Two properties of M\"untz spaces

TL;DR

This paper investigates M"untz spaces within $C[0,1]$, demonstrating they contain asymptotically isometric copies of $c_0$ and have duals with octahedral geometry, revealing new structural properties.

Contribution

It establishes two novel geometric properties of M"untz spaces: containing asymptotically isometric copies of $c_0$ and having octahedral dual spaces.

Findings

M"untz spaces contain asymptotically isometric copies of $c_0

Dual spaces of M"untz spaces are octahedral

Reveals new geometric structure of M"untz spaces

Abstract

We show that M\"{u}ntz spaces, as subspaces of $C[0,1]$, contain asymptotically isometric copies of $c_0$ and that their dual spaces are octahedral.

Equations

$$m\underset{n}{\sup }|t_n|\le \Vert \sum _n{t}_n{x}_n\Vert \le M\underset{n}{\sup }|t_n|,$$

$$\underset{n\to \infty }{\lim }\Vert x_n\Vert =M.$$

$$\Vert x+y\Vert \ge (1-\epsilon )(\Vert x\Vert +1)\text{mbox}forallx\in F.$$

$$S(x^{\ast },\epsilon ):=\{x\in B_X:x^{\ast }(x)>1-\epsilon ,\epsilon >0,x^{\ast }\in S_{X^{\ast }}\}.$$

$$p(x)=x^{\alpha }-x^{\beta },$$

$$p_k(x)=x^{{\lambda }_k}-x^{{\lambda }_{k+1}}.$$

$$\Vert p_k\Vert \ge r_k(x_k)=\frac{1}{2}-(\frac{1}{2}{)}^2=\frac{1}{4}.$$

$${\bar{x}}_k\ge 1/({\lambda }_{k+1}-{\lambda }_k{)}^{1/({\lambda }_{k+1}-{\lambda }_k)}=:y_k.$$

$$C=\sum _{j=1}^n{\mu }_{j}S(x_j^{\ast },{\epsilon }_j),$$

$$h_k^{j+}$$

$$h_k^{j-}$$

$$\frac{1}{1+\epsilon }\sum _{j=1}^n{\mu }_j{h}_k^{j\pm }\in C,$$

$$\Vert \frac{1}{1+\epsilon }\sum _{j=1}^n{\mu }_j{h}_k^{j+}-\frac{1}{1+\epsilon }\sum _{j=1}^n{\mu }_j{h}_k^{j-}\Vert \ge \frac{1}{1+\epsilon }\left(\sum _{j=1}^n{\mu }_j[h_k^{j+}(x_k)-h_k^{j+}(x_k)]\right)=\frac{2}{1+\epsilon }.$$

$$\frac{p_k(x)}{\Vert p_k\Vert }>\epsilon$$

$$\underset{u,v\in (a_k,b_k)}{\sup }|g^j(u)-g^j(v)|$$

$$|h_k^{j+}(x)|=\left|g^j(x)+(1-g^j(x_k))\frac{p_k(x)}{\Vert p_k\Vert }\right|\le |g^j(x)|+2\epsilon \le 1+2\epsilon .$$

$$|h^{j+}(x)|$$

$$<\left|g^j(x)+(1-g^j(x))\frac{p_k(x)}{\Vert p_k\Vert }\right|+\epsilon .$$

$$\left|g^j(x)+(1-g^j(x))\frac{p_k(x)}{\Vert p_k\Vert }\right|\le g^j(x)+(1-g^j(x))=1.$$

$$\left|g^j(x)+(1-g^j(x))\frac{p_j(x)}{\Vert p_k\Vert }\right|\le |g^j(x)|\le 1.$$

$$f_1=(1-1/2)\frac{p_{k_1}}{\Vert p_{k_1}\Vert }.$$

$$x\in I_2\iff p_{k_2}(x)$$

$$x\in I_2\Rightarrow p_{k_1}(x)$$

$$f_2=(1-1/2^2)\frac{p_{k_2}}{\Vert p_{k_2}\Vert }.$$

$$x\in I_3\iff p_{k_3}(x)$$

$$x\in I_3\Rightarrow p_{k_j}(x)$$

$$f_3=(1-1/2^3)\frac{p_{k_3}}{\Vert p_{k_3}\Vert }.$$

$$m\underset{n}{\sup }|t_n|\le \Vert \sum _n{t}_n{f}_n\Vert \le M\underset{n}{\sup }|t_n|$$

