset \{1,\ldots, i_0\}$ and ${\mathcal {G}}(x,m,t)$ is a finite set, so $b$ is a minimum. Now choose $A\in {\mathcal {G}}(x,m,t)$ and $n\in A$ with $|\mathbf {x}_n^*(x)|=b$. If $b=0$, choose $j\in {\rm supp}(x)\setminus A$. Then $0=b\ge t |\mathbf {x}_j^*(x)|>0$, a contradiction.

Proof. To prove (i), choose a finitely supported $z$ so that $\|x-P_D(x)-z\|< (1+\|P_D\|)^{-1}\epsilon$, and define $y:=z-P_D(z)+P_D(x)$. We have

\begin{align*} \|x\!-y\|\!\le\! \|x\!-P_D(x)\!-z\|\!+\!\|P_D(z\!-x\!+\!P_D(x)\|\!\le\! (1\!+\!\|P_D\|)\|x-P_D(x)-z\|\!<\!\epsilon. \end{align*}

To prove (ii), set ${\mathcal {B}}^*=(\mathbf {x}^*_i)_{i\in \mathbb {N}}$ the dual basis of ${\mathcal {B}}$. We may assume that $x$ has infinite support, so the hypotheses of lemma 3.13 hold for $m=m_0$. Let

\begin{align*} B:=\bigcup_{A\in {\mathcal{G}}(x,m_0,t)}A, \quad b:=\min_{i\in B}|\mathbf{x}_i^*(x)|. \end{align*}

First note that if $m_0>1$, $1\le m< m_0$ and $A\in {\mathcal {G}}(x,m,t)$, then $A\cup A_1\in {\mathcal {G}}(x,m_0,t)$ for every $A_1\in {\mathcal {G}}(x-P_A(x),m_0-m,1)$. Hence, $A\subset B$.

Since ${\mathcal {B}}^*$ is weak star null, there is $i_0\in \mathbb {N}$ such that $|\mathbf {x}_i^*(x)|< 2^{-1}tb$ for all $i\ge i_0$. Set

\begin{align*} D:=\{1,\ldots,i_0\}, \quad\epsilon_1:=\frac{t \min\{\epsilon, 1, b\}}{2(1+\lambda')}, \end{align*}

and let $y$ be obtained by an application of (i) to $x$, $D$ and $\epsilon _1$. To see that $y$ has the desired properties, first note that for each $n\in D$ and each $k\not \in D$,

(3.3)\begin{equation} |\mathbf{x}_k^*(y)|\le |\mathbf{x}_k^*(x-y)|+|\mathbf{x}_k^*(x)|\le \lambda'\|x-y\|+2^{{-}1}tb< tb\le t|\mathbf{x}_{n}^*(x)|=t|\mathbf{x}_{n}^*(y)|.\end{equation}

Since $i_0>m_0$, it follows that ${\mathcal {G}}(y,m,t)\subset D$ for all $1\le m\le m_0$. Now fix $1\le m\le m_0$, and choose $A\in {\mathcal {G}}(x,m,t)$, $n\in A$ and $k\not \in A$. If $k\in D$, then

\begin{align*} |\mathbf{x}_n^*(y)|=|\mathbf{x}_n^*(x)|\ge t |\mathbf{x}_k^*(x)|=t|\mathbf{x}_k^*(y)|, \end{align*}

whereas if $k\not \in D$, then $t |\mathbf {x}_k^*(y)|\le |\mathbf {x}_{n}^*(y)$ by (3.3). Therefore, $A\in {\mathcal {G}}(y,m,t)$.

Similarly, choose $A\in {\mathcal {G}}(y,m,t)$, $n\in A$ and $k\not \in A$. Since $n\in D$, the case $k\in D$ is handled as before but changing the roles of $x$ and $y$, whereas if $k\not \in D$, then $t|\mathbf {x}_k^*(x)|< b\le |\mathbf {x}_n^*(y)|=|\mathbf {x}_n^*(x)|$. We conclude that $A \in {\mathcal {G}}(x,m,t)$.

Proof. Pick $x\in {\mathbb {X}}\setminus \{0\}$ and $m\in \mathbb {N}$. If $|{\rm supp}(x)|< m$, then ${\rm supp}(x) \subset A$ for every $A \in {\mathcal {G}}(x,m,t)$, and there is nothing to prove. Otherwise, for every $n\in \mathbb {N}$, by lemma 3.14 there is $x_n\in {\mathbb {X}}$ with finite support such that $\|x-x_n\|\le n^{-1}$ and ${\mathcal {G}}(x_n,m,t)={\mathcal {G}}(x,m,t)$. By hypothesis, for each $n$ there are $A_n\in {\mathcal {G}}(x,m,t)$ and $y_n\in {\mathbb {X}}$ with ${\rm supp}(y_n)\subset A_n$ such that

\begin{align*} \|x_n-y_n\|\le \mathbf{C}\inf_{\substack{z\in {\mathbb{X}}\\ |{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(A_n)}}\|x_n-z\|. \end{align*}

By lemma 3.13, ${\mathcal {G}}(x,m,t)$ is finite. Thus, passing to a subsequence, we may assume $A_n=A$, a fixed set. Passing to a further subsequence, we may also assume that there is $y$ supported in $A$ such that $\|y_n-y\|\le n^{-1}$ for all $n$. Now pick $z\in {\mathbb {X}}$ with $w({\rm supp}(z))\le w(A)$. By remark 3.3, for each $n\in \mathbb {N}$ there is $z_n\in {\mathbb {X}}$ with finite support contained in ${\rm supp}(z)$ such that $\|z_n-z\|\le n^{-1}$. Hence,

\begin{align*} \|x-y\|& \le2n^{{-}1}+\|x_n-y_n\|\le 2n^{{-}1}+\mathbf{C}\|x_n-z_n\|\\& \le \mathbf{C}\|x-z\|+2n^{{-}1}(1+\mathbf{C}), \forall n\in \mathbb{N}. \end{align*}

As this holds for every $n\in \mathbb {N}$, we get $\|x-y\|\le \mathbf {C}\|x-z\|$. Now the proof is completed by taking infimum over all such $z$.

