Proof. The first statement is straightforward, and thus we only provide a proof of the second one. Let Y and Z be complemented subspaces of X, which are isomorphic to X. Let $P:X\to Y$ and $Q:X\to Z$ be the associated projections, and $A:X\to Y$ and $B:X\to Z$ the associated isomorphisms satisfying $\|A\|\|A^{-1}P\| \leq C$ , $\|B\|\|B^{-1}Q\|\leq D$ . $\|A^{-1}PT - S\| \leq \varepsilon $ and $\|B^{-1}QSB - R\| \leq \delta $ .

We define $\tilde P = AQA^{-1}P$ and $\tilde A = AB$ . Then $\tilde P$ is a projection onto $\tilde A[X]$ and $\|\tilde P\|\|\tilde A^{-1}P\| \leq CD$ . We obtain

$$ \begin{align*} \|B^{-1}Q(A^{-1}PTA)B - B^{-1}QSB\| \leq \|B^{-1} Q\|\|B\|\|A^{-1}PTA - S\|\leq D \varepsilon \end{align*} $$

and thus $\|B^{-1}QA^{-1}PTAB - R\| \leq D\varepsilon +\delta $ . Finally, observe that

$$ \begin{align*}\tilde A^{-1}\tilde P = B^{-1}A^{-1}AQA^{-1}P = B^{-1}QA^{-1}P\end{align*} $$

and thus $\|\tilde A^{-1}\tilde PT\tilde A - R\|\leq D\varepsilon + \delta $ .

Proof. Let Y be a subspace of X that is isomorphic to X and complemented in X, with associated projection and isomorphism $P,A:X\to Y$ , so that $\|A^{-1}P\|\|A\|\leq C$ and so that there exists a scalar $\lambda $ with $\|(A^{-1}P)TA - \lambda I\| \leq \varepsilon $ . If $|\lambda | \geq 1/2$ , then

$$\begin{align*}\big\|\underbrace{\lambda^{-1}A^{-1}PTA}_{=:B} - I\big\| \leq 2\varepsilon <1\end{align*}$$

and thus $B^{-1}$ exists with $\|B^{-1}\| \leq 1/(1-2\varepsilon )$ . We obtain that if $S = B^{-1}\lambda ^{-1}A^{-1}P$ , then $STA = I$ and $\|S\|\|A\| \leq 2C/(1-2\varepsilon )$ . If, on the other hand $|\lambda | <1/2$ , then, because $\|A^{-1}P(I-T)A - (1-\lambda )I\|\leq \varepsilon $ , we achieve the same conclusion for $I-T$ instead of T.

If $X = Y\oplus Z$ and $Q:X\to Y$ is a projection, then we deduce that either Y or Z contains a complemented subspace isomorphic to X. To see that we can assume that for some scalar $\lambda $ , with $|\lambda |\ge 1/2$ , Q is a C-projectional factor with error $\varepsilon \in (0,1/2)$ of $\lambda I$ . Otherwise, we replace Q by $I-Q$ . From what we have proved so far, we deduce that there are operators $S,A :X\to X$ so that $SQA = I$ . Then $W = QA(X)$ is a subspace of Y that is isomorphic to X. It is also complemented via the projection $R = (S|_W)^{-1}S:X\to W$ . So we obtain that Y is a complemented subspace of X and X is isomorphic to complemented subspace of Y. Since in addition X satisfies the accordion property, it follows from Pełczyński’s famous classical argument from [Reference Pełczyński30] that $X\simeq Y$ . Similarly, if $(I-Q)$ is a factor of the identity, we deduce $X\simeq Z$ .

Proof. Note that the sequence $(B_k)_{k=1}^\infty $ is a partition of $[0,1)$ , and for $k\in \mathbb {N}$ , $B_k$ is the set in $[0,1]$ of measure $2^{-k}$ on which $\theta _k h_{I_k}$ takes the value $-1$ . Let $f = a_0 h_\emptyset + \sum _{k=1}^n \theta _ka_k|I_k|^{-1}h_{I_k}$ . For $k\in \mathbb {N}$ , put $b_k = a_k$ if $k\leq n$ and $b_k = 0$ otherwise. For each $k\in \mathbb {N}$ , the function f is constant on $B_k$ and in fact for $s\in B_k$ , we have

$$ \begin{align*} f(s) =b_0 + \sum_{j=1}^{k-1}|I_j|^{-1}b_j - |I_k|^{-1}b_k = b_0 + \sum_{j=1}^{k-1}2^{j-1}b_j - 2^{k-1}b_k =: c_k. \end{align*} $$

Therefore, for any $m=1,2,\ldots n$ ,

(23) $$ \begin{align} \left\|f \chi_{\bigcup_{j=m}^n B_j} \right\|_{L_1} = \sum_{k=m}^\infty |X_k|, \end{align} $$

where for each $k\in \mathbb {N}$ ,

$$\begin{align*}X_k = \frac{c_k}{2^{k}} = \frac{1}{2^{k}}b_{0} + \sum_{j=1}^{k-1}\frac{2^{j-1}}{2^{k}}b_j - \frac{1}{2}b_k.\end{align*}$$

