Smooth norms in dense subspaces of $\ell _p(\Gamma )$ and operator ranges

Abstract

For $1\le p<\infty$, we prove that the dense subspace $\mathcal {Y}_p$ of $\ell _p(\Gamma )$ comprising all elements y such that $y \in \ell _q(\Gamma )$ for some $q \in (0,p)$ admits a $C^{\infty }$-smooth norm which locally depends on finitely many coordinates. Moreover, such a norm can be chosen as to approximate the $\left\| \cdot \right\| _p$-norm. This provides examples of dense subspaces of $\ell _p(\Gamma )$ with a smooth norm which have the maximal possible linear dimension and are not obtained as the linear span of a biorthogonal system. Moreover, when $p>1$ or $\Gamma$ is countable, such subspaces additionally contain dense operator ranges; on the other hand, no non-separable operator range in $\ell _1(\Gamma )$ admits a $C^1$-smooth norm.

1 Introduction

The present paper is a continuation of the research of the authors [4, 5, 12], dedicated to the study of smoothness in (incomplete) normed spaces. The main question that we face in this ongoing project is the following: given a Banach space $\mathcal {X}$ and $k\in \mathbb {N}\cup \{\infty ,\omega \}$ is there a dense subspace $\mathcal {Y}$ of $\mathcal {X}$ such that $\mathcal {Y}$ admits a $C^k$-smooth norm? (By definition, $C^\omega$-smooth means analytic; however, the case $k=\omega$ will not be considered in this article.) Such a line of research can be traced back at least to the papers [9, 20] from the early Nineties, where it is proved that every separable Banach space admits a dense subspace with a $C^\infty$-smooth norm. In particular, for a separable normed space $\mathcal {X}$ the existence of a $C^1$-smooth norm does not imply that $\mathcal {X}^*$ is separable, a result that is possibly surprising at first sight. Our goal in [4, 5] was to push such a theory to the non-separable context and it culminated in the main result of [5] asserting that every Banach space with a fundamental biorthogonal system has a dense subspace with a $C^\infty$-smooth norm (let us also refer to the same paper for a more thorough introduction to the subject).

As it turns out, in most of the above results, the dense subspace $\mathcal {Y}$ of $\mathcal {X}$ is the linear span of a certain biorthogonal system in $\mathcal {X}$ (the unique exception being [4, Theorem 3.1] where an analytic norm is constructed in the dense subspace $\ell _\infty ^F=\textrm{span}\{{\varvec{1}}_A:A\subseteq \mathbb {N}\}$ of $\ell _\infty$). In particular, when $\mathcal {X}$ is separable, the subspace $\mathcal {Y}$ has countable dimension (namely, it is the linear span of a countable set). In this paper we focus on the classical (long) sequence spaces and we show that it is possible to go beyond this limitation; in particular, we build $C^\infty$-smooth norms on dense subspaces that are ‘large’ in a sense that we specify below. More precisely, the following is our main result (for the necessary notation, we refer the reader to Sect. 1.1 below).

Main Theorem

Let $1\le p <\infty$ and $\Gamma$ be any infinite set. Then

$$\begin{aligned} \mathcal {Y}_p{:}{=}\Big \{y\in \ell _p(\Gamma ):\Vert y\Vert _q< \infty \text { for some } q\in (0,p) \Big \}= \bigcup _{0<q<p} \ell _q(\Gamma ) \end{aligned}$$

is a dense subspace of $\ell _p(\Gamma )$ which admits a $C^{\infty }$-smooth and LFC norm that approximates the $\left\| \cdot \right\| _p$-norm.

Plainly, when p is an even integer, the canonical norm of $\ell _p(\Gamma )$ is $C^\infty$-smooth, hence the $C^\infty$-smoothness part of the theorem is obvious. On the other hand, if $p\notin 2\mathbb {N}$, $\ell _p(\Gamma )$ does not have any $C^{\big \lceil p \big \rceil }$-smooth norm ($\big \lceil p \big \rceil$ denotes the ceiling of p) [10, p. 295]. Moreover, no $\ell _p(\Gamma )$ has an LFC norm [18], so in order to obtain an LFC norm in the main theorem it is indeed necessary to pass to the subspace $\mathcal {Y}_p$. Finally, recall that $c_0(\Gamma )$ has a $C^\infty$-smooth and LFC norm [10, p. 284], for which reason we do not consider $c_0(\Gamma )$ in our result. Also, notice that we don’t consider $\ell _\infty (\Gamma )$ in our result simply because the corresponding subspace $\mathcal {Y}_\infty$ is not dense in $\ell _\infty (\Gamma )$.

