1 Introduction
Let $(X,\\cdot \)$ be a Banach space with dual space $(X^*,\\cdot \_{*})$ , let $(e_i)$ be a seminormalized basic sequence in X, and let $\alpha = \sum _{i=1}^n a_i e_i$ and $\beta = \sum _{i=1}^m b_i e_i$ , where $b_m \ne 0$ , belong to $\operatorname {span}((e_i))$ . Define
and $\alpha \otimes 0 = 0$ . Note that $(\operatorname {span}((e_i)),\otimes )$ is a noncommutative semigroup with identity element $e_1$ . However, the semigroup multiplication is not continuous.
If $(e_i)$ is equivalent to the unit vector basis (u.v.b.) of $c_0$ or $\ell _p$ ( $1 \le p < \infty $ ), then there exists $K>0$ such that for all $\alpha , \beta \in \operatorname {span}((e_i))$ ,
The following is the main open question related to this note.
Question 1.1 Let $(e_i)$ be a seminormalized basis for X satisfying (1.2). Is $(e_i)$ equivalent to the u.v.b. of $c_0$ or $\ell _p$ for some $1 \le p < \infty $ ?
We prove below that Question 1.1 has a positive answer when $(e_i)$ is either bidemocratic, almost greedy, or invariant under spreading. However, we do not know the answer in general or for other natural classes of bases such as the class of unconditional bases.
Suppose that $(e_i)$ is a symmetric basis for X and that both X and $[(e^*_i)]$ have a unique symmetric basic sequence up to equivalence. Altshuler [Reference Altshuler3] proved that $(e_i)$ is equivalent to the u.v.b. of $c_0$ or $\ell _p$ . This theorem was recently extended to subsymmetric bases [Reference Casazza, Dilworth, Kutzarova and Motakis6]. In Section 3, we adapt Altshuler’s proof to give an answer to Question 1.1 for bidemocratic bases (see Section 2 for the definition of democratic and bidemocratic bases) by imposing on $(e_i)$ a condition that is formally weaker than (1.2). In Section 4, we provide a new characterization of $\ell _1$ which implies a solution to Question 1.1 under additional assumptions. The proof uses Szemerédi’s theorem on arithmetic progressions. In Section 5, we use the latter characterization to provide an answer to Question 1.1 for the class of democratic bases.
The final section contains results about subsequences. We prove that Question 1.1 has a positive solution if $(e_i)$ is invariant under spreading or, more generally, if every subsequence of $(e_i)$ satisfies (1.2). Using infinite Ramsey theory, we also prove a related characterization of basic sequences which are saturated by subsequences equivalent to the u.v.b. of $c_0$ or $\ell _p$ .
2 Notation and terminology
We use standard Banach space theory notation and terminology as in [Reference Lindenstrauss and Tzafriri13]. We also require some terminology and results from the theory of greedy bases which we briefly review here. For further information on this topic, we refer the reader to [Reference Albiac and Kalton2, Reference Temlyakov15].
Let $(e_i)$ be a seminormalized basic sequence in X. For finite $A \subset \mathbb {N}$ , let $\lambda (A) = \\sum _{i \in A} e_i\$ . The fundamental function $(\Phi (n))$ of $(e_i)$ is defined by
We say that $(e_i)$ is democratic with constant $\Delta $ if $\Phi (A) \le \Delta \lambda (A)$ for all finite $A \subset \mathbb {N}$ . Democratic bases were introduced in [Reference Konyagin and Temlyakov12] in order to prove the celebrated characterization of greedy bases as unconditional and democratic.
We say that $(e_i)$ is unconditional for constant coefficients if there exists $C>0$ such that, for all finite $A \subset \mathbb {N}$ and all choices of signs $\pm $ ,
Now suppose that $(e_i)$ is a Schauder basis for X with biorthogonal functionals $(e_i^*) \subset X^*$ . Let $(\Phi ^*(n))$ be the fundamental function of $(e_i^*)$ . We say that $(e_i)$ is bidemocratic if
(We write $a_n \asymp b_n$ if there exists $C>0$ such that $a_n/C \le b_n \le Ca_n$ for all $n\in \mathbb {N}$ .) It is known that if $(e_i)$ is bidemocratic, then both $(e_i)$ and $(e_i^*)$ are democratic and unconditional for constant coefficients [Reference Dilworth, Kutzarova, Knealton and Temlyakov8, Proposition 4.2]. Moreover, every subsymmetric basis is bidemocratic [Reference Lindenstrauss and Tzafriri13, Proposition 3.a.6].