Putting $X_0 = 0$ , a calculation yields that for all $k\in \mathbb {N}$ ,

(24) $$ \begin{align} X_k = \frac{1}{2}X_{k-1} + \frac{1}{2}\left(b_{k-1} - b_k\right). \end{align} $$

Applying the triangle inequality to equations (23) and (24), we conclude

$$ \begin{align*} \| f\|_{L_1}&=\sum_{k=m}^\infty |X_k|=\sum_{k=1}^\infty 2|X_k|-|X_{k-1}|\\ &\le \sum_{k=1}^\infty |2X_k -X_{k-1}|=\sum_{k=1}^\infty|b_k-b_{k-1}|, \end{align*} $$

which yields the upper bound of equation (21). The lower bound is proved with a similar computation. To obtain equation (22), we deduce from equation (24)

$$ \begin{align*} &\sum_{k=m+1}^\infty |X_k| \ge\frac12 \sum_{k=m+1}^\infty |b_k-b_{k-1}|-\frac12\sum_{k=m} |X_k| \end{align*} $$

and therefore

$$ \begin{align*} & \frac32 \sum_{k=m+1}^\infty |X_k| +\frac12 |X_{m}| \ge \frac12 \sum_{k=m+1}^\infty |b_k-b_{k-1}|, \end{align*} $$

which yields

$$ \begin{align*} & \| f \chi_{\bigcup_{j=m}^n B_j}\|_{L_1} = \sum_{k=m}^\infty |X_k| \ge \sum_{k=m+1}^\infty |X_k| +\frac13|X_m| \ge \frac13 \sum_{k=m+1}^\infty |b_k-b_{k-1}| \end{align*} $$

and proves equation (22).

Proof. By equation (19), D is always well defined on the linear span of the set $\mathcal {X} = \{|I|^{-1}\chi _I:I\in \mathcal {D}\}$ . In fact, the closed convex symmetric hull of $\mathcal {X}$ is the unit ball of $L_1$ . We deduce that $\|D\| = \sup \{\|Df\|:f\in \mathcal {X}\}$ , under the convention that $\|D\| = \infty $ if and only if D is unbounded. Fix $f = |I|^{-1}\chi _I\in \mathcal {X}$ . Use equation (19) to write

$$\begin{align*}f = |I_{k_0}|^{-1}\chi_{I_{k_0}} = \sum_{k=0}^{k_0-1}\theta_k|I_k|^{-1}h_{I_k},\ that is,\ Df = \sum_{k=0}^{k_0-1}a_k\theta_k|I_k|^{-1}h_{I_k}.\end{align*}$$

Extend $(I_k)_{k=0}^{k_0}$ to a branch $(I_k)_{k=0}^\infty $ . By equation (21), we have

(27) $$ \begin{align} \frac{1}{3}\big(\sum_{k=1}^{k_0-1}|a_{I_k} - a_{I_{k-1}}| + |a_{I_{k_0-1}}|\big) \leq \|Df\|_{L_1} \leq \sum_{k=1}^{k_0-1}|a_{I_k} - a_{I_{k-1}}| + |a_{I_{k_0-1}}|. \end{align} $$

By the triangle inequality, $\|Df\|_{L_1} \leq \sum _{k=1}^{\infty }|a_{I_k} - a_{I_{k-1}}| + \lim _k|a_{I_{k}}| \leq {\left \vert \kern -0.25ex\left \vert \kern -0.25ex\left \vert D \right \vert \kern -0.25ex\right \vert \kern -0.25ex\right \vert }$ . The lower bound is achieved by taking in equation (27) all $f\in \mathcal {X}$ .

Notation. For every disjoint collection $\Delta $ of $\mathcal {D}^+$ and $\theta \in \{-1,1\}^\Delta $ , we denote $h_\Delta ^\theta = \sum _{J\in \Delta }\theta _Jh_J$ . If $\theta _J = 1$ for all $J\in \Delta $ , we write $h_\Delta = \sum _{j\in \Delta }h_J$ . For a finite disjoint collection $\Delta $ of $\mathcal {D}$ , we denote $\Delta ^* = \cup \{I:I\in \Delta \}$ .

Proof of Theorem 2.9

Let us sketch the proof of the first statement. Let $(a_I)_{I\in \mathcal {D}^+}$ be the entries of D. For every $I\in \mathcal {D}$ , denote by $Q_I$ the Haar multiplier that has entries 1 for all $J\subset I$ and zero for all others. Then ${\left \vert \kern -0.25ex\left \vert \kern -0.25ex\left \vert Q_I \right \vert \kern -0.25ex\right \vert \kern -0.25ex\right \vert } = 1$ . First, note that for every $\varepsilon>0$ , there exists $I_0\in \mathcal {D}^+$ so that ${\left \vert \kern -0.25ex\left \vert \kern -0.25ex\left \vert DQ_{I_0} - a_{I_0}Q_{I_0} \right \vert \kern -0.25ex\right \vert \kern -0.25ex\right \vert }\leq \varepsilon $ . Otherwise, we could easily deduce ${\left \vert \kern -0.25ex\left \vert \kern -0.25ex\left \vert D \right \vert \kern -0.25ex\right \vert \kern -0.25ex\right \vert } = \infty $ . Construct a dilated and renormalised faithful Haar system $(\tilde h_I)_{I\in \mathcal {D}^+}$ with closed linear span Z in the range of $Q_{I_0}$ , and let $P:L_1\to Z$ be the corresponding norm-one projection and $A:L_1\to Z$ be an onto isometry. Then $\|A^{-1}PDA - a_{I_0}I\| \leq \varepsilon $ .

