Proof For $n \ge 1$ , let $\lambda (n) = \|\sum _{i=1}^n e_i\|$ and $\lambda ^*(n) =\|\sum _{i=1}^n e^*_i\|_*$ . Note that if $\alpha = \sum _{i=1}^n e_i$ and $k \ge 1$ , then $\alpha ^k = \sum _{i=1}^{n^k}e_i$ . So (2.2) applied to $ \alpha $ yields

(3.1) $$ \begin{align} [\lambda(n)]^k \le K^k \lambda(n^k). \end{align} $$

Similarly (2.2) applied to $\alpha ^* = \sum _{i=1}^n e_i^*$ gives

$$ \begin{align*} [\lambda^*(n)]^k \le K^k \lambda^*(n^k). \end{align*} $$

Since $(e_i)$ is bidemocratic, there exists $C>0$ such that

$$ \begin{align*} \frac{n}{\lambda(n)} \le \lambda^*(n) \le C \frac{n}{\lambda(n)}. \end{align*} $$

Hence,

$$ \begin{align*}\left[\frac{n}{\lambda(n)}\right]^k \le CK^k \frac{n^k}{\lambda(n^k)},\end{align*} $$

i.e.,

(3.2) $$ \begin{align} \lambda(n^k) \le CK^k[\lambda(n)]^k. \end{align} $$

By the proof (see page 60) of [Reference Lindenstrauss and Tzafriri13, Theorem 2.a.9], (3.1) and (3.2) imply that $\lambda (n) \asymp n^{1/p}$ for some $1 \le p \le \infty $ . Since $(e_i)$ is bidemocratic, it follows that $\Phi (n) \asymp n^{1/p}$ and $\Phi ^*(n) \asymp n^{1/q}$ , where $q=p/(p-1)$ . By Remark 2.1, if $p=1$ or $p=\infty $ , then $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ or $c_0$ , respectively.

So suppose that $1<p<\infty $ . Consider $\alpha = \sum _{i=1}^m a_i e_i \in \operatorname {span}((e_i))$ . By (2.2), $\|\alpha \|^n \le K^n \|\alpha ^n\|$ for each $n \ge 1$ . Note that

$$ \begin{align*}\alpha^n = \sum_{i_1+\dots +i_m =n} a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m}\bigg(\sum_{j \in A(i_1,\dots,i_m)} e_j\bigg),\end{align*} $$

where

$$ \begin{align*}|A(i_1,\dots,i_m)| = \binom{n}{i_1 \dots i_m}.\end{align*} $$

Hence,

$$ \begin{align*} \|\alpha^n\| &\le \sum_{i_1+\dots+ i_m =n} |a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m}| \lambda(|A(i_1,\dots,i_m)|)\\ & \le \sum_{i_1+\dots+ i_m =n} |a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m}| \Phi(|A(i_1,\dots,i_m)|)\\&\le C \sum_{i_1+\dots+ i_m =n} |a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m}| \binom{n}{i_1 \dots i_m}^{1/p} \end{align*} $$

(for some $C>0$ )

$$ \begin{align*} &\le C(n+1)^{m/q}\bigg(\sum_{i_1+\dots+ i_m =n} |a_1^{i_1}a_2^{i_2}\cdots a_m^{i_m}|^p \binom{n}{i_1 \dots i_m}\bigg)^{1/p} \end{align*} $$

by Hölder’s inequality. Hence,

$$ \begin{align*}\|\alpha\|^n \le K^n C (n+1)^{m/q} \bigg(\sum_{i=1}^m |a_i|^p\bigg)^{n/p}.\end{align*} $$

Taking the nth root and then the limit as $n \rightarrow \infty $ gives

(3.3) $$ \begin{align} \left\|\sum_{i=1}^m a_i e_i\right\| \le K \bigg(\sum_{i=1}^m |a_i|^p\bigg)^{1/p}.\end{align} $$

Since $(e_i^*)$ also satisfies (2.2) and $\Phi ^*(n) \asymp n^{1/q}$ , the same argument gives

(3.4) $$ \begin{align} \left\|\sum_{i=1}^m a_i e^*_i\right\|_* \le K \bigg(\sum_{i=1}^m |a_i|^q\bigg)^{1/q}. \end{align} $$

Hence, by duality, $(e_i)$ is equivalent to the u.v.b. of $\ell _p$ .

Proof Iteration of (3.5) yields $\|\alpha \|^n \le K^n \|\alpha ^n\|$ and $\|\alpha ^{*}\|_*^n \le K^n \|\alpha ^{*n}\|_*$ when n is a power of $2$ . This is enough to prove (3.3) and (3.4) and conclude the proof as above.

