TITLE: Some question on Harish-Chandra height function
QUESTION [2 upvotes]: Let $G$ be a connected reductive group over a number field $F$ and fix a minimal parabolic subgroup $P_0$ of $G$. Let $K$ be a fixed good maximal compact subgroup of $G(\mathbb{A}_F)$ such that $G=PK$ for all standard parabolic subgroup $P=UM$. (here $U,M$ are the unipotent radical and Levi of $P$.
Then Harish-chandra defined a height function $H_G:G(\mathbb{A}_F) \to \mathcal{a}_G$. (here $\mathcal{a}_G$ is the real vector space spanned by coroots of $G$. For the precise definition of Height function, please see 16 page of the Arthur's book 'An introduction to the trace formula'.)
For a standard parabolic subgroup $P \subset G$, Harish-Chandra height function $H_P : G(\mathbb{A_F}) \to \mathbb{a}_P$ is defined by
$$H_P(g)=H_{M_P}(m)$$ where $g=umk$ where $u \in N_P, m\in M_P, k\in K$ and $G=N_PM_PK$ is the Iwasawa decomposition of $P$.
Let $G^1(\mathbb{A}_F)=\{g \in G(\mathbb{A}_F) \ | \ H_G(g)=0\}$.
Then I have two questions;

Is this true that $U_P(\mathbb{A}_F) \subset G^1(\mathbb{A}_F)$ for all standard parabolic subgroups $P \subset G$?
For an element $g \in G(\mathbb{A}_F)^1$, we decompose $g=umk$ using Iwasawa decomposition with respect to $P$ as above. Then $u$ should be $1$ and $m \in M_P(\mathbb{A}_F)^1$?

I think that both are true but I can't prove it.
Any comments are greatly welcome!

REPLY [1 votes]: Is it true that $U_P(\mathbb A) \subseteq G(\mathbb A_F)^1$ for standard parabolic subgroups $P \subseteq G$?

Yes. Let $x \in U_P(\mathbb A)$ be written as $x = umk$ where $u \in U_P(\mathbb A), m \in M_P(\mathbb A)$ and $k \in K$. Clearly $u = x, m = 1$ and $k = 1$. Then $H_P(x) = H_{M_P}(m) = 0$ so $x \in G(\mathbb A)^1$.
Edit: If $H_P(g) = 0$ then $H_G(g) = 0$.
Observe that if $P_1 \subseteq P_2$ then $M_{P_1} \subseteq M_{P_2}$ and the restriction homomorphism $X(M_{P_2}) \to X(M_{P_1})$ is injective [Clay notes $\S 5$]. Thus $H_{P_2}(g) = 0$, which is equivalent to $\chi(g) = 0 \ \forall \chi \in X(M_{P_2})$ is true whenever $\chi(g) = 0 \ \forall \chi \in X(M_{P_1})$, which is equivalent to $H_{P_1}(g) = 0$.

On the other hand, the second question is not true. For a quick counterexample, take $G = SL(2), P$ to be the minimal parabolic subgroup of upper triangular matrices and $g = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Note that $G(\mathbb A) = G(\mathbb A)^1$ and $g \in U_P(\mathbb A)$ but $g \neq 1$.


TITLE: Corollary of Lifting Lemma for Algebraic Closures
QUESTION [1 upvotes]: If $L$ is algebraic extension of a field $K$ and $F$ is an algebraically closed field. Then for any nonzero homomorphism $j: K \to F$, there exist $i: L \to F$ such that $i|_K = j$

(note $i|_K$ is the restriction of $i$ to K)
Now apparently the following is an immediate corollary:

For any nonzero homomorphisms $j:K \to L$, $i:K \to F$ with L algebraic over $j(K)$ and F algebraically closed, there exists $h: L \to F$ so that $h(j(a)) = i(a)$ for all $a\in K$.

If my notes are correct then the hint for this exercise is use Zorn, I did not see it, so I decided to think a little for myself. So in short I am wondering if this proof is correct or am I missing anything.
Proof:
Using the setup we can get a homomorphism $h': j(K) \to F$, where $h' = i \circ j'^{-1}$ and $j': K \to j(K)$, $j'(a) = j(a)$ for all $j\in K$.
So applying the theorem to h' we get our h.
$\forall a \in K$, $h(j(a)) = h'(j(a)) = i(a)$
Any feedback is greatly appreciated

REPLY [1 votes]: Yeah, if I'm not mistaken, the proof is right. Zorn's lemma is supposed to be used only for the main theorem above, not the corollary.
I think the proof should have a few more clarifications at some points, so I'd write

Since $j$ is injective, the map $j':K\rightarrow j(K)$ defined by $j'(a)=j(a)$ is an ismorphism, so then we form the map $h'=i\circ (j')^{-1}: j(K)\rightarrow F$. Also, $L/j(K)$ is algebraic. Applying the theorem to $h'$ we get our $h$. For all $a\in K$, $h(j(a)) = h'(j(a)) = i(a)$.


TITLE: How to prove that f(w)=0
QUESTION [1 upvotes]: Let I=[a,b] , let f:$\rightarrow$R is continuous on I,and assume f(a) < 0 and f(b)>0 .let W={x$\in$I:f(x)<0},and let w:=supW .prove that f(w)=0.
My work: since w=supW then either w$\in$W or w is a limit point of W. If w is in W then this question does not make any sense so w is a limit point of W and f(w)$\geq$0. Since w is a limit point there exist a sequence ($x_n$) in W such that it converges to w. But after that I cannot proceed.

REPLY [0 votes]: $$w=\sup W\implies w=\lim_{n\to+\infty}w_n\;\;$$
with $ w_n\in W$.
$$f(w_n)<0\implies \lim_{n\to+\infty}f(w_n)\le 0$$
$$\implies f(\lim_{n\to+\infty}w_n)\le 0$$
$$\implies \color{red}{f(w)\le 0}$$
On the other hand, for $ n$ great enough, $ f(w+\frac 1n)\ge 0$
because $w+\frac 1n \notin W$
thus
$$\lim_{n\to+\infty}f(w+\frac 1n)\ge 0$$
$$\implies f(\lim_{n\to+\infty}(w+\frac 1n))\ge 0$$
$$\implies \color{red}{f(w)\ge 0}$$


TITLE: Derivative of Hadamard product
QUESTION [5 upvotes]: What is the derivative of Hadamard product of two matrices with respect to one of them?
I.e. what is $D(AB)$ with respect to $A$?

REPLY [5 votes]: Consider the 6th-order tensor ${\mathbb M}$, whose components ${\mathbb M}_{ijklmn}$ are unity if $\,(i=k=m)$ and $(j=l=n),\,$ but zero otherwise.
This tensor can be used, along with the Frobenius Product (:), to rewrite the Hadamard Product ($\circ$) between two matrices as
$$\eqalign{
 B\circ A &= B:{\mathbb M}:A \cr\cr
}$$
This makes the finding the gradient (a 4th-order tensor) especially simple
$$\eqalign{
 F &= A\circ B = B\circ A = B:{\mathbb M}:A \cr\cr
dF &= B:{\mathbb M}:dA \cr\cr
\frac{\partial F}{\partial A} &= B:{\mathbb M} \cr
}$$


TITLE: How can we prove the identity $\sum_{r=0, r~ even}^{k}\binom{k}{r}\binom{r}{r/2}2^{k-r}=\binom{2k}{k}$?
QUESTION [2 upvotes]: We know that $$\sum_{r=0}^{k}{\binom{k}{r}}^2=\binom{2k}{k}$$ Can we prove such identity for the above case. I can check the identity for some numerical values. For example $k=4$ case satisfies it.

REPLY [0 votes]: We seek to prove the identity
$$\sum_{q=0}^{\lfloor n/2\rfloor}
{n\choose 2q} {2q\choose q} 2^{n-2q} = {2n\choose n}.$$
We have
$${n\choose 2q} {2q\choose q}
= \frac{n!}{(n-2q)! \times q! \times q!}
= {n\choose q} {n-q\choose n-2q}.$$
The LHS is
$$\sum_{q=0}^{\lfloor n/2\rfloor}
{n\choose q} {n-q\choose n-2q} 2^{n-2q}
= 2^n [z^n] (1+z)^n
\sum_{q=0}^{\lfloor n/2\rfloor}
{n\choose q} z^{2q} (1+z)^{-q} 2^{-2q}.$$
Here the coefficient extractor enforces the upper limit of the sum and
we obtain
$$2^n [z^n] (1+z)^n
\sum_{q\ge 0}
{n\choose q} z^{2q} (1+z)^{-q} 2^{-2q}
\\ = 2^n [z^n] (1+z)^n
\left(1+\frac{z^2}{4(1+z)}\right)^n
= 2^n [z^n] \left(1+z+\frac{z^2}{4}\right)^n
\\ = 2^n [z^n] \left(1+\frac{z}{2}\right)^{2n}
= 2^n {2n\choose n} \frac{1}{2^n}
= {2n\choose n}.$$
This sum also appeared at the following MSE link.
Addendum. In hindsight we may simplify this starting from
$${2n\choose n} = [z^n] (1+z)^{2n}
= [z^n] (1+2z+z^2)^n
= [z^n] \sum_{q=0}^n {n\choose q} z^q (2+z)^q
\\ = \sum_{q=0}^n {n\choose q}
[z^{n-q}] (2+z)^q
= \sum_{q=0}^n {n\choose q} {q\choose n-q} 2^{2q-n}
\\ = \sum_{q=0}^n {n\choose q} {n-q\choose q} 2^{n-2q}
= \sum_{q=0}^n {n\choose q} {n-q\choose n-2q} 2^{n-2q}.$$
This is what the first answer to appear used.


text:
\begin{document}
\title[Structure of the Lipschitz free $p$-spaces $\F_p(\Int^d)$ and $\F_p(\Rea^d)$]
{Structure of the Lipschitz free $p$-spaces $\F_p(\Int^d)$ and $\F_p(\Rea^d)$ for $0<p\le 1$}

\author[F. Albiac]{Fernando Albiac}
\address{Department of Mathematics, Statistics, and Computer Sciences--InaMat$^2$ \\
Universidad P\'ublica de Navarra\\
Campus de Arrosad\'{i}a\\
Pamplona\\
31006 Spain}
\email{fernando.albiac@unavarra.es}

\author[J. L. Ansorena]{Jos\'e L. Ansorena}
\address{Department of Mathematics and Computer Sciences\\
Universidad de La Rioja\\
Logro\~no\\
26004 Spain}
\email{joseluis.ansorena@unirioja.es}

\author[M. C\'uth]{Marek C\'uth}
\address{Faculty of Mathematics and Physics, Department of Mathematical Analysis\\
Charles University\\
186 75 Praha 8\\
Czech Republic}
\email{cuth@karlin.mff.cuni.cz}

\author[M. Doucha]{Michal Doucha}
\address{Institute of Mathematics\\
Czech Academy of Sciences\\
\v Zitn\'a 25\\
115 67 Praha 1\\
Czech Republic}
\email{doucha@math.cas.cz}

\begin{abstract} Our aim in this article is to contribute to the theory of Lipschitz free $p$-spaces for $0<p\le 1$ over the Euclidean spaces $\Rea^d$ and $\Int^d$. To that end, on one hand we show that $\F_p(\Rea^d)$ admits a Schauder basis for every $p\in(0,1]$, thus generalizing the corresponding result for the case $p=1$ by H\'ajek and Perneck\'a \cite{HP2014}*{Theorem 3.1} and answering in the positive a question that was raised in \cite{AACD2019}. Explicit formulas for the bases of both $\F_p(\Rea^d)$ and its isomorphic space $\F_p([0,1]^d)$ are given. On the other hand we show that the well-known fact that $\F(\Int)$ is isomorphic to $\ell_{1}$ does not extend to the case when $p<1$, that is, $\F_{p}(\Int)$ is not isomorphic to $\ell_{p}$ when $0<p<1$.
\end{abstract}

\thanks{F. Albiac acknowledges the support of the Spanish Ministry for Economy and Competitivity under Grant MTM2016-76808-P for \emph{Operators, lattices, and structure of Banach spaces}. F. Albiac and J.~L. Ansorena acknowledge the support of the Spanish Ministry for Science, Innovation, and Universities under Grant PGC2018-095366-B-I00 for \emph{An\'alisis Vectorial, Multilineal y Aproximaci\'on}. M.~C\'uth has been supported by Charles University Research program No. UNCE/SCI/023. M. Doucha was supported by the GA\v{C}R project EXPRO 20-31529X and RVO: 67985840.}

\maketitle

\section{Introduction and background}
\noindent Suppose $0<p\le 1$. Given a pointed $p$-metric space $\MM$ it is possible to construct a unique $p$-Banach space $\F_p(\MM)$ in such a way that $\MM$ embeds isometrically in $\F_{p}(\MM)$ via a canonical map denoted $\delta_\MM$, and for every $p$-Banach space $X$ and every Lipschitz map $f\colon \MM\to X$ with Lipchitz constant $\Lip(f)$ that maps the base point $0$ in $\MM$ to $0\in X$ extends to a unique linear bounded map $T_f\colon \F_p(\MM)\to X$ with $\Vert T_f\Vert =\Lip(f)$. The space $\F_p(\MM)$ is known as the \emph{Lipschitz free $p$-space} over $\MM$.
This class of $p$-Banach spaces provides a canonical linearization process of Lipschitz maps between $p$-metric spaces: any Lipschitz map $f$ from a $p$-metric space $\MM_1$ to a $p$-metric space $\MM_{2}$ which maps the base point in $\MM_1$ to the base point in $\MM_2$ extends to a continuous linear map $L_f\colon \F_p(\MM_1)\to \F_p(\MM_2)$ with $\Vert L_f\Vert \le \Lip(f)$.

