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We completely characterize $1$-complemented subspaces and norm-one projections in complex spaces $\ell_p(\ell_q)$ for $1 \leq p, q < \infty$. Finally we give a full description of the subspaces that are spanned by a family of disjointly supported vectors and which are $1$-complemented in (real or complex) % %page 2 Orlicz or Lorentz sequence spaces. In particular if an Orlicz or Lorentz space $X$ is not isomorphic to $\ell_p$ for some $1 \leq p < \infty$ then the only subspaces of $X$ which are $1$-complemented and disjointly supported are the closed linear spans of block bases with constant coefficients. \end{abstract} \maketitle \section{Introduction} %page 3 Projections and norm one projections have been studied by many authors. The question about the form of a (norm-one) projection and the structure of its range arises naturally not only in geometry of Banach spaces, but also in approximation theory, spectral theory, ergodic theory; see, e.g., the surveys \cite{ChP,Doust} for more detailed discussions of applications. The difficulty in studying $1$-complemented subspaces of spaces with $1$-unconditional bases arises from the following classical fact due to Lindenstrauss \cite{L72} (cf. also \cite[Theorem~3.b.1]{LT1}) \begin{theorem}\lb{nohope} Every space $Y$ with a $1$-unconditional basis is $1$-complemented in some symmetric space $X$. \end{theorem} Thus it seems hopeless to give any characterization of $1$-complemented subspaces of, even symmetric, spaces with $1$-unconditional bases. The only class of spaces where the full characterization of $1$-complemented subspaces was available are the classical spaces $\ell_p$ and $c_0$. Namely it is well known that every subspace of a Hilbert space is $1$-complemented (with the unique orthogonal projection) and in $\ell_p$, for $p \neq 2, \infty$, we have the following result: %page 4 \begin{theorem}[{\cite{Ando}, cf.\ also \cite[Theorem 2.a.4]{LT1}}] \label{lp} Let $F \subset \ell_p$, where $1 \leq p < \infty$, $p \neq 2$. Then $F$ is $1$-complemented if and only if \begin{itemize} \item[(a)] $F$ is isometric to $\ell_p^{\dim F}$, \end{itemize} \noindent or \begin{itemize} \item[(b)] $F$ is spanned by a family of mutually disjoint vectors. \end{itemize} \end{theorem} It is clear that Theorem \ref{lp}(a) cannot be extended to other spaces. Namely Lindberg \cite{L73} showed a class of Orlicz functions $\phi$ (for necessary definitions see Section~\ref{pre}) so that there exists a $1$-complemented subspace $F$ in $\ell_\phi$ such that $F$ is not even isomorphic to $\ell_\phi$. Altshuler, Casazza and B.~L.~Lin \cite{ACL} showed a similar example in the class of Lorentz sequence spaces $\ell_{w,p}$. % %page 5 However, both of these examples were spanned by a family of mutually disjoint vectors; in fact they were closed linear spans of a block basis with constant coefficients. Also the symmetric space $X$ constructed in Theorem~\ref{nohope} was such that $Y$ was isometrically isomorhic to a closed linear span of a block basis with constant coefficients. It is well known that all such spans are $1$-complemented in any symmetric space ({\cite[Theorem 3.a.4]{LT1}}), so in fact all of those examples satisfy condition $(b)$ of Theorem~\ref{lp}. In this paper we prove that indeed Theorem~\ref{lp}(b) can be extended to a large class of $1$-complemented subspaces of complex spaces with $1$-unconditional basis. % %page 6 Namely, if $X$ is a complex, strictly monotone sequence space with a $1$-unconditional basis, $Y \subset X$ is $1$-complemented in $X$, and $Y$ does not contain a band isometric to $\ell_2^2$, then $Y$ is spanned by a family of disjointly supported vectors (see Corollary~\ref{cor}). It is clear that our restrictions on $X$ and $Y$ are necessary (see Remark after Corollary~\ref{cor} and examples in Section~\ref{secl2}). The above-mentioned assumption on $Y$ is satisfied, for example, in all spaces $X$ that do not have a $1$-complemented subspace isometric to $\ell_2^2$. We discuss it in greater detail in Section~\ref{secl2}. In Theorem \ref{main} we also describe the form of general $1$-complemented subspaces of complex strictly monotone spaces. Our method of proof % %page 7 cannot be extended to real sequence spaces. We use in particular the fact that every $1$-complemented subspace of a complex space with $1$-unconditional basis also has a $1$-unconditional basis. The analogous fact is false in real spaces \cite{Le79,BFL} (see \cite{R86} for the disscusion in special real spaces). As a consequence of Theorem \ref{main} we obtain a complete characterization of $1$-comple\-mented subspaces of complex $\ell_p(\ell_q)$, where $11$. For the careful analysis and properties of Hilbert components of various spaces we refer to \cite{KW} and to expository papers \cite{F83,R83}. Here we just want to recall some properties which will be used in our arguments. First, recall that if $X$ has 1-unconditional basis $\{e_i\}_{i\in I}$ then each basis element is hermitian. Moreover Kalton and Wood proved the following: \begin{theorem}[{\cite[Theorem 6.5]{KW}}] \label{KW} Let $X$ be a Banach space with a normalized $1$-uncon\-ditional basis. Then $x \in X$ is hermitian in $X$ if and only if \begin{itemize} \item[(i)] $\|y\|_X = \|y\|_2$ for all $y \in X$ with $\supp y \subset \supp x$, and \item[(ii)] for all $y, z \in X$ with $\supp y \cup \supp z \subset \supp x$ and for all $v \in X$ with $\supp v \cap \supp x = \varnothing$ if $\|y\|_X = \|z\|_X$ then $\|y+v\|_X = \|z+v\|_X$. \end{itemize} \end{theorem} %page 12 For our main result we will need the following two facts. \begin{proposition}[{\cite[Lemma~5.2]{KW}}] \lb{kw} Suppose that $x, y$ are hermitian elements in $X$. Denote by $x^*$ a norming functional for $x$. If $x^*(y)\neq 0$ then $\Span \{x, y\} \subset h(X)$. \end{proposition} \begin{proposition}[{\cite[Lemma~4]{F83}}] \lb{yherm} Suppose that $X$ has 1-unconditional basis $\{e_i\}_{i\in I}$ and let $P: X\lra X$ be norm one projection with range of $P$ equal to $Y$. Then for all $i\in I$, $Pe_i$ is a hermitian element in $Y$. \end{proposition} We will also frequently use the following well-known fact: \begin{proposition} \label{conj} Let $X$ be a Banach space with a $1$-unconditional basis. Suppose that $Y \subset X$ is $1$-complemented and a norm-one projection $P:X \to Y$ is given by $$ P(x) = \sum_i y_i^{*}(x) y_i $$ where $Y = \overline\Span\{y_i\}$ and $y_i^{*}$ is norming for $y_i$ for all $i$. Then for any $y \in Y$ there exists $y^{*}$ norming for $y$ and constants $K_i$ so that $$ y^{*} = \sum_i K_i y_i^*. $$ \end{proposition} Moreover, we have \begin{proposition}[Calvert \cite{Cal75}] \label{calvert} Let $X$ be a strictly convex reflexive Banach space with strictly convex dual $X^*$. Let $J:X\lra X^*$ be the duality map; $\|Jx\| = \|x\|, \ Jx(x) = \|x\|^2.