Proof. To prove (i), fix $0<\epsilon <1$, and choose $j_0\in \mathbb {N}$ so that

\begin{align*} w_{j_0}^{{-}1}\le \inf_{j\in \mathbb{N}}w_j^{{-}1}+\epsilon. \end{align*}

Given $A$ and $\varepsilon$ as in the statement, define the possibly empty set

\begin{align*} A_1:=\{i\in A: w_i<2^{{-}1}w_{j_0} \}. \end{align*}

We first estimate the norm $\|\mathbf{1}_{\varepsilon,A\setminus A_1}\|$:

(3.4)\begin{align} \|\mathbf{1}_{\varepsilon,A\setminus A_1}\|& \le \lambda |A\setminus A_1|\le 2\lambda w_{j_0}^{{-}1}w(A)\le 2\lambda (1+\epsilon)\max\{\inf_{j\in \mathbb{N}}w_j^{{-}1},1\}\max\{w(A),1\}\nonumber\\ & \le \frac{1}{3}(1+\epsilon)\mathbf{C}_1\max\{w(A),1\}. \end{align}

If $A_1=\varnothing$, as $\epsilon$ is arbitrary there is nothing else to prove. Else, to estimate the norm $\|\mathbf{1}_{\varepsilon,A_1}\|$ choose a partition of $A_1$ as follows: first, pick a set $A_{1,1}\subset A_1$ of maximum cardinality such that $w(A_{1,1})\le w_{j_0}$. If $A_{1,1}\not =A_1$, then pick $A_{1,2}\subset A_1\setminus A_{1,1}$ of maximum cardinality such that $w(A_{1,2})\le w_{j_0}$, and so on. By this procedure, we get a partition of $A_1$ into finitely many sets $\{A_{1,k}\}_{1\le k\le k_1}$ with $w(A_{1,k})\le w_{j_0}$ for all $1\le k\le k_1$. If $k_1>1$, then by construction, for every $1\le k\le k_1-1$, there is $i\in A_1\setminus A_{1,k}$ such that $w(A_{1,k})+w_i>w_{j_0}$, which implies that $w(A_{1,k})>2^{-1}w_{j_0}$. Thus,

\begin{align*} w(A_1)> \sum\limits_{k=1}^{k_1-1}w(A_{1,k})\ge 2^{{-}1}(k_1-1)w_{j_0}, \end{align*}

so

(3.5)\begin{equation} k_1\le 2w(A_1)w_{j_0}^{{-}1}+1\le 2w(A)w_{j_0}^{{-}1}+1.\end{equation}

For each $1\le k\le k_1$, define

\begin{align*} z_k:=(1+\epsilon)s^{{-}1}\mathbf{x}_{j_0}+\mathbf{1}_{\varepsilon,A_{1,k}}. \end{align*}

As ${\mathcal {G}}(z_k,1,s)=\{\{j_0\}\}$, there is $b_k\in \mathbb {K}$ such that

\begin{align*} \|z_k-b_k\mathbf{x}_{j_0}\|\le \mathbf{C}\inf_{\substack{|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w_{j_0}}}\|z_k-z\|\le \mathbf{C}(1+\epsilon)s^{{-}1}\|\mathbf{x}_{j_0}\|. \end{align*}

Hence, by the triangle inequality,

\begin{align*} \|\mathbf{1}_{\varepsilon,A_{1,k}}\|& \le \|z_k-b_k\mathbf{x}_{j_0}\|+\|((1+\epsilon)s^{{-}1}-b_k)\mathbf{x}_{j_0}\|\\& \le (1+\epsilon)s^{{-}1}\mathbf{C}\|\mathbf{x}_{j_0}\|+ |(1+\epsilon)s^{{-}1}-b_k|\|\mathbf{x}_{j_0}\|. \end{align*}

Let ${\mathcal {B}}^*=(\mathbf {x}^*_i)_{i\in \mathbb {N}}$ be the dual basis of ${\mathcal {B}}$. Since

\begin{align*} |(1+\epsilon)s^{{-}1}-b_k|=|\mathbf{x}_{j_0}^{*}(z_k-b_k\mathbf{x}_{j_0})|\le \|\mathbf{x}_{j_0}^{*}\|\|z_k-b_k\mathbf{x}_{j_0}\|\le \mathbf{C}\|\mathbf{x}_{j_0}^{*}\|(1+\epsilon)s^{{-}1}\|\mathbf{x}_{j_0}\| \end{align*}

we obtain

\begin{align*} \|\mathbf{1}_{\varepsilon,A_{1,k}}\|\le \mathbf{C}(1+\epsilon)s^{{-}1}(1+\|\mathbf{x}_{j_0}\|\|\mathbf{x}_{j_0}^{*}\|)\|\mathbf{x}_{j_0}\|. \end{align*}

Using again the triangle inequality and (3.5), we get

\begin{align*} \|\mathbf{1}_{\varepsilon,A_1}\|& \le (2w(A)w_{j_0}^{{-}1}+1) \mathbf{C}(1+\epsilon)s^{{-}1}(1+\|\mathbf{x}_{j_0}\|\|\mathbf{x}_{j_0}^{*}\|)\|\mathbf{x}_{j_0}\|\\ & \le 2\mathbf{C}(1+\epsilon)s^{{-}1}(1+\lambda \lambda ')\lambda \max\{2w_{j_0}^{{-}1}w(A),1\}\\ & = \frac{2}{3} \mathbf{C}_1 (1+\epsilon)(\max\{2\inf_{j\in \mathbb{N}} w_j^{{-}1},1\})^{{-}1}\max\{2w_{j_0}^{{-}1}w(A),1\}\\ & \le \frac{2}{3}\mathbf{C}_1(1+\epsilon)\max\{w(A),1\}. \end{align*}

Given that $\epsilon$ is arbitrary, the proof of (i) is completed combining the above inequality with (3.4).

Now suppose (i) is false, and choose $x^*\in S_{{\mathbb {X}}^*}$ for which the result does not hold. Then, there is $A\subset \mathbb {N}$ finite with $w(A)>1$ such that

\begin{align*} |x^*(\mathbf{x}_j)|>\mathbf{C}_1 w_j,\quad\forall j\in A. \end{align*}

Define $\varepsilon \in {\mathcal{E}}_A$ by

\begin{align*} \varepsilon_j:=\frac{|x^*(\mathbf{x}_j)|}{x^*(\mathbf{x}_j)}, \quad \forall j\in A. \end{align*}

As $w(A)>1$, using (i) we get

\begin{align*} \mathbf{C}_1w(A)\ge \|\mathbf{1}_{\varepsilon,A}\|\ge |x^*(\mathbf{1}_{\varepsilon,A})|=\sum_{j\in A}|x^*(\mathbf{x}_j)|>\sum_{j\in A}\mathbf{C}_1 w_j=\mathbf{C}_1 w(A), \end{align*}

a contradiction.

To prove (iii), fix $x\in {\mathbb {X}}$, $m\in \mathbb {N}$ and $A\subset \mathbb {N}$ with $|A|\le m$. By (i),

\begin{align*} \|P_A(x)\|\le \|x\|_{\infty}\max_{\varepsilon\in {\mathcal{E}}_A}\|\mathbf{1}_{\varepsilon,A}\|\le \lambda '\mathbf{C}_1\max\{w(A),1\}\|x\|. \end{align*}

so the proof is completed by taking supremum.