Let us now discuss the novelty of the result. First of all, for every $1\le p<\infty$, the dense subspace $\mathcal {Y}_p$ of $\ell _p(\Gamma )$ has the same linear dimension of $\ell _p(\Gamma )$, hence it is as large as possible in the linear sense. In particular, in the case of $\ell _p$, i.e., when $\Gamma$ is countable, we obtain a dense subspace of dimension continuum. This is in sharp contrast with the results in [4, 5, 9, 20], where the dense subspaces had dimension equal to the density character of the Banach space $\mathcal {X}$. Comparing this result with [9], it seems conceivable to conjecture that for every separable Banach space $\mathcal {X}$ there is a dense subspace $\mathcal {Y}$ of dimension continuum and with a $C^\infty$-smooth norm. We leave the validity of such a conjecture as an open problem.

Moreover, if ${\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$ is a norm on $\ell _p(\Gamma )$ that coincides with the $C^\infty$-smooth and LFC one on $\mathcal {Y}_p$, then it is standard to verify (see, e.g., the proof of [5, Corollary 3.5]) that the norm ${\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$ is $C^\infty$-smooth and LFC at every point of $\mathcal {Y}_p$ (as a function on $\ell _p(\Gamma )$). In lineability terms, this assertion can be restated as stating that the set of points in $\ell _p(\Gamma )$ where the norm ${\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$ is $C^\infty$-smooth and LFC is maximal densely lineable in $\ell _p(\Gamma )$. Let us refer to [1, 2, 17] and the references therein for information on lineability in Banach spaces and for the relevant definitions.

Finally, there is a second sense in which the subspace $\mathcal {Y}_p$ can be considered to be ‘large’, which is connected to the notion of operator ranges, [3, 7, 8]. Recall that a normed space $\mathcal {Y}$ is an operator range if there are a Banach space $\mathcal {Z}$ and a surjective bounded linear operator $T:\mathcal {Z}\rightarrow \mathcal {Y}$ (and, up to passing to a suitable quotient, one can assume that T is injective); in other words, $\mathcal {Y}$ is the linear (injective) image of a Banach space. Notice that operator ranges bear a certain form of completeness, since, for instance, they satisfy the Baire Category Theorem, even if in a finer linear topology. When $p>1$, it is clear that $\mathcal {Y}_p$ contains a dense operator range, since the Banach space $\ell _1(\Gamma )$ injects in $\mathcal {Y}_p$. Hence, when $p>1$, our theorem also implies the existence of a dense operator range in $\ell _p(\Gamma )$ that admits a $C^\infty$-smooth and LFC norm. On the other hand, the situation is different when $p=1$ (and $\Gamma$ is uncountable): indeed, it is a folklore result, essentially due to Rosenthal [19], that every non-separable operator range in $\ell _1(\Gamma )$ contains an isomorphic copy of $\ell _1(\omega _1)$, hence it admits no $C^1$-smooth norm (see Proposition 3.1). This result is extremely relevant to the topic of the paper, since it is one of the few instances where an incomplete normed space is proved not to admit any $C^1$-smooth norm. Moreover it is also the first occurrence where $\ell _1(\Gamma )$ behaves worse than $\ell _p(\Gamma )$ ($1<p<\infty$) for what concerns the existence of smooth norms on dense subspaces. Such behaviour was to be expected, but it was not present in the literature so far; quite surprisingly, there are in fact instances of the opposite situation (see the discussion concerning Theorem A(iii) in [4]).

1.1 Definitions and notation

All the spaces that we consider in the paper are real normed spaces. If $\mathcal {X}$ is a normed space, then the norm $\left\| \cdot \right\|$ of $\mathcal {X}$ is said to be $C^k$-smooth if its k-th Fréchet derivative exists and it is continuous at every point of $\mathcal {X}\setminus \{0\}$. When this holds for every $k \in \mathbb {N}$, the norm is $C^\infty$-smooth. The norm $\left\| \cdot \right\|$ locally depends on finitely many coordinates (is LFC, for short) on $\mathcal {X}$ if for each $x\in \mathcal {X}\setminus \{0\}$ there exist an open neighbourhood $\mathcal {U}$ of x, functionals $\varphi _1,\dots ,\varphi _k \in \mathcal {X}^*$, and a function $G:\mathbb {R}^k\rightarrow \mathbb {R}$ such that

$$\begin{aligned} \Vert y\Vert = G\big ( \langle \varphi _1,y\rangle ,\dots , \langle \varphi _k,y\rangle \big ) \qquad \text {for every } y\in \mathcal {U}. \end{aligned}$$

We say that a norm $\left\| \cdot \right\|$ on $\mathcal {X}$ can be approximated by norms with a certain property P if, for every $\varepsilon >0$, there is an equivalent norm ${\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$ on $\mathcal {X}$ with property P and such that $(1-\varepsilon ){\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }\le \left\| \cdot \right\| \le (1+\varepsilon ){\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$. For further information concerning smoothness we refer to the monographs [6, 10].