Proof We may assume that $\|e_i\| \le 1$ for all $i \ge 1$ . By assumption, there exist an infinite $M \subseteq \mathbb {N}$ and $\delta>0$ such that

$$ \begin{align*}\underset{\pm}{\operatorname{Ave}}\left\|\sum_{i=1}^m \pm e_i\right\|>\delta m\end{align*} $$

for all $m \in M$ . By Elton’s “ $\ell _1^n$ theorem” [Reference Elton9], there exist $\delta _1>0$ and $c>0$ such that for each $m \in M$ there exists $A_m \subset \{1,2,\dots ,m\}$ , with $|A_m| \ge \delta _1 m$ , such that for all scalars $(a_i)_{i \in A_m}$ ,

$$ \begin{align*}c \sum_{i \in A_m} |a_i| \le \left\|\sum_{i \in A_m} a_i e_i\right\| \le \sum_{i \in A_m} |a_i|.\end{align*} $$

Let $k \in \mathbb {N}$ . By Szemerédi’s theorem [Reference Szemerédi14], when m is sufficiently large $A_m$ contains an arithmetic progression of length k, $\{n_1, n_1+d, n_1+2d,\dots , n_1+(k-1)d\}$ . Let $n_1 = bd + r$ , where $1 \le r \le d$ . Fix scalars $(a_i)_{i=1}^k$ and $\varepsilon>0$ , and set

$$ \begin{align*}\alpha = \sum_{i=1}^k a_i e_{b+i}\quad\text{and}\quad \beta_\varepsilon=e_r + \varepsilon \sum_{i=r+1}^d e_i.\end{align*} $$

Note that

$$ \begin{align*}\alpha\otimes \beta _\varepsilon= \sum_{i=1}^k a_i e_{n_1+(i-1)d} + \varepsilon y\end{align*} $$

for some $y \in \operatorname {span}((e_i))$ . Thus, applying (4.1),

$$ \begin{align*} c\sum_{i=1}^k |a_i| - \varepsilon \|y\| \le \|\alpha \otimes \beta_\varepsilon\| \le K \|\alpha\| \|\beta_\varepsilon\| \le K(1 + (d-r)\varepsilon)\|\alpha\|. \end{align*} $$

Taking the limit as $\varepsilon \rightarrow 0$ , we get

$$ \begin{align*}\|\alpha\| \ge \frac{c}{K} \sum_{i=1}^k |a_i|.\end{align*} $$

Let $n = \lfloor k/3 \rfloor $ . There exists $s \in \mathbb {N}$ such that $[(s-1)n +1,sn] \subset [b+1,b+k]$ . We may assume that $a_n \ne 0$ . Then, applying (4.1) again,

$$ \begin{align*}\left\|\sum_{i=1}^n a_ie_i\right\| \ge \frac{1}{K} \left\|e_s \otimes \sum_{i=1}^n a_i e_i\right\|=\frac{1}{K}\left\|\sum_{i=1}^n a_i e_{(s-1)n+i}\right\| \ge \frac{c}{K^2}\sum_{i=1}^n |a_i|.\end{align*} $$

Since k (and hence n) are arbitrary, it follows that $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ .

Proof By considering $\alpha = \sum _{i=1}^m e_i$ and $\beta {\kern-1.2pt}={\kern-1.2pt} \sum _{i=1}^n e_i$ , (1.2) implies that, for all $m,n {\kern-1.2pt}\in{\kern-1.2pt} \mathbb {N}$ ,

(5.1) $$ \begin{align} \frac{1}{K} \lambda(m) \lambda(n) \le \lambda(mn) \le K \lambda(m) \lambda(n). \end{align} $$

Hence, by [Reference Lindenstrauss and Tzafriri13], $\lambda (n) \asymp n^{1/p}$ for some $1 \le p \le \infty $ . Since $(e_i)$ is democratic, it follows that $\Phi (n) \asymp n^{1/p}$ .

If $(\Phi (n))$ is bounded, then, since $(e_i)$ is unconditional for constant coefficients, there exists $C>0$ such that $\|\sum _{i=1}^n \pm e_i\| \le C$ for all $n \ge 1$ and all choices of signs. Hence, $(e_i)$ is equivalent to the u.v.b. of $c_0$ .

If $\phi (n) \asymp n$ , then by the democratic assumption and unconditionality for constant coefficients, it follows that

$$ \begin{align*}\underset{\pm}{\operatorname{Ave}}\left\|\sum_{i=1}^n \pm e_i\right\| \asymp n.\end{align*} $$

Hence, by Theorem 4.1, $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ .

So suppose that $1<p<\infty $ . The sequence $(n^{1/p})$ has the “upper regularity property” for $1<p<\infty $ (see [Reference Dilworth, Kutzarova, Knealton and Temlyakov8, p. 586]). So $(e_i)$ is quasi-greedy and democratic and $(\Phi (n))$ has the upper regularity property. Hence, by [Reference Dilworth, Kutzarova, Knealton and Temlyakov8, Proposition 4.4], $(e_i)$ is bidemocratic. In particular, $\Phi ^*(n) \asymp n^{1/q}$ , where $q=p/(p-1)$ .