$ Then a closed linear subspace $Y$ of $X$ is $1$-complemented in $X$ \wtw $J(Y)$ is a linear subspace of $X^*$. \end{proposition} %page 13 Finally we recall a few definitions (see \cite{LT1}). We say that a Banach space $X$ with $1$-unconditional basis is \emph{ strictly monotone\/} if $\|x+y\| > \|x\|$ for all $x, y \geq 0$ with $y \neq 0$. An \emph{Orlicz function} $\phi$ is a convex non-decreasing function $\phi: [0,\infty)\longrightarrow [0,\infty]$ with $\phi(0) = 0$ and $\phi(1) = 1$ or $\infty$. To any Orlicz function $\phi$ we associate the \emph{Orlicz space} $\ell_\phi$ of all sequences of scalars $x = (x_i)_i$ such that $$ \sum^\infty_{i=1} \phi \Bigl(\frac{|x_i|}{\rho}\Bigr) < \infty \quad \text{ for some } \rho > 0, $$ with the norm $$ \|x\|_\phi = \inf\biggl\{\rho > 0 : \sum^\infty_{i=1} \phi \Bigl(\frac{|x_i|}{\rho}\Bigr) < 1\biggr\}. $$ Let $1 \leq p < \infty$ and let $w=\{w_i\}_{i \in I}$, where $I=\N$ or $I = \{1,\dots, d\}$, be a % %page 14 non-increasing sequence such that $w_1=1$ and $w_i \geq 0$ for all $i$. The Banach space of all sequences of scalars $x = (x_i)_{i \in I}$ for which $${\disp \|x\|_{w,p} = \sup\limits_{\sigma \in \cal{P}(I)}\biggl(\sum_{i \in I} |x_{\sigma(i)}|^p w_i\biggr)^{\frac{1}{p}} < \infty,} $$ where $\cal{P}(I)$ is the set of all permutations of $I$, is denoted $\ell_{w,p}$ and it is called a \emph{Lorentz sequence space} (another notation frequently used in the literature is $d(w,p))$. Notice that \begin{itemize} \item $\ell_\phi$ is strictly monotone if and only if $\phi(t) > 0$ for all $t > 0$ and $\phi(t) < \infty $ for all $t \leq 1$. \item $\ell_{w,p}$ is strictly monotone if and only if $w_i > 0$ for all $i \in I$. \item $\ell_p(\ell_q)$ is strictly monotone if and only if $p,q \neq \infty$. \end{itemize} %page 15 For any $ 1 \leq p, q \leq \infty$ we denote be $\ell_p(\ell_q)$ the space of sequences of scalars $x = (x_{ij})_{i \in I, j \in J}$ such that $$ \|x\|_{\ell_p(\ell_q)} = \bigl\| ( \| (x_{ij})_{j \in J}\|_{\ell_q})_{i \in I}\bigr\|_{\ell_p} < \infty. $$ We follow standard notations as defined in \cite{LT1} and this is also where we refer the reader for all undefined terms. % %page 16 \section{General form of contractive projections} We are now ready to present our main theorem. \begin{theorem}\lb{main} Suppose that $X$ is a complex strictly monotone sequence space with 1-unconditional basis $\{e_i\}$ and $X\ne \ell_2$ and let $P$ be the projection of norm 1 in $X$. Let $\{H_\g\ : \g\in\G\}$ be the collection of Hilbert components of $Y = PX$. Then $H_\g's$ are disjointly supported as elements of $X$. \end{theorem} \begin{proof} %{Proof of Theorem~\ref{main}} By Proposition~\ref{yherm} all $\{Pe_i\}_{i\in I}$ are hermitian elements of $Y$. Let $i,j$ be such that $Pe_i$ and $Pe_j$ are not in the same Hilbert component of $Y$. Assume that there exists $k$ such that $k\in \supp Pe_i\cap \supp Pe_j$. If $Pe_k\neq 0$, then $P^*e_k^*$ is a norming functional for $Pe_k$ in $Y$ and $(P^*e_k^*,Pe_i)=(e_k^*,Pe_i)\neq 0. $ Thus, by Proposition~\ref{kw}, $Pe_k, Pe_i$ are in the same Hilbert component of $Y$. Similarly $Pe_k, Pe_j$ are in the same Hilbert component of $Y$. But then $Pe_i, Pe_j$ are in the same Hilbert component of $Y$ contrary to our assumption. Thus $Pe_k = 0$. Now suppose that $Pe_i = \sum_{l\in S} \alpha_l e_l +\alpha_k e_k$ for some $\al_l, \al_k \neq 0$, where $S= \supp Pe_i\setminus \{k\}$. Since $P$ is a projection and $P(e_k)=0$ we get \begin{align*} \sum_{l\in S} \alpha_l e_l +\alpha_k e_k &=Pe_i = P(Pe_i) = P\biggl( \sum_{l\in S} \alpha_l e_l\biggr) + \alpha_k P(e_k) =\\ &=P\biggl( \sum_{l\in S} \alpha_l e_l\biggr) \ . \end{align*} Hence $S\ne \emptyset$ and by strict monotonicity of $X$ $$\Big\| P\biggl( \sum_{l\in S} \alpha_l e_l\biggr)\Big\| = \Big\| \sum_{l\in S} \alpha_l e_l + \alpha_k e_k\Big\| > \Big\| \sum_{l\in S} \alpha_l e_l \Big\|$$ which contradicts the fact that $\|P\|=1$. Thus if $Pe_i,Pe_j$ are not in the same Hilbert component of $Y$ then they are disjoint. \end{proof} \begin{corollary}\lb{cor} Suppose that $X$ is a complex strictly monotone sequence space with 1-unconditional basis $\{e_i\}$ and $X\ne \ell_2$ and let $P$ be the projection of norm 1 in $X$. Suppose that $Y = PX\subset X$ has no nontrivial Hilbert components. Then there exist disjointly supported elements $\{y_j\}_{j=1}^m$ ($m= \dim PX \le\infty$) which span $Y = PX$. Moreover, for all $x\in X$, $$Px = \sum_{j=1}^m y_j^*(x)y_j,$$ where $\{y_j^*\}_{j=1}^m\subset X^*$ satisfy $\|y_j\| = \|y_j^*\| = y_j^*(y_j) = 1$ for all $j$. \end{corollary} \begin{proof} By Proposition~\ref{yherm} all $\{Pe_i\}_{i\in I}$ are hermitian elements in $Y$. By our assumption all Hilbert components of $Y$ are one-dimensional so if $Pe_i,Pe_j$ are linearly independent then they belong to different Hilbert components of $Y$. Therefore, by Theorem~\ref{main}, if $Pe_i,Pe_j$ are linearly independent then they are disjoint and $Y$ can be presented as $\hbox{span}\{Pe_i :i\in I\}$ where $I$ is a collection of such indices $i,j$ that $Pe_i,Pe_j$ are mutually disjoint. Then $y_i = {Pe_i}/{\|Pe_i\|}$ for all $i\in I$, and for each $x\in X$ we have $Px = \sum_{i\in I} C_i y_i$, where $C_i\in \bbC$ are uniquely determined by $x$. Clearly $y_i^*(x) \defeq C_i(x)$ satisfies the conclusion of the theorem. Notice also that $\supp y_i^* = \supp y_i$ for all $i\in I$. \end{proof} \begin{remarks} \begin{enumerate} \item Notice that the assumption about $X$ being strictly monotone is important. Indeed, Blatter and Cheney \cite{BCh} (see also \cite{B88}) showed examples of $1$-complemented hyperplanes in $\ell^3_\infty$ that are not spanned