If $x:\Gamma \rightarrow \mathbb {R}$ and $p\in (0,\infty )$, we write as usual

$$\begin{aligned} \Vert x\Vert _p{:}{=}\left( \sum _{\gamma \in \Gamma } |x(\gamma )|^p\right) ^{1/p} \quad \text {and}\quad \ell _p(\Gamma ){:}{=}\Big \{x:\Gamma \rightarrow \mathbb {R}:\Vert x\Vert _p<\infty \Big \}. \end{aligned}$$

When $p\ge 1$, $\ell _p(\Gamma )$ is obviously a Banach space with the norm $\left\| \cdot \right\| _p$, while for $p\in (0,1)$ it is only a quasi-Banach space and $\left\| \cdot \right\| _p$ is a quasi-norm. When $0<q<p<\infty$, we sometimes write $\ell _q(\Gamma ) \subseteq \ell _p(\Gamma )$, by which we mean the inclusion of the corresponding underlying vector spaces (which is also a continuous injection of (quasi-)Banach spaces).

We write $\mathbb {N}$ for the set of positive natural numbers and $\mathbb {N}_0{:}{=}\mathbb {N}\cup \{0\}$. We write $\omega _1$ for the smallest uncountable ordinal. We denote by |A| the cardinality of a set A. Given a set $\Gamma$, we use the standard set-theoretic notation $[\Gamma ]^{<\omega } {:}{=}\{A\subseteq \Gamma :|A|<\infty \}$.

2 Proof of the main theorem

In this section we provide the proof of our main theorem. The argument is inspired by the proof of [4, Theorem 4.1]; in particular, the formula for the norm and the scheme of the argument are essentially the same. Nevertheless, the similarity between the two proofs is more formal than substantial due to the crucial part of the proof (Claim 2.1 below, that corresponds to [4, Claim 4.2]) which is rather different in the two papers: indeed, in the present paper we require some new ingredients, including combinatorial ones and finer estimates.

In what follows, we occasionally write $a_k \sim b_k$ to mean that the scalar sequences $(a_k)_{k=1}^\infty$ and $(b_k)_{k=1}^\infty$ are asymptotic (i.e., $a_k/b_k\rightarrow 1$ as $k\rightarrow \infty$).

Proof of the Main Theorem

Let us start by choosing some parameters. Let $(\delta _k)_{k=0}^{\infty } \subseteq \mathbb {R}$ be a decreasing sequence with

$$\begin{aligned} \delta _k \sim \frac{1}{\log k} \qquad \text {as } k\rightarrow \infty . \end{aligned}$$

(2.1)

Next, let $(\theta _k)_{k=0}^{\infty } \subseteq \mathbb {R}$ be another decreasing sequence such that

$$\begin{aligned} \theta _k \searrow 0 \ \ \ \text{ and } \ \ \ \frac{1 + \delta _{k+1}}{1 + \delta _k} < 1 - 2 \theta _{k+1} \ \text{ for } \text{ every } \ k \ge 0. \end{aligned}$$

(2.2)

Additionally, fixed $\varepsilon >0$, the sequences are chosen so that

$$\begin{aligned} \frac{1+\theta _1}{1-\theta _1}\cdot (1+\delta _1)^2\le 1+\varepsilon . \end{aligned}$$

(2.3)

Next, we fix some notation we will be using throughout the proof. For $A\in [\Gamma ]^{<\omega }$, we identify $\ell _p(A)$ with the finite-dimensional subspace of $\ell _p(\Gamma )$ comprising all vectors with support contained in A. Moreover, when $x\in \ell _p(\Gamma )$, we define $Ax\in \ell _p(A)$ as

$$\begin{aligned} (Ax)(\gamma )\,{:}{=}{\left\{ \begin{array}{ll} x(\gamma ) &{} \text {if}\ \gamma \in A, \\ 0 &{} \text {if}\ \gamma \notin A. \end{array}\right. } \end{aligned}$$