$$\underset{n\to \infty }{\lim }\Vert f_n\Vert =M$$

$$\Vert \sum _{n\in \mathbb{N}}{t}_n{f}_n\Vert$$

$$\ge 1-\frac{1}{2^N}-\sum _{n\ne N}|f_n(x_N)|$$

$$>1-\frac{1}{2^N}-\frac{1}{4}\ge \frac{1}{4}.$$

$$\sum _n{t}_n{f}_n(x)\le \sum _n{f}_n(x)\le \sum _n\frac{1}{2^{2n}}\le \frac{1}{3}.$$

Full text

1. Introduction

Let $\Lambda=(\lambda_{k})_{k=0}^{\infty}$ be a strictly increasing sequence of non-negative real numbers and let $M(\Lambda)=\operatorname{\overline{span}}\{t^{\lambda_{k}}\}_{k=0}^{\infty}\subset C[0,1]$ where $C[0,1]$ is the space of real valued continuous functions on $[0,1]$ endowed with the $\max$-norm. We will call $M(\Lambda)$ a Müntz space provided $\sum_{k=1}^{\infty}1/\lambda_{k}<\infty$. The name is justified by Müntz’ wonderful discovery that if $\lambda_{0}=0$ then $M(\Lambda)=C[0,1]$ if and only if $\sum_{k=1}^{\infty}1/\lambda_{k}=\infty$.

It is well known that $C[0,1]$ contains isometric copies of $c_{0}$ (see e.g. [2, p. 86] how to construct them) and that its dual space is isometric to an $L_{1}(\mu)$ space for some measure $\mu$. The aim of this paper is to demonstrate that Müntz spaces inherit quite a bit of structure from $C[0,1]$ in that they always contain asymptotically isometric copies of $c_{0}$, and that their dual spaces are always octahedral. (An $L_{1}(\mu)$ space is octahedral. See below for an argument.) Let us proceed by recalling the definitions of these two concepts and put them into some context.

Definition 1.1**.**

[4, Theorem 2] A Banach space $X$ is said to contain an asymptotically isometric copy of $c_{0}$ if there exist a sequence $(x_{n})_{n=1}^{\infty}$ in $X$ and constants $0<m<M<\infty$ such that for all sequences $(t_{n})_{n=1}^{\infty}$ with finitely many non zero terms

[TABLE]

and

[TABLE]

R. C. James proved a long time ago (see [9]) that $X$ contains an almost isometric copy of $c_{0}$ as soon at is contains a copy of $c_{0}$. Note that containing an asymptotically isometric copy of $c_{0}$ is a stronger property, see e.g. [4, Example 5].

Definition 1.2**.**

A Banach space $X$ is said to be octahedral if for any finite-dimensional subspace $F$ of $X$ and every $\varepsilon>0$, there exists $y\in S_{X}$ with

[TABLE]

This concept was introduced by Godefroy in [6], and there also the following result can be found on page 12 (see also [8] for a proof of it):

Theorem 1.3**.**

Let $X$ be a Banach space. Then $X^{*}$ is octahedral if and only if every finite convex combination of slices of $B_{X}$ has diameter 2.

By a slice of $B_{X}$ we mean a set of the form

[TABLE]

When we show that the dual of Müntz spaces are octahedral we will use Theorem 1.3 and establish the equivalent property stated there. Note that an $L_{1}(\mu)$ space is octahedral. Indeed, the bidual of such a space can be written $L_{1}(\mu)^{**}=L_{1}(\mu)\oplus_{1}X$ for some subspace $X$ of $L_{1}(\mu)^{**}$ (see e.g. [7, IV. Example 1.1]). From here the octahedrality of $L_{1}(\mu)$ is a straightforward application of the Principle of Local Reflexivity.