by disjointly supported vectors. \item Also the assumption about $Y$ cannot be removed. We discuss it in greater detail in the next section. \item As mentioned in Introduction, Calvert and Fitzpatrick \cite{CF86} showed that if every subspace of the form described in Corollary~\ref{cor} is $1$-complemented in $X$ then $X$ is isometric to $\ell_p$, for some $p$, $1 \leq p < \infty$, or to $c_0$. \end{enumerate} \end{remarks} As a consequence of Corollary~\ref{cor} we can express $1$-complemented subspaces as an intersection of hyperplanes of special simple form (see \cite{BP88} for analogous representation in $\ell_p$). \begin{corollary}\lb{ker} Let $X$ and $Y$ be as described in Corollary~\ref{cor}. Then $Y$ can be presented as intersection of kernels of functionals $f_j$, such that $\card(\supp f_j)\le 2$ for all $j$. \end{corollary} %page 17 \section{$1$-complemented copies of $\ell^2_2$} \label{secl2} In this section we discuss in what situation it is possible that a space $X$ has a $1$-complemented subspace $Y$ with nontrivial Hilbert components. This clearly reduces to the question of characterizing under what conditions $X$ can have a $1$-complemented subspace $F$ that is isometric to $\ell^2_2$. % %page 18 The question that arises here is:\par\nobreak\medskip \begin{quote} \it Is it possible that a space $X$ with only $1$-dimensional Hilbert components has a $1$-complemented subspace isometric to $\ell^2_2$? \end{quote} \medbreak One quickly realizes that the answer is yes: \begin{example}\lb{exl2lq} Consider the space $X= \ell_2(\ell_1)$ and for $x=(x_{ij})_{i,j} \in \ell_2(\ell_1)$ let $Px = (x_{i1})_i$. Then $PX$ is isometric to $\ell^2_2$. Further there exists an orthogonal projection $Q$ of $PX$ onto a span of any collection of orthogonal vectors from $PX = \ell_2$. \end{example} In Lemma \ref{l2cor} we show that if $\ell_p(\ell_q)$ has a $1$-complemented subspace $F$ that is isometric to $\ell^2_2$ then either $q=2$ and $F$ is contained in a Hilbert component of $\ell_p(\ell_q)$ or $p=2$ and $F$ is similar to the range of $QP$ in Example~\ref{exl2lq}. Below we present two more examples: a $1$-complemented copy of $\ell^2_2$ in a Lorentz space and in a Orlicz space. We do not know a full characterization of spaces $X$ that have $1$-complemented copies of $\ell^2_2$, but we suspect that if $X$ is a Lorentz or Orlicz space then $X$ has to be very similar to the examples presented below. % %page 19 \begin{example} Consider Lorentz space $\ell_{w,2}$ with weight $ w = (1,1, w_3, w_4, \dots)$. Then $\Span \{e_1, e_2\} \subset \ell_{w,2}$ is isometric to $\ell^2_2$ and clearly it is $1$-complemented. \end{example} \begin{example}\lb{ex2} Consider 4 dimensional Orlicz space $\ell_\varphi$ where $$\varphi (t) = \left\{ \begin{array}{ll} t^2&\mbox{if }\ 0\le t\le a\\ (1+a)t-a&\mbox{if }\ a\le t\le1\end{array} \right.$$ and $\sqrt{2/3} a$. Then, by \eqref{ex22} $$\sum_{i\ne j} \frac{|x_i|^2}{\alpha^2} < 1-a^2\ .$$ Hence $$\left| \frac{x_1+x_2+x_3}{\alpha}\right| \ge \frac{|x_j|}{\alpha} - \sum_{\textstyle \genfrac{}{}{0pt}{}{i=1}{i\ne j}}^3 \frac{|x_i|}{\alpha} \ge a-\sqrt2\ \sqrt{1-a^2} > 0$$ since $a> \sqrt{2/3}$. But this contradicts the fact that $x\in F = \ker (e_1^* + e_2^* +e_3^*)$. Thus $\frac{|x_j|}{\alpha} \le a$ for $j=1,2,3$, so $$1= \sum_{i=1}^4 \varphi \left( \frac{|x_i|}{\alpha}\right) = \sum_{i=1}^3 \frac{|x_i|^2}{\alpha^2} +\varphi \left(\frac{|x_4|}{\alpha}\right) \ .$$ Equation \eqref{ex21} and Theorem~\ref{KW} immediately imply that $e_1-e_2$ and $e_1-e_3$ are hermitian in $F$ and belong to the same Hilbert component of $F$, but clearly $F$ is not isometric to $\ell_2$. Moreover \eqref{ex21} implies that $F$ is 1-complemented in $\ell_\varphi$. Indeed, define $P:\ell_\varphi\to F$ by: \begin{equation*} \begin{split} P(x_1,x_2,&x_3,x_4) = \\ &\left( x_1 -\frac13 (x_1+x_2+x_3), x_2-\frac13 (x_1+x_2+x_3) , x_3 -\frac13 (x_1+x_2+x_3),x_4\right) . \end{split} \end{equation*} Notice that $Q:\ell_2^3 \to \ell_2^3$ defined by $$Q(x_1,x_2,x_3) = \left( x_1 -\frac13 (x_1+x_2+x_3), x_2-\frac13 (x_1+x_2+x_3) , x_3 -\frac13 (x_1+x_2+x_3)\right)$$ is the norm one projection on this Hilbert space. Let $x\in\ell_\f$. Denote $y=Px$ and $\|Px\|_\varphi =\beta$. Then, by \eqref{ex21} \begin{eqnarray*} 1&=&\sum_{i=1}^3 \frac{|y_i|^2 }{\beta^2} + \varphi \left(\frac{|y_4|}{\beta} \right) = \frac{\|(y_1,y_2,y_3)\|_2^2 }{ \beta^2} +\varphi \left( \frac{|y_4|}{\beta}\right) \\ &=& \frac{\|Q(x_1,x_2,x_3)\|_2^2 }{\beta^2} +\varphi\left( \frac{|y_4|}{\beta}\right) \le \frac{\|(x_1,x_2,x_3)\|_2^2}{ \beta^2} +\varphi \left( \frac{|x_4|}{\beta}\right) \\ &=& \frac{|x_1|^2}{\beta^2} + \frac{|x_2|^2}{\beta^2} + \frac{|x_3|^2}{\beta^2} +\varphi\left( \frac{|x_4|}{\beta}\right)\le\\ &\le& \sum_{i=1}^4 \varphi \left( \frac{|x_i|}{\beta}\right)\ . \end{eqnarray*} Thus $\|x\|_\varphi \ge \beta = \|Px\|_\varphi$, i.e. $\|P\|\le1$. \end{example} %1.5 hours more %page 20 We finish this section with a lemma characterizing $1$-complemented subspaces $F$ in $X$ that are isometric to $\ell^2_2$ in terms of norming functionals. % %page 21 \begin{lemma} \label{l2} If $\Span \{x,y\} \subset X$ is $1$-complemented in $X$ and $\Span\{x,y\}$ is isometric to $\ell^2_2$, then there exist $x^*$ norming for $x$ and $y^*$ norming for $y$ such that, for all $a, b \in \C$ with $|a|^2 + |b|^2 = 1$, the functional $\bar{a} x^* + \bar{b} y^*$ is norming for $ax + by$. \end{lemma} \begin{proof}%[Proof of Lemma \ref{l2}] Let $P: X \to \Span\{x,y\}$ be a norm-one projection. Then there exist $x^*$ norming for $x$ and $y^* $ norming for $y$ such that $$ P(u) = x^*(u)x + y^*(u)y \quad \text{ for all } u \in X $$ and $x^*(y) = y^*(x)= 0$. Moreover, $\Span \{x^*, y^*\} \subset X^*$ is isometric to $\ell^2_2$. Thus $$ (\bar{a} x^* + \bar{b} y^*)(ax + by) = a \bar{a} x^*(x) + \bar{b} b y^*(y)= |a|^2 + |b|^2 = \|\bar{a} x^* + \bar{b} y^*\|_{x^*} \|ax + by\|_x. \eqno\qed $$ \let\qed\relax \end{proof} %page 22 \section{$1$-complemented subspaces of $\ell_p(\ell_q)$} \label{seclplq} The main results of this section (Theorems \ref{lplq} and \ref{l2lq}) completely characterize $1$-com\-ple\-men\-ted subspaces of the complex space $\ell_p(\ell_q)$, $ 1 < p, q < \infty$, $p \neq q$. To formulate the theorem