Now suppose that ${\mathcal {B}}$ is $\mathbf {K}$-$\mathbf {w}$-disjoint superdemocratic. Then all of the steps of the above proof hold with the only modification consisting in replacing $\mathbf {C}_1$ with $\mathbf {K}_1$, except for the bounds for $\|\mathbf{1}_{\varepsilon, A\setminus A_{1}}\|$ and $\|\mathbf{1}_{\varepsilon, A_{1}}\|$; we give bounds for these norms as follows: First,

\begin{align*} \|\mathbf{1}_{\varepsilon, A_{1,k}}\|\le \mathbf{K}\|\mathbf{x}_{j_0}\|\le \mathbf{K}\lambda, \quad \forall 1\le k\le k_1. \end{align*}

Thus, using (3.5),

\begin{align*} \|\mathbf{1}_{\varepsilon, A_{1}}\|\le k_1\mathbf{K}\lambda \le \mathbf{K}\lambda (1+2w(A)w_{j_0}^{{-}1})\le \frac{1}{2}\mathbf{K}_1(1+2\epsilon)\max\{w(A),1\}. \end{align*}

On the other hand, arguing as in the proof of (3.4) we obtain

\begin{align*} \|\mathbf{1}_{\varepsilon,A\setminus A_1}\|\le 2\lambda (1+\epsilon)\max\{1,\inf_{j\in \mathbb{N}}w_j^{{-}1}\}\max\{w(A),1\}\le \frac{1}{2}(1+\epsilon)\mathbf{K}_1\max\{w(A),1\}, \end{align*}

and the result follows by the above inequalities.

If ${\mathcal {B}}$ is $\mathbf {K}$-$\mathbf {w}$-disjoint almost greedy, it is $\mathbf {K}$-$\mathbf {w}$-disjoint superdemocratic by lemma 3.9 and [Reference Berná16, theorem 1.5].

Finally, if ${\mathcal {B}}$ is $\mathbf {K}$-$\mathbf {w}$-disjoint democratic, by convexity we have

\begin{align*} \|\mathbf{1}_{\varepsilon, A_{1,k}}\|\le 2\kappa \sup_{B\subset A_{1,k}}\|\mathbf{1}_{B}\| \le 2\kappa \mathbf{K}\|\mathbf{x}_{j_0}\|\le 2\kappa \mathbf{K}\lambda, \quad 1\le k\le k_1, \end{align*}

and the rest of the proof is the same as that of the $\mathbf {K}$-$\mathbf {w}$-disjoint superdemocratic case.

Proof. Let $c:=\min _{i\in A}|\mathbf {x}_i^*(x)|$. Clearly we may assume $c>0$. Since ${\mathcal {B}}^*$ is weak star null, there is $n_0\in \mathbb {N}$ such that for each $n\ge n_0$, every set in ${\mathcal {G}}(x,n,s)$ contains $A$. Let

\begin{align*} n_1:=\min\{{n\in \mathbb{N}}: \exists B\in {\mathcal {GS}}(x,n,s): B\supset A\}, \end{align*}

and choose $B \in {\mathcal {GS}}(x,n_1,s)$ containing $A$. If $B=A$, there is nothing to prove. Otherwise, since $n_1>1$, we can choose $D \in {\mathcal {GS}}(x,n_1-1,s)$. By the minimality of $n_1$, it follows that

\begin{align*} A\not\subset D. \end{align*}

Hence, for all $j\in D$,

(3.7)\begin{equation} |\mathbf{x}_j^*(x)|\ge s c.\end{equation}

Thus, if there exists $j_0\in D\setminus B$, it follows that for all $j\in B$,

\begin{align*} |\mathbf{x}_j^*(x)|\ge s |\mathbf{x}_{j_0}^*(x)|\ge s^{2} c. \end{align*}

On the other hand, if $D\subseteq B$, given that $A\not \subset D$ and $A\subseteq B$, there is $i_1\in A$ such that

\begin{align*} B=D\cup \{ i_1\}, \end{align*}

which implies that (3.7) also holds for all $j\in B$.

Proof. To prove (i), choose $0<\epsilon <1$ and let $(\mathbf {x}_{i_{k_j}})_{j\in \mathbb {N}}$ be a subsequence given by an application of corollary 2.7 to $(\mathbf {x}_{i_k})_{k\in \mathbb {N}}$ and $({\mathbb {X}}, \mathbf {M}, \epsilon )$. Fix $x$, $m$, $t$ and $A$ as in the statement, and $y$ such that $w({\rm supp}(y))\le w(A)$ and ${\rm supp}(y)\cap A=\varnothing$. We assume first that both $x$ and $y$ have finite support, and we may also assume $x\not =P_A(x)$, so

\begin{align*} a:=\min_{i\in A}|\mathbf{x}_i^*(x)| \end{align*}

is positive. Pick $i_0>{\rm supp}(x)\cup {\rm supp}(y)$, and set $\mathbb {F}:=[ \mathbf {x}_i: 1\le i\le i_0]$. We will consider two cases:

Case 1. Suppose that there is a set $E\subset \{i_{k_j}\}_{j\ge s_{\mathbb {F},\epsilon }}$ such that $|E|\le 8$ and $2w(A)\le w(E)$. Define

\begin{align*} z_1:=x-P_A(x)+a(1+\epsilon) t^{{-}1}s^{{-}1}\mathbf{1}_{E }. \end{align*}

Notice that $|\mathbf {x}_i^*(z_1)|=|\mathbf {x}_i^*(x)|\le t^{-1}a$ for all $i\not \in E$, so ${\mathcal {G}}(z_1,|E|,s)=\{E\}$. Hence, there is $z_2\in {\mathbb {X}}$ with ${\rm supp}(z_2) \subset E$ such that

\begin{align*} \|z_1-z_2\|\le \mathbf{C} \inf_{\substack{|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(E)}}\|z_1-z\|. \end{align*}

Given that $w({\rm supp}(y))+w(A)\le w(E)$, we have

\begin{align*} \|z_1-z_2\|& \le \mathbf{C}\|z_1+P_A(x)-y\|\le \mathbf{C}\|x-y\|+\mathbf{C} a(1+\epsilon)t^{{-}1}s^{{-}1}\|\mathbf{1}_E\|\\ & \le \mathbf{C}\|x-y\|+8\mathbf{C} a(1+\epsilon)t^{{-}1}s^{{-}1} \lambda . \end{align*}

Pick any $i\in A$. Since $A\cap {\rm supp}(y)=\varnothing$, we have

\begin{align*} a\le |\mathbf{x}_i^*(x-y)|\le \lambda '\|x-y\|. \end{align*}

Thus,

\begin{align*} \|z_1-z_2\|\le (\mathbf{C}+8\mathbf{C}(1+\epsilon)t^{{-}1}s^{{-}1} \lambda \lambda ')\|x-y\|. \end{align*}