Lipschitz free $p$-spaces were introduced in \cite{AlbiacKalton2009}, where they were used to provide for every for $0<p<1$ a couple of \emph{separable} $p$-Banach spaces which are Lipschitz-isomorphic without being linearly isomorphic. These spaces constitute a new family of $p$-Banach spaces which are easy to define but whose geometry is difficult to grasp. This task was undertaken by the authors in \cite{AACD2018} and continued in the articles \cites{AACD2019,AACD2020}.

Within this subject, it is specially interesting and challenging to understand the structure of Lipschitz free $p$-spaces over subsets of the metric space $\Rea$ (or more generally over $\Rea^d$ for $d\in\Nat$) endowed with the Euclidean distance (see \cite{AACD2018}*{Comments at the end of section \S5}). Although the papers \cites{AACD2019,AACD2020} do not focus on this kind of Lipschitz free $p$-spaces, they contain results that apply in particular to them. Let us gather the most significant advances concerning the geometry of Lipschitz free $p$-spaces over Euclidean spaces achieved in those papers. Some of them are explicitly stated in the articles \cites{AACD2019,AACD2020} (in which case we provide the reference), while others are straightforward consequences of more general results. In the list below we assume $0<p\le 1$.

\begin{enumerate}[label={\textbf{(A.\arabic*)}}, leftmargin=*,widest=iii]
\item\label{two:net} For every $d\in\Nat$ and every net $\NN$ in $\Rea^d$, $\F_p(\NN)\simeq\F_p(\Int^d)$ (\cite{AACD2019}*{Proposition~3.6}).

\item\label{two:basis} The space $\F_p([0,1])$ has a Schauder basis (\cite{AACD2019}*{Theorem~5.7}).

\item For every $d\in\Nat$, the space $\F_p(\Int^d)$ has a Schauder basis (\cite{AACD2019}*{Theorem~5.3}).

\item\label{three:isomorphismsRd} $\F_p(\Rea^d)\simeq \F_p([0,1])^d\simeq \F_p(\Rea_+^d)\simeq \F_p(S^d)\simeq \ell_p(\F_p(\Rea^d))$ for every $d\in\Nat$ (\cite{AACD2020}*{Theorem~4.15, Corollary~4.17 and Theorem~4.21}).

\item\label{three:isomorphismsNd} $\F_p(\Int^d)\simeq \F_p(\Nat^d) \simeq \ell_p(\F_p(\Int^d))$ for every $d\in\Nat$ (\cite{AACD2020}*{Theorems 5.8 and 5.12}).

\item\label{three:local} The spaces $\F_p(\Int^d)$ and $\F_p(\Rea^d)$, despite being non-isomorphic, have the same local structure (\cite{AACD2020}*{Corollary 5.14}).

\item\label{three:complemeted} For every $d\in\Nat$ there is a constant $C=C(p,d)$ such that for every $\MM\subset\NN\subset\Rea^d$, the space $\F_p(\MM)$ is $C$-complemented in $\F_p(\NN)$ (\cite{AACD2020}*{Corollary 5.3}).

\item\label{three:piproperty} For every $0<p\le 1$, every $d\in\Nat$ and every $\MM\subset\Rea^d$, $\F_p(\MM)$ has the $\pi$-property (\cite{AACD2020}*{Corollary 5.3}).

\item\label{three:lp} For every $\MM\subset\Rea^d$ infinite, there is $\NN\subset\MM$ such that $\F_p(\NN) \simeq \ell_p$ (\cite{AACD2020}*{Theorem 3.2}).

\end{enumerate}

Combining \ref{three:complemeted} with \ref{three:isomorphismsRd}, and using Pe{\l}czy\'nski's decomposition method (see, e.g., \cite{AlbiacKalton2016}*{Theorem 2.2.3}), yields that $\F_p(\NN)\simeq \F_p(\Rea^d)$ whenever the subset $\NN$ of $\Rea^d$ has non-empty interior. So, the research on Lipschitz free $p$-spaces over subsets of Euclidean spaces reduces to the following two main topics (with non-empty intersection).

\begin{enumerate}[label={\textbf{(Q.\alph*)}}, leftmargin=*]
\item\label{QA} The study of the geometry of the spaces $\F_p(\Rea^d)$.

\item\label{QB} To contrast the spaces $\F_p(\NN)$ for different subsets $\NN\subset\Rea^d$ with empty interior.
\end{enumerate}

As far as the topic \ref{QA} is concerned, the main question suggested by the previous work on the subject is whether \ref{two:basis} extends to higher dimensions. Within the scale of approximation properties, the existence of a Schauder basis is the most demanding one. Thus, this question also connects with \ref{three:piproperty}, which states in particular that $\F_p([0,1]^d)$ has the $\pi$-property.

The approximation properties of Lipschitz free spaces (i.e., Lipschitz free $p$-spaces for $p=1$) has attracted a lot of attention from the specialists from the upsurge of interest in the theory back 2003. We refer the reader to \cites{GodefroyKalton2003,Dalet2015,GodefroyOzawa2014,LP2013,AmbrosioPuglisi2016,PS2015,Dalet2015b,Godefroy2015,Godefroy2015b,WojFonf2008} for a non exhaustive list of papers containing positive answers to this question within different frameworks. In contrast, determining whether a given Lipschitz free space has a Schauder basis has shown to be a somewhat elusive task. To the best of our knowledge, the list of papers that achieve positive answers in this direction reduces to \cites{HP2014,HN2017,CuthDoucha2016,DoKa2020}.

As far as Lipschitz-free spaces over Euclidean spaces is concerned, it is known that $\F(\Rea^d)$ has a Schauder basis (see \cite{HP2014}*{Theorem 3.1}). One of the goals of this paper is to see whether analogous statements hold for the more general case of Lipschitz-free $p$-spaces for $p\in(0,1]$. In this spirit, here we extend this result by proving that $\F_p(\Rea^d)$ admits a Schauder basis for every $p\in(0,1]$, thus answering in the positive \cite{AACD2019}*{Question 6.5}. Moreover, exact formulas for the basis of both $\F_p(\Rea^d)$ and its isomorphic space $\F_p([0,1]^d)$ are provided. Section~\ref{Sect:Basis} will be devoted to take care of this (see Theorem~\ref{thm:basis01d} and Theorem~\ref{thm:basisRd}).

As for the topic \ref{QB}, it is natural to initiate the study with uniformly separated subsets. Note that if $\MM\subset \Rea^d$ is uniformly separated, then it is contained in a net $\NN$. Consequently, by \ref{two:net}, \ref{three:lp} and \ref{three:complemeted}, $\ell_p$ is complemented in $\F_p(\MM)$ and $\F_p(\MM)$ is complemented in $\F_p(\Int^d)$. Hence, if $\F_p(\Int^d)$ were isomorphic to $\ell_p$, applying Pe{\l}czy\'nski's decomposition method would give $\ell_p \simeq \F_p(\MM)$. So, our first task should be to determine whether $\F_p(\Int^d)$ is isomorphic to $\ell_p$ or not. It is known (\cite{NaorSchechtman2007}) that, for $d\ge 2$, $\F(\Int^d)$ is not isomorphic to $\ell_1$. Taking envelopes yields $\F_p(\Int^d)\not\simeq\ell_p$ for any $0<p\le 1$ and $d\ge 2$ (see \cite{AACD2018}*{Corollary 4.2}). On the other hand, it is known and easy to prove, that $\F(\Int)\simeq\ell_1$. More generally, we have $\F(\MM)\simeq\ell_1$ whenever $\MM$ is the closure of a zero-measure subset of $\Rea$ (see \cite{CuthDoucha2016}). This result was extended to $0<p<1$ by the authors replacing the Euclidean distance $|\cdot|$ on $\mathbb R$ with its anti-snowflaking $|\cdot|^{1/p}$. That is, we have $\F_p(\Int,|\cdot|^{1/p})\simeq\ell_p$ for every $0<p\le 1$. A question that implicitly arose from \cite{AACD2018} is whether the same holds for the Euclidean distance, i.e., whether $\F_p(\Int)$ is isomorphic to $\ell_p$ or not for $0<p<1$. Section~\ref{sect:PpNotlp} is devoted to providing a negative answer to this problem.  Note that  $\F_p(\Int,|\cdot|^{\alpha})\simeq\ell_p$ for $0<\alpha<1$ and $0<p\le 1$ (see \cite{AACD2020}*{comments preceding Question 8}). So, this new result exhibits a somewhat surprising discontinuity in the behavior of the Lipschitz free $p$-spaces over the family of $p$-metric spaces $(\Int,|\cdot|^\alpha)$
for $0<\alpha\le 1/p$.

\subsection{Terminology}
Throughout this article we use standard facts and notation from quasi-Banach spaces and Lipschitz free $p$-spaces over quasimetric spaces as can be found in \cite{AACD2018}. Nonetheless, we will record the notation that is most heavily used. A \emph{quasi-norm} on a vector space $X$ over the real field $\Rea$ is a map $\Vert \cdot\Vert\colon X\to[0,\infty)$ satisfying $\Vert x\Vert>0$ unless $x=0$, $\Vert t\, x\Vert=|t| \, \Vert x\Vert$ for all $t\in\Rea$ and all $x\in X$, and
\begin{equation}\label{eq:qn}
\Vert x+y\Vert\le \kappa (\Vert x\Vert +\Vert y\Vert), \quad x,\, y\in X.
\end{equation}
for some constant $\kappa\ge 1$.
The optimal constant such that \eqref{eq:qn} will be called the \emph{modulus of concavity} of $X$.

Let $0<p\le 1$. If $\Vert \cdot\Vert$ fulfils the condition
\[
\Vert x+y\Vert^p \le \Vert x\Vert^p+\Vert y\Vert^p, \quad x,\, y\in X,
\]
we say that $\Vert \cdot\Vert$ is a \emph{$p$-norm.} Any $p$-norm is a quasi-norm with modulus of concavity at most $2^{1/p-1}$. A quasi-norm induces a Hausdorff vector topology on $X$. If $X$ is a complete topological vector space, we say that $(X,\Vert \cdot\Vert)$ is a \emph{quasi-Banach space}. A $p$-Banach space will be a quasi-Banach space equipped with a $p$-norm.

A quasi-Banach space $X$ is said to have the \emph{bounded approximation property} (BAP for short) if there exists a net $(T_\alpha)_{\alpha\in\A}$ consisting of finite-rank linear operators with
\begin{equation*}
\sup_{\alpha\in\A} \Vert T_\alpha\Vert <\infty
\end{equation*}
that converges to $\Id_X$ uniformly on compact sets. If, moreover, each operator $T_\alpha$ is a projection we say that $X$ has the \emph{$\pi$-property}.