The proof is now concluded as in Proposition 3.2. More precisely, the left-hand inequality of (1.2) gives an upper p-estimate for $(e_i)$ and (2.3) gives an upper q-estimate for $(e_i^*)$ .

Proof As remarked above, $(e_{2i}-e_{2i-1})$ is IS and unconditional. Because, for every choice of scalars $a_1,\ldots ,a_n$ ,

$$\begin{align*}\bigg(\sum_{i=1}^na_ie_i\bigg)\otimes(e_2-e_1) = \sum_{i=1}^na_i(e_{2i}-e_{2i-1}),\end{align*}$$

(1.2) yields that $(e_i)_i$ is equivalent to $(e_{2i}-e_{2i-1})$ . Therefore, $(e_i)_i$ is almost greedy and, thus, by Theorem 5.1, the result follows.

Proof This follows at once from (1.2) along with the observation that

$$ \begin{align*}e_{m+1} \otimes \bigg(\sum_{i=1}^n a_i e_i \bigg)= \sum_{i=1}^n a_i e_{mn+i}.\end{align*} $$

Proof of Theorem 6.2

By Theorem 6.1, it suffices to prove that $(e_i)$ is IS. Let $(y_i)$ be a subsequence of $(e_i)$ which generates a spreading model $(Y, \|\cdot \|_Y)$ with basis $(s_i)$ . We define a subsequence $(x_i)$ of $(e_i)$ inductively. For $1 \le i \le 3$ , let $x_i = e_i$ . For the inductive step, suppose that $n \ge 1$ and that $(x_i)_{i=1}^{3^n}$ have been defined with $x_i = e_{N(i)}$ , where $(N(i))_{i=1}^{3^n}$ is strictly increasing. Choose $m \in \mathbb {N}$ with $m3^n>N(3^n)$ and define $x_{3^n + i} = e_{m3^n +i}$ for $1 \le i \le 3^n$ . Thus, $x_i$ has now been defined for $1 \le i \le 2\cdot 3^n$ .

Now choose $p> (m+1)3^n$ such that

(6.1) $$ \begin{align} \frac{1}{2}\left\|\sum_{i=1}^{3^n} a_i s_i\right\|_Y \le \left\|\sum_{i=1}^{3^n} a_i y_{p+i}\right\| \le 2 \left\|\sum_{i=1}^{3^n} a_i s_i\right\|_Y \end{align} $$

for all scalars $(a_i)_{i=1}^{3^n}$ . This is possible because $(y_i)$ generates the spreading model with basis $(s_i)$ . Define $x_{2\cdot 3^n+i} = y_{p+i}$ for $1 \le i \le 3^n$ . This completes the inductive definition of $x_i = e_{N(i)}$ for $1 \le i \le 3^{n+1}$ . Note that

$$ \begin{align*}N(3^n+1) = m3^n+1>N(3^n)\end{align*} $$

and

$$ \begin{align*}N(2\cdot 3^n + 1) \ge p+1)> (m+1)\cdot 3^n = N(2 \cdot 3^n).\end{align*} $$

Hence, $(N(i))_{i=1}^{3^{n+1}}$ is strictly increasing as desired.

By assumption, $(x_i)$ satisfies (1.2) for some $K>0$ . Hence, by Lemma 6.3, $(x_{3^n+i})_{i=1}^{3^n} = (e_{m\cdot 3^n+i})_{i=1}^{3^n}$ is uniformly equivalent to $(e_i)_{i=1}^{3^n}$ . Again, by Lemma 6.3, $(x_{3^n+i})_{i=1}^{3^n}$ is uniformly equivalent to $(x_{2\cdot 3^n+i})_{i=1}^{3^n}$ , which in turn is uniformly equivalent to $(s_i)_{i=1}^{3^n}$ by (6.1). So $(e_i)_{i=1}^{3^n}$ is uniformly equivalent to $(s_i)_{i=1}^{3^n}$ , i.e., $(e_i)$ is equivalent to $(s_i)$ as desired.

Proof (a) $\Rightarrow $ (b) is obvious. Suppose (b) holds. Let $(f_i)$ be any subsequence of $(e_i)$ . Let

$$ \begin{align*}B = \{ (n_k)_{k=1}^\infty \in [\mathbb{N}]^\omega \colon (f_{n_k})_{k=1}^\infty \,\text{satisfies (1.2) for some }K>0\}.\end{align*} $$

Then B is easily seen to be a Borel set. Hence, by the infinite Ramsey theorem of Galvin and Prikry [Reference Galvin and Prikry11], there exists $(n_k)_{k=1}^\infty \in [\mathbb {N}]^\omega $ such that either every subsequence of $(n_k)_{k=1}^\infty $ belongs to B or every subsequence belongs to the complement of B. The latter contradicts (b). By Theorem 6.2, the former implies that $(f_{n_k})$ is equivalent to the u.v.b. of $c_0$ or $\ell _p$ .