In other words, we also denote by A the canonical projection from $\ell _p(\Gamma )$ onto $\ell _p(A)$. Since $\ell _p(A)$ is finite-dimensional, $C^\infty$-smooth norms are dense in $\ell _p(A)$, hence there exists a $C^\infty$-smooth norm $\left\| \cdot \right\| _{s,A}$ on $\ell _p(A)$ that $\frac{1}{1 + \theta _{|A|}}$-approximates the $\left\| \cdot \right\| _p$-norm ($\left\| \cdot \right\| _{s,A}$ is also trivially LFC). In particular, for every $y\in \mathcal {Y}$ we have

$$\begin{aligned} \frac{1}{1 + \theta _{|A|}} \Vert Ay\Vert _p \le \Vert Ay\Vert _{s,A} \le \Vert Ay\Vert _p. \end{aligned}$$

(2.4)

Finally, in several computations it will be more convenient to use, instead of $\left\| \cdot \right\| _p$, the following auxiliary equivalent norm on $\mathcal {Y}$

$$\begin{aligned} \nu (x) {:}{=}\sup _{|A|<\infty } (1 + \delta _{|A|})^2 \Vert Ax\Vert _p \ \ \ (x \in \mathcal {Y}). \end{aligned}$$

The norm $\nu$ approximates the norm $\left\| \cdot \right\| _p$ on $\mathcal {Y}$; more precisely,

$$\begin{aligned} \left\| \cdot \right\| _p \le \nu \le (1+\delta _1)^2 \left\| \cdot \right\| _p \end{aligned}$$

(2.5)

We now come to the crucial part of the proof, where we prove the following ‘strong maximum’ result.

Claim 2.1

Let $x \in \mathcal {Y}$ be such that $\nu (x) \le 1$. Then there exist an open neighbourhood $\mathcal {O}_{x}$ of x and a finite collection of subsets ${\mathfrak {F}}_{x} \subseteq [\Gamma ]^{<\omega }$ such that, for each $y \in \mathcal {O}_{x}$ and each $A \in [\Gamma ]^{<\omega } {\setminus } {\mathfrak {F}}_{x}$, we have

$$\begin{aligned} (1+\delta _{|A|})^2 \Vert Ay\Vert _{s,A} \le 1-\theta _{|A|}. \end{aligned}$$

(2.6)

Proof of Claim 2.1

In fact, we prove a stronger estimate that only involves the point x. Indeed, we show that there exist a finite collection of subsets ${\mathfrak {F}}_x \subseteq [\Gamma ]^{<\omega }$ and $k_0 \in \mathbb {N}$ such that, for every $A \in [\Gamma ]^{<\omega } \setminus {\mathfrak {F}}_x$, we have

$$\begin{aligned} (1+\delta _{|A|})^2 \Vert Ax\Vert _p \le {\left\{ \begin{array}{ll} \displaystyle 1-2\theta _{|A|}, &{} \text{ if } \ |A|\le k_0,\\ \displaystyle 1-2\theta _{k_0+1}, &{} \text{ if } \ |A|>k_0. \end{array}\right. } \end{aligned}$$

(2.7)

Let us show first that (2.7) is indeed stronger than (2.6). Suppose that we have proved (2.7). Define the following open neighbourhood of x:

$$\begin{aligned} \mathcal {O}_x\,{:}{=}\Big \{y\in \mathcal {Y}:\nu (y-x)< \theta _{k_0+1} \Big \}. \end{aligned}$$

Then, for every $A \in [\Gamma ]^{<\omega } \setminus {\mathfrak {F}}_x$ and $y \in \mathcal {O}_x$, we have that

$$\begin{aligned} (1 + \delta _{|A|})^2 \Vert Ay\Vert _{s,A}&{\mathop {\le }\limits ^{(2.4)}} ( 1 + \delta _{|A|})^2 \Vert Ay\Vert _p\\&\le (1 + \delta _{|A|})^2 \Vert Ax\Vert _p + (1 + \delta _{|A|})^2 \Vert A(y-x)\Vert _p \\&\le (1 + \delta _{|A|})^2 \Vert Ax\Vert _p + \nu (y - x)< (1 + \delta _{|A|})^2 \Vert Ax\Vert _p + \theta _{k_0+1}. \end{aligned}$$

If $|A|\le k_0$, we use (2.7) and continue from the above inequalities

$$\begin{aligned} (1+\delta _{|A|})^2 \Vert Ay\Vert _{s,A} {\mathop {\le }\limits ^{(2.7)}} 1 - 2 \theta _{|A|} + \theta _{k_0+1}\le 1 - \theta _{|A|}. \end{aligned}$$

Similarly, if $|A| > k_0$,

$$\begin{aligned} (1+\delta _{|A|})^2 \Vert Ay\Vert _{s,A}\le 1-\theta _{k_0+1}\le 1-\theta _{|A|}. \end{aligned}$$