Remark 2.1 As is well known, if $(e_i)$ is bidemocratic and $(\Phi (n))$ is bounded, then there exists $C>0$ such that $\\sum _{i1}^n \pm e_i\ \le C$ for all $n\ge 1$ and for all choices of signs. Hence $(e_i)$ is equivalent to the u.v.b. of $c_0$ . On the other hand, if $(e_i)$ is bidemocratic and $\Phi (n) \asymp n$ , then $(\Phi ^*(n))$ is bounded. Hence, $(e^*_i)$ is equivalent to the u.v.b. of $c_0$ . By duality, $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ .
Example 2.2 (1) For $1<p<\infty $ , let $(e_i)$ be the u.v.b. of pconvexified Tsirelson space $T_p$ [Reference Figiel and Johnson10]. Then $(e_i)$ is democratic and unconditional and $\Phi (n) \asymp n^{1/p}$ . Since $(n^{1/p})$ has the upper regularity property (see [Reference Dilworth, Kutzarova, Knealton and Temlyakov8, p. 586]), it follows from [Reference Dilworth, Kutzarova, Knealton and Temlyakov8, Proposition 5.4] that $(e_i)$ is bidemocratic. Moreover, from the special properties of block bases in $T_p$ [Reference Casazza and Shura7, Corollary II.5], it follows that there exists $K>0$ such that for all $\alpha , \beta \in \operatorname {span}((e_i))$ ,
(2) Let $(e_i)$ be a subsymmetric basis and suppose that all subsymmetric block bases of $(e_i)$ are equivalent to $(e_i)$ . Altshuler [Reference Altshuler4] constructed the first symmetric example of this type (other than the u.v.b. of $c_0$ or $\ell _p$ ). A second symmetric example was constructed in [Reference Casazza and Shura7]. Recently, the first example of a subsymmetric basis of this type, which in addition is not symmetric, was constructed in [Reference Casazza, Dilworth, Kutzarova and Motakis6].
It was proved in [Reference Casazza, Dilworth, Kutzarova and Motakis6, Lemma 17] that if $(e_i)$ is any subsymmetric basis of this type, then there exists $K>0$ such that for all $\alpha ^*, \beta ^* \in \operatorname {span}((e^*_i))$ ,
Remark 2.3 Suppose that $(e_i)$ is a basis that satisfies (1.2) with constant K. Then, there exists a constant $\tilde K$ such that for every $n\in \mathbb {N}$ and $\alpha \in \operatorname {span}((e_i))$ , $\alpha ^* \in \operatorname {span}((e^*_i))$ ,
Note that the left inequality follows simply by iterating (1.2). For the right one, assume without loss of generality that $(e_i)$ is normalized and monotone. First, make the following easy observation. If $a = \sum _{i=1}^ma_ie_i$ and $a^* = \sum _{i=1}^mb_ie_i^*$ with $a_m\neq 0$ and $b_m\neq 0$ , then, for all $n\in \mathbb {N}$ , $a^{*n}(a^n) = (a^*(a))^n$ .
Next, take an $a^* = \sum _{i=1}^mb_ie_i^*$ and $n\in \mathbb {N}$ . By monotonicity, find a normone $a = \sum _{i=1}^\ell a_ie_i$ with $\ell \leq m$ such that $a^*(a) = \a^*\_*$ . Assume first that $\ell < m$ . For $\varepsilon> 0$ , let $a_\varepsilon = a + \varepsilon e_m$ and note that $a^*(a_\varepsilon ) \geq (1\varepsilon )\a^*\_*$ , while $\a_\varepsilon \ \leq 1 + \varepsilon $ . Therefore,
and let $\varepsilon {\to } 0$ . If $\ell =m$ , the argument is slightly simpler and does not require the perturbation of a.
3 Bidemocratic bases
The main theorem of this section is the following.
Theorem 3.1 Suppose that $(e_i)$ is a bidemocratic basis for X that satisfies (1.2). Then $(e_i)$ is equivalent to the u.v.b. of $c_0$ or $\ell _p$ .
The above is an immediate consequence of the next proposition and Remark 2.3.
Proposition 3.2 Suppose that $(e_i)$ is a bidemocratic basis for X that satisfies (2.2). Then $(e_i)$ is equivalent to the u.v.b. of $c_0$ or ${\ell }_p$ .
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
Similarly (2.2) applied to $\alpha ^* = \sum _{i=1}^n e_i^*$ gives
Since $(e_i)$ is bidemocratic, there exists $C>0$ such that
Hence,
i.e.,
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 $q=p/(p1)$ . 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
where
Hence,
(for some $C>0$ )
by Hölder’s inequality. Hence,
Taking the nth root and then the limit as $n \rightarrow \infty $ gives
Since $(e_i^*)$ also satisfies (2.2) and $\Phi ^*(n) \asymp n^{1/q}$ , the same argument gives
Hence, by duality, $(e_i)$ is equivalent to the u.v.b. of $\ell _p$ .