For the second part, we will use that the Rademacher sequence $(r_n)_n$ (i.e., $r_n = \sum _{L\in \mathcal {D}_n}h_L$ , for $n\in \mathbb {N}$ )) is weakly null in $L_1$ and $w^*$ -null in $(L_1)^* \equiv L_\infty $ . Using this fact, we inductively construct a faithful Haar system $(\tilde h_I)_{I\in \mathcal {D}^+}$ so that for each $I\neq J$ , we have

$$\begin{align*}\big|\big\langle\tilde h_I,T\big(|J|^{-1}\tilde h_J\big)\big\rangle\big|\leq \varepsilon_{(I,J)},\end{align*}$$

where $(\varepsilon _{(I,J)})_{(I,J)\in \mathcal {D}^+}$ is a prechosen collection of positive real numbers with $\sum \varepsilon _{(I,J)} \leq \varepsilon $ . This is done as follows. If we have chosen $\tilde h_I$ for $\iota (I) = 1,\ldots ,k-1$ . Let $I\in \mathcal {D}^+$ with $\iota (I) = k$ , and let $I_0$ be the predecessor of I: that is, either $I = I_0^+$ or $I = I_0^-$ . Let us assume $I = I_0^+$ . We then choose the next function $\tilde h_I$ among the terms of a Rademacher sequence with support $[\tilde h_\emptyset \tilde h_{I_0} = 1]$ . Denote by Z the closed linear span of $(\tilde h_I)_{I\in \mathcal {D}^+}$ , and take the canonical projection $P:L_1\to Z$ as well as the onto isometry $A:L_1\to Z$ given by $A h_I = \tilde h_I$ . Consider the operator $S=A^{-1}PTA:L_1\to L_1$ , and note that for all $I\neq J$ , we have $|\langle h_I, S(|J|^{-1}h_J) \rangle | = |\langle \tilde h_I, T(|J|^{-1}\tilde h_J )\rangle | \leq \varepsilon _{(I,J)}$ . It follows that the entries $a_I = \langle h_I, S(|I|^{-1}h_I )\rangle $ define a bounded Haar multiplier D and $\|S - D\| \leq \varepsilon $ : that is, T is a 1-projectional factor with error $\varepsilon $ of D.

Proof. By the first condition in Definition 2.12, we have

$$ \begin{align*}\big\|\sum_{I\in \mathcal D_n} \!a_I\chi_{\pi(I)} \big\|\!=\!\big\|\sum_{I\in \mathcal D_n} \!a_I\chi_{I} \big\| \end{align*} $$

for all $n\in \mathbb {N}$ , all permutations $\pi $ on $\mathcal D_n$ and all scalar families $(a_I:I\in \mathcal D_n)$ .

To show the first inequality in (a), let $n\in \mathbb {N}$ , $f=\sum _{I\in \mathcal D_n} a_I \chi _I\in Z$ , and let $\pi :\mathcal D_n\to \mathcal D_n $ , be cyclic (i.e., $\{\pi ^r(I):r=1,2\ldots ,2^n\}=\mathcal D_n$ for $I\in \mathcal D_n$ ). Then

$$ \begin{align*}\|f\|=\big\| |f|\big\| \ge \frac1{2^n} \Big\|\sum_{r=1}^{2^n} \sum_{I\in \mathcal D_n} |a_I| \chi_{\pi^r(I)}\Big\|=\frac1{2^n}\Big|\sum_{I\in \mathcal D_n} |a_I|\Big|=\|f\|_{L_1}.\end{align*} $$

The second inequality in (a) follows from the observation that for each $n\in \mathbb {N}$ , the family $(\chi _I: I \in \mathcal D_n)$ is $1$ -unconditional.

We identify each $g\in Z$ with the bounded functional $x^*_g$ , defined by $x^*_g(f)=\int _0^1 f g$ , and we denote the dual norm by $\|\cdot \|_{*}$ . From this representation it is clear that $\|\cdot \|_*$ also satisfies the first condition in Definition 2.12. Since $\|1_{[0,1)]}\|=1$ , and since for all $f\in Z$ , $ \int f \le \|f\|_1\le \|f\|$ , we deduce that the second condition in Definition 2.12 holds true for the norm $\|\cdot \|_*$ .