conveniently we will need some notation. For all $x$ in $\ell_p(\ell_q)$ we will write $x=\sum^n_{i=1}x_i$, where $n = \dim \ell_p \leq \infty$, and $x_i \in \ell_q$ for all $i \leq n$. We will also write $$ x= \sum^n_{i=1} \sum^m_{j=1} x_{ij} e_{ij}, $$ where $m = \dim \ell_q \leq \infty$ and $e_{ij}$ are standard basis elements in $\ell_p(\ell_q)$. % %page 23 We will distinguish two types of support of $x$: \begin{itemize} \item the usual support $$ \supp x = \{ (i,j) \subset \{1, \dots,n\} \times \{1, \dots, m\}: x_{ij} \neq 0\}, $$ which we will sometimes call the \emph{scalar support} of $x$, and \item the \emph{vector support} $$ \vsupp x = \{i \subset \{1, \dots, n\} : \ell_q \ni x_i \neq \vec{0} \}. $$ \end{itemize} Thus we will have notions of disjointness of elements in $\ell_p(\ell_q)$ in the scalar and vector senses. % %page 24 \begin{theorem} \label{lplq} Let $1 < p, q < \infty$ with $p \neq q, 2$. Consider the complex space $\ell_p(\ell_q)$. Then $Y \subset \ell_p(\ell_q)$ is $1$-complemented if and only if there exist $\{v^i\}^{\dim Y}_{i=1} \subset \ell_p(\ell_q)$ so that $$ Y = \overline\Span \{v^i\}^{\dim Y}_{i=1}, $$ and for all $i\ne j \le \dim Y$ one of the following conditions holds: \begin{itemize} \item[(a)] $\vsupp v^i \cap \vsupp v^j = \varnothing $, \end{itemize} \noindent or \begin{itemize} \item[(b)] $ \vsupp v^i = \vsupp v^j $, $\|v^i_k\|_q=\|v^j_k\|_q$ for all $k\in \vsupp v^i,$ and \begin{itemize} \item[(b1)] if $q\ne2$ then $\supp v^i \cap \supp v^j = \varnothing $, \item[(b2)] if $q=2$ then $ v^i_k, v^j_k$ are orthogonal for each $k\in \vsupp v^i.$ \end{itemize} \end{itemize} \end{theorem} The structure of 1-complemented subspaces of $\ell_2(\ell_q)$ can be somewhat more complicated as illustrated in Example~\ref{exl2lq}. We have the following: \begin{theorem} \label{l2lq} Let $1 < q < \infty$ with $q \neq 2$. Consider the complex space $\ell_2(\ell_q) = \ell^n_2(\ell^m_q)$, $n,m \in \bbN\cup\{\infty\}$. Then $Y \subset \ell_2(\ell_q)$ is $1$-complemented if and only if there exist $\{v^i\}^{\dim Y}_{i=1} \subset \ell_p(\ell_q)$ so that $$ Y = \overline\Span \{v^i\}^{\dim Y}_{i=1}, $$ and for all $i\ne j \le \dim Y$ one of the conditions $(a)$ or $(b)$ of Theorem~\ref{lplq} holds or \begin{itemize} \item[(c)] for each $k\in \vsupp v^i \cap \vsupp v^j$ there exists a constant $C_k\in\bbC$ with $$v^i_k = C_k v^j_k$$ and for every map $\s:\{1,\dots,n\} \lra \{1,\dots,m\}$ such that $(k,\s(k))\in\supp v^i \cup \supp v^j$ whenever $k\in \vsupp v^i \cup \vsupp v^j$, the vectors $$s_\s(v^i) =\left(\|v^i_k\|_q \frac{v^i_{k\s(k)}}{|v^i_{k\s(k)}|}\right)_{k\le n} \in \ell_2, \text{ and } s_\s(v^j) =\left(\|v^j_k\|_q \frac{v^j_{k\s(k)}}{|v^j_{k\s(k)}|}\right)_{k\le n} \in \ell_2$$ (with the convention $0/0 =0$) are orthonormal. \end{itemize} \end{theorem} \begin{corollary} \lb{ge} If $Y \subset \ell_p(\ell_q)$, $1 < p, q < \infty$, $p \neq q$, is $1$-complemented in $\ell_p(\ell_q)$ \wtw $Y$ is isometric to $\sum\oplus_p Y_i$, where each $Y_i$ is isometrically isomorphic to $\ell_q^{d_i}$, $d_i = \dim Y_i \in \bbN \cup\{\infty\}$. \end{corollary} \begin{proof}[Proof of Corollary~\ref{ge}] The ``only if'' part follows immediately from Theorems~\ref{lplq} and \ref{l2lq}, and the ``if'' part when $q\ne2$ is a simple consequence of \cite[Lemma~6]{Kol91} and Theorems~\ref{lplq} and \ref{l2lq}. When $q=2$ the ``if'' part follows from Lemma~\ref{l2cor}. \end{proof} For the proof of Theorems~\ref{lplq} and \ref{l2lq} we use Theorem \ref{main} and the following lemmas: \begin{lemma} \label{disj} Suppose that $x^1,x^2$, $\|x^1\|=\|x^2\|=1$, are disjointly supported (in the scalar sense). Then $\Span\{x^1,x^2\}$ is $1$-complemented in $\ell_p(\ell_q)$, with $1< p, q <\infty$, $p \neq q$, \wtw one of the following conditions holds: \begin{itemize} \item[(a)] $\vsupp x^1 \cap \vsupp x^2 = \varnothing $, in this case $F$ is isometric to $\ell_p^2$, \end{itemize} \noindent or \begin{itemize} \item[(b)] $\vsupp x^1 = \vsupp x^2$ and $\|x^1_i\|_q=\|x^2_i\|_q$ for all $i\in \vsupp x^1,$ in this case $F$ is isometric to $\ell_q^2$, \end{itemize} \end{lemma} \begin{lemma} \label{l2cor} Suppose that $F \subset \ell_p(\ell_q) $ is isometric to $\ell^2_2$ and is $1$-complemented in $\ell_p(\ell_q),$ $1\le p, q <\infty$, $p \neq q$. Then \begin{itemize} \item[(a)] $F$ is spanned by disjointly supported vectors (in the scalar sense), \item[(b)] $p=2$, \end{itemize} \noindent or \begin{itemize} \item[(c)] $q=2$ and there exists a surjective isometry $U$ of $\ell_p(\ell_q)$ such that $UF$ is spanned by disjointly supported vectors (in the scalar sense). \end{itemize} \end{lemma} \begin{remark} It follows immediately from Lemma~\ref{disj} that if Lemma~\ref{l2cor}$(a)$ holds then $p=2$ or $q=2$. \end{remark} \begin{lemma} \label{l2iff} Let $x, y \in \ell^n_2(\ell^m_q)$, $n,m \in \bbN\cup\{\infty\}$, $1\le q <\infty$, $q \neq 2$, $\|x\|=\|y\|=1$. Then $F= \Span\{x,y\}\subset\ell_2(\ell_q)$ is isometrically isomorphic to $\ell_2^2$ and $1$-complemented in $\ell_2(\ell_q)$ \wtw for each $i\in \vsupp x \cap \vsupp y$ there exists a constant $C_i\in\bbC$ with $$x_i = C_i y_i$$ and for every map $\s:\{1,\dots,n\} \lra \{1,\dots,m\}$ such that $(i,\s(i))\in\supp x \cup \supp y$ whenever $i\in \vsupp x \cup \vsupp y$, the vectors $$s_\s(x) =\left(\|x_i\|_q \frac{x_{i\s(i)}}{|x_{i\s(i)}|}\right)_{i\le n} \in \ell_2, \text{ and } s_\s(y) =\left(\|y_i\|_q \frac{y_{i\s(i)}}{|y_{i\s(i)}|}\right)_{i\le n} \in \ell_2$$ (with the convention $0/0 =0$) are orthonormal. \end{lemma} \begin{proof}[Proof of Theorems~\ref{lplq} and \ref{l2lq}] Since $\ell_p(\ell_q)$ is complex, if $Y$ is $1$-complemented in $\ell_p(\ell_q),$ then $Y$ has a 1-unconditional basis $ \{v^i\}^{\dim Y}_{i=1}$. Consider $v^i, v^j$ for some $i\ne j \le \dim Y$. If $v^i, v^j$ belong to different Hilbert components of $Y$, then by Theorem~\ref{main}, $v^i$ and $v^j$ are scalarly disjoint and conditions $(a), (b)$ of Theorem~\ref{lplq} follow from Lemma~\ref{disj} (since it is clear that $\Span\{v^i, v^j\}$ is 1-complemented in $Y$ and therefore in $\ell_p(\ell_q)$). If $v^i, v^j$ are in the same Hilbert component of $Y$, then $\Span\{v^i, v^j\}$ is isometric to $\ell^2_2$ and when $p\ne 2$ Lemma~\ref{l2cor} reduces