The following corollary replaces (2.2) by a weaker condition but adds an assumption about the fundamental function.
Corollary 3.3 Let $1 < p < \infty $ . Suppose that $(e_i)$ is a bidemocratic basis for X such that $\Phi (n) \asymp n^{1/p}$ . Suppose also that there exists $K>0$ such that for all $\alpha \in \operatorname {span}((e_i))$ and $\alpha ^* \in \operatorname {span}((e^*_i))$ ,
Then $(e_i)$ is equivalent to the u.v.b. of $\ell _p$ .
4 A characterization of $\ell _1$
Theorem 4.1 Let $(e_i)$ be a seminormalized basis for X. Suppose that there exists $K>0$ such that, for all $\alpha ,\beta \in \operatorname {span}((e_i))$ ,
and also that
Then $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ .
Remark 4.2 Note that (1.2) implies that $\lambda (n) \asymp n^{1/p}$ for some $1 \le p \le \infty $ (see (5.1)). If $p=1$ and $(e_i)$ is unconditional for constant coefficients, then (4.2) is satisfied. So Theorem 4.1 gives a positive answer to Question 1.1 in this case.
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
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 $(a_i)_{i \in A_m}$ ,
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, $\{n_1, n_1+d, n_1+2d,\dots , n_1+(k1)d\}$ . Let $n_1 = bd + r$ , where $1 \le r \le d$ . Fix scalars $(a_i)_{i=1}^k$ and $\varepsilon>0$ , and set
Note that
for some $y \in \operatorname {span}((e_i))$ . Thus, applying (4.1),
Taking the limit as $\varepsilon \rightarrow 0$ , we get
Let $n = \lfloor k/3 \rfloor $ . There exists $s \in \mathbb {N}$ such that $[(s1)n +1,sn] \subset [b+1,b+k]$ . We may assume that $a_n \ne 0$ . Then, applying (4.1) again,
Since k (and hence n) are arbitrary, it follows that $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ .
Corollary 4.3 Suppose that $(e_i)$ satisfies (4.1) and is unconditional for constant coefficients and that $\lambda (n) \asymp n$ . Then $(e_i)$ is equivalent to the u.v.b. of $\ell _1$ .
Remark 4.4 Corollary 4.3 does not admit an $\ell _p$ version for $p>1$ . To see this, let $q = p/(p1)$ . As observed in Example 2.2, the u.v.b. $(e_i)$ of $T_q^*$ is unconditional and bidemocratic with $\phi (n) \asymp n^{1/p}$ . Moreover, (2.1) dualizes due to the special property of block bases of $T_q$ [Reference Casazza and Shura7, Corollary II.5]. In particular, $(e_i)$ satisfies (4.1). Thus, Corollary 4.3 does not admit an $\ell _p$ version for $p>1$ .
5 Democratic bases
Democratic bases are more general than bidemocratic bases. In this section, we prove a characterization of the u.v.b. of $\ell _p$ or $c_0$ among democratic bases.
The characterization involves the notion of quasigreedy basis which we now recall. Let $(e_i)$ be a basis for X. For $x \in X$ and $\delta>0$ , let
Then $(e_i)$ is quasigreedy if there exists $A>0$ such that for all $x \in X$ and for all $\delta>0$ , $\\mathcal {G}_\delta (x)\ \le A \x\$ . Quasigreedy bases were introduced in [Reference Konyagin and Temlyakov12] in connection with the Thresholding Greedy Algorithm. While unconditional bases are quasigreedy, there are also important examples of conditional quasigreedy bases [Reference Albiac, Ansorena, Dilworth and Kutzarova1, Reference Konyagin and Temlyakov12, Reference Wojtaszczyk16]. However, quasigreedy bases are always unconditional for constant coefficients [Reference Wojtaszczyk16]. We refer to [Reference Dilworth, Kutzarova, Knealton and Temlyakov8] for the definition of an almost greedy basis. Almost greediness of a basis is proved in [Reference Dilworth, Kutzarova, Knealton and Temlyakov8] to be equivalent to being quasigreedy and democratic.
Theorem 5.1 Let $(e_i)$ be a quasigreedy democratic (i.e., almost greedy) basis for X which satisfies (1.2). Then $(e_i)$ is equivalent to the u.v.b. of $c_0$ or $\ell _p$ .