Given that

\begin{align*} \|x-P_A(x)\|\le (\mathbf{M}+\epsilon)\|z_1-z_2\|, \end{align*}

it follows that

(3.8)\begin{equation} \|x-P_A(x)\|\le \mathbf{C} (\mathbf{M}+\epsilon) (1+8(1+\epsilon) t^{{-}1}s^{{-}1} \lambda \lambda ')\|x-y\|.\end{equation}

Case 2. Suppose there is ${j_0}>s_{\mathbb {F},\epsilon }$ such that

(3.9)\begin{equation} w(A)>4w_{i_{k_{j}}},\quad \forall j\ge {j_0}.\end{equation}

Choose $l_1, l_2\ge 3$ so that

\begin{align*} & w(\{i_{k_{2({j_0}+d)-1}},i_{k_{2({j_0}+d)}}: 1\le d< l_1\})\\& \quad \le w(A)\le w(\{i_{k_{2({j_0}+d)-1}},i_{k_{2({j_0}+d)}}: 1\le d\le l_1\});\\ & w(\{i_{k_{2({j_0}+l_1+d)-1}},i_{k_{2({j_0}+l_1+d)}}: 1\le d< l_2\})\\& \quad \le w(A)\le w(\{i_{k_{2({j_0}+l_1+d)-1}},i_{k_{2({j_0}+l_1+d)}}: 1\le d\le l_2\}). \end{align*}

Set

\begin{align*} & E_{1,1}:=\{i_{k_{2({j_0}+d)-1}}: 1\le d< l_1\}; \quad E_{1,2}:=\{i_{k_{2({j_0}+d)}}: 1\le d< l_1\};\\ & E_{2,1}:=\{i_{k_{2({j_0}+l_1+d)-1}}: 1\le d< l_2\}; \quad E_{2,2}:=\{i_{k_{2({j_0}+l_1+d)}}: 1\le d< l_2\};\\ & E_{3,1}:=\{i_{k_{2({j_0}+l_1)-1}},i_{k_{2({j_0}+l_1+l_2)-1}}\}; \quad E_{3,2}:=\{i_{k_{2({j_0}+l_1)}},i_{k_{2({j_0}+l_1+l_2)}}\};\\ & E:=\bigcup_{b=1}^{3}\bigcup_{d=1}^{2}E_{b,d}. \end{align*}

It follows from our choices and (3.9) that

(3.10)\begin{equation} \max_{1\le b\le 3}w(E_{b,1}\cup E_{b,2})\le w(A)\le\frac{w(E)}{2}. \end{equation}

Now define

\begin{align*} z_3:=x-P_A(x)-as^{{-}1}t^{{-}1}(1+\epsilon)\left(\sum_{b=1}^{3}\mathbf{1}_{E_{b,1}} -\mathbf{1}_{E_{b,2}}\right). \end{align*}

As before, we have $|\mathbf {x}_i^*(z_3)|=|\mathbf {x}_i^*(x-P_A(x))|\le a t^{-1}$ for all $i\not \in E$, so ${\mathcal {G}}(z_3,|E|,s)=\{E\}$. Hence, there is $z_4$ with ${\rm supp}(z_4)\subset E$ such that

\begin{align*} \|z_3-z_4\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w(E)\\|{\rm supp}(z)|<\infty}}\|z_3-z\|. \end{align*}

Since $w({\rm supp}(y))+w(A)\le 2w(A) \le w(E)$, we have

(3.11)\begin{align} \|x-P_A(x)\|& \le (\mathbf{M}+\epsilon)\|z_3-z_4\|\le \mathbf{C}(\mathbf{M}+\epsilon) \|z_3+P_A(x)-y\|\nonumber\\ & \le \mathbf{C}(\mathbf{M}+\epsilon)\|x-y\|+ \mathbf{C}(\mathbf{M}+\epsilon)as^{{-}1}t^{{-}1}(1+\epsilon)\|\sum_{b=1}^{3}\mathbf{1}_{E_{b,1}} -\mathbf{1}_{E_{b,2}}\|. \end{align}

To estimate $\|\sum _{b=1}^{3}\mathbf{1}_{E_{b,1}} -\mathbf{1}_{E_{b,2}}\|$, set

\begin{align*} z_5:=x-y+a s^2(1-\epsilon)(\mathbf{1}_{E_{1,1}}-\mathbf{1}_{E_{1,2}}). \end{align*}

By lemma 3.21, there is a set $D\supset A$ with $D\in {\mathcal {GS}}(z_5,|D|,s)$ such that

\begin{align*} \min_{j\in D}|\mathbf{x}_j^*(z_5)|\ge s^2\min_{i\in A}|\mathbf{x}_i^*(z_5)|=s^2\min_{i\in A}|\mathbf{x}_i^*(x)|=s^2a, \end{align*}

which implies that $D\subset \{1,\dots,i_0\}$. Choose $z_6$ with ${\rm supp}(z_6)\subset D$ so that

\begin{align*} \|z_5-z_6\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w(D)\\|{\rm supp}(z)|<\infty}}\|z_5-z\|. \end{align*}

Given that $z_6\in \mathbb {F}$, using (3.10) and the fact that $w(A)\le w(D)$ we infer that

\begin{align*} \|a s^2(1-\epsilon)(\mathbf{1}_{E_{1,1}}-\mathbf{1}_{E_{1,2}})\|& \le \|z_5-z_6\|+\|x-y-z_6\|\le (2+\epsilon)\|z_5-z_6\|\\ & \le (2+\epsilon)\mathbf{C}\|z_5-a s^2(1-\epsilon)(\mathbf{1}_{E_{1,1}}-\mathbf{1}_{E_{1,2}})\|\\ & = (2+\epsilon)\mathbf{C}\|x-y\|. \end{align*}

The same argument gives

\begin{align*} \|a s^2(1-\epsilon)(\mathbf{1}_{E_{b,1}}-\mathbf{1}_{E_{b,2}})\|\le (2+\epsilon)\mathbf{C}\|x-y\,|\qquad\forall 2\le b\le 3. \end{align*}

Combining these inequalities with (3.11), by the triangle inequality we get

(3.12)\begin{equation} \|x-P_A(x)\|\le \mathbf{C}(\mathbf{M}+\epsilon)(1+3(2+\epsilon)\mathbf{C}(1+\epsilon)(1-\epsilon)^{{-}1}t^{{-}1}s^{{-}3})\|x-y\|. \end{equation}