We do not know of much research in the direction of our results. But we would like to mention a paper of P. Petráček ([10]), where he demonstrates that Müntz spaces are never reflexive and asks whether they can have the Radon-Nikodým property. Since the Radon-Nikodým property implies the existence of slices of arbitrarily small diameter, we now understand that Müntz spaces rather belong to the “opposite world” of Banach spaces.

2. Results

Definition 2.1**.**

We will say that a strictly increasing sequence of non-negative real numbers $(\lambda_{k})_{k=0}^{\infty}$ has the *Rapid Increase Property (RIP) * if $\lambda_{k+1}\geq 2\lambda_{k}$ for every $k\geq 0$.

We will call a function of the form

[TABLE]

where $0\leq\alpha<\beta,$ a spike function.

Remark 2.2*.*

If $\alpha>0$ it should be clear that any spike function $p$ satisfies $p(0)=p(1)=0,$ attains its norm on a unique point $x_{p}$, is strictly increasing on $[0,x_{p}],$ and strictly decreasing on $[x_{p},1].$

We will need the following result below.

Lemma 2.3**.**

Let $(\lambda_{k})_{k=0}^{\infty}$ be an RIP sequence and $(p_{k})_{k=0}^{\infty}$ the sequence of corresponding spike functions $p_{k}(x)=x^{\lambda_{k}}-x^{\lambda_{k+1}}$. Then $\inf_{k}\|p_{k}\|\geq 1/4.$ Moreover, the sequence $(p_{k}/\|p_{k}\|)_{k=1}^{\infty}$ converges to [math] weakly in $M(\Lambda)$.

Proof.

We want to find the norm of the spike function defined by

[TABLE]

Observe that $r_{k}(x):=x^{\lambda_{k}}-x^{2\lambda_{k}}\leq p_{k}(x)$ for all $x\in[0,1]$. Now, by standard calculus, $r_{k}$ attains its maximum at $x_{k}$ where ${x_{k}}^{\lambda_{k}}=\frac{1}{2}$. Thus

[TABLE]

As $(p_{k})_{k=1}^{\infty}$ converges pointwise to [math] and $\inf_{k}\|p_{k}\|\geq 1/4,$ the sequence $(p_{k}/\|p_{k}\|)_{k=1}^{\infty}$ converges pointwise to [math] and thus weakly to [math] as it is bounded. ∎

Remark 2.4*.*

By standard calculus one can show that the point at which $p_{k}$ in Lemma 2.3 obtains its norm is $\bar{x}_{k}=(\lambda_{k}/\lambda_{k+1})^{1/(\lambda_{k+1}-\lambda_{k})}$. For sufficiently large $\lambda_{k}$ (e.g. for $\lambda_{k}>3$) it is straightforward to show that

[TABLE]

$(y_{k})$ is strictly monotone, and converges to 1.

Theorem 2.5**.**

The dual of any Müntz space is octahedral.

Proof.

Let $M(\Lambda)$ be a Müntz space. Let

[TABLE]

where $\sum_{j=1}^{n}\mu_{j}=1,\mu_{j}>0$, and $S(x_{j}^{*},\varepsilon_{j}),1\leq j\leq n$, is a slice of $B_{M(\Lambda)}.$ We will show that the diameter of $C$ is 2 (cf. Theorem 1.3). To this end, start with some $f\in C$ and write $f=\sum_{j=1}^{n}\mu_{j}g^{j}$, where $g^{j}\in S(x_{j}^{*},\varepsilon_{j})$. Let $(\lambda_{k})_{k=0}^{\infty}$ be an RIP subsequence of $\Lambda$ and put

[TABLE]

where $(p_{k})_{k=0}^{\infty}$ is the sequence of spike functions corresponding to $(\lambda_{k})_{k=0}^{\infty}$ and $x_{k}$ the (unique) point where $p_{k}$ attains its norm. We will prove that, for any $\varepsilon>0$, there exists a $K=K(\varepsilon)$ such that whenever $k\geq K$ we have $\frac{1}{1+\varepsilon}h_{k}^{j+},\frac{1}{1+\varepsilon}h_{k}^{j-}\in S(x_{j}^{*},\varepsilon_{j})$ for every $1\leq j\leq n$. Then, clearly

[TABLE]

and

[TABLE]

for all $k\geq K.$ Since $\varepsilon$ is arbitrary, we can thus conclude that $C$ has diameter 2.