A \emph{Schauder basis} of a quasi-Banach space $X$ is a sequence $(x_n)_{n=1}^\infty$ in $X$ such that for every $x\in X$ there is a unique sequence $(a_n)_{n=1}^\infty$ with $x=\sum_{n=1}^\infty a_n\, x_n$. Associated to the Schauder basis $(x_n)_{n=1}^\infty$ the partial-sum projections $P_m\colon X\to X$, $m\in\Nat$, given by
\[
x=\sum_{n=1}^\infty a_n\, x_n\mapsto P_m(x)= \sum_{n=1}^m a_n\, x_n,
\]
are uniformly bounded. Therefore, if a quasi-Banach space $X$ has a Schauder basis, then it has both the BAP and the $\pi$-property. Conversely, given a sequence $(P_m)_{m=1}^\infty$ of linear maps from $X$ into $X$ such that $\sup_m \Vert P_m\Vert<\infty$, $\cup_{m=1}^\infty P_m(X)$ is dense in $X$, $\dim(P_m(X))=m$, and $P_m\circ P_n =P_{\min\{n,m\}}$ for all $n$, $m\in\Nat$, there is a Schauder basis whose associated projections are $(P_m)_{m=1}^\infty$. Namely, if for each $n\in\Nat$ we pick an arbitrary non-zero vector $x_n$ in the one-dimensional space $P_n(X)\cap \Ker(P_{n-1})$, then $(x_n)_{n=1}^\infty$ is such a Schauder basis.

We say that a quasi-Banach space $X$ is \emph{$K$-complemented} in $Y$ if there are bounded linear maps $S\colon X\to Y$ and $P\colon Y\to X$ with $P\circ S=\Id_X$ and $\Vert S\Vert\, \Vert T\Vert\le K$. If, moreover, $S\circ P=\Id_Y$, the spaces $X$ and $Y$ are said to be \emph{$K$-isomorphic.} In the case when $S$ is the inclusion map, that is, $P$ is a projection onto $X$, we say that $X$ is a $K$-complemented subspace of $Y$. If the constant $K$ is irrelevant, we simply drop it from the notation. We can define similar notions replacing the quasi-Banach spaces $X$ and $Y$ with metric, or quasi-metric, spaces $\MM$ and $\NN$, and replacing bounded linear maps with Lipschitz maps. In the ``metric case'', we will say that $\MM$ is a Lipschitz retract of $\NN$ with constant $K$, or that $\MM$ and $\NN$ are Lipschitz isomorphic with distortion at most $K$, respectively.

A subset $\NN$ of a metric space $(\MM,d)$ is said to be \emph{uniformly separated} if
\[
\inf\{ d(x,y)\colon x,y\in\NN, x\not=y\}>0.
\]
A \emph{net} is a uniformly separated set $\NN$ with $\sup_{x\in\MM} d(x,\NN) <\infty$.

We use the symbol $\Nat_*$ to denote the set consisting of all non-negative integers, i.e., $\Nat_*=\Nat\cup\{0\}$.  Given $n\in\Nat$ we will put $\Nat_n=\Int\cap[0,n]$.

\section{The $p$-Banach space $\F_p(\Int^d)$ is not isomorphic to $\ell_p$}\label{sect:PpNotlp}
\noindent Once we conjecture that two quasi-Banach spaces are not isomorphic, the best strategy for substantiating our guess is to come up with a feature that tells apart them. We find this wished-for property within the theory of locally complemented subspaces and $\LL_p$-spaces developed by Kalton in \cite{Kalton1984}. A subspace $X$ of a quasi-Banach space $Y$ is said to be \emph{locally $K$-complemented} in $Y$ for some $K>0$, if for every finite-dimensional space $V\subset Y$ and every $\varepsilon>0$ there is $P\colon V\to X$ with
\begin{equation*}
\Vert P\Vert \le K \text{ and }\Vert P(x)-x\Vert\le\varepsilon\Vert x\Vert, \quad x\in V\cap X.
\end{equation*}
If the constant $K$ is irrelevant, we say that $X$ is locally complemented in $Y$.

A quasi-Banach space $X$ is a \emph{$\LL_p$-space,} $0<p<\infty$, if it is isomorphic to a locally complemented subspace of $L_p(\mu)$ for some measure $\mu$.
This notion was introduced in \cite{Kalton1984} with the aim of developing an attractive definition of a $\LL_p$-space for $0<p<1$ as it is not clear whether $L_p$($0<p<1$) would satisfy the analogue of the classical Lindenstrass-Pe{\l}czy\'{n}ski definition. For $p=1$, the above definition is equivalent to the classical one, while for $1<p<\infty$ the only exception to the equivalence is that, with the definition used in this paper, Hilbert spaces become $\LL_p$-spaces.

\begin{Lemma}\label{lem:slpdf}
Let $0<p<\infty$, let $X$ be a quasi-Banach space and let $(V_\alpha)_{\alpha\in A}$ be an increasing net consisting of finite-dimensional subspaces of $X$ with $\overline{\cup_{\alpha\in A} V_\alpha}=X$. Suppose that there is $K\in(0,\infty)$ such that for every $\alpha\in A$ and every $\varepsilon>0$ there is $S\colon V_\alpha\to \ell_p$ and $T \colon \ell_p \to X$ with
\[
\Vert T\Vert\,\Vert S\Vert \le K \text{ and }\Vert T(S(x))-x\Vert\le\varepsilon\Vert x\Vert, \quad x\in V_\alpha.
\]
Then $X$ is an $\LL_p$-space.
\end{Lemma}

\begin{proof}
Let $V$ be a finite-dimensional subspace of $X$ and let $\varepsilon>0$. Set
\[
\varepsilon_0:=\min\left\{1, \frac{\varepsilon}{\kappa+2\kappa^2} \right\}.
\]
A standard argument yields $\alpha\in A$ and $J\colon V\to V_\alpha$ with $\Vert J(x)-x\Vert\le\varepsilon_0\Vert x\Vert$ for all $x\in V$. By hypothesis, there are $S\colon V_\alpha\to\ell_p$ and $T\colon \ell_p\to X$ such that $\Vert T\Vert \,\Vert S\Vert \le K$ and $\Vert T(S(x))-x\Vert \le \varepsilon_0\Vert x\Vert$ for all $x\in V_\alpha$. We have
\[
\Vert J \Vert =\Vert J -\Id_V+\Id_V\Vert
\le \kappa (\varepsilon_0+1)\le 2\kappa.
\]
Hence $\Vert T\Vert \,\Vert S\circ J\Vert \le \Vert T\Vert \,\Vert S\Vert \,\Vert J\Vert \le 2\kappa K$.
If $x\in V$,
\begin{align*}
\Vert T(S(J(x)))-x \Vert
&\le \kappa( \Vert T(S(J(x)))-J(x)\Vert + \Vert J(x)-x\Vert)\\
&\le \kappa (\varepsilon_0 \Vert J(x)\Vert+ \varepsilon_0 \Vert x \Vert )\\
&\le \kappa \varepsilon_0 ( 2 \kappa \Vert x\Vert + \Vert x \Vert )
=\varepsilon \Vert x\Vert.
\end{align*}
Appealing to \cite{Kalton1984}*{Theorem 6.1} finishes the proof.
\end{proof}

It is known that complemented subspaces inherit the property of being $\LL_p$-spaces (see \cite{Kalton1984}*{Proposition 3.3}). For the sake of clarity, we state and prove the  quantitative version of this result, which is the one we will need.
\begin{Lemma}\label{lem:FDSLp}Let $X$ be an $\LL_p$-space and let $\lambda\in[1,\infty)$. There is a constant $K=K(X,\lambda)$ such that for every  quasi-Banach space $Y$ which is $\lambda$-complemented in $X$, every finite-dimensional subspace $V$ of $Y$, and every $\varepsilon>0$ there are bounded linear maps $S\colon V\to \ell_p$ and $T\colon\ell_p\to Y$ with $\Vert S \Vert \, \Vert T\Vert \le K$ and $\Vert T\circ S-\Id_V\Vert\le\varepsilon$.
\end{Lemma}

\begin{proof}Let $J\colon Y\to X$ and $P\colon X\to Y$ be such that $\Vert J \Vert \, \Vert P\Vert \le \lambda$ and $P\circ J=\Id_Y$. By \cite{Kalton1984}*{Theorem 6.1} there are $S\colon J(V)\to \ell_p$ and $T\colon\ell_p\to X$ with $\Vert S \Vert \, \Vert T\Vert \le K_0$ and $\Vert T\circ S-\Id_{J(V)}\Vert\le\varepsilon/\lambda$, where $K_0\in[1,\infty)$ depends only on $X$. We have
\[
\Vert S\circ J|_V\Vert\,\Vert P\circ T\Vert
\le \Vert S\Vert\, \Vert J\Vert\,\Vert P\Vert\, \Vert T\Vert
\le \lambda K_0
\] and
\[
\Vert P\circ T\circ S\circ J|_V-\Id_V\Vert =\Vert P\circ(T\circ S-\Id_{J(V)})\circ J\Vert \le \varepsilon.\qedhere
\]
\end{proof}

Given a quasi-Banach space $X$ and $0<q\le 1$ the $q$-Banach envelope of $X$ is a pair $(\Env[q]{X},E_{X,q})$, where $\Env[q]{X}$ is a $q$-Banach space and $E_{X,q}\colon X\to \Env[q]{X}$ is a linear contraction, defined by the following universal property: for every bounded linear map $T\colon X\to Y$, where $Y$ is a $q$-Banach space, there is a unique linear map $T'\colon \Env[q]{X}\to Y$ with $\Vert T'\Vert\le \Vert T\Vert$ and $T=T'\circ E_{X,q}$. If $X$ and $Y$ are quasi-Banach spaces and $T\colon X\to Y$ is linear and bounded, there is a unique bounded linear map $\Env[q]{T}\colon \Env[q]{X}\to \Env[q]{Y}$ such that $\Env[q]{T}\circ E_{X,q}=E_{Y,q}\circ T$. Moreover, $\Vert \Env[q]{T}\Vert\le \Vert T\Vert$. For background on envelopes, see e.g.\@ \cite{AABW2019}*{\S 9}.

Banach envelopes inherit BAP. For further reference, we record this result.

\begin{Lemma}\label{lem:BAPEnv}
Let $X$ be a quasi-Banach space with the BAP. Then $\Env[q]{X}$ has the BAP for $0<q\le 1$.
\end{Lemma}

Given $0<p\le 1$, a subset $\C$ of a vector space $V$ is said to be \textit{absolutely $p$-convex} if for any $x$ and $y\in \C$ and any scalars $\lambda$ and $\mu$ with $\ |\lambda|^p+|\mu|^p \le 1$ we have $\lambda \, x + \mu\, y \in \C$. We will denote by $\co_p(Z)$ the \textit{$p$-convex hull} of $Z\subset V$, i.e., the smallest absolutely $p$-convex set containing $Z$.

\begin{Lemma}\label{lem:compacthull}
Let $Z$ be a $q$-Banach space and $K\subset Z$ be relatively compact. Then the absolutely $q$-convex hull $\co_q(K)$ of $K$ is relatively compact.
\end{Lemma}

\begin{proof}
Since the map $(t,x)\mapsto tx$ is continuous and the unit sphere of the scalar field is compact, we can suppose that $t x\in K$ whenever $x\in K$ and $|t|=1$. If suffices to prove that $\co_q(K)$ possesses a finite $\varepsilon$-net for every $\varepsilon>0$. Let $\NN_0$ be a finite $(2^{-1/q}\varepsilon)$-net for $K$. Then
\[
\NN_1=\left\{\sum_{k=1}^m a_k x_k\colon x_k\in\NN_0,\, a_k\ge 0,\, \sum_{k=1}^m a_k^q\le 1\right\}
\]
is a $(2^{-1/q}\varepsilon)$-net for $\co_q(K)$