Let $(h_n)$ be the Haar basis linearly ordered in the usual way, meaning that if $m<n$ , then either $\mathrm {supp}(h_n)\subset \mathrm {supp}(h_m)$ or $\mathrm {supp}(h_n)\cap \mathrm {supp}(h_m)=\emptyset $ . The claim of condition (c) follows from the fact that if $f=\sum _{j=1}^n a_j h_j\in Z$ , then for any scalar $a_{n+1}$ , the absolute values of the functions $f+a_{n+1} h_{n+1}$ and $f-a_{n+1} h_{n+1}$ have the same distribution and their average is f.

Let $n\!\in \!\mathbb {N}$ and $I\!\in \!\mathcal D_n$ using for $k\!>\!n$ cyclic permutations on $\{ J\!\in \!\mathcal D_k, J\!\subset \!I\}$ . We deduce that $\sup _{f\in Z,\|f\|\le 1} \int _I f$ is attained for $f=\chi _I/\|\chi _I\|$ and thus $\|\chi _I\|\cdot \|\chi _I\|_*=2^{-n}$ . Since, second, for each n, $(\chi _I: I \in \mathcal D_n)$ is an orthogonal family, we deduce (d).

Since faithful Haar systems have the same joint distribution, we deduce the first part of (e). Since by (b), this is also true with respect to the dual norm, we deduce the second part of (e).

Proof. It is immediate that $P\otimes Q$ is a norm-one projection onto $Z(W)$ and that $A\otimes B$ is a norm-one map with dense image. It also follows that $A\otimes B$ is 1-1 on $L_1\otimes X$ . One way to see this is to identify $L_1\otimes X$ and $Z\otimes W$ with spaces of bilinear forms on $(L_1)^*\times X^*$ and $Z^*\times W^*$ , respectively. To conclude that $A\otimes B$ is an isometry and that $Z(W)=Z\otimes _{\pi } W$ , take u in $L_1\otimes X$ . Note that $v:=(A\otimes B)(u)$ is in $Z\otimes W\subset L_1\otimes X$ , and write $v = \sum _{i=1}^nf_i\otimes x_i$ , where $f_1,\ldots ,f_n\in L_1$ and $x_1,\ldots ,x_n\in X$ . We will see that $\sum _{i=1}^n\|f_i\|\|x_i\| \geq \|u\|$ , which will imply the conclusion by the definition of $\|v\|$ . Indeed, $v = (P\otimes Q)(v) = \sum _{i=1}^n(Pf_i)\otimes (Qx_i)$ and

$$ \begin{align*} \|v\| &\geq \sum_{i=1}^n\|Pf_i\|\|Qx_i\| = \sum_{i=1}^n\|A^{-1}Pf_i\|\|B^{-1}Qx_i\|\\ & \geq \big\|\underbrace{\sum_{i=1}^n\big(A^{-1}Pf_i\big)\otimes\big(B^{-1}Qx_i\big)}_{=:y}\big\|. \end{align*} $$

It is immediate that $(A\otimes B)(y) = v$ and thus $y = u$ .

Proof. By approximating $\mathscr {A}$ in measure by closed sets from the inside, we can assume that $\mathscr {A}$ is closed. For $k\in \mathbb {N}$ , let $A_k = \cup \{I:I\in \mathcal {A}\cap \mathcal {D}_k\}$ and

$$\begin{align*}\mathscr{A}_k =\{ (I_n)_{n=0}^\infty \in [\mathcal D^+]: I_k\in\mathcal D_k, I_k\subset A_k\},\end{align*}$$

that is, $\mathscr {A}_k$ is the set of all sequences $(I_n)_{n=0}^\infty $ in $[\mathcal D^+]$ such that the k’th entry $I_k$ is a subset of $A_k$ . Then it follows that $\mathscr {A}=\bigcap _k \mathscr {A}_k$ , and letting $A = \cap _k A_k$ , we deduce that

$$ \begin{align*}|A|= \lim_{k\to\infty}|A_k|= \lim_{k\to\infty} |\mathscr{A}_k |= |\mathscr{A}|.\end{align*} $$

But also, for any $J\notin \mathcal {A}$ , we have $J\cap A=\emptyset $ . It follows that for any $f\in L_1$ with $f|_{A^c} = 0$ and $J\notin \mathcal {A}$ , we have $\langle h_J, f\rangle = 0$ and thus $f\in Y_{\mathscr {A}}$ . In particular, the restriction operator $R_A:L_1\to L_1$ is a 1-projection onto a subspace that is isometrically isomorphic to $L_1$ .

Notation. Let X be a Haar system space. For $L\in \mathcal {D}^+$ , we denote

1. (i) $q^L = q^{(\nu _Lh_L)}:L_1(X)\to L_1$ ,

2. (ii) $j^L = j^{(\mu _Lh_L)}:L_1\to L_1(X)$ , and

3. (iii) $P^L = j^Lq^L:L_1(X)\to L_1(X)$ .

Note that for any $k\in \mathbb {N}$ , $\|\sum _{\{L:\iota (L)\leq k\}}P^L\| = 1$ . This is because this operator coincides with $I\otimes P^{[\iota \leq k]}$ , where $P^{[\iota \leq k]}:X\to X$ is the basis projection onto $(\mu _Lh_L)_{\iota (L)\leq k}$ (this is easy to verify on vectors of the form $u =h_I\otimes h_L$ whose linear span is dense in $L_1(X)$ ). We may therefore state the following.