As $\epsilon$ is arbitrary, a combination of (3.8) and (3.12) gives (i) for $x, y\in {\mathbb {X}}$ with finite support. Now choose again $x,m, t, A$ and $y$ as in the statement, $x\not =P_A(x)$, fix $\delta >0$, and let $\mathbf {K}:=\mathbf {C}\mathbf {M} \max \{1+8 t^{-1}s^{-1} \lambda \lambda ',1+6\mathbf {C} t^{-1}s^{-3} \}$. By lemma 3.14, there is $x_1\in {\mathbb {X}}$ with finite support such that $A\in {\mathcal {G}}(x_1,m,t)$, $P_A(x_1)=P_A(x)$ and $\|x-x_1\|\le \delta$. Also, remark 3.3 gives $y_1\in {\mathbb {X}}$ with finite support such that ${\rm supp}(y_1)\subset {\rm supp}(y)$ and $\|y-y_1\|\le \delta$. Applying the result for vectors with finite support, we obtain

\begin{align*} \|x-P_A(x)\|& \le\|x_1-x\|+\|x_1-P_A(x_1)\|\le \delta+\mathbf{K}\|x_1-y_1\|\\ & \le \delta+\mathbf{K}\|x-y\|+\mathbf{K}\|x_1-x\|+\mathbf{K}\|y-y_1\|\le 3\delta+\mathbf{K}\|x-y\|. \end{align*}

Since $\delta$ is arbitrary, the proof of (i) is complete.

To prove (ii), choose $0<\epsilon <1$, let $(\mathbf {x}_{i_{k_j}})_{j\in \mathbb {N}}$ be as in the proof of (i), and let $A$, $\varepsilon$ and $x$ be as in the statement, and suppose $x$ has finite support. Choose $i_0>A\cup {\rm supp}(x)$, and set

\begin{align*} \mathbb{F}:=[\mathbf{x}_i: 1\le i\le i_0]. \end{align*}

We will consider again two cases:

Case 1. Suppose there are $j_2>j_1> s_{\mathbb {F},\epsilon }$ such that

\begin{align*} w(A)\le w_{i_{k_{j_1}}}+w_{i_{k_{j_2}}}. \end{align*}

Set

\begin{align*} E:=\{i_{k_{j_1}}, i_{k_{j_2}}\} \quad \text{and} \quad z_1:=\mathbf{1}_{\varepsilon, A} +(1+\epsilon) s^{{-}1}\mathbf{1}_{E }. \end{align*}

Since ${\mathcal {G}}(z_1,2,s)=\{E\}$, there is $z_2\in \mathbb {N}$ with ${\rm supp}(z_2)\subset E$ such that

\begin{align*} \|z_1-z_2\|\le \mathbf{C} \inf_{\substack{|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(E)}}\|z_1-z\|. \end{align*}

Hence,

\begin{align*} \|\mathbf{1}_{\varepsilon,A}\|\le (\mathbf{M}+\epsilon)\|z_1-z_2\|\le (\mathbf{M}+\epsilon) \mathbf{C} (1+\epsilon) s^{{-}1}\|\mathbf{1}_{E}\|. \end{align*}

Given that

\begin{align*} \|\mathbf{1}_E\|\le 2\lambda \le 2\lambda \lambda '\|x\|, \end{align*}

we obtain

(3.13)\begin{equation} \|\mathbf{1}_{\varepsilon,A}\|\le 2(\mathbf{M}+\epsilon)\mathbf{C}(1+\epsilon)s^{{-}1}\lambda \lambda '\|x\|. \end{equation}

Case 2. Suppose that there is ${j_0}>s_{\mathbb {F},\epsilon }$ such that

\begin{align*} w(A)\ge 2 w(i_{k_{j}}),\quad \forall j\ge {j_0}, \end{align*}

and choose $l_1\ge 2$ so that

\begin{align*} & w\big(\{i_{k_{2({j_0}+d)-1}},i_{k_{2({j_0}+d)}}: 1\le d< l_1\}\big)\le w(A)\\& \quad \le w\big(\{i_{k_{2({j_0}+d)-1}},i_{k_{2({j_0}+d)}}: 1\le d\le l_1\}\big). \end{align*}

Define

\begin{align*} & E_{1,1}:=\{i_{k_{2({j_0}+d)-1}}: 1\le d< l_1\}; \quad E_{1,2}:=\{i_{k_{2({j_0}+d)}}: 1\le d< l_1\};\\ & E_{2,1}:=\{i_{k_{2({j_0}+l_1)-1}} \}; \quad E_{2,2}:=\{i_{k_{2({j_0}+l_1)}}\};\\ & E:=\bigcup_{b=1}^{2}\bigcup_{d=1}^{2}E_{b,d}. \end{align*}

Note that $\max \{w(E_{1,1}\cup E_{1,2}),w(E_{2,1}\cup E_{2,2})\}\le w(A)\le w(E)$. Let

\begin{align*} B:=\{i\in \mathbb{N}: |\mathbf{x}_i^*(x)|\ge 1\}\quad\text{and}\quad z_1:=x + s^2(1-\epsilon)(\mathbf{1}_{E_{1,1}}-\mathbf{1}_{E_{1,2}}). \end{align*}

By lemma 3.21, there is a set $D\supset B$ with $D\in {\mathcal {GS}}(z_1,|D|,s)$ and a vector $z_2$ with ${\rm supp}(z_2)\subset D$ such that

\begin{align*} \min_{j\in D}|\mathbf{x}_j^*(x)|\ge s^2, \end{align*}

and

\begin{align*} \|z_1-z_2\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w(D)\\|{\rm supp}(z)|<\infty}}\|z_1-z\|. \end{align*}

Since $D\subset \{1,\dots,i_0\}$, considering that $w(D)\ge w(B)\ge w(A)\ge w(E_{1,1})+w(E_{1,2})$, we have

\begin{align*} \| s^2(1-\epsilon)(\mathbf{1}_{E_{1,1}}-\mathbf{1}_{E_{1,2}})\|& \le \|z_1-z_2\|+\|x-z_2\|\le (2+\epsilon)\|z_1-z_2\|\\ & \le (2+\epsilon)\mathbf{C}\|x\|. \end{align*}

The same argument gives

\begin{align*} \| s^2(1-\epsilon)(\mathbf{1}_{E_{2,1}}-\mathbf{1}_{E_{2,2}})\|\le (2+\epsilon)\mathbf{C}\|x\|. \end{align*}

To finish the proof, define

\begin{align*} z_3:= \mathbf{1}_{\varepsilon, A}+ s^{{-}1}(1+\epsilon)\left(\sum_{b=1}^{2}\mathbf{1}_{E_{b,1}} -\mathbf{1}_{E_{b,2}}\right)\!. \end{align*}

Since ${\mathcal {G}}(z_3,|E|,s)=\{E\}$, there is $z_4$ with ${\rm supp}(z_4)\subset E$ such that

\begin{align*} \|z_3-z_4\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\ w({\rm supp}(z))\le w(E)\\|{\rm supp}(z)|<\infty}}\|z_3-z\|. \end{align*}

Therefore,

\begin{align*} \|\mathbf{1}_{\varepsilon, A}\|& \le (\mathbf{M}+\epsilon)\|z_3-z_4\|\le \mathbf{C}(\mathbf{M}+\epsilon) s^{{-}1}(1+\epsilon)\left\|\sum_{b=1}^{2}\mathbf{1}_{E_{b,1}} -\mathbf{1}_{E_{b,2}}\right\|\\ & \le 2\mathbf{C}^2(\mathbf{M}+\epsilon) s^{{-}3}(1+\epsilon)(1-\epsilon)^{{-}1}(2+\epsilon)\|x\|. \end{align*}

Now the proof is completed combining the above result with (3.13), and letting $\epsilon$ tend to zero.