Enumerate $\NN_0=\{ y_j \colon 1\le j \le n\}$, and let $s\in\Nat$ be such that $ 2n \sup_j\Vert y_j\Vert^q \le \varepsilon^q 2^{sq}$. We will conclude the proof by showing that
\[
\left\{2^{-s} \sum_{j=1}^n b_j y_j\colon b_j\in\Nat_*,\,
\sum_{j=1}^n b_j^q\le 2^{sq}\right\}
\]
is a (finite) $(2^{-1/q}\varepsilon)$-net for $\NN_1$. Let $x\in\NN_1$. There is $(a_j)_{j=1}^n$ in $[0,\infty)$ such that $x=\sum_{j=1}^n a_j y_j$ and $\sum_{j=1}^n a_j^q\le 1$. For each $j=1$, \dots, $n$, let $b_j\in\Nat_*$ be such that $b_j\le 2^s a_j<b_j+1$. Then,
\[
\left\Vert x - 2^{-s} \sum_{j=1}^n b_j y_j\right\Vert^q\le
n 2^{-sq} \sup_j \Vert y_j\Vert^q \le \frac{\varepsilon^q}{2}.\qedhere
\]
\end{proof}

\begin{Theorem}\label{thm:compactEnv}
Let $T\colon X\to Y$ be a compact linear operator between quasi-Banach spaces $X$ and $Y$. Then the operator $\Env[q]{T}$ is compact for any $0<q\le 1$.
\end{Theorem}

\begin{proof}
Since the space of compact operators forms an ideal, we can assume that $Y$ is $q$-Banach, so that $\Env[q]{T}$ is a map from $\Env[q]{X}$ into $Y$ with $\Env[q]{T}\circ E_{q,X}=T$. Then, by construction,
\begin{align*}
\overline{\Env[q]{T}(B_{\Env[q]{X}})}
&= \overline{\Env[q]{T}(\co_q(E_{q,X}(B_X)))}\\
&= \overline{\co_q(\Env[q]{T}(E_{q,X}(B_X)))}\\
&=\overline{\co_q(T(B_X))},
\end{align*}
which is compact by Lemma~\ref{lem:compacthull}.
\end{proof}

Let $X$ be a subspace of a quasi-Banach space $Y$. We say that $X$ has the \emph{compact extension property} (CEP for short) in $Y$ if every compact operator $T\colon X\to Z$, where $Z$ is a quasi-Banach space, extends to a compact operator $\widetilde T\colon Y\to Z$. Let $0<q\le 1$. We say that $X$ has the \emph{compact extension property for $q$-Banach spaces} ($q$-CEP for short) in $Y$ if the compact extension property holds when $Z$ is a $q$-Banach space. If, moreover, we can ensure that $\Vert \widetilde T \Vert\le K \Vert T\Vert$ for some constant $K\in[1,\infty)$ (depending on $X$, $Y$ and $q$), we say that $X$ has the $q$-CEP in $Y$ with constant $K$. Notice that, by the Aoki-Rolewicz theorem, $X$ has the CEP in $Y$ if and only if $X$ has the $q$-CEP in $Y$ for every $0<q\le 1$. We have the following results in this respect.

\begin{Theorem}\label{thm:Lindens}
Let $X$ be a subspace of a quasi-Banach space $Y$. Suppose that $X$ has the $q$-CEP in $Y$ for some $0<q\le 1$. Then there is a constant $K\in[1,\infty)$ for which $X$ has the $q$-CEP in $Y$ with constant $K$.
\end{Theorem}

\begin{proof}
Although the proof of \cite{Lind1964}*{Theorem 2.2} was done for Banach spaces, the arguments therein apply to our case without any important modifications.
\end{proof}

\begin{Theorem}\label{thm:CEPvsLC}
Let $0<q\le 1$ and $Y$ be a $q$-Banach space with the BAP. Let $X$ be a closed subspace of $Y$. The following are equivalent.
\begin{enumerate}[label={(\roman*)}, leftmargin=*, widest=iii]
\item\label{thm:CEPvsLCitem1} $X$ is locally complemented in $Y$.
\item\label{thm:CEPvsLCitem2} $X$ has the BAP and the CEP in $Y$.
\item\label{thm:CEPvsLCitem3} $X$ has the BAP and the $q$-CEP in $Y$.
\end{enumerate}
\end{Theorem}

\begin{proof}
Although \cite{Kalton1984}*{Theorem 5.1} only establishes the equivalence between \ref{thm:CEPvsLCitem1} and \ref{thm:CEPvsLCitem2}, the very same proof gives that \ref{thm:CEPvsLCitem3} implies \ref{thm:CEPvsLCitem1}. So, taking into account Theorem~\ref{thm:Lindens}, \ref{thm:CEPvsLCitem3} is equivalent to \ref{thm:CEPvsLCitem1} and \ref{thm:CEPvsLCitem2}.
\end{proof}

\begin{Theorem}\label{lem:lcenv}
Let $X$ be a subspace of a quasi-Banach space $Y$. Denote by $J$ the inclusion of $X$ into $Y$. Suppose that $X$ has the BAP and that $X$ has the $q$-CEP in $Y$ for some $0<q\le 1$. Then $\Env[q]{J}$ is an isomorphic embedding and $\Env[q]{J}(\Env[q]{X})$ has the $q$-CEP in $\Env[q]{Y}$.
\end{Theorem}

\begin{proof}
Use Theorem~\ref{thm:Lindens} to pick $K$ such that $X$ has the $q$-CEP in $Y$ with constant $K$. Let $Z$ be a $q$-Banach space and let $T\colon\Env[q]{X}\to Z$ be a compact operator. Since $T\circ E_{X,q}$ is also compact, the compact extension property and Theorem~\ref{thm:compactEnv} yield a compact operator $S\colon \Env[q]{Y}\to Z$ with $\Vert S \Vert \le K \Vert T\Vert$ and such that the diagram
\[
\xymatrix{Y \ar[r]^{E_{Y,q}} & \Env[q]{Y} \ar[dr]^{S} & \\
X \ar[r]_{E_{X,q}} \ar[u]^{J} & \Env[q]{X} \ar[r]_{T} & Z}
\]
commutes. Since $E_{X,q}(X)$ is dense in $\Env[q]{X}$, merging this commutative diagram with
\[
\xymatrix{Y \ar[r]^{E_{Y,q}} & \Env[q]{Y} \\
X \ar[r]_{E_{X,q}} \ar[u]^{J} & \Env[q]{X} \ar[u]_{\Env[q]{J}}}
\]
yields that the diagram
\[
\xymatrix{\Env[q]{Y} \ar[dr]^{S} & \\
\Env[q]{X} \ar[r]_{T} \ar[u]^{\Env[q]{J}} & Z}
\]
commutes. It remains to show that $\Env[q]{J}$ is an isomorphic embedding. To that end, use Lemma~\ref{lem:BAPEnv} to pick $C$ such that $\Env[q]{X}$ has that BAP with constant $C$. Let $x\in \Env[q]{X}$ and $\varepsilon>0$. There is a linear operator $T\colon \Env[q]{X}\to \Env[q]{X}$ with finite-dimensional range such that $\Vert x-T(x)\Vert \le \varepsilon$ and $\Vert T\Vert \le C$. Since $T$ is compact, there is $S\colon \Env[q]{Y}\to \Env[q]{X}$ such $S\circ \Env[q]{J}=T$ and $\Vert S\Vert \le C K$. We have
\[
\Vert x \Vert^q \le \Vert x-T(x)\Vert^q + \Vert S (\Env[q]{J}(x))\Vert^q
\le \varepsilon^q+ C^q K^q \Vert \Env[q]{J}(x)\Vert^q.
\]
Letting $\varepsilon$ tend to zero we obtain $\Vert x \Vert \le CK \Vert \Env[q]{J}(x)\Vert$.
\end{proof}

\begin{Theorem}\label{thm:EnvSLpIsSLq}
Let $0<p\le q \le 1$ and $X$ be a separable $\LL_p$-space with the BAP. Then $\Env[q]{X}$ is isomorphic to a locally complemented subspace of $\ell_q$ and has a Schauder basis.
\end{Theorem}

\begin{proof}
By \cite{Kalton1984}*{Theorem 6.4}, $X$ is isomorphic to a locally complemented subspace $Y$ of $\ell_p$, and $Y$ has a Schauder basis. By Theorem~\ref{thm:CEPvsLC}, $Y$ has CEP in $\ell_p$. Hence, by Theorem~\ref{lem:lcenv}, $\Env[q]{X}$ is isomorphic to a subspace $Z$ of $\ell_q$ and has the $q$-CEP in $\ell_q$. As it is clear that $\Env[q]{Y}$ (and so  $\Env[q]{X}$) has a Schauder basis, applying once again Theorem~\ref{thm:CEPvsLC} completes the proof.
\end{proof}

The following straightforward consequence of Theorem~\ref{thm:EnvSLpIsSLq} partially solves \cite{AAW2020}*{Question 4.18}.
\begin{Corollary}\label{cor:EnvLpisLq}
Let $0<p\le q \le 1$ and $X$ be a separable $\LL_p$-space with the BAP. Then $\Env[q]{X}$ is a $\LL_q$-space.
\end{Corollary}

We are ready to prove the main results of this section.
\begin{Theorem}\label{thm:FpMNotSLp}Let $\MM$ be a $p$-metric space, $0<p<1$. Suppose that there is a constant $C$ such that $\Nat_n$ is a Lipschitz retract of $\MM$ with constant $C$ for all $n\in\Nat$. Then $\F_p(\MM)$ is not an $\LL_p$-space. In particular, $\F_p(\MM)\not\simeq\ell_p$.
\end{Theorem}

\begin{Theorem}\label{thm:FpMNotSLpB}
Let $\MM$ be a metric space. Suppose that there is a constant $K$ such that, for all $n\in\Nat$, $\Nat_n$ Lipschitz-isomorphically embeds in $\MM$ with distortion at most $K$. Then $\F_p(\MM)$ is not an $\LL_p$-space. In particular, $\F_p(\MM)\not\simeq\ell_p$.
\end{Theorem}

\begin{proof}[Proof of Theorems~\ref{thm:FpMNotSLp} and \ref{thm:FpMNotSLpB}]
Since $\Rea$ is a doubling metric space, there is $D\ge 1$ such that every subset of $\Rea$, in particular $\Nat_n$ for all $n\in\Nat$, is a doubling metric space with doubling constant $D$. Thus, under the assumptions in Theorem~\ref{thm:FpMNotSLpB}, applying \cite{AACD2020}*{Theorem 5.1} yields a constant $K$ such that $\F_p(\Nat_n)$ is $K$-complemented in $\F_p(\MM)$ for all $n\in\Nat$.
By \cite{AACD2018}*{Lemma 4.19}, this holds under the assumptions in Theorem~\ref{thm:FpMNotSLp} as well. Since
\[
\Dy_k=\{ x\in [-k,k] \colon 2^k x\in\Int\}
\]
is a doubling metric space with constant $D$ for all $k\in\Nat$, applying again \cite{AACD2020}*{Theorem 5.1} yields a constant $C$ such that the linearization of the inclusion of $\Dy_k$ into $\Rea$ is a $C$-isomorphic embedding for all $k\in\Nat$. Taking into account that $\Dy_k$ is Lipschitz isomorphic to $\Nat_{k 2^{k+1}}$ with distortion $1$ we infer the existence of a constant $K_1$ such that the finite-dimensional subspace
\[
V_k:=\spn( \delta_\Rea(x) \colon x \in \Dy_k)
\]
of $\F_p(\Rea)$ is $K_1$-complemented in $\F_p(\MM)$ for all $k\in\Nat$.

Suppose by contradiction that $\F_p(\MM)$ is an $\LL_p$-space. Then, by Lemma~\ref{lem:FDSLp}, there is a constant $K_2$ such that for all $k\in\Nat$ and $\varepsilon>0$ there are
linear bounded maps $S \colon V_k\to \ell_p$ and $T \colon\ell_p\to V_k\subset\F_p(\Rea)$ with $\Vert S \Vert \, \Vert T \Vert \le K_2$ and $\Vert T\circ S - \Id_{V_k} \Vert \le \varepsilon$.
Since the set $\Dy$ consisting of all dyadic rationals is dense in $\Rea$,
\[
\cup_{k=1}^\infty V_k=\spn(\delta_\Rea(x) \colon x\in\Dy)
\]
is a dense subspace of $\F_p(\Rea)$. Applying Lemma~\ref{lem:slpdf} yields that $\F_p(\Rea)$ is an $\LL_p$-space. Since $\F_p(\Rea)$ has the BAP (see \ref{three:piproperty}) combining Theorem~\ref{thm:EnvSLpIsSLq} with  \cite{AACD2018}*{Proposition 4.20} yields that $\F(\Rea)$ is isomorphic to a subspace of $\ell_1$. Using that $\F(\Rea)$ is isometric to $L_1(\Rea)$ and that $\ell_2$ is a subspace of $L_1(\Rea)$, we obtain that $\ell_2$ is isomorphic to a subspace of $\ell_1$, an absurdity.