Notation. Let X be a Haar system space, and let $T:L_1(X)\to L_1(X)$ be a bounded linear operator. For $L,M\in \mathcal {D}^+$ , we denote $T^{(L,M)} = T^{(\nu _Lh_L,\mu _Mh_M)}$ (recall from Proposition 2.13 that the scalars $\mu _M$ and $\nu _L$ are positive and chosen so that $\mu _Mh_M$ is normalised in $X^*$ and $\nu _L h_L$ is normalised in X), so that for every $u\in L_1(X)$ , we have

(29) $$ \begin{align} Tu &= \sum_{L\in\mathcal{D}^+}P^LT\Big(\sum_{M\in\mathcal{D}^+}P^Mu\Big) = \sum_{L\in\mathcal{D}^+}\sum_{M\in\mathcal{D}^+}j^LT^{(L,M)}q^Mu\nonumber\\ &= \sum_{L\in\mathcal{D}^+}\sum_{M\in\mathcal{D}^+}\Big(T^{(L,M)}(q^Mu)\Big)\otimes(\mu_Lh_L). \end{align} $$

For $L\in \mathcal {D}^+$ , we denote $T^L = T^{(L,L)}$ .

Proof. The ‘in particular’ part follows from the separability of $L_1$ and the fact that the set in question if bounded by $\|T\|\sup _{(x^*,x)\in A\times B}\|x^*\|\|x\|$ .

Fix a sequence $(x_n^*,x_n)\in A\times B$ . Assume that $(T^{(x_n^*,x_n)}f)_n$ is not uniformly integrable. Then after passing to a subsequence, there exist $\delta>0$ and a sequence of disjoint measurable subsets $(A_n)_n$ of $[0,1)$ so that for all $n\in \mathbb {N}$ , we have

$$\begin{align*}\delta\leq \Big|\int_{A_n}\big(T^{(x_n^*,x_n)}f\big)(s)ds\Big| = \big|\langle \chi_{A_n},T^{(x_n^*,x_n)}f\rangle\big| = \big|\langle \chi_{A_n}\otimes x_n^*, T(f\otimes x_n)\rangle\big|.\end{align*}$$

For every $n\in \mathbb {N}$ , define the scalar sequence $\xi _n = (\xi _n(m))_m$ given by $\xi _n(m) = \langle \chi _{A_m}\otimes x_m^*,T(f\otimes x_n)\rangle $ . Then for every $m_0\in \mathbb {N}$ , we have that for appropriate scalars $(\zeta _m)_{m=1}^N$ of modulus one,

(30) $$ \begin{align} \sum_{m=1}^{m_0}|\xi_n(m)| &= \Big|\big\langle\sum_{m=1}^{m_0}\chi_{A_m}\otimes \zeta_mx_m^*,T(f\otimes x_n)\big\rangle\Big|\\\nonumber & \leq \underbrace{\Big\|\sum_{m=1}^{m_0}\chi_{A_{m}}\otimes (\zeta_m x^*_{m})\Big\|}_{= \max_{1\leq m\leq m_0}\|x^*_{m}\|}\|T\|\|f\| \|x_n\|\nonumber\\ &\leq \|T\|\|f\|\sup_{(x^*,x)\in A\times B}\|x^*\|\|x\|.\nonumber \end{align} $$

By Rosenthal’s Lemma 3.1, there exists an infinite subset $N = \{n_j:j\in \mathbb {N}\}$ of $\mathbb {N}$ so that for all $i_0\in \mathbb {N}$ , we have $\sum _{j\neq i_0}|\xi _{n_{i_0}}(n_j)| \leq \delta /2$ . After relabelling, for all $n_0\in \mathbb {N}$ , we have

$$\begin{align*}\sum_{m\neq n_0}|\xi_{n_0}(m)|\leq \delta/2.\end{align*}$$

We now show that $(x_n)_n$ is equivalent to the unit vector basis of $\ell _1$ . Fix scalars $a_1,\ldots ,a_N$ . For appropriate scalars $\theta _1,\ldots ,\theta _N$ of modulus 1, we have

(31) $$ \begin{align} \sum_{n=1}^Na_n\theta_n\langle \chi_{A_{n}}\otimes x^*_{n},T(f\otimes x_{n})\rangle \geq \delta\sum_{n=1}^N|a_n|. \end{align} $$

Put

$$ \begin{align*} \Lambda & = \Big|\big\langle\sum_{m=1}^N\chi_{A_{m}}\otimes (\theta_m x^*_{m}),T\big(\sum_{n=1}^Nf\otimes a_nx_{n}\big)\big\rangle\Big|\\ &\leq \underbrace{\Big\|\sum_{m=1}^N\chi_{A_{m}}\otimes (\theta_m x^*_{m})\Big\|}_{= \max_{1\leq m\leq N}\|x^*_{m}\|}\|T\|\|f\| \Big\|\sum_{n=1}^Na_nx_n\Big\|\\ &\leq \|T\|\|f\|\sup_{x^*\in A}\|x^*\|\Big\|\sum_{n=1}^Na_nx_n\Big\|. \end{align*} $$