Therefore to prove Claim 2.1 all we need to do is to prove (2.7). In order to do so, let us consider the sequence $(\phi _k(x))_{k=1}^{\infty }$ defined by

$$\begin{aligned} \phi _k(x){:}{=}\, (1+\delta _k) \sup _{|A|=k} \Vert Ax\Vert _p \qquad (k\in \mathbb {N}). \end{aligned}$$

Note that the supremum above is actually attained. Indeed, if $(\gamma _j)_{j=1}^{\infty } \subseteq \Gamma$ is an injective sequence such that ${{\,\textrm{supp}\,}}(x) \subseteq \{\gamma _j\}_{j=1}^{\infty }$ and $\big (|x(\gamma _j)| \big )_{j=1}^{\infty }$ is non-increasing, it is clear that

$$\begin{aligned} \phi _k(x)=(1+\delta _k)\cdot \left( \sum _{j=1}^k |x(\gamma _j)|^p \right) ^{1/p} \qquad \text {for every}\ k\in \mathbb {N}. \end{aligned}$$

Claim 2.2

There exists $k_0\in \mathbb {N}$ such that $\phi _{k+1}(x)\le \phi _k(x)$ for every $k \ge k_0$.

Proof of Claim 2.2

Clearly, $\phi _{k+1}(x)\le \phi _k(x)$ is equivalent to

$$\begin{aligned} (1+\delta _{k+1})^p\cdot \left( \sum _{j=1}^k |x(\gamma _j)|^p+ |x(\gamma _{k+1})|^p \right) \le (1+ \delta _k)^p \cdot \sum _{j=1}^k |x(\gamma _j)|^p \end{aligned}$$

which is in turn equivalent to

$$\begin{aligned} |x(\gamma _{k+1})|^p\le \frac{(1+\delta _k)^p - (1+\delta _{k+1})^p}{(1+\delta _{k+1})^p} \cdot \sum _{j=1}^k |x(\gamma _j)|^p. \end{aligned}$$

(2.8)

Thus, it is enough to check that (2.8) is true for all large enough k. By using first-order Taylor expansions and (2.1) we readily get

$$\begin{aligned} (1+\delta _k)^p - (1+\delta _{k+1})^p \sim \frac{p}{k (\log k)^2}. \end{aligned}$$

As the sequence $\left( \sum _{j=1}^k |x(\gamma _j)|^p \right) _{n=1}^{\infty }$ is non-decreasing and bounded, for some constant $C > 0$ we thus have that

$$\begin{aligned} \frac{(1+\delta _k)^p - (1+\delta _{k+1})^p}{ (1+\delta _{k+1})^p} \cdot \sum _{j=1}^k |x(\gamma _j)|^p \sim C \cdot \frac{1}{k(\log k)^2}. \end{aligned}$$

(2.9)

We now estimate the left-hand side of (2.8). Since $x \in \mathcal {Y}$, there exists $q < p$ such that $x \in \ell _q(\Gamma )$. By definition of the sequence $(\gamma _j)_{j=1}^\infty$, for every $k \in \mathbb {N}$, we have that¹

$$\begin{aligned} \Vert x\Vert _q^q \ge \sum _{j=1}^k |x(\gamma _j)|^q \ge k |x(\gamma _k)|^q. \end{aligned}$$

In other words, there exists ${\tilde{C}} > 0$ such that

$$\begin{aligned} |x(\gamma _{k+1})|^p \le {\tilde{C}} \cdot \frac{1}{k^{p/q}} \end{aligned}$$

(2.10)

Comparing (2.9) and (2.10) and recalling that $\frac{p}{q}>1$, it is clear that (2.8) is true for k sufficiently large, which proves Claim 2.2 as desired. $\square$

We now return to the proof of Claim 2.1. Note first that $(1+\delta _k)\phi _k(x)\le \nu (x)$, for every $k\in \mathbb {N}$. Let $k_0\in \mathbb {N}$ be as in Claim 2.2. Then, for every set $A\in [\Gamma ]^{<\omega }$ with $|A|>k_0$, we have

$$\begin{aligned} (1+\delta _{|A|})^2 \Vert Ax\Vert _p\le & {} (1 + \delta _{k_0+1})(1 + \delta _{|A|}) \Vert Ax\Vert _p \\\le & {} \frac{1 + \delta _{k_0+1}}{1 +\delta _{k_0}} \cdot (1 + \delta _{k_0}) \phi _{|A|}(x)\\\le & {} \frac{1 + \delta _{k_0+1}}{1 +\delta _{k_0}} \cdot (1 + \delta _{k_0}) \phi _{k_0}(x) \\\le & {} \frac{1 + \delta _{k_0+1}}{1 +\delta _{k_0}} \cdot \nu (x) \le \frac{1 + \delta _{k_0+1}}{1 +\delta _{k_0}} {\mathop {<}\limits ^{(2.2)}} 1 - 2 \theta _{k_0+1}. \end{aligned}$$