For the last part of the statements, we just need to note that $\ell_p$ is an $\LL_p$-space.
\end{proof}

\begin{Corollary}\label{thm:notEllp}
Let $\MM$ be a metric space containing a subset which is Lipchitz isomorphic either to $[0,1]$ or to $\Nat$. Then $\F_p(\MM)$ is not a $\LL_p$-space for any $0<p<1$.
\end{Corollary}

\begin{proof}
Just notice that $\{ k/n \colon k \in\Nat_n\}\subset[0,1]$ is Lipschitz isomorphic to $\Nat_n$ with distortion $1$, and apply Theorem~\ref{thm:FpMNotSLpB}.
\end{proof}

\begin{Corollary}
Let $X$ be a $p$-Banach space ($0<p<1$) with nontrivial dual. Then $\F_p(X)$ is not a $\LL_p$-space.
\end{Corollary}

\begin{proof}By Corollary~\ref{thm:notEllp}, $\F_p(\Rea)$ is not a $\LL_p$-space.  Since, by assumption, $\Rea$ is a complemented subspace of $X$, by \cite{AACD2018}*{Lemma 4.19}  it follows that $\F_p(\Rea)$ is a complemented subspace of
$\F_p(X)$. Consequently, $\F_p(X)$ is a not an $\LL_p$-space either.
\end{proof}

\section{Schauder bases in $\F_p([0,1]^d)$ and $\F_p(\Rea^d)$}\label{Sect:Basis}
\noindent
The basic idea for building Schauder bases for $\F_p([0,1]^d)$ and $\F_p(\Rea^d)$ comes, on one hand, from \cite{LP2013}, where the authors present a canonical way of extending linearly Lipschitz functions on $d$-dimensional hypercubes, and on the other hand from \cite{AACD2019}, where a method for building linear projections on $\F_p([0,1])$ is given.

Fix $d\in\Nat$. Given $R>0$ and $w\in\Int^d$, we denote by $Q^d_{w,R}$, the hypercube having edge-length $R$ and with vertices in the points
\[
V^d_{w,R}=\{ R w + R\varepsilon \colon \varepsilon\in\{0,1\}^d\},
\]
That is, if $w=(w_i)_{i=1}^d$,
\[
Q^d_{w,R}=\co[V^d_{w,R}]=\prod_{i=1}^d [R w_i, Rw_i+R].
\]
For $R>0$ fixed, the set of hypercubes
\[
\QQ^d_R= \{ Q^d_{w,R} \colon w\in\Int^d\}
\]
tessellates the space $\Rea^d$. If $Q\in\QQ^d_R$, we denote by $\VV(Q)$ its set of vertices, that is $\VV(Q^d_{w,R})=V^d_{w,R}$ for every $w\in\Int^d$. We have
\[
\bigcup_{Q\in \QQ^d_R} \VV(Q)=\VV^d_R:=\{ R w \colon w\in\Int^d\}.
\]

We shall define a fuzzy pull back of $\Rea^d$ into the set of vertices $\VV^d_R$. Given $x\in[0,1]$ and $w\in\Int$ we set
\[
x^{(w)}
=\begin{cases} x & \text{ if } w=1, \\ 1-x & \text{ if } w=0, \\ 0 & \text{ if } w\in\Int\setminus\{0,1\}.\end{cases}
\]
Given $x=(x_i)_{i=1}^d\in[0,1]^d$ and $w=(w_i)_{i=1}^d\in\Int^d$ we put
\[
x^{(w)}=\prod_{i=1}^d x_i^{(w_i)}.
\]

\begin{Lemma}\label{lem:FPB}
Let $d\in\Nat$ and $R>0$. There is a mapping
\[
\Lambda=(\Lambda(v,\cdot))_{v\in\VV^d_R} \colon \Rea^d \to [0,1]^{\VV^d_R}
\]
such that $\Lambda(Ru,Rw+Rx) = x^{(u-w)}$ for every $x\in [0,1]^d$ and $u,w\in\Int^d$.
\end{Lemma}

\begin{proof}
Since the function $x\mapsto Rw+Rx$ maps $[0,1]^d$ onto $Q^d_{w,R}$, if such a function $\Lambda$ exists, it is unique. By dilation, it suffices to consider the case $R=1$. If $\Lambda$ is as desired in the one-dimensional case, then
\[
\Lambda(u,\cdots)=\Lambda(u_1,\cdot)\otimes \cdots \otimes\Lambda(u_i,\cdot)\otimes \cdots \otimes \Lambda(u_d,\cdot), \quad u=(u_i)_{i=1}^d\in\Int^d,
\]
is as desired in the $d$-dimensional case. Hence, we can also assume that $d=1$. To prove the result in this particular case, we must check that, given $w\in\Int$, the function given for $x\in\Rea$ by
\[
x\mapsto (x-w)^{(u-w)} \text{ if } u\in\Int \text{Â and }Â w\le x \le w+1
\]
is well-defined. Suppose that $w\le x \le w+1$ and $v\le x \le v+1$ with $v$, $w\in\Int$. Assume without lost of generality that $v<w$. Then $x=w=v+1$. Since $x-w=0$, we have $ (x-w)^{(u-w)}=0$ unless $u-w=0$, in which case $ (x-w)^{(u-w)}=1$. Since $x-v=1$, we have $(x-v)^{(u-v)}=0$ unless $u-v=1$, in which case $(x-v)^{(u-v)}=1$. Since $u-w=u-v-1$ for every $u\in\Int$, we are done.
\end{proof}

\begin{Definition}\label{def:FPB}
Given $R>0$ and $d\in\Nat$, we define
\[\Lambda_R^d=(\Lambda_{R}^d(v,\cdot))_{v\in\VV^d_R}\]
as the function provided by Lemma~\ref{lem:FPB}.
\end{Definition}

If $d=1$ we simply put $\Lambda_R=\Lambda^1_R$ and $\VV_R=\VV^1_R$. Given a finite set $A\subset \Nat$ we can carry out the above construction replacing the set $\{1,\dots,d\}$ with the set $A$. We will denote by $\VV^A_R$ the corresponding set of vertices and by $\VV^A_R$ the corresponding function defined as in Definition~\ref{def:FPB}.

Let us give an auxiliary lemma followed by some properties of the function $\Lambda_R^d$.

\begin{Lemma}\label{lem:ineq}
Let $x=(x_i)_{i=1}^d$, $y=(y_i)_{i=1}^d\in[0,1]^d$. Then
\[
\left| \prod_{i=1}^d x_i - \prod_{i=1}^d y_i\right|\le \vert x-y\vert_1.
\]
\end{Lemma}

\begin{proof}
We proceed by induction on $d$. For $d=1$ the result is obvious. Assume that $d\in\Nat$ and that the result holds for $d-1$. Then
\begin{align*}
\left |\prod_{i=1}^d x_i - \prod_{i=1}^d y_i\right|
&\le \left|\prod_{i=1}^d x_i - y_d \prod_{i=1}^{d-1} x_i\right| + \left| y_d \prod_{i=1}^{d-1} x_i - \prod_{i=1}^d y_i\right| \\
&= |x_d-y_d| \prod_{i=1}^{d-1} x_i + y_d \left|\prod_{i=1}^{d-1} x_i - \prod_{i=1}^{d-1} y_i\right| \\
&\le |x_d-y_d| + \sum_{i=1}^{d-1} |x_i-y_i| =\vert x-y\vert_1.\qedhere
\end{align*}
\end{proof}

\begin{Lemma}\label{prop:FPB}
Let $d\in\Nat$ and $S\ge R>0$ with $S/R\in\Int$. We have:
\begin{enumerate}[label={(\roman*)}, leftmargin=*,widest=iii]
\item\label{prop:FPB:1} $\Lambda^d_R(v,x)=0$ if $x\in Q\in\QQ^d_R$ and $v\notin \VV(Q)$.
\item\label{prop:FPB:2} $\Lambda^d_R(v,u)=\delta_{u,v}$ for every $u$, $v\in \VV^d_R$.
\item\label{prop:FPB:3} $\sum_{v\in\VV_R} \Lambda^d_R(v,x)=1$ for every $x\in\Rea^d$.
\item\label{prop:FPB:4} If there is $Q\in\QQ^d_R$ such that $x$, $y\in Q$, then
\[|\Lambda^d_R(v,x)-\Lambda^d_R(v,y)|\le R^{-1} \vert x-y\vert_1\] for every $v\in\VV^d_R$.
\item\label{prop:FPB:6}Let $(A,B)$ be a partition of $\{1,\dots,d\}$. Then
\[
\Lambda^d_R((u,v),(x,y))=\Lambda^A_R(u,x)\Lambda^B_R(v,y)
\]
for every $u\in\VV^A_R$, $v\in\VV^B_R$, $x\in\Rea^A$ and $y\in\Rea^B$.
\item\label{prop:FPB:5} $\Lambda^d_S(v,x)=\sum_{u\in\VV^d_R} \Lambda^d_S(v,u)\Lambda^d_R(u,x)$ for every $x\in\Rea^d$ and $v\in\VV^d_S$.
\end{enumerate}
\end{Lemma}

\begin{proof}
\ref{prop:FPB:1} is clear from the definition. \ref{prop:FPB:2} follows from the equality $0^{(0)}=1$. A straightforward induction on $d$ yields
\[
\sum_{w\in\Int^d} x^{(w)}=1, \quad x\in[0,1]^d,
\]
and \ref{prop:FPB:3} is clear from this identity. \ref{prop:FPB:4} is a consequence of Lemma~\ref{lem:ineq}. \ref{prop:FPB:6} is clear from the definition. In light of (v), in order to prove \ref{prop:FPB:5} it suffices to consider the case $d=1$. Given $x\in\Rea$ there are $u_0$, $u_1\in\VV_S$ and $v_0$, $v_1\in\VV_R$ with $u_0\le v_0 \le x<v_1\le u_1$, and we have $v_1=v_0+R$ and $u_1=u_0+S$. Suppose that $u\in\VV_S\setminus\{u_0,u_1\}$. Then $\Lambda_S(u,v)=0$ for $v\in\{v_0,v_1\} $. Since $\Lambda_R(v,x)=0$ for $v\in\VV_R\setminus\{ v_0,v_1\} $ we have
\[
\Gamma(u,x):=\sum_{v\in\VV_R} \Lambda_S(u,v)\Lambda_R(v,x)=0=\Lambda_S(u,x).
\]
Hence, considering also the symmetry $x\mapsto -x$, if suffices to prove the result in the case when $u=u_1$. We have
\begin{align*}
\Gamma(u_1,x)
&=\Lambda_S(u_1,v_0)\Lambda_R(v_0,x) + \Lambda_S(u_1,v_1)\Lambda_R(v_1,x)\\
&=\eta(x):=\frac{v_0-u_0}{S} \frac{v_1-x}{R} + \frac{v_1-u_0}{S} \frac{x-v_0}{R}.
\end{align*}
Since $\eta(u_0)=0$ we have $\eta(y)=\gamma(y-u_0)$ for all $y\in\Rea$, where
\[
\gamma=\left(-\frac{v_0-u_0}{SR}+\frac{v_1-u_0}{SR} \right)=\frac{1}{S}.
\]
Since $\Lambda_S(u_1,x)=(x-u_0)/S$ we are done.
\end{proof}