Also,

$$ \begin{align*} \Lambda & = \Big|\sum_{n=1}^Na_n\theta_n\langle \chi_{A_{n}}\otimes x^*_{n},T(f\otimes x_{n})\rangle + \sum_{n=1}^Na_n\sum_{m\neq n}\theta_m\langle \chi_{A_{m}}\otimes x^*_{m},T(f\otimes x_{n})\rangle\Big|\\ &\geq \delta\sum_{n=1}^N|a_n| - \sum_{n=1}^N|a_n|\sum_{m\neq n}|\xi_n(m)| \geq \delta/2\sum_{n=1}^N|a_n|. \end{align*} $$

Thus, $\|\sum _{n=1}^Na_nx_n\| \geq c\sum _{n=1}^N|a_n|$ , where $c = \delta /(2\|T\|\|f\|\sup _{x^*\in A}\|x^*\|)$ .

Notation. For $n\in \mathbb {N}$ , we denote by $P_{(\leq n)}:L_1\to L_1$ the norm-one canonical basis projection onto $\langle \{h_I:I\in \mathcal {D}^n\}\rangle $ . We also denote $P_{(>n)} = I - P_{(\leq n)}$ .

Proof. Let $(B_k)$ be the partition of $[0,1)$ defined by $B_k=I_k\setminus I_{k+1}$ , $k\in \mathbb {N}$ . We conclude from the inequality in equation (22) in Proposition 2.5 that:

1. (i) for every $k \leq \ell _n \leq \ell \in \mathbb {N}$ and $s\in B_k$ , we have $f_n^\ell (s) = f^{\ell _n}_n(s)$ and

2. (ii) for every $m\leq \ell _n\in \mathbb {N}$ , we have

   $$\begin{align*}\int_{\cup_{k=m}^{\ell_n} B_k}|f_n^{\ell_n}(s)|ds \geq \frac{1}{3}\sum_{k=m+1}^{\ell_n} |a_k - a_{k-1}|.\end{align*}$$

Put $A_n = \cup _{i=k_n}^{\ell _n}B_i$ . The conclusion follows directly from (i) and (ii).

Proof of Theorem 3.4

Put $\mathscr {T} = \{T^{(x^*,x)}:(x^*,x)\in A\times B\}$ . Take a family $(D_T)_{T}$ as in Definition 3.3. For each $T\in \mathscr {T}$ , we have

(32) $$ \begin{align} \|(T - D_T)P_{(>k)}\| &\leq \sup_{S\in\mathscr{T}}\Big(\big\|(S - D_S)P_{(>k)}\big\|\Big) =\varepsilon_k. \end{align} $$

(33) $$ \begin{align} \|P_{(>k)}TP_{(\leq k)}\| &\leq \underbrace{\big\|P_{(>k)}D_TP_{(\leq k)}\big\|}_{=0} + \big\|P_{(>k)}(T - D_T)P_{(\leq k)}\big\|\nonumber\\ &\leq \big\|P_{(>k)}(T - D_T)\|\nonumber\\ &\leq \sup_{S\in\mathscr{T}}\Big(\big\|P_{(>k)}(S- D_S)\big\|\Big) \!=\!\delta_k. \end{align} $$

Both $(\varepsilon _k)_k$ and $(\delta _k)_k$ tend to zero. For each $T\in \mathscr {T}$ , denote by $(a^T_I)_{I\in \mathcal {D}^+}$ the entries of $D_T$ .

Claim: Fix $\sigma = (I_k)_{k=1}^\infty \in [\mathcal {D}^+]$ and $r>0$ . Then there exists $k_0\in \mathbb {N}$ so that for all $T\in \mathscr {T}$ , we have $\sum _{k=k_0}^\infty |a^T_{I_k} - a^T_{I_{k-1}}| \leq r$ .

We will assume that the claim is true and proceed with the rest of the proof. For every $N,k_0,\in \mathbb {N}$ , let

$$\begin{align*}\mathcal{A}_{N,k_0} = \Big\{\sigma = (I_k)_{k=1}^\infty\in[\mathcal{D}^+]: \sup_{T\in\mathscr{T}}\sum_{k=k_0}^\infty|a^T_{I_k} - a^T_{I_{k-1}}| \leq 2^{-N}\Big\},\end{align*}$$

which is a closed subset of $[\mathcal {D}^+]$ , and by the claim, we have $\cup _{k_0}\mathscr {A}_{N,k_0} = [\mathcal {D}^+]$ for all $N\!\in \!\mathbb {N}$ . We may therefore pick a strictly increasing sequence of natural numbers $(k_N)$ so that for each N, we have $|\mathscr {A}_{N,k_N}| \geq 1-\eta /2^N$ . We put $\mathscr {A} = \cap _N\mathscr {A}_{N,k_N}$ , and we demonstrate that this is the desired set.