Proof. The statement for the case of an $r$-norming dual basis follows from proposition 2.3, so we need to prove (i) and (ii). To prove the former, fix $x$, $m$, $t$ and $A$ as in the statement, and $y$ with $w({\rm supp}(y))\le w(A)$. We may assume $x\not =P_A(x)$, and we will also assume that both $x$ and $y$ have finite support. Set

\begin{align*} a:=\min_{i\in A}|\mathbf{x}_i^*(x)|, \end{align*}

choose $i_0>{\rm supp}(x)\cup {\rm supp}(y)$, and set $\mathbb {F}:=[ \mathbf {x}_i: 1\le i\le i_0]$. Given that $(w_{i_k})_{k\in \mathbb {N}}\in \texttt {c}_0\setminus \ell _1$, there is a sequence of finite sets of positive integers $(A_k)_{k\in \mathbb {N}}$ such that for all $k\in \mathbb {N}$, $i_0< A_k< A_{k+1}$ and

(3.14)\begin{equation} \frac{w(A)}{3}\le w(A_k)\le \frac{w(A)}{2}. \end{equation}

By hypothesis $\mathbf {M}_{fs}((\mathbf{1}_{A_k})_{k\in \mathbb {N}},{\mathbb {X}})\le \mathbf {M}$. To simplify our notation, we may assume that $(\mathbf{1}_{A_k})_{k\in \mathbb {N}}$ is already a separating sequence for $({\mathbb {X}},\mathbf {M},\epsilon )$. By proposition 3.18, $(\mathbf{1}_{A_k})_{k\in \mathbb {N}}$ is bounded. Since it is also $\lambda '^{-1}$-uniformly discrete, we may choose $0<\epsilon <1$ and apply lemma 2.6 to $(\mathbf{1}_{A_k})_{k\in \mathbb {N}}$. Again, we assume that $(\mathbf{1}_{A_k})_{k\in \mathbb {N}}$ is already the subsequence given by the lemma. Set

\begin{align*} \mathbb{L}:=\{k\in 5\mathbb{N}: k> \max\{r_{\mathbb{F},(\mathbf{1}_{A_n})_{n\in \mathbb{N}},\epsilon}, j_{\mathbb{F},(\mathbf{1}_{A_n})_{n\in \mathbb{N}},\epsilon}, \}\}, \end{align*}

and for every $k\in \mathbb {L}$, define

\begin{align*} z_{1,k}& :=x-P_A(x)+s^{{-}1}t^{{-}1}a(1+\epsilon)\sum_{l=1}^{3}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}});\\ z_{2,k,l}& :=x-y-s^2(1-\epsilon)a(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}}), \quad\forall 1\le l\le 3. \end{align*}

Fix $k$ as above and $1\le l\le 3$. By lemma 3.21, there is a set $D\supset A$ such that $D\in {\mathcal {GS}}(z_{2,k,l},|D|,s)$ and

\begin{align*} \min_{j\in D}|\mathbf{x}_j^*(z_{2,k,l})|\ge s^2\min_{j\in A}|\mathbf{x}_j^*(z_{2,k,l})|=s^2\min_{j\in A}|\mathbf{x}_j^*(x)|=s^2 a. \end{align*}

It follows that $D\subset \{1,\ldots,i_0\}$, so there is $z_{3,k,l}\in \mathbb {F}$ such that

\begin{align*} \|z_{2,k,l}-z_{3,k,l}\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(D)}}\|z_{2,k,l}-z\|. \end{align*}

As $D\supset A$, using the above inequality and (3.14) we get

\begin{align*} \|z_{2,k,l}-z_{3,k,l}\|\le \mathbf{C}\|x-y\|. \end{align*}

Hence,

(3.15)\begin{align} \|s^2(1-\epsilon)a(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})\|& \le \|z_{2,k,l}-z_{3,k,l}\|+\|x-y-z_{3,k,l}\|\nonumber\\ & \le (2+\epsilon) \|z_{2,k,l}-z_{3,k,l}\|\le \mathbf{C}(2+\epsilon)\|x-y\|. \end{align}

Now we consider $z_{1,k}$: set

\begin{align*} B_k:=\bigcup_{l=1}^{3}A_{2k+2l-1}\cup A_{2k+2l}. \end{align*}

Notice that ${\mathcal {G}}(z_{1,k},|B_k|,s)=\{B_k\}$. Hence, there is $z_{4,k}$ with ${\rm supp}(z_{4,k})\subset B_k$ such that

\begin{align*} \|z_{1,k}-z_{4,k} \|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(B_k)}}\|z_{1,k}-z\|. \end{align*}

By (3.14), $2w(A)\le w(B_k)$. Since $w({\rm supp}(y))\le w(A)$, this gives

(3.16)\begin{align} \|z_{1,k}-z_{4,k} \|& \le \mathbf{C} \|z_{1,k}+P_A(x)-y\| \nonumber\\ & \le \mathbf{C}\|x-y\|+\mathbf{C} a(1+\epsilon)s^{{-}1}t^{{-}1}\left\|\sum_{l=1}^{3}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})\right\|. \end{align}

Note that the sequence

\begin{align*} \left(u_k:=a(1+\epsilon)s^{{-}1}t^{{-}1}\sum_{l=1}^{3}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})-z_{4,k}\right)_{k\in \mathbb{L}} \end{align*}

has the FDSP with constant $\le \mathbf {M}$. Indeed, if $u_k=0$ for infinitely many values of $k$, the sequence has this property with constant $1$, whereas if this is not the case, there is $k_0\in \mathbb {L}$ such that the subsequence beginning in $k_0$ is a block basis of ${\mathcal {B}}$, so we have this bound by hypothesis. In particular, it follows that there is $k\in \mathbb {L}$ such that

\begin{align*} \|x-P_A(x) \|& \le (\mathbf{M}+\epsilon) \left \| \vphantom{\sum_{l=1}^{3}} x-P_A(x) \right. \\ & \quad + \left. a(1+\epsilon)s^{{-}1}t^{{-}1}\sum_{l=1}^{3}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})-z_{4,k} \right \|\\ & = (\mathbf{M}+\epsilon)\|z_{1,k}-z_{4,k}\|, \end{align*}

which, when combined with (3.15), (3.16) and the triangle inequality gives

\begin{align*} \|x-P_A(x)\|\le \mathbf{C}(\mathbf{M}+\epsilon)(1+3\mathbf{C}(1+\epsilon)(1-\epsilon)^{{-}1}(2+\epsilon)t^{{-}1}s^{{-}3})\|x-y\|. \end{align*}

As $\epsilon$ is arbitrary, this completes the proof of (i) for $x$ and $y$ with finite support, and the general case is proven by the argument given in the proof of theorem 3.22.