Although the previous auxiliary results are stated in terms of the $\ell_1$-norm, in this section we will consider $\Rea^d$ and its subsets equipped with the supremum norm $\|\cdot\|_\infty$.
\begin{Theorem}\label{thm:LipFPB}
Let $d\in\Nat$ and $0<p\le 1$. There is a constant $C=C(p,d)$ such that for every $R>0$ and every $\RR\subset \QQ^d_R$, if we denote $K=\cup_{Q\in\RR} Q$ and $V=\cup_{Q\in\RR} \VV(Q)$, and we choose an arbitrary point of $V$ as base point of both metric spaces, there is a $C$-Lipschitz map $r=r_{K,V}\colon K\to \F_p(V)$ such that
\[
r(x) = \sum_{v\in V} \Lambda^d_R(v,x) \, \delta_V(v), \quad x\in K.
\]
\end{Theorem}

\begin{proof}Let $Q\in\RR$ and $x$, $y\in Q$. Pick $u\in \VV(Q)$. By Lemma~\ref{prop:FPB},
\begin{align*}
\Vert r(x)-r(y)\Vert^p
&= R \left\Vert \sum_{v\in \VV(Q)\setminus\{ u\} } (\Lambda^d_R(v,x) - \Lambda^d_R(v,y)) \frac{\delta_V(v)-\delta_V(u)}{\vert v-u\vert_\infty} \right\Vert^p \\
&\le R \sum_{v\in \VV(Q)\setminus\{ u\} } | \Lambda^d_R(v,x) - \Lambda^d_R(v,y) |^p \\
&\le (2^d-1) \vert x-y\vert_1^p.
\end{align*}
Let $x=(x_i)_{i=1}^d\in K$ and $y=(y_i)_{i=1}^d\in K$. Pick $u=(u_i)_{i=1}^d$ and $w=(w_i)_{i=1}^d\in\Int^d$ such that $x\in Q^d_{R,u}$ and $y\in Q^d_{R,w}$. Define
\[
F=\{i\in\{1,\dots,d\} \colon u_i=w_i\}.
\]
For each $i\in G=\{1,\dots,d\}\setminus F$ there is $m_i\in\{u_i,u_i+1\}$ and $n_i\in\{w_i,w_i+1\}$ such that
\[
|y_i-x_i|=|y_i-Rn_i|+|Rn_i-Rm_i|+|Rm_i-x_i|.
\]
Suppose that $n_i=m_i$ for every $i\in G$. Define $z=(z_i)_{i=1}^d$ by
\[
z_i=\begin{cases} x_i & \text{ if } i\in F, \\ Rn_i=Rm_i & \text{ if } i\in G. \end{cases}
\]
We have $z\in Q^d_{R,u} \cap Q^d_{R,w}$. Consequently,
\begin{align*}
\Vert r(x)-r(y)\Vert^p
&\le \Vert r(x)-r(z)\Vert^p+\Vert r(z)-r(y)\Vert^p\\
&\le (2^d-1)( \vert x-z\vert_1^p+\vert z-y\vert_1^p)\\
&\le 2^{1-p} (2^d-1) (\vert x-z\vert_1+\vert z-y\vert_1)^p\\
&= 2^{1-p} (2^d-1) \vert x-y\vert_1^p.
\end{align*}
Suppose that $m_i\not=n_i$ for some $i$. Define $x'=(x_i')_{i=1}^d$ and $y'=(y_i')_{i=1}^d$ by
\[
x_i'=\begin{cases} Ru_i=Rw_i & \text{ if } i\in F, \\ Rm_i & \text{ if } i\in G, \end{cases}
\quad
y_i'=\begin{cases} Ru_i=Rw_i & \text{ if } i\in F, \\ R n_i & \text{ if } i\in G.
\end{cases}
\]
We have $x'\in V^d_{R,u}$, $y'\in V^d_{R,w}$, and $1\le \mu:=\max_{i\in G} |m_i-n_i|$. Hence
\begin{align*}
\Vert r(x)&-r(y)\Vert^p\\
&\le \Vert r(x)-r(x')\Vert^p+ \Vert r(x')-r(y')\Vert^p+ \Vert r(y')-r(y)\Vert^p\\
&\le (2^d-1)\vert x-x'\vert_1^p+\vert x'-y'\vert_\infty^p + (2^d-1)\vert y-y'\vert_1^p\\
&\le 2 (2^d-1) d R^p + \mu^p R^p \\
&\le (1 + 2 d (2^d-1) ) \mu^p R^p\\
&\le (1 + 2 d (2^d-1) ) \vert x-y\vert_\infty.
\end{align*}
This way the result holds with $C(p,d)= (1 + 2 d (2^d-1) )^{1/p}$.
\end{proof}

Next, we show that $\F_p(\Rea^d)$ has a Schauder basis for every $d\in\Nat$.
\begin{Lemma}\label{lem:17}
Let $\MM$ and $\NN$ be quasi-metric spaces. For $i=1$, $2$, let $f_i\colon\MM_i\subset\MM\to \NN$ be a Lipschitz function. Assume that
\begin{enumerate}[label={(\roman*)}, leftmargin=*,widest=iii]
\item $\MM=\MM_1\cup \MM_2$,
\item $f_1|_{\MM_1\cap \MM_2}=f_2|_{\MM_1\cap \MM_2}$, and
\item\label{it:17:3} There is a constant $C$ such that $d(x_1,\MM_1\cap \MM_2)\le Cd(x_1,x_2)$ for every $(x_1,x_2)\in\MM_1\times\MM_2$.
\end{enumerate}
Then, the map $f\colon\MM\to\NN$ defined by $f|_{\MM_i}=f_i$ for $i=1$, $2$ is Lipschitz. Moreover, if $k_\NN$ and $k_\MM$ are the quasi-metric constanst of $\NN$ and $\MM$, respectively, we have
\[
\Lip(f) \le k_\NN(C+k_\MM + Ck_\MM)\max_{i=1,2} \Lip(f_i).
\]
\end{Lemma}

\begin{proof}
Put $L=\max_{i=1,2} \Lip(f_i)$. Let $(x_1,x_2)\in\MM_1\times\MM_2$ and pick $x\in\MM_1\cap \MM_2$. Since $f_1(x)=f_2(x)$ we have
\begin{align*}
d_\NN(f_1(x_1), f_2(x_2))
&\le k_\NN ( d_\NN(f_1(x_1), f_1(x))+d_\NN(f_2(x), f_2(x_2)))\\
&\le k_\NN L ( d_\MM(x_1,x)+ d_\MM(x,x_2))\\
&\le k_\NN L ( (1+k_\MM) d_\MM(x_1,x)+ k_\MM d_\MM(x_1,x_2)).
\end{align*}
Since the element $x$ can be chosen so that $d_\MM(x_1,x)$ is arbitrarily close to $d_\MM(x_1,\MM_1\cap\MM_2)$, using the assumption \ref{it:17:3} yields the desired result.
\end{proof}

The following lemma exhibits a situation in which Lemma~\ref{lem:17} is useful that will occur several times throughout this section,.
\begin{Lemma}\label{lem:18}
Let $\MM$ and $\NN$ be quasi-metric spaces. For $i=1$, $2$, let $f_i\colon\MM_i\subset\MM\to \NN$ be a Lipschitz function. Assume that $\MM=\MM_1\cup \MM_2$ and that $f_1|_{\MM_1\cap \MM_2}=f_2|_{\MM_1\cap \MM_2}$. Suppose that there are constants $\lambda>0$ and $C\ge 1$ such that
\begin{enumerate}[label={(\roman*)},widest=ii]
\item $\MM$ is $\lambda$-separated, and
\item $d(x_1,\MM_1\cap \MM_2)\le C$ for every $x_1\in\MM_1$.
\end{enumerate}
Then, the map $f\colon\MM\to\NN$ defined by $f|_{\MM_i}=f_i$ for $i=1$, $2$ is Lipschitz. Moreover, if $k_\NN$ and $k_\MM$ are the quasi-metric constanst of $\NN$ and $\MM$, respectively,
\[
\Lip( f) \le k_\NN \left( \frac{C}{\lambda} ( 1+k_\MM) +k_\MM \right) \max_{i=1,2} \Lip (f_i).
\]
\end{Lemma}

\begin{proof}Let $(x_1,x_2)\in\MM_1\times\MM_2$ with $x_1\not=x_2$. We have
\[
d(x_1,\MM_1\cap \MM_2)\le \frac{C}{\lambda} d(x_1,x_2).
\]
Hence, the result follows from Lemma~\ref{lem:17}.
\end{proof}

\begin{Theorem}[cf. \cite{AACD2019}*{Theorem 5.7}]\label{thm:basis01d}
Let $0<p\le 1$ and $d\in\Nat$. Then $\F_p([0,1]^d)$ has a Schauder basis. In fact, if $V_n=[0,1]^d \cap 2^{-n}\Int^d$ for all $n\in\Nat_*$ and $V_{-1}=\{0\}$, and we put $\alpha(x)=n$ if $x\in V_n\setminus V_{n-1}$, then, any arrangement $(f(x_j))_{j=1}^\infty$ of the family
\[
f(x)=\delta_{[0,1]^d}(x)-\sum_{v\in V_{n-1}} \Lambda^d_{2^{-n+1}}(v,x) \delta_{[0,1]^d}(v) \quad n\in\Nat_*, \, x\in V_n\setminus V_{n-1}
\]
such that $(\alpha(x_j))_{j=1}^\infty$ is non-decreasing is a Schauder basis of $\F_p([0,1]^d)$.
\end{Theorem}

\begin{proof}
Set $\delta=\delta_{[0,1]^d}$ and $\delta_n=\delta_{V_n}$ for $n\in\Nat_*$.
By Theorem~\ref{thm:LipFPB}, there exist a constant $C$ and linear maps $T_n\colon \F_p([0,1]^d)\to \F_p(V_n)$ such that $\Vert T_n\Vert\le C$ and
\[
T_n(\delta(x)) = \sum_{v\in V_n} \Lambda^d_{2^{-n}}(v,x) \, \delta_{n}(v), \quad x\in [0,1]^d.
\]
Let $m$, $n\in\Nat_*$ with $m\le n$. Since $V_m\subset V_n$, we can consider the canonical map $L_{m,n}\colon \F_p(V_m)\to\F_p(V_n)$ associated to the inclusion. Consider also the canonical map $L_n\colon \F_p(V_n)\to\F_p([0,1]^d)$ associated to the inclusion of $V_n$ into $[0,1]^d$. Applying Proposition~\ref{prop:FPB}~\ref{prop:FPB:5} yields
\begin{itemize}[leftmargin=*]
\item $T_m\circ L_{n}\circ T_n=T_m$ and $T_n \circ L_m\circ T_m=L_{m,n}\circ T_m$.
\end{itemize}
Moreover, since $\cup_{n=0}^\infty V_n$ is dense in $[0,1]^d$,
\begin{itemize}[leftmargin=*]
\item if $X_n= L_n(T_n(\F_p([0,1]^d)))$, $\cup_{n=0}^\infty X_n$ is dense is $\F_p([0,1]^d)$.
\end{itemize}
Since, with the convention $L_{-1}\circ T_{-1}=0$,
\[
Y_n=\{ x -L_{n-1}(T_{n-1}(x)) \colon x\in X_n\},
\]
and $X_n=\spn\{\delta(x) \colon x \in V_n\}$, the family of nonzero vectors
\[
\BB_n:=(f (x))_{x\in V_n\setminus V_{n-1}}
\]
generates the space $Y_n$. We shall prove that $\BB_n$ is an unconditional basis of $Y_n$ with uniformly bounded unconditional basis constant. Let $F\subset V_n\setminus V_{n-1}$. We define $r_{n,F}\colon V_{n} \to \F_p([0,1]^d)$ by
\[
r_{n,F}(x)=\begin{cases} L_{n-1}\circ T_{n-1}(\delta(x)) &\text{ if } x\in V_{n}\setminus F, \\ \delta(x) &\text{ if } x\in F. \end{cases}
\]
If $z\in V_{n-1}$, then $L_{n-1}( T_{n-1}(\delta(z)))=L_{n-1}(\delta_{n-1}(z))=\delta(z)$. For every $x\in V_n$ there is $z\in V_{n-1}$ such that $\vert x-z\vert_\infty=2^{-n}$, and $V_n$ is $2^{-n}$-separated. By Lemma~\ref{lem:18}, $r_{n,F}$ is $C_1$-Lipschitz with $C_1=(1+2^{1/p})2^{1/p-1}C$. We infer that there is $U_{n,F}\colon \F_p(V_{n})\to \F_p([0,1]^d)$ such that $\Vert U_{n,F}\Vert\le C_1$ and $U_{n,F}\circ \delta_{n}=r_{n,F}$. Put $Q_{n,F}=U_{n,F}\circ T_n|_{Y_n}$. We have $\Vert Q_{n,F}\Vert \le C C_1$ and
\[
Q_{n,F}(f(x))=\begin{cases}
f(x) & \text{ if } x\in F, \\
0 & \text{ if } x\in (V_n\setminus V_{n-1})\setminus F.
\end{cases}
\]
Thus, the mappings $(Q_{n,F})_{F\subset V_n\setminus V_{n-1}}$ are commuting projections satisfying $Q_{n,F}(Y_n)=\spn\{f(x)\colon x\in F\}$ and $Q_{n,F}^{-1}(0)=\spn\{f(x)\colon x\in (V_n\setminus V_{n-1})\setminus F\}$, which implies that $\B_n$ is unconditional basis of $Y_n$ with basis constant at most $CC_1$.
\end{proof}

We have explicitly constructed a Schauder basis of $\F_p([0,1]^d)$. Using the isomorphism $\F_p([0,1]^d)\simeq\F_p(\Rea^d)$ we infer that $\F_p(\Rea^d)$ has a Schauder basis, but now our proof is not constructive. So, it is worth to mention that an argument slightly more involved than the one we have used to prove Theorem~\ref{thm:basis01d} yields an explicit Schauder basis of $\F_p(\Rea^d)$.