To produce the $K=K(\varepsilon)$ above, note that $h_{k}^{j\pm}$ converges to $g^{j}$ pointwise, and thus weakly since the sequences are bounded. As $U_{j}:=\{x\in M(\Lambda):x_{j}^{\ast}(x)>1-\varepsilon\}$ is weakly open, each sequence $(h_{k}^{j\pm})_{k=0}^{\infty}$ enters $U_{j}$ eventually. Since there are only a finite number of sets $U_{j}$, this entrance is uniform. So, what is left to prove is that for $\varepsilon>0$ there exists $K$ such that $\|h_{k}^{j\pm}\|\leq 1+\varepsilon$ whenever $k\geq K$.

Now, let $\varepsilon>0$. Combining Remark 2.2, Remark 2.4, that $(p_{k}/\|p_{k}\|)_{k=1}^{\infty}$ converges pointwise to [math], and the continuity of $g^{j}$, we can find $K\in\mathbb{N}$ such that for all $k\geq K$ there are points $0<a_{k}<x_{k}<b_{k}<1$ such that

[TABLE]

We will see that this $K$ does the job for the given $\varepsilon>0$: Let $k\geq K$ and suppose $x\not\in(a_{k},b_{k})$. Then

[TABLE]

If $x\in(a_{k},b_{k})$, observe that

[TABLE]

Now, if $g^{j}(x)\geq 0,$ then

[TABLE]

If $g^{j}(x)<0$ and $g^{j}(x)+(1-g^{j}(x))p_{k}(x)/\|p_{k}\|\geq 0,$ then

[TABLE]

If $g^{j}(x)<0$ and $g^{j}(x)+(1-g^{j}(x))p_{k}(x)/\|p_{k}\|<0,$ then

[TABLE]

In any case we have for $k\geq K$ and $x\in[0,1]$ that $|h_{k}^{j+}(x)|\leq 1+2\varepsilon.$ The argument that $\|h_{k}^{j-}\|\leq 1+\varepsilon$ is similar. ∎

Theorem 2.6**.**

Müntz spaces contain asymptotically isometric copies of $c_{0}$.

Proof.

We will construct a sequence $(f_{n})_{n=1}^{\infty}\subset M(\Lambda)$ and pairwise disjoint intervals $I_{n}=(a_{n},b_{n})\subset[0,1]$ such that for all $n\in\mathbb{N}$

• (i)

$f_{n}(x)\geq 0$ for all $x\in[0,1],$

• (ii)

$\|f_{n}\|=1-1/2^{n},$

• (iii)

$b_{n}<a_{n+1}$

• (iv)

$f_{n}(x)>1/2^{2n}\Leftrightarrow x\in I_{n},$

• (v)

$f_{n}(x)<1/2^{2m}$ whenever $m\geq n$ and $x\in I_{m}.$

To this end choose a subsequence of $\Lambda$ with the RIP (which is possible as $\sum_{k=1}^{\infty}1/\lambda_{k}<\infty$). For simplicity denote also this subsequence by $(\lambda_{k})_{k=0}^{\infty}$. Let $(p_{k})_{k=1}^{\infty}$ be its corresponding sequence of spike functions, and let $x_{k}$ be the (unique) point in $(0,1)$ where $p_{k}$ obtains its maximum.

Now, start by letting $k_{1}=1$ and put

[TABLE]

Using continuity and properties of $p_{1}$, we can find an interval $I_{1}=(a_{1},b_{1})$ such that $0<a_{1}<b_{1}<1$ and $f_{1}(x)>\frac{1}{2^{2}}\Leftrightarrow x\in I_{1}.$ By construction $f_{1}$ satisfies the conditions (i) - (iv).