It remains to prove that there exists a finite subset ${\mathfrak {F}}_x$ of $[\Gamma ]^{<\omega }$ such that, for each $A \in [\Gamma ]^{<\omega } {\setminus } {\mathfrak {F}}_x$ with $|A| \le k_0$, we have that $(1 + \delta _{|A|})^2 \Vert Ax\Vert _p \le 1 - 2 \theta _{|A|}$. Suppose that this is not the case. Then, there exists a sequence of mutually distinct sets $(A_j)_{j=1}^{\infty } \subseteq \Gamma$ with $|A_j|\le k_0$ for every $j\in \mathbb {N}$ and such that

$$\begin{aligned} (1 + \delta _{|A_j|})^2 \Vert A_j x\Vert _p > 1 - 2 \theta _{|A_j|}. \end{aligned}$$

(2.11)

Up to passing to a subsequence, we can assume that there is $k\in \mathbb {N}$ such that $|A_j|=k$ for every $j\in \mathbb {N}$. Hence, since all the sets $A_j$ have the same cardinality, we can apply the Delta System Lemma for countable families (see, for instance, [16, p. 167]). Therefore, there exist a subsequence of $(A_j)_{j=1}^{\infty }$, still denoted by $(A_j)_{j=1}^{\infty }$, and a set $\Delta \in [\Gamma ]^{<\omega }$ (possibly empty) such that $A_i\cap A_j=\Delta$ for every $i\ne j$. Notice that $|\Delta | \le k-1$ because the sets $A_j$ are mutually distinct. Now, since the elements of the sequence $(A_j \setminus \Delta )_{j=1}^{\infty }$ are disjoint and $x \in \ell _p(\Gamma )$, we have that

$$\begin{aligned} \Vert A_j x\Vert _p^p = \Vert \Delta x\Vert _p^p + \Vert (A_j\setminus \Delta )x\Vert _p^p \rightarrow \Vert \Delta x\Vert _p^p \qquad \text {as} \ j\rightarrow \infty . \end{aligned}$$

Thus, taking the limit when $j \rightarrow \infty$ in (2.11), we get that

$$\begin{aligned} (1+\delta _k)^2 \Vert \Delta x\Vert _p \ge 1-2\theta _k \end{aligned}$$

(2.12)

which will yield a contradiction. Indeed, since $(1+\delta _{|\Delta |})^2 \Vert \Delta (x)\Vert _p \le \nu (x) \le 1$ and $|\Delta | \le k-1$, (2.12) yields

$$\begin{aligned} 1\!-\!2\theta _k\!\le \! \frac{(1+\delta _k)^2}{(1\!+\! \delta _{|\Delta |})^2} \cdot (1\!+\!\delta _{|\Delta |})^2 \Vert \Delta x\Vert _p\!\le \! \left( \frac{1 + \delta _k}{1 + \delta _{k-1}} \right) ^2 \!\le \!\frac{1 + \delta _k}{1 + \delta _{k-1}} {\mathop {<}\limits ^{(2.2)}} 1 \!-\! 2 \theta _k \end{aligned}$$

which is absurd. This concludes the proof of Claim 2.1. $\square$

From this point on, we just have to glue together the ingredients in the standard way. Even if the argument is the same as in [4, Theorem 4.1], we give the details for the sake of being self-contained.

For every $n\in \mathbb {N}_0$, let $\rho _n:\mathbb {R}\rightarrow [0,\infty )$ be a $C^{\infty }$-smooth, even, and convex function such that $\rho _n\equiv 0$ on $[0,1- \theta _n^2]$ and $\rho _n(1)=1$. Then, for every $n\in \mathbb {N}_0$, we have that $\rho _n(t) \le 1$ if and only if $|t|\le 1$. Define $\Psi :\mathcal {Y}\rightarrow [0,\infty ]$ by

$$\begin{aligned} \Psi (x){:}{=}\sum _{|A|<\infty } \rho _{|A|} \Big ((1+\delta _{|A|})^2 \cdot (1+\theta _{|A|})\cdot \Vert Ax\Vert _{s,A} \Big ) \qquad (x \in \mathcal {Y}). \end{aligned}$$

(2.13)