The proof of (ii) is similar: fix $A$, $\varepsilon$ and $x$ as in the statement, set

\begin{align*} B:=\{i\in \mathbb{N}: |\mathbf{x}_i^*(x)|\ge 1\}, \end{align*}

choose $0<\epsilon <1$ and $i_0>A\cup {\rm supp}(x)$. Now choose a sequence of sets of positive integers $(A_k)_{k\in \mathbb {N}}$ so that for all $k$, $i_0< A_k< A_{k+1}$ and

(3.17)\begin{equation} \frac{w(A)}{4}\le w(A_k) \le \frac{w(A)}{2}.\end{equation}

As before, we assume that $(\mathbf{1}_{A_k})_{k\in \mathbb {N}}$ is already a separating sequence for $({\mathbb {X}},\mathbf {M},\epsilon )$, and that we have applied lemma 2.6. Set

\begin{align*} \mathbb{F}& :=[ \mathbf{x}_i: 1\le i\le i_0];\\ \mathbb{L}& :=\{k\in 5\mathbb{N}: k >\max\{r_{\mathbb{F},(\mathbf{1}_{A_n})_{n\in \mathbb{N}},\epsilon}, j_{\mathbb{F},(\mathbf{1}_{A_n})_{n\in \mathbb{N}},\epsilon},\} \}. \end{align*}

For each $k\in \mathbb {L}$, define

\begin{align*} z_{1,k}& :=\mathbf{1}_{\varepsilon,A}+s^{{-}1} (1+\epsilon)\sum_{l=1}^{2}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}});\\ z_{2,k,l}& :=x-s^2(1-\epsilon) (\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}}),\quad\forall 1\le l\le 2. \end{align*}

Fix $k\in \mathbb {L}$ and $1\le l\le 2$. By lemma 3.21, there is a set $D\supset B$ such that $D\in {\mathcal {GS}}(z_{2,k,l},|D|,s)$ and

\begin{align*} \min_{j\in D}|\mathbf{x}_j^*(z_{2,k,l})|\ge s^2\min_{j\in B}|\mathbf{x}_j^*(z_{2,k,l})|=s^2\min_{j\in B}|\mathbf{x}_j^*(x)|\ge s^2, \end{align*}

which implies that $D\subset \{1,\dots,i_0\}$. Hence, there is $z_{3,k,l}\in \mathbb {F}$ such that

\begin{align*} \|z_{2,k,l}-z_{3,k,l}\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(D)}}\|z_{2,k,l}-z\|. \end{align*}

As $D\supset B$, using the above inequality, (3.17) and the fact that $w(A)\le w(B)$ we get

\begin{align*} \|z_{2,k,l}-z_{3,k,l}\|\le \mathbf{C}\|x\|. \end{align*}

Thus, the property of lemma 2.6 gives

(3.18)\begin{align} \|s^2(1-\epsilon)(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})\|& \le \|z_{2,k,l}-z_{3,k,l}\|+\|x-z_{3,k,l}\|\nonumber\\ & \le (2+\epsilon) \|z_{2,k,l}-z_{3,k,l}\|\le \mathbf{C}(2+\epsilon)\|x\|. \end{align}

For fixed $k\in \mathbb {L}$, set

\begin{align*} B_k:=\bigcup_{l=1}^{2}A_{2k+2l-1}\cup A_{2k+2l}. \end{align*}

Note that ${\mathcal {G}}(z_{1,k},|B_k|,s)=\{B_k\}$. Thus, there is $z_{4,k}$ with support contained in $B_k$ such that

\begin{align*} \|z_{1,k}-z_{4,k} \|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\|{\rm supp}(z)|<\infty\\ \mathbf{w}({\rm supp}(z))\le w(B_k)}}\|z_{1,k}-z\|. \end{align*}

By (3.17), $w(A)\le w(B_k)$, so

(3.19)\begin{equation} \|z_{1,k}-z_{4,k} \|\le \mathbf{C} \|z_{1,k} -\mathbf{1}_{\varepsilon,A} \|=\mathbf{C} (1+\epsilon)s^{{-}1} \left \|\sum_{l=1}^{2}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})\right \|.\end{equation}

As before, the sequence

\begin{align*} \left((1+\epsilon)s^{{-}1}\sum_{l=1}^{2}(\mathbf{1}_{A_{2k+2l-1}}-\mathbf{1}_{A_{2k+2l}})-z_{4,k}\right)_{k\in \mathbb{L}} \end{align*}

has the FDSP with constant $\le \mathbf {M}$. In particular, there is $k\in \mathbb {L}$ such that

\begin{align*} \|\mathbf{1}_{\varepsilon,A}\|\le (\mathbf{M}+\epsilon)\|z_{1,k}-z_{4,k}\| \end{align*}

which, when combined with (3.18), (3.19) and the triangle inequality gives

\begin{align*} \|\mathbf{1}_{\varepsilon,A}\|\le 2\mathbf{C}^2(\mathbf{M}+\epsilon)(1+\epsilon)(1-\epsilon)^{{-}1}(2+\epsilon)s^{{-}3}\|x\|. \end{align*}

As $\epsilon$ is arbitrary, the proof is complete.

Proof. As this is not a quantitative result, we will not keep track of the constants (even so, note that the right-hand side of the inequality in (ii) was estimated in proposition 3.18).

Note that (ii) follows from (i), proposition 3.18 and the fact that $\|x\|_{\infty }\le \lambda '\|x\|$ for all $x\in {\mathbb {X}}$; (iii) follows from that fact together with (i) and (ii), whereas (iv) follows from (ii). Thus, we only need to prove (i). Moreover, by lemma 3.14, it is sufficient to prove (i) for $x$ with finite support.

Let ${\mathcal {B}}=( \mathbf {x}_i)_{i\in \mathbb {N}}$. Since $(v_k)_{k\in \mathbb {N}}$ is not weakly null, passing to a subsequence we may assume there is $\epsilon >0$ and $x^*\in S_{X^*}$ such that

\begin{align*} |x^*(v_k)|\ge \epsilon, \quad\forall k\in\mathbb{N}. \end{align*}

For each $k\in \mathbb {N}$, set $A_k:={\rm supp}(v_k)$, and define $c_1:=\max \{1,\| (\|v_k\|_{\infty })_{k\in \mathbb {N}}\|_{\infty }\}$.