\begin{Theorem}\label{thm:basisRd}Let $0<p\le 1$ and $d\in\Nat$. Given an increasing sequence of natural numbers $(k_n)_{n=-1}^\infty$, put
\begin{align*}
V_n&=\{x \in \Rea^d \colon \vert x \vert_\infty \le k_n\} \cap 2^{-n}\Int^d, \text{ and}\\
W_n&=\{x \in \Rea^d \colon \vert x \vert_\infty \le k_{n-1}\} \cap 2^{-n}\Int^d\\
\end{align*}
if $n\in\Nat_*$, and set $V_{-1}=\{0\}$. Define
\[
s_n(x)=\left( \min \left\{ -2^{-n}+ \vert x \vert_\infty, |x_i| \right\}\sgn(x_i)\right)_{i=1}^d, \, x=(x_i)_{i=1}^d\in V_n\setminus W_n,
\]
and put $\eta(x)=(n,\vert x\vert_\infty-k_{n-1})$ if $x\in V_n\setminus W_n$ and $\eta(x)=(n,0)$ if $x\in W_n\setminus V_{n-1}$. Define $f(x)$ for $x \in \cup_{n=0}^\infty V_n$ by
\[
f(x)=\begin{cases} \delta_{\Rea^d}(x) - \delta_{\Rea^d}(s_n(x)) & \text{if } x \in V_n\setminus W_n, \\
\delta_{\Rea^d}(x)-\sum_{v\in V_{n-1}} \Lambda^d_{2^{-n+1}}(v,x) \delta_{\Rea^d}(v)
& \text{if } x\in W_n\setminus V_{n-1}.
\end{cases}
\]
Let $(x_j)_{j=1}^\infty$ be an arrangement of $\cup_{n=0}^\infty V_n$ such that $(\eta(x_j))_{j=1}^\infty$ is non-decreasing with respect to the lexicographical order. Then $(f(x_j))_{j=1}^\infty$ is a Schauder basis of $\F_p(\Rea^d)$.
\end{Theorem}

\begin{proof}
Given $t\in(0,\infty)$, for $d=1$ define $r_t\colon\Rea\to\Rea$ by
\[
r_t(x)=\min\left\{ 1, \frac{t}{|x|}\right\} x, \quad x\in\Rea.
\]
In general, we define $r_t\colon\Rea^d \to \Rea^d$ by
\[
r_t((x_i)_{i=1}^d)=(r_t(x_i))_{i=1}^d.
\]
The map $r_t$ is $1$-Lipschitz and its range is $B_t:=\{ x \in\Rea ^d \colon \Vert x\Vert_\infty\le t\}$. Let $ S_t\colon \F_p(\Rea^d)\to \F_p(B_t)$ be obtained by linearization of $r_t$. Given $R>0$ such that $t/R\in\Nat$, denote $V_{t,R}=\VV^d_R\cap B_t$, and let $T_R\colon \F_p(\Rea^d)\to \F_p(\VV^d_R) $ and $ T_{t,R}\colon \F_p(B_t)\to \F_p(V_{t,R})$ be the linear maps provided by Theorem~\ref{thm:LipFPB}. Notice that Theorem~\ref{thm:LipFPB} yields a constant $C$ be such that $\Vert T_{t,R}\Vert\le C$ for every $t>0$ and $R$ with $t/R\in\Nat$. Let $L_{\MM,\NN}$ denote the linearization of the inclusion of $\MM$ into $\NN$, and put $L_{R}=L_{\VV^d_R,\Rea^d}$, $L_{t,R}=L_{V_{t,R},B_t}$ and $L'_{t,R}=L_{V_{t,R},\Rea^d}$. We shall prove that
\begin{equation}\label{eq:commutes}
L_{t,R}\circ T_{t,R}\circ S_t = S_t\circ L_{R}\circ T_R, \quad \frac{t}{R}\in\Nat.
\end{equation}
Set $\delta=\delta_{\Rea^d}$, $\delta_t=\delta_{B_t}$, and $\delta_{t,R}=\delta_{V_{t,R}}$.

By Proposition~\ref{prop:FPB}~\ref{prop:FPB:6} a similar argument works for the general case, hence for notational ease we will deal with the case $d=1$. Let $x\in\Rea$. In the case when $|x|\le t$, we have $r_t(x)=x$ and $\Lambda_R(v,x)=0$ unless $|v|\le t$, in which case $r_t(v)=v$. Consequently,
\begin{align*}
S_t(L_{R}(T_R(\delta(x))))
&=\sum_{v\in\VV_R} \Lambda_R(v,x)\delta_t(r_t(v))\\
&=\sum_{v\in\VV_R} \Lambda_R(v,x)\delta_t(v)\\
&=L_{t,R}(T_{t,R}(\delta_t(x)))\\
&=L_{t,R}(T_{t,R}(\delta_t(r_t(x)))).
\end{align*}
In the case when $|x|>t$ we have $\Lambda_R(v,x)=0$ unless $|v|\ge t$ and $\sgn(v)=\sgn(x)$, in which case $r_t(v)=r_t(x)$. Then we have
\begin{align*}
S_t(L_{R}(T_R(\delta(x))))
&=\sum_{v\in\VV_R} \Lambda_R(v,x)\delta_t(r_t(v))\\
&=\sum_{v\in\VV_R} \Lambda_R(v,x)\delta_t(r_t(x))\\
&=\delta_t(r_t(x))\\
&=L_{t,R}(T_{t,R}(\delta_t(r_t(x)))).
\end{align*}
Since $\delta_t(r_t(x))=S_t(\delta(x))$ and $\{ \delta(x) \colon x\in\Rea\}$ spans $\F_p(\Rea)$, \eqref{eq:commutes} holds. Note that the range of $U_{t,R}:= T_{t,R}\circ S_t$ is $\spn\{\delta_{t,R}(x) \colon x\in V_{t,R}\}$. Put $P_{t,R}:=L'_{t,R}\circ T_{t,R}\circ S_t$. By \eqref{eq:commutes}, for every $x\in\Rea^d$ we have
\begin{equation}\label{eq:commutes3}
P_{t,R}(\delta(x)) = \sum_{v\in\VV^d_R} \Lambda^d_R(v,x)\delta(r_t(v)), \quad \frac{t}{R}\in\Nat.
\end{equation}
Thus, for every $v\in \VV^d_R$ we have $P_{t,R}(\delta(v)) = \delta(r_t(v))$, which in combination with \eqref{eq:commutes3} implies that $P_{t,R}$ is a projection with range equal to $\spn\{\delta(x)\colon x\in V_{t,R}\}$. We infer that
\begin{equation}\label{eq:commutes2}
P_{t,R} \circ P_{t',R'} = P_{t',R'} \circ P_{t,R}= P_{t,R}, \quad
\frac{R}{R'}, \frac{t'}{R'}, \frac{t}{R}\in\Nat, \, t'\ge t>0.
\end{equation}
Indeed, $P_{t',R'} \circ P_{t,R}= P_{t,R}$ follows from the fact that the range of $P_{t,R}$ is contained in the range of $P_{t',R'}$ and we have
\begin{align*}
P_{t,R}\circ P_{t',R'} & = L'_{t,R}\circ T_{t,R}\circ S_t\circ L_{B_{t'},\Rea^d}\circ S_{t'}\circ L_{R'}\circ T_{R'}\\
& = L'_{t,R}\circ T_{t,R}\circ S_t\circ L_{R'}\circ T_{R'}\\
&= P_{t,R},
\end{align*}
where in the first equality we used \eqref{eq:commutes}, in the second we used the observation that $S_t\circ L_{B_{t'},\Rea^d}\circ S_{t'} = S_t$ and the third equality follows from Proposition~\ref{prop:FPB}~\ref{prop:FPB:5}.

Hence, if for $n\in\Nat_*$ we put
\[
P_{2n}=P_{k_n,2^{-n}} \text{ and }
P_{2n-1}= P_{k_{n-1},2^{-n}},
\]
we have $P_j\circ P_{j'}=P_{j'}\circ P_j=P_j$ whenever $-1\le j\le j'$. Therefore, there is Schauder decomposition $(Y_j)_{j=-1}^\infty$ of $\F_p(\Rea^d)$ whose associated projections are $(P_j)_{j=-1}^\infty$. Moreover, for all $n\in\Nat_*$ the range of $P_{2n}$ is $\spn\{\delta(x) \colon x\in V_n\}$ and the range of $P_{2n-1}$ is $\spn\{\delta(x) \colon x\in W_n\}$.

Similar arguments as in the proof of Theorem~\ref{thm:basis01d} show that there is a constant $C=C(p,d)$ such that
\[
(f(x))_{x\in W_n\setminus V_{n-1}}
\]
is a $C$-unconditional basis of this $Y_{2n-1}$ for all $n\in\Nat_*$. Let $n\in\Nat_*$. Note that $r_t(x)=r_t(s_n(x))$ for all $t$ with $2^n t\in\Nat$ and all $x\in V_n\setminus W_n$ with $\Vert x \Vert_\infty>t$. In particular,
\[
r_{k_{n-1}}(x)=r_{k_{n-1}}(s_n(x)), \quad x\in V_n\setminus W_n.
\]
Therefore, $P_{2n-1}(f(x))=0$ for every $x\in V_n\setminus W_n$, which in turn implies that $f(x)\in Y_{2n}$. An inductive argument yields that for every $x\in V_n\setminus W_n$,
\[
g(x):=\delta(x)-\delta(r_{k_{n-1}}(x))
\]
is a linear combination of the family (of nonzero vectors)
\[
\BB_n=(f(y))_{y\in V_n\setminus W_n}.
\]
Since, by an argument similar to that used in the proof of Theorem~\ref{thm:basis01d}, $(g(x))_{x\in V_n\setminus W_n}$ generates $Y_{2n}$, $\BB_n$ also generates $Y_{2n}$. Let $F\subset V_n\setminus W_n$ be such that
\[
V_{t-2^{-n},2^{-n}}\setminus W_n
\subset F
\subset V_{t, 2^{-n}} \setminus W_n
\]
for some $t\in(k_{n-1}, k_n]\cap 2^{-n}\Int$. Set $\alpha=(t,2^{-n})$ and let $\widehat{\alpha}$ be the ``predecesor'' of $\alpha$ given by $\widehat{\alpha}=(t-2^{-n},2^{-n})$. Define $r_{\alpha,F}\colon V_{\alpha}\to \Rea^d$ by
\[
r_{\alpha,F}(x)=\begin{cases} r_{t-2^{-n}}(x) &\text{ if } x\in V_{\alpha} \setminus F, \\ x &\text{ if } x\in F. \end{cases}
\]
If $x\in V_{\widehat{\alpha}}$ then $r_{t-2^{-n}}(x)=x$. Given $x\in V_\alpha$ there is $z\in V_{\widehat{\alpha}}$ with $\vert x-z\vert_\infty=2^{-n}$, and $V_{\alpha}$ is $2^{-n}$-separated. Hence, by Lemma~\ref{lem:18}, $r_{\alpha,F}$ is $C_p$-Lipschitz, where $C_p=(1+2^{1/p})$. Let $U_{\alpha,F}\colon \F_p(V_\alpha)\to \F_p(\Rea^d)$ be the linear map defined by $\delta_\alpha(x) \mapsto \delta(x)$ if $x\in F$ and $\delta_\alpha(x) \mapsto \delta (r_{t-2^{-n}}(x))$ if $x\in V_{\alpha}\setminus F$. Set $U_\alpha= T_\alpha\circ S_t$. We infer that, if
\[
Q_{\alpha,F}=U_{\alpha,F}\circ U_\alpha|_{Y_{2n}},
\]
then $\Vert Q_{\alpha,F} \Vert\le C C_p$. Note that $U_\alpha(\delta(x))=\delta_\alpha(x)$ for all $x\in V_\alpha$ and that $U_\alpha(\delta(x))=U_\alpha(\delta(s_n(x)))$ for all $x\in V_n\setminus V_\alpha$. If $x\in V_\alpha$, then $s_n(x)\in V_{\widehat{\alpha}}$, and so $Q_{\alpha,F}(f(x))=f(x)$ for every $f\in F$. If $x\in V_\alpha\setminus V_{\widehat{\alpha}}$, then $s_n(x)=r_{t-2^{-n}}(x)$. Consequently, $Q_{\alpha,F}(f(x))=0$ for every $x\in V_\alpha\setminus F$. Finally, we deduce that $Q_{\alpha,F}(f(x))=0$ for every $x\in V_n\setminus V_{\alpha}$. We infer that, if $(x_j)_{j=1}^{|V_n|-|W_n|}$ is an arrangement of $V_n\setminus W_n$ with $(|x_j|)_{j=1}^{|V_n|-|W_n|}$ non-decreasing, then $(f(x_j))_{j=1}^{|V_n|-|W_n|}$ is a Schauder basis of $Y_{2n-1}$ with basis constant at most $CC_p$.
\end{proof}