our considerations to the case of disjointly supported vectors (in the scalar sense), where we apply Lemma~\ref{disj}. When $p=2$ we apply Lemma~\ref{l2iff}. \end{proof} %page 26 \begin{proof}[Proof of Lemma \ref{disj}] Notice that when $z\in\ell_p(\ell_q)$ then the norming functional $z^*$ for $z$ is given by \begin{equation}\lb{zn} z^* = \frac1{\|z\|^{p-1}} \sum_i \|z_i\|_q^{p-q}\sum_j |z_{ij}|^{q-1} \sgn(\overline{z_{ij}}) e_{ij}^*. \end{equation} By Proposition~\ref{calvert}, $\Span\{x^1,x^2\}$ is 1-complemented in $\ell_p(\ell_q)$ \wtw for all $a^1, a^2 \in \bbC$ there exist $K_s,$ ($= K_s(a^1, a^2)$), $s=1, 2$, in $\bbC$ so that \begin{equation}\lb{cal} (a^1x^1 + a^2x^2)^* = K_1x^{1*} + K_2x^{2*}. \end{equation} This is equivalent, by disjointness of $x^1,x^2$ and \eqref{zn}, to the existence of constants $K_s,$ ($= K_s(a^1, a^2)$), $s=1, 2$, such that for all $(i,j)\in\supp x^s$: \begin{equation*} \begin{split} \frac{1}{\|a^1x^1 + a^2x^2\|^{p-1}} &\|a^1x^1_i + a^2x^2_i\|_q^{p-q} |a^sx^s_{ij}|^{q-1} \sgn(\overline{a^sx^s_{ij}}) \\ &= K_s(a^1, a^2) \|x^s_i\|_q^{p-q} |x^s_{ij}|^{q-1} \sgn(\overline{x^s_{ij}}), \end{split} \end{equation*} which is further equivalent to the existence of constants $C_s,$ ($= C_s(a^1, a^2)= K_s(a^1, a^2)\cdot \|a^1x^1 + a^2x^2\|^{p-1}$), $s=1, 2$, such that for all $i\in\vsupp x^s$: \begin{equation*} \frac{\|a^1x^1_i + a^2x^2_i\|_q^{p-q}}{\|x^s_i\|_q^{p-q}} = \left(|a^s| + |a^t| \frac{\|x^t_i\|_q^{q}}{\|x^s_i\|_q^{q}} \right)^{\frac{p-q}{q}} = C_s(a^1, a^2), \end{equation*} where $t\ne s$ and $\{t,s\} =\{1,2\}$. Since $p\ne q$ this is equivalent to the conditions that for all $(i,j)\in\supp x^s$: \begin{equation*} \frac{\|x^t_i\|_q}{\|x^s_i\|_q} = \frac{\|x^t_j\|_q}{\|x^s_j\|_q}. \end{equation*} This means that either $\vsupp x^1 \cap \vsupp x^2 = \varnothing $ or $\vsupp x^1 = \vsupp x^2$ and $\|x^1_i\|_q=\|x^2_i\|_q$ for all $i\in \vsupp x^1$ (since $\|x^1\|=\|x^2\|$). The fact that $\Span\{x^1,x^2\}$ is isometric to $\ell_p^2$ or $\ell_q^2$, resp., follows immediately. \end{proof} \begin{proof}[Proof of Lemma \ref{l2cor}] \mbox{ } First we prove that if Part $(a)$ does not hold then $p=2$ or $q=2$. The proof follows the standard technique of showing that $\ell_2$ can be isometrically embedded in $\ell_p$ only when $p$ is an even integer (see \cite{LV}) and was suggested to me by Alex Koldobsky. Assume that $F= \Span \{x,y\}$, where $\|x\| = \|y\| = 1$ and for all $a, b \in \C$ $$ \|ax + by\| = (|a|^2 + |b|^2)^{\frac{1}{2}}. $$ Then \begin{equation} \label{diff} (|a|^2 + |b|^2 )^{\frac{p}{2}} = \|ax + by\|^p = \sum^n_{i=1} \biggl(\sum^m_{j=1} |ax_{ij} + by_{ij}|^q\biggr)^{\frac{p}{q}}. \end{equation} For each $i \leq n$ let $S_i = \{ j\le m \ : \ (x_{ij}, y_{ij})\ne (0,0)\}= \{ j\le m \ : (i,j)\in \supp x\cup \supp y\}$. We define an equivalence relation $R_i$ on the set $S_i$ by the condition that %page 33 $(j_1, j_2) \in R_i $ if and only if the pairs $(x_{ij_1}, y_{ij_1})$ and $(x_{ij_2}, y_{ij_2})$ are proportional. Let $\Lambda_i$ denote the set of equivalence classes for $R_i$ and let $J= \{i\le n \ : \ \card(\La_i) = 1\}$ and $M = \{i\le n \ : \ \card(\La_i) > 1\}$. For each $i\in J$ let $j_i$ be the representative of the equivalence class of $R_i$ (i.e. $j_i \in S_i$) and for $i\in M$ let $\{j_\alpha\}_{\alpha \in \Lambda_i}$ be the set of representatives of each equivalence class. Then there exist positive constants $\{K_i\}_{i\in J}$, $\{C_{i,\alpha}\}_{i\in M, \al\in \La_i}$ so that \eqref{diff} can be written as \begin{equation} \label{diff1} (|a|^2 + |b|^2)^{\frac{p}{2}} = \sum_{i\in J} K_i |a x_{ij_i } + b y_{ij_i } |^p+ \sum_{i\in M} \biggl(\sum_{\alpha \in \Lambda_i} C_{i,\alpha} |a x_{ij_\alpha } + b y_{ij_\alpha } |^q\biggr)^{\frac{p}{q}}, \end{equation} where the pairs $(x_{ij_\alpha}, y_{ij_\alpha})_{\alpha \in \Lambda_i}$ are mutually linearly independent for each $i\in M$. If there exists $i\in M$ and $ \beta \in \Lambda_i$ with $x_{ij_\beta} \cdot y_{ij_\beta} \neq 0$, then there exist $a_0, b_0 \in \C$ with \begin{align*} a_0 x_{ij_\beta} + b_0 y_{ij_\beta} &= 0\\\noalign{\vskip-6pt} \intertext{and}\noalign{\vskip-6pt} a_0 x_{ij_\alpha} + b_0 y_{ij_\alpha} &\neq 0 \quad \text{ for all } \alpha \neq \beta, \al\in\La_i. \end{align*} %page 34 Fix $b = b_0$ and differentiate \eqref{diff1} with respect to $a$ along the real axis at $a_0$. The left-hand side of \eqref{diff1} can be differentiated (in this fashion) infinitely many times. But, if $q$ is not an even integer then the $([q] + 1)$-st derivative of $|ax_{ij_\beta}+b_0 y_{ij_\beta}|^q$ does not exist at $a_0$. Hence, if $q$ is not even, $x_{ij} \cdot y_{ij} = 0$ for all $i\in M,\ j\in S_i$. In particular, this implies that $\card(\La_i) = 2$ for all $i\in M$. % i.e., $x$ and $y$ are (scalarly) disjoint. Similarly, if there exists $i\in J$ with $x_{ij_i} \cdot y_{ij_i} \neq 0$, then there exist $a_0, b_0 \in \C$ with $ a_0 x_{ij_\beta} + b_0 y_{ij_\beta} = 0 $, we fix $b = b_0$ and differentiate \eqref{diff1} with respect to $a$ along the real axis at $a_0$. As before, the left-hand side of \eqref{diff1} can be differentiated (in this fashion) infinitely many times. But, if $p$ is not an even integer then the $([p] + 1)$-st derivative of $|ax_{ij_i}+b_0 y_{ij_i}|^p$ does not exist at $a_0$. Hence, if $p$ is not even, $x_{ij} \cdot y_{ij} = 0$ for all $i\in J,\ j\in S_i$. Further, by Lemma~\ref{l2}, if $x^* , y^*$ denote norming functionals for $x, y$, respectively, then $\Span \{x^* , y^*\}$ is isometric to $\ell_2^2$ and is 1-complemented in $(\ell_p (\ell_q ))^* = \ell_{p'}(\ell_{q'})$, where $1/p + 1/p' = 1=1/q+ 1/q'$. By \eqref{zn} relations analogous to $R_i$ defined using coefficients of $x^*$ and $y^*$ are identical as the original relations $R_i$. Hence, if $q'$ is not even, then $x^*_{ij} \cdot y^*_{ij} = 0$ for all $i\in M,\ j\in S_i$ and if $p'$ is not even, then $x^*_{ij} \cdot y^*_{ij} = 0$ for all $i\in J,\ j\in S_i$. Thus, by \eqref{zn}, if $q$, $q'$, $p$, $p'$ are not all even integers, that is, $q \ne 2$ and $p \ne 2$ then $x$ and $y$ are (scalarly) disjoint. \end{proof} We postpone the proof of Part~$(c)$, because the proof of Lemma~\ref{l2iff} is the direct continuation of just presented argument and uses the same notation. \begin{proof}[Proof