Let $x \in \mathcal {Y}$ with $\nu (x) \le 1$ and take $\mathcal {O}_x$ and ${\mathfrak {F}}_x$ as in Claim 2.1. If $y\in \mathcal {O}_x$ and $A \in [\Gamma ]^{<\omega } {\setminus } {\mathfrak {F}}_x$, we have that

$$\begin{aligned} (1 + \delta _{|A|})^2 (1 + \theta _{|A|}) \Vert Ay\Vert _{s,A} {\mathop {\le }\limits ^{(2.6)}} (1 - \theta _{|A|})(1 + \theta _{|A|}) = 1 - \theta _{|A|}^2 \end{aligned}$$

which implies that $\rho _{|A|} \Big ((1+\delta _{|A|})^2 \cdot (1+\theta _{|A|})\cdot \Vert Ay\Vert _{s,A} \Big ) = 0$. Hence, on the set $\mathcal {O}_x$, only the finitely many summands with $A\in {\mathfrak {F}}_x$ are different from 0. Also, note that each summand in (2.13) is $C^\infty$-smooth on $\mathcal {Y}$ (each summand vanishes in a neighbourhood of 0, so $\Psi$ is also differentiable there). Therefore, $\Psi$ is (real-valued and) $C^\infty$-smooth and LFC on the open set $\mathcal {O}$ defined by

$$\begin{aligned} \mathcal {O}\,{:}{=}\bigcup \Big \{\mathcal {O}_x:x \in \mathcal {Y}, \ \nu (x)\le 1 \Big \}. \end{aligned}$$

Next, we note that the convex and symmetric set $\{\Psi <1\}$ is contained in $\{\nu \le 1\} \subseteq \mathcal {O}$ (hence it is also open, as $\Psi$ is continuous on $\mathcal {O}$). Indeed, if $\Psi (x)<1$, then $\rho _{|A|} \Big ((1+\delta _{|A|})^2 \cdot (1+\theta _{|A|})\cdot \Vert Ax\Vert _{s,A} \Big ) \le 1$ for every $A \in [\Gamma ]^{<\omega }$. So the properties of the functions $\rho _n$ give

$$\begin{aligned} (1 + \delta _{|A|})^2 \Vert Ax\Vert _p {\mathop {\le }\limits ^{(2.4)}} (1 + \delta _{|A|})^2(1 + \theta _{|A|}) \Vert Ax\Vert _{s,A} \le 1 \end{aligned}$$

and $\nu (x)\le 1$. Moreover, the set $\{\Psi \le 1-\theta _1\}$ is closed in $\mathcal {Y}$ by the lower semi-continuity of $\Psi$. Hence, a standard consequence of the Implicit Function Theorem (see [4, Lemma 2.5] or [10, Chapter 5, Lemma 23]) implies that the Minkowski functional ${\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$ of $\{\Psi \le 1-\theta _1\}$ is an equivalent $C^\infty$-smooth and LFC norm on $\mathcal {Y}$.

It remains to check that ${\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }$ approximates $\left\| \cdot \right\| _p$. For this aim, we first show that $\Psi (x)=0$ whenever $\nu (x)\le \frac{1-\theta _1}{1+\theta _1}$. Indeed, if $x\in \mathcal {Y}$ satisfies $\nu (x)\le \frac{1-\theta _1}{1+\theta _1}$, then for every $A \in [\Gamma ]^{<\omega }$, we have that

$$\begin{aligned} (1 + \delta _{|A|})^2 (1 + \theta _{|A|}) \Vert Ax\Vert _{s,A}&{\mathop {\le }\limits ^{(2.4)}} (1 + \delta _{|A|})^2(1 + \theta _{|A|}) \Vert Ax\Vert _p \\&\le \frac{1 - \theta _1}{1 + \theta _1} \cdot (1 + \theta _{|A|}) \le 1 - \theta _{|A|}^2. \end{aligned}$$

Hence $\Psi (x) = 0$. Combining this inclusion with the inclusion $\{\Psi \le 1\} \subseteq \{\nu \le 1\}$, which was proved above, we have that

$$\begin{aligned} \nu \le {\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }\le \frac{1+\theta _1}{1-\theta _1} \nu . \end{aligned}$$

Together with (2.5) we finally reach the conclusion that

$$\begin{aligned} \left\| \cdot \right\| _p\le {\left| \hspace{-1.0625pt}\left| \hspace{-1.0625pt}\left| \cdot \right| \hspace{-1.0625pt}\right| \hspace{-1.0625pt}\right| }\le \frac{1+\theta _1}{1- \theta _1} \cdot (1+\delta _1)^2 \left\| \cdot \right\| _p {\mathop {\le }\limits ^{(2.3)}}(1+\varepsilon ) \left\| \cdot \right\| _p. \end{aligned}$$

$\square$

3 Operator ranges in $\ell _{1}(\Gamma )$

In this short section, we discuss the problem of whether $\ell _p(\Gamma )$ (for $1\le p<\infty$) contains a dense operator range with a $C^\infty$-smooth and LFC norm. As we mentioned already in the Introduction, operator ranges bear a certain form of completeness, that is not shared by all normed spaces (for example, recall the standard fact that there is no complete norm on a normed space of dimension less than continuum). For this reason, building a smooth norm on an operator range is more complicated than building one on a general normed space.