To show that $\{TP_{\mathscr {A}}:T\in \mathscr {T}\}$ is relatively compact with respect to the operator norm, we fix $\varepsilon>0$ and $(T_n)_n$ in $\mathscr {T}$ . For each $n\in \mathbb {N}$ , denote $D_n = D_{T_n}$ . We will find $M\in [\mathbb {N}]^{\infty }$ so that for all $n,m\in M$ , we have $\|T_nP_{\mathscr {A}} - T_mP_{\mathscr {A}}\| \leq 11\varepsilon $ . Fix $N\in \mathbb {N}$ so that $2^{-N}\leq \varepsilon $ , $\varepsilon _{k_N}\leq \varepsilon $ , and $\delta _{k_N}\leq \varepsilon $ . For each $n\in \mathbb {N}$ , write

$$ \begin{align*} T_n = D_nP_{(>k_N)} + \underbrace{(T_n-D_n)P_{(>k_N)}}_{=:A_n}+\underbrace{P_{(>k_N)}T_nP_{(\leq k_N)}}_{=:B_n} + \underbrace{P_{(\leq k_N)}T_nP_{(\leq k_N)}}_{=:C_n}. \end{align*} $$

Then we have $\|A_n\| \leq \varepsilon _{k_N}\leq \varepsilon $ and $\|B_n\| \leq \delta _{k_N}\leq \varepsilon $ . By passing to a subsequence of $(T_n)$ , we may assume that for all $n,m\in \mathbb {N}$ , we have (letting $a_I^n=a^{T_n}_I$ )

(34) $$ \begin{align} \sum_{\left\{\substack{I\in\mathcal{D}^+\\|I|\geq 1/2^{k_N+1}}\right\}}|a_I^n - a_I^m| \leq \varepsilon. \end{align} $$

Since the $C_n$ are bounded elements of a finite-dimensional space, we can also assume that $\|C_n - C_m\| \leq \varepsilon $ , for $m,n\in \mathbb {N}$ . Therefore, for $n,m\in \mathbb {N}$ , we have

$$ \begin{align*} \|T_nP_{\mathscr{A}} - T_mP_{\mathscr{A}}\| \leq \underbrace{\| D_nP_{(>k_N)}P_{\mathscr{A}} - D_mP_{(>k_N)}P_{\mathscr{A}} \|}_{=:\Lambda} + 5\varepsilon. \end{align*} $$

Luckily, the remaining quantity $\Lambda $ is the norm of a Haar multiplier on $L_1$ , and we know how to compute this. If for $\sigma = (I_k)_{k=0}^\infty \in \mathscr {A}$ , we put

$$\begin{align*}\Lambda_\sigma &= \sum_{k=k_N+1}^\infty|(a^n_{I_k}-a^m_{I_k}) - (a^n_{I_{k-1}}-a^m_{I_{k-1}})| + |a^n_{I_{k_N}} - a^m_{I_{k_N}}| + \lim_k|a^n_{I_k}-a^m_{I_k}|\\&\leq 2\underbrace{\sum_{k=k_N+1}^\infty|(a^n_{I_k}-a^m_{I_k}) - (a^n_{I_{k-1}}-a^m_{I_{k-1}})|}_{\leq 2/2^N\leq 2\varepsilon} + 2\underbrace{|a^n_{I_{k_N}} - a^m_{I_{k_N}}|}_{\leq \varepsilon} \leq6\varepsilon. \end{align*}$$

Then by Remark 2.8, $\Lambda = \sup _{\sigma \in \mathscr {A}}\Lambda _\sigma $ and thus $\|T_nP_{\mathscr {A}} - T_mP_{\mathscr {A}}\|\leq 11\varepsilon $ .

We now provide the required proof of the claim. We fix $\sigma = (I_k)_{k=0}^\infty $ , with associated signs $(\theta _k)_{k=0}^\infty $ . Let us assume that the conclusion fails. Then we may find $(T_n)_n = (T^{(x^*_n,x_n)})_n$ in $\mathscr {T}$ , each $T_n$ is associated with a $D_n$ (each $D_n$ has entries $(a_I^n)_{I\in \mathcal {D}^+}$ ) and $k_1 < \ell _1<k_2<\ell _2<\cdots $ so that for all $n\in \mathbb {N}$

$$ \begin{align*} \sum_{k=k_n+1}^{\ell_n}|a^{n}_{I_k} - a^{n}_{I_{k-1}}|\geq r. \end{align*} $$

Pick $k_0\in \mathbb {N}$ so that $\varepsilon _{k_0}\leq r/12$ . For $k,n\in \mathbb {N}$ define $b_k^n = 0$ if $k\leq k_0$ and $b_k^n = a_{I_k}^{n}$ if $k>k_0$ . If we additionally assume that $k_1>k_0$ , then for all $n\in \mathbb {N}$ , we have

(35) $$ \begin{align} \sum_{k=k_n+1}^{\ell_n}|b_k^n - b^n_{k-1}|\geq r. \end{align} $$