of Lemma~\ref{l2iff}] \mbox{ } The ``only if'' part: By the first part of the proof of Lemma~\ref{l2cor}, since $q \ne 2$, $x_{ij} \cdot y_{ij} = 0$ for all $i\in M,\ j\in S_i$. We will show that $M = \varnothing$. Since $\Span\{x,y\} = \ell_2^2$ we get, by Lemma~\ref{l2}, that for all $a, b \in \C$ with $|a|^2 + |b|^2 = 1$, the functional $\bar{a} x^* + \bar{b} y^*$ is norming for $ax + by$. That is, by \eqref{zn}, \begin{equation*} \begin{split} \sum_i \|ax_i&+by_i\|_q^{2-q}\sum_j |ax_{ij}+by_{ij}|^{q-1} \sgn(\overline{ax_{ij}+by_{ij}}) e_{ij}^* = \\ &\bar a \sum_i \|x_i\|_q^{2-q}\sum_j |x_{ij}|^{q-1} \sgn(\overline{x_{ij}}) e_{ij}^* + \bar b \sum_i \|y_i\|_q^{2-q}\sum_j |y_{ij}|^{q-1} \sgn(\overline{y_{ij}}) e_{ij}^*. \end{split} \end{equation*} Thus if $i\in M$, by disjointness of $x_i, y_i$, for each $j$ with $(i,j)\in \supp x$ we have \begin{equation*} \|ax_i+by_i\|_q^{2-q} |ax_{ij}|^{q-1} \sgn(\overline{ax_{ij}}) = \bar a \|x_i\|_q^{2-q} |x_{ij}|^{q-1} \sgn(\overline{x_{ij}}). \end{equation*} Hence, since $q\ne 2$, $$ \|ax_i+by_i\|_q = | a | \|x_i\|_q = \|ax_i\|_q. $$ Since $x_i$ and $y_i$ are disjoint and $\ell_q$ is strictly monotone we conclude that $y_i = 0$. But this implies that $\card(\La_i) = 1$, which contradicts the fact that $i\in M$. Thus $M= \varnothing$. Hence $J= \vsupp x\cup \vsupp y$ and for each $i\in \vsupp x \cap \vsupp y$ there exists a constant $C_i\in\bbC$ with $$x_i = C_i y_i.$$ Now \eqref{diff}, \eqref{diff1} and definition of $J$ imply that for every map $\s:\{1,\dots,n\} \lra \{1,\dots,m\}$ such that $(i,\s(i))\in\supp x \cup \supp y$ whenever $i\in \vsupp x \cup \vsupp y$ we have (with the convention $0/0 =0$): \begin{equation*} \begin{split} |a|^2 + |b|^2 &= \|ax + by\|^2 = \sum_{i\in J} \biggl(\sum_{j\in S_i} |ax_{ij} + by_{ij}|^q\biggr)^{\frac{2}{q}}\\ &= \sum_{i\in J} \left(\sum_{j\in S_i} \left(\frac{|x_{ij}|}{|x_{i\s(i)}|}|ax_{i\s(i)} + by_{i\s(i)}|\right)^q\right)^{\frac{2}{q}}\\ &= \sum_{i\in J} \left(\sum_{j\in S_i} \frac{|x_{ij}|^q}{|x_{i\s(i)}|^q}\right)^{\frac{2}{q}}|ax_{i\s(i)} + by_{i\s(i)}|^2\\ &= \sum_{i\in J} \frac{\|x_{i}\|_q^2}{|x_{i\s(i)}|^2} |ax_{i\s(i)} + by_{i\s(i)}|^2\\ &= \sum_{i\in J} \left|a\frac{\|x_{i}\|_q}{|x_{i\s(i)}|}x_{i\s(i)} + b\frac{\|y_{i}\|_q}{|y_{i\s(i)}|}y_{i\s(i)}\right|^2. \end{split} \end{equation*} Thus the vectors $$s_\s(x) =\left(\|x_i\|_q \frac{x_{i\s(i)}}{|x_{i\s(i)}|}\right)_{i\le n} \in \ell_2, \text{ and } s_\s(y) =\left(\|y_i\|_q \frac{y_{i\s(i)}}{|y_{i\s(i)}|}\right)_{i\le n} \in \ell_2$$ (with the convention $0/0 =0$) are orthonormal. The ``if'' part: It is clear from above calculations that if $x, y$ are of the form described in Lemma~\ref{l2iff} then $\Span\{x, y\}$ is isometrically isomorphic to $\ell_2^2.$ Further, by \eqref{zn} we have for all $a, b\in\bbC$: \begin{equation*} \begin{split} (ax &+by)^* = \frac1{(|a|^2 + |b|^2 )^{\frac12}} \sum_i \|ax_i+by_i\|_q^{2-q}\sum_j |ax_{ij}+by_{ij}|^{q-1} \sgn(\overline{ax_{ij}+by_{ij}}) e_{ij}^* \\ &=\frac1{(|a|^2 + |b|^2 )^{\frac12}} \biggl(\sum_{i\in\vsupp x\setminus\vsupp y} \|ax_i\|_q^{2-q}\sum_j |ax_{ij}|^{q-1} \sgn(\overline{ax_{ij}}) e_{ij}^* \\ &\hspace{5mm} + \sum_{i\in\vsupp x\cap\vsupp y} \|ax_i+bC_ix_i\|_q^{2-q}\sum_j |ax_{ij}+bC_ix_{ij}|^{q-1} \sgn(\overline{ax_{ij}+bC_ix_{ij}}) e_{ij}^* \\ &\hspace{5mm} + \sum_{i\in\vsupp y\setminus\vsupp x} \|by_i\|_q^{2-q}\sum_j |by_{ij}|^{q-1} \sgn(\overline{by_{ij}}) e_{ij}^* \biggr) \end{split} \end{equation*} \begin{equation*} \begin{split} &=\frac1{(|a|^2 + |b|^2 )^{\frac12}} \biggl(\bar a \sum_{i\in\vsupp x\setminus\vsupp y} \|x_i\|_q^{2-q}\sum_j |x_{ij}|^{q-1} \sgn(\overline{x_{ij}}) e_{ij}^* \\ &\hspace{5mm} + \overline{a+bC_i}\sum_{i\in\vsupp x\cap\vsupp y} \|x_i\|_q^{2-q}\sum_j |x_{ij}|^{q-1} \sgn(\overline{x_{ij}}) e_{ij}^* \\ &\hspace{5mm} + \bar b \sum_{i\in\vsupp y\setminus\vsupp x} \|y_i\|_q^{2-q}\sum_j |y_{ij}|^{q-1} \sgn(\overline{y_{ij}}) e_{ij}^* \biggr)\\ &=\frac1{(|a|^2 + |b|^2 )^{\frac12}} ( \bar a x^* + \bar b y^*). \end{split} \end{equation*} Thus, by Lemma~\ref{calvert}, $\Span\{x, y\}$ is 1-complemented. \end{proof} \begin{proof}[Proof of Lemma~\ref{l2cor}$(c)$] Assume that $q=2$. First notice that, if $z \in \ell_p (\ell_2 )$ and $\|z\| = 1$, then the norming functional $z^*$ for $z$ is given by \begin{equation} \label{zstar} z^*=\sum_{i=1}^n\|z_i\|_2^{p-2} \cdot \sum _{j=1}^m \overline{z_{ij}}e_{ij} \end{equation} Let $F = \Span \{x, y\}$, where $\|x\| = \|y\| = 1$ and $\|ax + by\| = (|a|^2 + |b|^2 )^{1/2}$ for all $a, b \in \C$. Let $U_i : \ell_m^2 \to \ell_m^2$ be a surjective isometry such that, for all $i \le n$, $$ U_ix_i = \|x_i\|_2 e_1 . $$ Define an isometry $U : \ell_p (\ell_2 ) \to \ell_p (\ell_2 )$ by $$ U ((z_i)^n_{i=1}) = (U_iz_i)^n_{i=1}. $$ Then $U F = \Span\{U x, U y\}$ is isometric to $\ell_2^2$ and is 1-complemented in $\ell_p (\ell_2 )$. Thus, by Lemma~\ref{l2}, for all $a, b \in \C$ with $|a|^2 + |b|^2 = 1$, we have $$ (a U x + bU y)^* = \bar a (U x)^* + \bar b(U y)^*. $$ Hence, by \eqref{zstar} and by the choice of $U$, we get for all $i \le n$ and all $j\geq 2$: % \begin{equation} \label{isol21} \begin{split} \|a(U x)_i + b(U y)_i\|^{p-2}_2 &\cdot \left(\bar a \cdot 0 + \bar b \cdot \overline{ (U y)_ {ij}}\right) = \\ &\bar a \cdot \|(U x)_i\|_2^{p-2} \cdot 0 + \bar b \cdot \|(Uy)_i\|_2^{p-2} \overline{(U y)_{ij}}. \end{split} \end{equation} Now for each $i$ with $\|(Ux)_i\|_2\ne 0$ we consider two cases: either \begin{enumerate} \item there exists $j\geq 2$ with $(U y)_ {ij}\ne0$, or \item $(U y)_i = ((U y)_{i1}, 0, 0,\dots,0)$. \end{enumerate} In case $(1)$ we have $$ \|a(U x)_i + b(U y)_i\|_2^{p-2} = \|(U y)i\|_2^{p-2} $$ whenever $|a|^2 + |b|^2 = 1$. In particular, since $p\ne 2$ and $\|(U x)_i\|_2 =\|(U y)_i\|_2$, we get $$ \left\| a \frac{(Ux)_i}{\|(Ux)_i\|_2}+ b \frac{(Uy)_i}{\|(Uy)_i\|_2}\right\|_2=1. $$ Thus ${(Ux)_i}/{\|(Ux)_i\|_2}$ and ${(Uy)_i}/{\|(Uy)_i\|_2}$ form an orthonormal basis for $\ell_2$. Since $(Ux)_i/\|(Ux)_i\|= (1, 0, 0,\dots)$, the vectors $(U x)_i$ and $(U y)_i$ are disjoint. In case $(2)$, i.e. when $(U y)_i = ((U y)_{i1}, 0, 0,\dots,0)$, then by \eqref{zstar} and by the form of $U$ we get for all $a, b$ with $|a|^2 + |b|^2 = 1$: \begin{equation} \label{isol22} \begin{split} |a(U