Proof By considering $\alpha = \sum _{i=1}^m e_i$ and $\beta {\kern1.2pt}={\kern1.2pt} \sum _{i=1}^n e_i$ , (1.2) implies that, for all $m,n {\kern1.2pt}\in{\kern1.2pt} \mathbb {N}$ ,
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
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 quasigreedy 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 $q=p/(p1)$ .
The proof is now concluded as in Proposition 3.2. More precisely, the lefthand inequality of (1.2) gives an upper pestimate for $(e_i)$ and (2.3) gives an upper qestimate for $(e_i^*)$ .
6 Subsequences
In this section, we use results from the theory of spreading models initiated by Brunel and Sucheston [Reference Brunel and Sucheston5]. We start with a brief review. A basic sequence $(e_i)$ is invariant under spreading (IS) if there exists $C>0$ such that
for all $n \ge 1$ , $i_1<\cdots <i_n$ , and scalars $(a_i)_{i=1}^n$ .
We say $(e_i)$ generates a spreading model if there exists a Banach space $(Y,\\cdot \_Y)$ with basis $(s_i)$ such that
for all $n\ge 1$ and scalars $(a_i)_{i=1}^n$ . It was proved in [Reference Brunel and Sucheston5] that every basic sequence has a subsequence $(x_i)$ which generates a spreading model. Note that $(s_i)$ is IS with $C=1$ . Moreover, $(s_{2i}s_{2i1})$ is suppression $1$ unconditional [Reference Brunel and Sucheston5]. Note that an IS sequence $(e_i)$ is equivalent to the spreading model $(s_i)$ generated by a subsequence of $(e_i)$ . Hence $(e_{2i}e_{2i1})$ is IS and unconditional.
Theorem 6.1 Suppose that $(e_i)$ is IS and satisfies (1.2). Then $(e_i)$ is equivalent to the u.v.b. of $c_0$ or $\ell _p$ .
Proof As remarked above, $(e_{2i}e_{2i1})$ is IS and unconditional. Because, for every choice of scalars $a_1,\ldots ,a_n$ ,
(1.2) yields that $(e_i)_i$ is equivalent to $(e_{2i}e_{2i1})$ . Therefore, $(e_i)_i$ is almost greedy and, thus, by Theorem 5.1, the result follows.
To state the next result, we first make some clarifying remarks to avoid a possible source of confusion. For any given seminormalized basic sequence $(e_i)$ , we defined a multiplication $\otimes $ on $\operatorname {span}((e_i))$ by (1.1). We must emphasize, however, that for a different choice of basic sequence, $(f_i)$ say, the corresponding multiplication will also be different. But to avoid cumbersome notation, we use the same symbol $\otimes $ for both. Likewise, when we say in the next result that all subsequences of $(e_i)$ satisfy (1.2), it is to be understood that the constant K in (1.2) will depend on the subsequence, i.e., we are not assuming a priori that there is a uniform K for all subsequences. (Of course, as the result shows, a uniform K does in fact exist.)
Theorem 6.2 Let $(e_i)$ be a seminormalized basic sequence. Suppose that every subsequence of $(e_i)$ satisfies (1.2). Then $(e_i)$ is equivalent to the u.v.b. of $c_0$ or $\ell _p$ .
The proof requires the following lemma.
Lemma 6.3 Let $(e_i)$ be a seminormalized basic sequence which satisfies (1.2). Then all sequences of the form $(e_{mn+i})_{i=1}^n$ ( $m,n \in \mathbb {N}$ ) are uniformly equivalent to $(e_i)_{i=1}^n$ , i.e., there exists $C>0$ such that for all $m,n \in \mathbb {N}$ and all scalars $(a_i)_{i=1}^n$ ,
Proof This follows at once from (1.2) along with the observation that
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
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
and
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.
We conclude with a characterization of basic sequences that are saturated by subsequences equivalent to the u.v.b. of $c_0$ or $\ell _p$ .
Let $[\mathbb {N}]^\omega $ denote the collection of increasing sequences $(n_k)_{k=1}^\infty $ of natural numbers endowed with the product topology.
Theorem 6.4 Let $(e_i)$ be a seminormalized basic sequence. The following are equivalent:

(a) Every subsequence of $(e_i)$ contains a further subsequence equivalent to the u.v.b. of $c_0$ or $\ell _p$ .

(b) Every subsequence of $(e_i)$ contains a further subsequence satisfying (1.2) for some $K>0$ .
Proof (a) $\Rightarrow $ (b) is obvious. Suppose (b) holds. Let $(f_i)$ be any subsequence of $(e_i)$ . Let
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 $\ell _p$ .