For each $n,\ell \in \mathbb {N}$ , put

$$\begin{align*}f_n^\ell = \sum_{k=0}^\ell b_k^n\theta_k|I_k|^{-1}h_{I_k} = D_nP_{(>k_0)}(\underbrace{|I_{\ell+1}|^{-1}\chi_{I_{\ell+1}}}_{=:\psi_\ell}).\end{align*}$$

By Lemma 3.6, we may find a sequence of $(A_n)_n$ of disjoint measurable sets so that for each $n\in \mathbb {N}$ , the sequence $(f_n^\ell (s))_{\ell \geq \ell _n}$ is constant for all $s\in A_n$ and $\|f_n^{\ell _n}|_{A_n}\|_{L_1} \geq r/3$ . For each $n\in \mathbb {N}$ fix $g_n$ in the unit sphere of $L_\infty $ with support in $A_n$ so that for all $\ell \geq \ell _n$

$$ \begin{align*} r/3&\leq \|f_n^{\ell_n}|_{A_n}\|_{L_1} = \big|\langle g_n ,f_n^{\ell}\rangle\big| = \big|\langle g_n, D_nP_{(>k_0)}(\psi_\ell)\rangle\big|\\ \nonumber &\leq \big|\langle g_n, T_nP_{(>k_0)}(\psi_\ell)\rangle\big| + r/12 = \big|\langle g_n\otimes x_n^*, T\big(\underbrace{(P_{(>k_0)}\psi_\ell)}_{\phi_\ell}\otimes x_n\big)\rangle\big| + r/12. \end{align*} $$

Note that for all $\ell \in \mathbb {N}$ , $\|\phi _\ell \|_{L_1}\leq 2$ . Then for all $n\in \mathbb {N}$ and $\ell \geq \ell _n$ ,

$$ \begin{align*} \big|\langle g_n\otimes x_n^*, T\big(\phi_\ell\otimes x_n\big)\rangle\big| \geq r/4. \end{align*} $$

Pick an $L\in [\mathbb {N}]^\infty $ so that for each $m,n\in \mathbb {N}$ , the limit

$$\begin{align*}\xi_n(m) := \lim_{\ell\in L}\langle g_m\otimes x_m^*, T(\psi_\ell\otimes x_n) \rangle \text{ exists}. \end{align*}$$

Because the sequence $(g_m)_m$ is disjointly supported, an identical calculation as in equation (30) yields that for all $n\in \mathbb {N}$ , we have

$$ \begin{align*}\sum_m|\xi_n(m)| \leq 2\|T\|\sup_{(x^*,x)\in C}\|x^*\|\|x\|.\end{align*} $$

Thus, by Rosenthal’s Lemma 3.1, we may pass to a subsequence and relabel so that for all $n_0\in \mathbb {N}$ , we have $\sum _{m\neq n_0}|\xi _{n_0}(m)| \leq r/8$ .

We will show that $(x_n)_n$ must be equivalent to the unit vector basis of $\ell _1$ , which would contradict our assumption and thus finish the proof. Fix scalars $a_1,\ldots ,a_N$ , and for $\ell \in L$ with $\ell \geq \ell _N$ pick appropriate scalars $\zeta ^\ell _1,\ldots ,\zeta ^\ell _N$ of modulus one so that we have

$$\begin{align*}\frac{r}{4}\sum_{n=1}^N|a_n| \leq \sum_{n=1}^N\langle g_n\otimes\big(\zeta^\ell_nx_n^*\big), T\big(\phi_\ell\otimes(a_nx_n)\big)\rangle \end{align*}$$

and put

$$ \begin{align*} \Lambda_\ell &= \big|\big\langle\sum_{m=1}^N g_n\otimes\big(\zeta^\ell_mx_m^*\big),\sum_{n=1}^NT\big(\phi_\ell\otimes(\sum_{n=1}^Na_nx_n)\big)\big\rangle\big|\\ &\leq \Big(2\|T\|\sup_{x^*\in A}\|x^*\|\Big)\big\|\sum_{n=1}^N a_nx_n\big\|. \end{align*} $$

But also,

$$ \begin{align*} \lim_{\ell\in L}\Lambda_\ell &= \lim_{\ell\in L}\Big| \sum_{n=1}^N\langle g_n\otimes\big(\zeta^\ell_nx_n^*\big), T\big(\phi_\ell\otimes(a_nx_n)\big)\rangle\\ &\quad + \sum_{n=1}^Na_n\sum_{m\neq n}\zeta_m^\ell \langle g_m\otimes x_m^*, T\big(\psi_\ell\otimes x_n\big) \rangle\Big|\\ &\geq \frac{r}{4}\sum_{n=1}^N|a_n| - \sum_{n=1}^N|a_n|\sum_{m\neq n}|\xi_n(m)| \geq \frac{r}{8}\sum_{n=1}^N|a_n|. \end{align*} $$

Therefore, $\|\sum _{n=1}^Na_nx_n\| \geq r/(16\|T\|\sup _{x^*\in A}\|x^*\|)\sum _{n=1}^N|a_n|$ .