When $p>1$ we have observed before that $\mathcal {Y}_p$ contains a dense operator range, since there is a continuous linear injection of $\ell _1(\Gamma )$ in $\mathcal {Y}_p$. Therefore, $\ell _p(\Gamma )$ contains a dense operator range with a $C^\infty$-smooth and LFC norm (by the Main Theorem). Here it is perhaps worth noting that, by the Closed Graph Theorem, $\ell _p(\Gamma )$ is not an operator range in $\ell _1(\Gamma )$, for $p<1$. (Indeed, if $T:\mathcal {Z}\rightarrow (\ell _p(\Gamma ),\left\| \cdot \right\| _1)$ is a continuous bijection, where $\mathcal {Z}$ is a Banach space, then $T:\mathcal {Z}\rightarrow (\ell _p(\Gamma ),\left\| \cdot \right\| _p)$ has closed graph. Hence, by the Closed Graph Theorem for Fréchet spaces, $T:\mathcal {Z}\rightarrow (\ell _p(\Gamma ),\left\| \cdot \right\| _p)$ would be continuous, hence an isomorphism, which is impossible.)

Thus when $p=1$ the situation is different and the result actually depends on the cardinality of $\Gamma$. If $\Gamma$ is countable, then $\mathcal {Y}_1$ still contains a dense operator range. In fact, it is sufficient to find a continuous linear injection of $\ell _\infty$ into $\mathcal {Y}_1$, which is simply given by the map $T:\ell _\infty \rightarrow \ell _{1/2}$ defined by

$$\begin{aligned} (x(j))_{j=1}^\infty \mapsto \left( 2^{-j}\cdot x(j)\right) _{j=1}^\infty . \end{aligned}$$

Hence, $\ell _1$ also contains a dense operator range with a $C^\infty$-smooth and LFC norm. This ceases to be true for uncountable $\Gamma$, as the next result shows. It is essentially due to Rosenthal [19] and, in a slightly weaker form, it can also be found in [13, Lemma 3.8]. Yet, a direct proof is so short that we give it here for the sake of completeness.

Proposition 3.1

Let $\mathcal {Y}$ be a non-separable operator range in $\ell _1(\Gamma )$. Then $\mathcal {Y}$ contains an isomorphic copy of $\ell _1(\omega _1)$; in particular, $\mathcal {Y}$ admits no Fréchet smooth norm.

Proof

Let $\mathcal {Z}$ be a Banach space and $T:\mathcal {Z}\rightarrow \ell _1(\Gamma )$ be a bounded linear operator such that $\mathcal {Y}=T[\mathcal {Z}]$. Then ${\overline{\mathcal {Y}}}$ is a non-separable closed subspace of $\ell _1(\Gamma )$, hence it contains an isomorphic copy of $\ell _1(\omega _1)$, [15, (5) on p.185]. Let $(y_\alpha )_{\alpha <\omega _1}$ be equivalent to the canonical basis of $\ell _1(\omega _1)$, fix $\varepsilon >0$, and take $z_\alpha \in \mathcal {Z}$ with $\Vert y_\alpha - Tz_\alpha \Vert <\varepsilon$. If $\varepsilon >0$ is sufficiently small, the sequence $(Tz_\alpha )_{\alpha <\omega _1}$ is also equivalent to the canonical basis of $\ell _1(\omega _1)$ (see, e.g., [11, Lemma 5.2], or [14, Example 30.12]). Moreover, up to passing to an uncountable subset of $\omega _1$ and relabelling, we can assume that $(z_\alpha )_{\alpha <\omega _1}$ is a bounded set. Consequently, the linear map $S:\textrm{span}\{Tz_\alpha \} _{\alpha<\omega _1} \rightarrow \textrm{span}\{z_\alpha \} _{\alpha <\omega _1}$ defined by $Tz_\alpha \mapsto z_\alpha$ ($\alpha <\omega _1$) is bounded. So $T\mathord {\upharpoonright }_{\overline{\textrm{span}} \{z_\alpha \} _{\alpha <\omega _1}}$ is an isomorphic embedding, and we are done. $\square$