\section{Open problems}
\noindent
By \cite{K86}*{Theorem 5.2}, there exists a subspace $Z$ of $\ell_p$ whose Banach envelope is isomorphic to $L_1$, so we would like to know whether $\F_p(\Rea)$ is different from the subspace $Z$, whose existence is guaranteed by a general abstract construction.

\begin{Question}
Let $p\in(0,1)$. Is $\F_p(\Rea)$ isomorphic to a subset of $\ell_p$?
\end{Question}

Once we have told apart $\F_p(\Nat)$ from $\ell_p$ for $p<1$, it is natural to go on with the topic \ref{QB} by trying to determine how many non-isomorphic Lipchitz free $p$-spaces one can obtain from subsets of $\Nat$. Note that if $\NN$ is a subset of $\Nat$ then $\F_p(\NN)$ is a complemented subspace of $\F_p(\Nat)$ by \ref{three:complemeted}. So, this problem connects with that of characterizing the complemented subspaces of $\F_p(\Nat)$. Let us illustrate the question with an example. Suppose that $\NN\subset\Nat$ contains arbitrarily long chains of consecutive integers.
Then, there is $(a_k)_{k=0}^\infty$ in $\Nat$ such that $a_k+\Nat_{2^k-1}\subset \NN$ and $2(a_{k}+2^{k}-1) \le a_{k+1}$ for all $k\in\Nat_*$. Set
\[
\NN_0= \bigcup_{k=0}^\infty a_k+\Nat_{2^k-1}.
\]
Combining \cite{AACD2019}*{Lemma 2.1} with \cite{AACD2020}*{Theorem 5.8} yields $\F_p(\NN_0)\simeq \F_p(\Nat)$. Then, by \ref{three:complemeted}, $\F_p(\NN)$ is complemented in $\F_p(\Nat)$ and, the other way around, $\F_p(\Nat)$ is complemented in $\F_p(\NN)$. Taking into account \ref{three:isomorphismsNd}, Pe{\l}czy\'nski's decomposition method yields $\F_p(\NN)\simeq\F_p(\Nat)$.

\begin{Question}Let $0<p<1$. Does there exist $\NN\subset\Nat$ such that $\F_p(\NN)$ is neither isomorphic to $\ell_p$ nor to $\F_p(\Nat)$?
\end{Question}

A well-known problem in Geometric Functional Analysis is whether $\F(\Nat^2)$ is isomorphic to $\F(\Nat^3)$ or, more generally, whether $\F(\Nat^d)$  is isomorphic to $\F(\Nat^{d+1})$ for $d\ge 2$.
As, by \cite{AACD2018}*{Proposition 4.20}, if  $\F_p(\Nat^d)\simeq\F_p(\Nat^{d+1})$ for some $p<1$ then $\F(\Nat^d)\simeq\F(\Nat^{d+1})$, investigating in more depth the geometry of $\F_p(\Nat^d)$ for $p<1$ and $d\ge 2$ could shed some light into this important question. Similarly,  telling apart $\F_p(\Rea^d)$ from $\F_p(\Rea^{d+1})$ for $p<1$ could be easier than telling apart $\F(\Rea^d)$ from $\F(\Rea^{d+1})$.

By \cite{AACD2020}*{Theorem 4.21} the latter  is equivalent to proving that Lipschitz $p$-free spaces over Euclidean spaces are not isomorphic to Lipschitz $p$-free spaces over its spheres.

\begin{Question}Let $0<p\le 1$ and $d\ge 2$. Is $\F_p(\Nat^d)$ is isomorphic to $\F_p(\Nat^{d+1})$?
\end{Question}

\begin{Question}Let $0<p\le 1$ and $d\ge 3$. Is $\F_p(\Rea^d)$ is isomorphic to  $\F_p(S_{\Rea^d})$?
\end{Question}

\end{document}

config: arxiv


TITLE: Computing quotients of group by elements of its lower exponent$-p$ central series
QUESTION [1 upvotes]: Let $G$ be a finite $p−$group of number of generators $d$ and exponent$−p$ class $c$, that is $c$ is the smallest integer satisfying $P_c(G)=1$ in the series
$$
G=P_0(G)≥...≥P_{i−1}(G)≥P_i(G)≥...
$$
Where $P_i(G)=[P_{i−1}(G),G]P_{i−1}(G)^p$.
1/ Can you show me how to calculate $G/P_i(G)$´s using GAP system?
2/ Can you show me how to compute $G/P_1(G)$ using abelianisation and row-echelonisation (by hand)?
Thanks in advance

REPLY [5 votes]: The series is in general called the $p$-central series, so the GAP command is PCentralSeries.
As for calculating it by hand, you might want to look at section 9.4.2 of Holt/Eick/O'Brien: Handbook of Computational Group Theory.


TITLE: How forces act on the two masses on a rotating table
QUESTION [0 upvotes]: Consider a system on the figure: two masses on a rotating turntable are connected by a string over a frictionless pulley.

The masses of the bodies are related as $m_A<m_B$. They are at distance $R$ from the center of the turntable.
The turntable is slowly accelerating, until it reaches some angular velocity $\omega_0$. After that the lower mass (B) slips outwards. The static friction coefficient between the blocks and the between the block and the table is $\mu$.
It is necessary to find $\omega_0$.
In order to do that, I'm writing the Newton's law for each body:

Body A: $\quad m_A{\bf \ddot r} = m_A{\bf g}+{\bf N}_A + {\bf T}+{\bf F}_A$
Body B: $\quad m_B{\bf \ddot r} = (m_A+m_B){\bf g}+{\bf N}_B + {\bf T}-{\bf F}_A+{\bf F}_B$

where ${\bf T}$ is for tension of the string acting on each body, ${\bf F}_{A}$ is friction between A and B, and ${\bf F}_B$ is friction between B and the turntable.
After that I shoul probably go to the cylindric coordinates $(r, \theta, z)$ and write the equations above component-by-component. However, I cannot understand what is the condition of B slipping outwards, and why is there a such a critical value $\omega_0$.
How do the forces of friction act in this case?
Please, help me to understand the physics behind this situation. Any kind of hint will be highly appreciated. Thank you!

REPLY [0 votes]: It is generally easiest assign the pieces and add them when you have all the pieces defined.  I see $\mu$ is missing from your equations, so something is wrong.  When the blocks start slipping is when the friction force is $\mu\ N$.  Prior to that it is lower.
The friction force on block A:
$\text{FrA}=-\mu\  \text{mA}\ g$
I am calling to the right positive and to the left negative.  Block A slips to the right and its friction force is opposite, so it is negative.
The friction force on Block B is positive.
$\text{FrB}=\mu\  (\text{mA}+\text{mB}) g$
Now find $\text{$\Sigma $F}=m\ a$ for each block. The acceleration for each block is its respective $\frac{m\ v^2}{r}=m\ r\ \omega^2$
Block A
$T+FrA=mA\ R\ \omega 0^2$
and block B
$T+FrB-FrA=mB\ R\ \omega 0^2$
Plugging in for the $FrA$ and $FrB$ from above and solving the two equations for $T$ and $\omega 0$, we get
$T=\frac{2 g\ \mu\ mA\ (mA+mB)}{mB-mA}$
$\omega 0=\sqrt{\frac{\mu\ g (3mA+mB)}{R (mB-mA)}}$


text:
\begin{document}

\subsection{Study of the tree structure}
\label{subsection:study_of_tree}

In general, for tree tensor networks the rank depends on the structure of the tree.
As already observed for large Markov chains using the tensor-train format, a change in the ordering of the automata within the tensor-train tree network can affect the rank, cf.~\cite{Masetti2019}.
In the following, we focus on the hierarchical Tucker format and study how the ordering of the automata in the leaves of the tree affects the low-rank approximability. We preserve the balanced binary structure of the tree because this is advantageous for parallelization, cf.~\cite{Grasedyck2017}.

We already studied the singular values of the matricizations corresponding to the canonical balanced tree, see~\cref{figure:sv_d=8_b=4}. We change only the arrangement of the automata in the leaves of the tree and compute the marginal distribution $\pPM$ again.
The decay of the singular values for each matricization is shown in~\cref{figure:sv_d=8_b=4_reverse} where the semi-logarithmic plot at each vertex displays the means of the singular values.
\begin{figure}[tbhp]
	\centering

	\includegraphics[width = \textwidth]{tikz/singular_values_modified_tree}

	\caption{\footnotesize{Mean of singular values of matricizations of $\pPM$ for a modified balanced tree and $100$ sample parameters with $d=8$ automata and block size $b=4$.}}
	\label{figure:sv_d=8_b=4_reverse}
\end{figure}
Comparing~\cref{figure:sv_d=8_b=4,figure:sv_d=8_b=4_reverse}, we observe that the choice of the canonical binary tree, i.e., the original ordering in the leaves, results in a significantly faster decay of the singular values close to the root.~\Cref{figure:sv_d=8_b=4_reverse} shows that the singular values closest to the root also have an exponential decrease, but at a much slower rate.
Note that blocks of size $b = 4$ indicate that the automata $\{\A_1, \A_2, \A_3, \A_4\}$ interact directly with one another and are thus highly correlated.
The same holds for the automata  $\{\A_5, \A_6, \A_7, \A_8\}$.
There are no direct interactions between automata of different blocks, i.e., these are only weakly correlated.
Hence, the modified balanced tree separates highly correlated automata, which explains the slower decline of the singular values in level $1$ (level $0$ being the root).
In contrast, the canonical balanced tree separates weakly correlated automata, leading to a faster decline of the singular values.
We will make similar observations when studying the approximation rank.
\end{document}


TITLE: Second derivative of a parametric equation with trig functions
QUESTION [1 upvotes]: I am solving a problem where I have to find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ given these parametric equations:
$ x = \cos t $
$ y = 3 \sin t $
I got $-3\operatorname{cot}t\;$ for the first derivative by finding $\frac{dy}{dt} / \frac{dx}{dt}$.
For the second derivative, I simply took the derivative of $-3\operatorname{cot}t$ to get $3\operatorname{csc}^2t$.
However, my teacher did this differently. He wrote that $\frac{d^2y}{dx^2} = \frac{-3(-\operatorname{csc}^2t)}{-\sin t} = -3\operatorname{csc}^3t$
I'd really appreciate it if someone could explain where I went wrong in my own solution.
Thanks $:)$

REPLY [1 votes]: You took the derivative with respect to $t$. Your teacher took the derivative with respect to $x$.
$$\frac{dy}{dx} = -3\cot t$$
$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{d}{dt}(-3\cot t)}{-\sin t} = \frac{3\csc^2 t}{-\sin t}$$


TITLE: Closure of the point spectrum.
QUESTION [1 upvotes]: I'm trying to prove the following about the closure of the point spectrum: $\overline \sigma_p \subseteq \sigma_c \cup \sigma_p$ where $\sigma_p$ is the point spectrum and $\sigma_c$ is the continuous spectrum. What I noticed so far is that the union of the closure of the point spectrum and the eigenvalues of infinity multiplicity forms the continuous spectra. But I think I'm missing something since I can't conclude.
Best Regards,
Lievet.

REPLY [2 votes]: I don't believe that your assertion is true. For example, let $H = \ell^2(\mathbb{N}) \times \ell^2(\mathbb{N})$, and define $A : H\rightarrow H$ by
$$
                  A(e_n,e_m)=(\frac{1}{n}e_n,e_{m+1}),\;\;\; n,m=1,2,3,\cdots.
$$
Then $A(e_n,0)=\frac{1}{n}(e_n,0)$, which means that $0\in\overline{\sigma_p(A)}$. However, the range of $A$ is not dense because $(0,e_1)\perp \mathcal{R}(A)$. Therefore $0\notin\sigma_c(A)$. And it is easy to check that $0\notin\sigma_p(A)$, meaning that $\mathcal{N}(A)=\{0\}$.