Note that $c_1$ is a well-defined positive number because ${\mathcal {B}}^*$ and $(v_k)_{k\in \mathbb {N}}$ are both bounded. For each $k\in \mathbb {N}$, choose $\varepsilon ^{(k)}\in {\mathcal{E}}_{A_k}$ so that

\begin{align*} \varepsilon_j^{(k)} x^*(\mathbf{x}_j)\ge 0, \quad \forall j\in A_k. \end{align*}

Note that $(\mathbf{1}_{\varepsilon ^{(k)},A_k})_{k\in \mathbb {N}}$ is bounded by proposition 3.18. Let $\epsilon _1:=\epsilon c_1^{-1}$. For each $k\in \mathbb {N}$, we have

(3.20)\begin{align} x^*(\mathbf{1}_{\varepsilon^{(k)},A_k})& =\sum_{j\in A_k}|x^*(\mathbf{x}_j)|\ge c_1^{{-}1}\sum_{j\in A_k}|\mathbf{x}_j^*(v_k)| |x^*(\mathbf{x}_j)|\ge c_1^{{-}1}|x^*\left(\sum_{j\in A_k}\mathbf{x}_j^*(v_j)\mathbf{x}_j\right)|\nonumber\\ & = c_1^{{-}1}|x^*(v_k)| \ge c_1^{{-}1}\epsilon =\epsilon_1>0. \end{align}

Set

\begin{align*} a=\liminf_{k\to \infty}w(A_k). \end{align*}

By hypothesis, $a$ is a nonnegative real number. We claim that $a>0$. Otherwise, there would be a subsequence $(\mathbf{1}_{\varepsilon ^{(k_j)},A_{k_j}})_{j\in \mathbb {N}}$ such that $(w(A_{k_j}))_{j\in \mathbb {N}}\in \ell _1$. By lemma 3.2, $((\mathbf {x}_i)_{i\in A_{k_j}})_{j\in \mathbb {N}}$ would be equivalent to the canonical unit vector basis of $\texttt {c}_{0}$, so all of its bounded block bases would be weakly null, contradicting (3.20). Thus, passing to a subsequence if necessary we may assume that

(3.21)\begin{equation} \frac{a}{2}\le w(A_k)\le 2a, \quad\forall k\in\mathbb{N}. \end{equation}

Let $\mathbf {M}:=\mathbf {M}_{fs}((\mathbf{1}_{\varepsilon ^{(k)},A_k})_{k\in \mathbb {N}},{\mathbb {X}})$. Applying proposition 2.3, again we may assume that $(\mathbf{1}_{\varepsilon ^{(k)},A_k})_{k\in \mathbb {N}}$ is already a separating sequence with the properties of definition 2.1 for $({\mathbb {X}},\mathbf {M},1)$.

Now fix $x\in {\mathbb {X}}$ with finite support, $A$ a finite nonempty subset of $\mathbb {N}$ and $\varepsilon \in {\mathcal{E}}_{A}$. We may assume that $A\subset {\rm supp}(x)$. If $w(A)\le 3a$, pick any $i\in A$. We have

(3.22)\begin{equation} \min_{i\in A}|\mathbf{x}_i^*(x)| w(A)\le 3a |\mathbf{x}_{i}^*(x)|\le 3a \lambda '\|x\|. \end{equation}

On the other hand, if $w(A)>3a$, define $b:=\min _{i\in A}|\mathbf {x}_i^*(x)|$, and set

\begin{align*} \mathbb{F}:=[\mathbf{x}_n: n\in {\rm supp}(x)]. \end{align*}

By (3.21), there is $B>j_{\mathbb {F},1}$ such that

(3.23)\begin{equation} \sum_{k\in B}w(A_k)\le w(A)\le 2 \sum_{k\in B}w(A_k). \end{equation}

Set

\begin{align*} z_1:=x+ \frac{bs^{2}}{2}\sum_{k\in B}\mathbf{1}_{\varepsilon^{(k)},A_k}. \end{align*}

By lemma 3.21, there is $D\supset A$ such that $D\in {\mathcal {GS}}(z_1,|D|,s)$ and

\begin{align*} \min_{j\in D}|\mathbf{x}_j^*(z_1)|\ge s^2 \min_{j\in A}|\mathbf{x}_j^*(z_1)|=s^2 \min_{j\in A}|\mathbf{x}_j^*(x)|=bs^2. \end{align*}

It follows that $D\subset {\rm supp}(x)$, so there is $z_2\in \mathbb {F}$ such that

\begin{align*} \|z_1-z_2\|\le \mathbf{C} \inf_{\substack{z\in {\mathbb{X}}\\|{\rm supp}(z)|<\infty\\ w({\rm supp}(z))\le w(D)}}\|z_{1}-z\|. \end{align*}

Thus, using (3.23) and the separating condition on $(\mathbf{1}_{\varepsilon ^{(k)},A_k})_{k\in \mathbb {N}}$ we deduce that

\begin{align*} \left\|\frac{b s^{2}}{2} \sum_{k\in B}\mathbf{1}_{\varepsilon^{(k)},A_k}\right\|\le \|z_1-z_2\|+\|x-z_2\|\le (\mathbf{M}+2)\|z_1-z_2\|\le (\mathbf{M}+2)\mathbf{C}\|x\|. \end{align*}

On the other hand, by (3.20), (3.21) and (3.23),

\begin{align*} \left\|\sum_{k\in B}\mathbf{1}_{\varepsilon^{(k)},A_k}\right\|\ge |x^*\left(\sum_{k\in B}\mathbf{1}_{\varepsilon^{(k)},A_k}\right)|\ge \epsilon_1 |B|\ge \epsilon_1 \sum_{k\in B}\frac{w(A_k)}{2a}\ge \frac{\epsilon_1}{4a} w(A). \end{align*}

Hence,

\begin{align*} \min_{i\in A}|\mathbf{x}_i^*(x)|w(A)=bw(A) \le 8 a \epsilon_1^{{-}1} s^{{-}2}\mathbf{C}(\mathbf{M}+2)\|x\|. \end{align*}

The proof is completed combining the above inequality and (3.22).

Proof. Note that if $\mathbf {w}\in \texttt {c}_0\setminus \ell _1$ and the hypotheses of proposition 3.24 do not hold, then by proposition 2.3 the hypotheses of theorem 3.23 do, with $\mathbf {M}=1$. Hence, it follows from theorems 3.22, 3.23, proposition 3.24 and lemma 3.2 that if ${\mathcal {B}}$ is $s$-$\mathbf {w}$-semi-greedy, it is truncation quasi-greedy and $\mathbf {w}$-superdemocratic. Then, by [Reference Berná, Dilworth, Kutzarova, Oikhberg and Wallis20, proposition 3.13], it has the $\mathbf {w}$-property (A).