To construct $f_{2}$ we use Lemma 2.3 and Remarks 2.2 and 2.4 to find $k_{2}\in\mathbb{N}$ and an interval $I_{2}=(a_{2},b_{2})$ with $b_{1}<a_{2}<b_{2}<1$ such that

[TABLE]

Let

[TABLE]

By construction $f_{1}$ now satisfies condition (v) for $m\leq 2$ and $f_{2}$ satisfies conditions (i) - (iv).

To construct $f_{3}$ we use Lemma 2.3 and Remarks 2.2 and 2.4 again to find $k_{3}\in\mathbb{N}$ and an interval $I_{3}=(a_{3},b_{3})$ with $b_{2}<a_{3}<b_{3}<1$ such that

[TABLE]

Let

[TABLE]

By construction $f_{1}$ and $f_{2}$ now satisfy condition (v) for $m\leq 3$ and $f_{3}$ satisfies conditions (i) - (iv). If we continue in the same manner we obtain a sequence $(f_{n})_{n=1}^{\infty}\subset M(\Lambda)$ and a sequence of intervals $I_{n}=(a_{n},b_{n})$ which satisfies the conditions (i) - (v).

Now we will show that $(f_{n})_{n=1}^{\infty}$ satisfies the requirements of Definition 1.1. To this end we need to find constants $0<m<M<\infty$ such that given any sequence $(t_{n})_{n=1}^{\infty}$ with finitely many non zero terms

[TABLE]

and

[TABLE]

We claim that (1) and (2) holds with $m=\frac{1}{4}$ and $M=1.$ First observe that we have $\lim_{n\rightarrow\infty}\|f_{n}\|=1$ immediately from the requirements, so (2) holds for $M=1$. In order to prove the two inequalities in (1), let $(t_{n})_{n=1}^{\infty}$ be an arbitrary sequence with finitely many non zero terms. First we will prove that $1/4\sup_{n}|t_{n}|\leq\left\|\sum_{n}t_{n}f_{n}\right\|$. We can assume by scaling that $\sup|t_{n}|=1$. Since $(t_{n})_{n=1}^{\infty}$ has finitely many non zero terms, its norm is attained at, say, $n=N$, i.e. $|t_{N}|=1$. Put $x_{N}=x_{k_{N}}$ where $x_{k_{N}}$ is the point where $p_{k_{N}}$ and thus $f_{N}$ attains its norm. Then

[TABLE]

We conclude that the left hand side of the inequality (1) holds. Now we will show the right hand side of this inequality holds, i.e. we want to prove that $|\sum_{n}t_{n}f_{n}(x)|\leq 1$ for all $x\in[0,1]$. Since $f_{n}\geq 0$ for all $n=1,2,\ldots,$ we may assume that every $t_{n}$ is positive. Now, if $x\not\in\cup_{n}(a_{n},b_{n})$, we have

[TABLE]

If, on the other hand $x\in(a_{n^{\prime}},b_{n^{\prime}})$ for some $n^{\prime}\in\mathbb{N},$ then

[TABLE]

These combined yields the right hand side of the inequality (1), so the proof is complete. ∎

A Banach space $X$ contains an asymptotically isometric copy of $\ell_{1}$ if it contains a sequence $(x_{n})_{n=1}^{\infty}$ for which there exists a sequence $(\delta_{n})_{n=1}^{\infty}$ in $(0,1)$, decreasing to [math], and such that

[TABLE]

for each finite sequence $(a_{n})_{n=1}^{m}$ in $\mathbb{R}$.

Merging ([5, Theorem 2]) and [1, Lemma 2.3] gives us that if either the Banach space $X$ contains an asymptotically isometric copy of $c_{0}$ or if $X^{\ast}$ is octahedral, then $X^{\ast}$ contains an asymptotically isometric copy of $\ell_{1}$. So, we have two ways of proving

Corollary 2.7**.**

$M(\Lambda)^{\ast}$* contains an asymptotically isometric copy of $\ell_{1}$.*

Moreover, we have

Corollary 2.8**.**

$M(\Lambda)^{\ast\ast}$* contains an isometrically isomorphic copy of $L_{1}[0,1]$.*

Proof.

This follows from Corollary 2.7 and [3, Theorem 2]. ∎