x)_{i1} + b(U y)_{i1}|^{p-2}&\cdot (\bar a (U x)_{i1} + \bar b\,\overline{(U y )_{i1}}) = \\ &\bar a (U x)^{p-2}_{i1} (U x)_{i1} + \bar b|(U y)_{i1}|^{p-2} \overline{ (U y)_{i1}}. \end{split} \end{equation} Let $a \in\R_+$ and $b = c e^{i\theta}$, where $c \in\R_+$, $a^2 + c^2 = 1$ and $\theta$ is such that $e^{-i\theta}\cdot \overline{ (U y)_{i1}} = |(U y)_{i1}|$. Set $\alpha = (U x)_{i1} > 0$, $\beta = |(U y)_{i1}|\geq 0$. Then \eqref{isol22} becomes \begin{equation} \label{isol23} (a\alpha + c\beta)^{p-1} = a\alpha^{p-1} + c\beta^{p-1} , \end{equation} and this equation holds for all $a, c \geq 0$ with $a^2 + c^2 = 1$. For any fixed $u, v \geq 0$ define a function $f_{u,v} : [0, 1] \to \R$ by $$ f_{u,v}(a) = au + \sqrt{ 1- a^2 } v. $$ It is not difficult to check that $f_{u,v}$ attains its maximum on $[0,1]$ at the point $a_0 = a_0 (u, v) = u (u^2 + v^2 )^{-1/2}$ and the maximum value of $f_{u,v}$ is equal to $M (u, v) = (u^2 + v^2 )^{1/2}$. Since equation \eqref{diff1} can be written as $$ (f_{\alpha,\beta}(a))^{p-1}=f_{\alpha^{p-1},\beta^{p-1}(a)}, $$ we have \begin{align*} a_0(\alpha,\beta)&=a_0(\alpha^{p-1},\beta^{p-1}),\\ M(\alpha,\beta)^{p-1}&=M(\alpha^{p-1},\beta^{p-1}). \end{align*} Thus $$ \biggl(\frac{\alpha}{\sqrt{\alpha^2+\beta^2}}\biggr)^{p-1}=\frac {\alpha}{\sqrt{\alpha^2+\beta^2}}. $$ Since $p\neq2$, we conclude that either $\alpha = 0$ or $\alpha (\alpha^2 + \beta^2 )^{-1/2} = 1$, i.e., $\beta = 0$. But $\alpha = (U x)_{i1}= \|(Ux)_i\|_2$ and $\beta = |(U y)_{i1}|= \|(Uy)_i\|_2$. Thus $(U x)_i = 0$ or $(U y)_i = 0$. Hence $U x$ and $U y$ are disjoint. \end{proof} \begin{remark} Lemmas \ref{disj}-\ref{l2iff} are all valid (with the presented proofs) both in the complex and real case. We suspect that Theorems~\ref{lplq} and \ref{l2lq}, too, are true in the real case, but our method of proof does not work then. \end{remark} %page 42 \section{$1$-complemented disjointly supported subspaces of Orlicz and Lorentz spaces} \label{gen} In this section we fully characterize subspaces of (real or complex) Orlicz and Lorentz spaces that are spanned by disjointly supported elements and $1$-complemented. In particular, it follows from Theorems \ref{orlicz} and \ref{glorentz} that in ``most'' Orlicz and Lorentz spaces the only $1$-complemented disjointly supported subspaces are those spanned by a block basis with constant coefficients (of some permutation of the original basis). % %page 43 \begin{theorem} \label{orlicz} Let $\ell_\phi$ be a (real or complex) Orlicz space and let $x,y \in \ell_\phi$, be disjoint elements such that $\|x\|_\phi = \|y\|_\phi = 1$ and $\Span \{x,y \} $ is $1$-complemented in $\ell_\phi$. Then one of three possibilities holds: \begin{itemize} \item[(1)] $\card (\supp x) < \infty$ and $|x_i| = |x_j|$ for all $i,j \in \supp x$; or \item[(2)] there exists $p$, $1 \leq p \leq \infty$, such that $\phi(t) = Ct^p$ for all $t \leq \|x\|_\infty$; or \item[(3)] there exists $p$, $1 \leq p \leq \infty$, and constants $C_1, C_2, \gamma \geq 0$ such that $C_2 t^p \leq\nobreak \phi(t) \leq C_1 t^p$ for all $t \leq \|x\|_\infty$ and such that, for all $j \in \supp x$, $$|x_j| = \gamma^{k(j)} \cdot \|x\|_\infty$$ for some $k(j) \in \Z$. %page 44 \end{itemize} \end{theorem} %page 45 For the proof of the theorem we will need the following (well-known?) lemma, whose proof is outlined in \cite{Z76}. For the convenience of the reader we provide its proof below. % %page 46 \begin{lemma} \label{z} Let $a > 0$ and suppose that $\phi: [0,a] \to \R $ is an increasing differentiable function with $\phi(0)=0$. Suppose that there exist $a < \alpha, \beta < 1$ so that, for all $u \leq a$, % \begin{equation} \label{mult} \phi(\alpha u) = \beta \phi(u). \end{equation} % Then there exist $p > 0$ and $C_1, C_2 > 0$ such that, for all $u \leq a$, $$ C_2 u^p \leq \phi(u) \leq C_1 u^p. $$ Moreover, if $\phi(u) \not\equiv C \cdot u^p$, there exists $\gamma > 0$ such that \eqref{mult} is satisfied (with the corresponding $\beta$) if and only if $\alpha = \gamma^k$ for some $k \in\Z$. \end{lemma} %page 47 \begin{proof}[Proof of Theorem \ref{orlicz}] Let $z \in \ell_\phi$. By \cite{GrzH} the norming functional $z^*$ of $z$ is given by: $$ z^*_i = \frac1C \sgn (\overline{z_i}) \phi' \Bigl(\frac{|z_i|}{\|z \|_\phi}\Bigr), $$ where $C$ is a constant depending on $z$. By Proposition \ref{conj}, for all $b \in \C$ there exist constants $K_1, K_2$ such that $$ (x + by)^* = K_1 x^* + K_2 y^*. $$ Since $x$ and $y$ are disjoint, there exists a constant $K=K(b)$ so that for all $i \in \supp x$ $$ \sgn (\overline{x_i}) \phi' \left(\frac{|x_i|}{\| x + by \|_\phi}\right ) = K \cdot \sgn (\overline{x_i})\phi' (|x_i|). $$ %page 48 %4 hours total Now for all $, 0 0$ so that for all $i \in \supp x$ $$ \phi'(|x_i|\cdot t) = C_t \phi'(|x_i|). $$ Hence for all $i,j \in \supp x$ and for all $t \leq 1$ $$ \frac{\phi'(|x_i|\cdot t)}{\phi'(|x_i|)} = \frac{\phi' (|x_j|\cdot t)}{\phi' (|x_j|)}. $$ Set $$ \beta = \frac{ \phi' (|x_i|)}{\phi' (|x_j|)}, \quad u= |x_j|\cdot t ,\quad \alpha = \frac{|x_i|}{|x_j|}. $$ In this notation we have $$\phi' (\alpha u)=\beta \phi'(u)$$ for all $u$ such that $0 \leq u \leq |x_j|$. Thus $$\phi(\alpha u) = \beta \alpha \phi(u)$$ for all $u$ such that $0 \leq u \leq |x_j|$. % %page 49 Let $j \in \supp x$ be such that $|x_j| = \|x\|_\infty$. If there exists $i \in \supp x $ with $|x_i| \neq |x_j| = \|x\|_\infty$ then, by Lemma \ref{z}, condition (2) or (3) holds. \end{proof} %page 50 \begin{proof}[Proof of Lemma \ref{z}] Let $p= \log_\alpha (\beta)$. Let $m_0 \in \Z$ be the smallest integer with $\alpha^{m_0} \leq a$. If $\alpha^{m_0} < u \leq a$ we have $$ \phi(u) \leq \phi(a) \leq \frac{\phi(a)}{\alpha^{m_0 p}} \cdot u^p $$ and $$ \phi(u) \geq \phi(\alpha^{m_0}) \geq \frac{\phi(\alpha^{m_0})}{a^p}\cdot u^p. $$ If $\alpha^{m+1} < u \leq \alpha^m$ for some $m \leq m_0$ we have $$ \thickmuskip=2mu plus 2mu \phi(u) \leq \phi(\alpha^m) = \beta\phi(\alpha^{m-1}) = \cdots =\beta^{m-m_0}\phi(\alpha^{m_0}) =\frac{\phi(\alpha^{m_0})}{\beta^{m_0+1}}(\alpha^p)^{m+1} \leq \frac{\phi(\alpha^{m_0})}{\beta^{m_0+1}} \cdot u^p $$ %page 51 and $$ \thickmuskip=1mu plus 2mu \phi(u) \geq \phi(\alpha^{m+1}) = \beta \phi(\alpha^m) = \dots =\beta^{m+1-m_0} \phi(\alpha^{m_0}) =\frac{\phi(\alpha^{m_0})}{\beta^{m_0-1}}\cdot(\phi^p)^m \geq \frac{\phi(\alpha^{m_0})}{\beta^{m_0-1}}u^p. $$ Set $ C_1 = \max \{ {\phi(a)}/{\alpha^{m_0 p}}, \ {\phi(\alpha^{m_0})}/{\beta^{m+1}}\} $ and $ C_2 = \min \{ {\phi(\alpha^{m_0})}/{a^p},\ {\phi(\alpha^{m_0})}/{\beta^{m_0-1}}\}. $ Then $$C_2 u^p \leq \phi(u) \leq C_1 u^p $$ for all $u$ with $0 \leq u \leq a$. Further define a function $h_\phi: (-\infty, \ln a] \to \R$ by $$ h_\phi(t) = \frac{d}{ds}(\ln(\phi(e^s)))\big|_{s=t}. $$ Then, by \eqref{mult}, $h_\phi(t+ \ln(\alpha)) = h_\phi(t)$ for all $ t \leq \ln a$. %page 52 Thus, since $\alpha \neq 1$, either \begin{itemize} \item $h_\phi$ is constant, that is, there exists a constant $K$ so that $\phi(u) = K \cdot u^p$ for all $ u \leq a$, or \item $h_\phi$ is periodic, that is, there exists $w$, with minimal $|w|$, so that $h_\phi(t+w) = h_\phi(t)$ for all $t \leq \ln a$. \end{itemize} Thus there exists $\gamma>0$ (namely $\gamma=e^w$) and $k \in \Z$ such that $\alpha = \gamma^k$. \end{proof} Our next theorem describes disjointly supported $1$-complemented subspaces of (real or complex) Lorentz sequence spaces. %page 53 \begin{theorem} \label{glorentz} Let $\ell_{w,p}$, with $1 < p < \infty$, be a real or complex Lorentz sequence space. Suppose that $\{ x_i \}_{i \in I}$ are mutually disjoint elements of $\ell_{w,p}$ such that $\card (I) \geq 2$ and $F = \overline{\Span} \{ x_i \}_{i \in I} $ is $1$-complemented in $\ell_{w,p}$. Suppose, moreover, that $w_\nu \neq 0$ for all $\nu \leq \Sigma \defeq \sum_{i \in I} \card (\supp x_i)\quad (\leq \infty)$. Then \begin{itemize} \item[$(a)$] $w_\nu = 1$ for all $ \nu \leq \Sigma$, \end{itemize} \noindent or \begin{itemize} \item[$(b)$] $ |x_{il}| = |x_{ik}|$ for all $i \in I$ and all $k, l \in \supp x_i$. \end{itemize} \end{theorem} %page 54 \begin{proof} With each element $z \in \ell_{w,p}$ we associate a decreasing sequence of positive numbers $(\tilde{z}_i)^{l(z)}_{i=1}$ and ``level sets'' $A_i(z)$ defined inductively as follows: $$ \def\eqalign#1{\null\,\vcenter{\openup\jot \ialign{\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil \crcr#1\crcr}}\,} \eqalign{ \tilde{z}_1 &= \| z \|_\infty, \cr \tilde{z_2} &= \max \{ |z_j|: j \in \N \setminus A_1(z) \},} \qquad \eqalign{A_1(z) &= \{j \in \N: |z_j| = \tilde{z}_1 \},\cr A_2(z) &= \{ j \in \N: |z_j| = \tilde{z}_2 \},} $$ and so on. Note that $l(z)$ is the largest integer such that $\tilde{z}_{l(z)} > 0$ and $\supp z = \bigcup^{l(z)}_{i=1} A_i(z)$. %page 55 For $i \leq l(z)$ introduce also $$s_0(z) = 0,\ \ s_i(z) = \sum^i_{j=1} \card (A_i(z)),$$ $$L_i(z) = \{ s_{i-1}(z) + 1, \dots, s_i(z) \} \subset \N,$$ and let $\delta_i: A_i(z) \to L_i(z)$ be a bijection. Finally, for any set $A \subset \N $ denote by $ \cal{P}(A)$ the set of all permutations of $A $. In this notation we can easily describe norming functionals $z^N$ for $z$. Namely, for each $ j $ with $ 1 \leq j \leq l(z) $, there exists a family of coefficients $ \{ \lambda_\sigma \}_{\sigma \in \cal{P}(L_j(z))}$ such that % %page 56 $ \lambda_\sigma \geq 0, \quad \sum_{\sigma \in \cal{P}(L_j)} \lambda_\sigma = 1 $ and $$ \bigl(|z^N_k|\bigr)_{k \in A_j(z)} = \left (\frac{\tilde{z}_j}{\| z \| }\right )^{p-1} \! \sum_{\sigma \in \cal{P}(L_j(z))} \lambda_\sigma (w_{\sigma(\delta_j (k))})_{k \in A_j(z)}. $$ In particular, we can compute the $\ell_1$-norm of $z^N$ restricted to a level set $A_j(z)$ \begin{equation} \label{l1n} \left\|\bigl(|z^N_k|\bigr)_{k \in A_j(z)}\right\|_{\ell_1} = \left (\frac{\tilde{z}_j}{\| z \| }\right )^{p-1} \! \sum_{n\in L_j(z)} w_n. \end{equation} Notice that the left hand side of \eqref{l1n} does not depend on the choice of the norming functional $z^N$ for $z$. % %page 57 Now assume that $ l(x_1) > 1 $. We will show that $ w_\nu= 1 $ for all natural numbers $\nu \leq \Sigma$, where $\Sigma \defeq \sum_{i \in I} \card (\supp x_i)$ as defined above. If $\nu \leq \Sigma$, there exists $n \leq l_1(x_1)$ such that $$ \nu- s_n(x_1) \leq \Sigma - \card (\supp x_1). $$ Further, there exist $ \mu \in \N$ and $ \{ j_i \}^{\mu}_{i=2} \subset \N $ such that $ j_i \leq l(x_i)$ for $ i=2, \dots, \mu $ and $$\sum^{\mu}_{i=2} s_{j_i}(x_i) =: M \geq \nu - s_N(x_1).$$ Choose $ \{ a_i \}^{\mu}_{i=1} \subset \R_+$ such that, %page 58 for all $i$ with $2 \leq i \leq \mu$, $$ a_1 (\widetilde{{x}_{1}})_{1} > a_i (\widetilde{{x}_{i}})_{j_i} > a_1 (\widetilde{{x}_{1}})_{2} $$ and $ a_1 (\widetilde{{x}_{1}})_{n} > a_i (\widetilde{{x}_{i}})_{j}$ for all $ j > j_i $. To shorten the notation, set $ x = \sum^{\mu}_{i=1} a_1 x_1 $. Then \begin{equation} \label{lov1} a_1 (\widetilde{{x}_{1}})_{1} = \tilde{x}_1\quad\hbox{and} \quad A_1(x) = A_1 (x_1). \end{equation} Moreover, there exists $ k \in \N$ with $2 < k \leq 1 + \sum^{\mu}_{i=2} j_i$ such that, for all $ \alpha $ satisfying $ 2 \leq \alpha \leq n $, \begin{equation} \label{lov3} a_1 (\widetilde{{x}_{1}})_{ \alpha } = \tilde{x}_{k +(\alpha -2)},\quad A_{k+(\alpha -2)}(x) = A_{\alpha}(x_1), \end{equation} and $$ s_{k-1}(x) = s_1 (x_1) +M. $$ Thus \begin{equation} \label{lov4} L_1(x) = L_1(x_1) = \{ 1, \dots, s_1(x_1) \} , \end{equation} %page 59 and, for all $ \alpha $ with $ 2 \leq \alpha \leq n $, \begin{equation} \label{lov5} L_{k + (\alpha -2)}(x) = M + L_\alpha (x_1) = \{ M + s_{\alpha -1}(x_1) + 1,\, \dots,\, M+ s_{\alpha} (x_1) \} . \end{equation} By Proposition \ref{conj} there exist norming functionals $ x^N$ for $x$ and $x_i^N$ for $x_i$, %where $i=1, \dots, \mu $, and constants $ K_i$, where $i=1, \dots, \mu $, such that $$ x^N = \sum^{\mu}_{i=1} K_i x_i^N. $$ Thus, by \eqref{l1n} and \eqref{lov1}, $$ \left ( \frac{{\tilde{x}_1}}{\|x\|}\right )^{p-1} \sum_{j \in L_1(x)} w_j = K_1 \cdot \left (\frac{a_1 (\widetilde{{x}_{1}})_{1}}{\| x_1 \|} \right )^{p-1}\sum_{j \in L_1(x_1)}w_j. $$ Hence, by \eqref{lov4}, \begin{equation} \label{lov6} K_1 \cdot\left( \frac{\|x\|}{\| x_1 \| }\right )^{p-1} = 1. \end{equation} %page 60 Moreover, by \eqref{l1n} and \eqref{lov3}, for all $\alpha$ with $ 2 \leq \alpha \leq n $ we get: $$ \left ( \frac{\tilde{x}_{k + (\alpha -2)}}{\|x\| }\right )^{p-1} \! \sum_{j \in L_{k + (\alpha-2)}(x)} w_j = K_1 \cdot \left ( \frac{a_1 (\widetilde{{x}_{1}})_\alpha}{\|x_1 \| }\right )^{p-1} \! \sum_{j \in L_\alpha (x_1)} w_j. $$ Hence, by \eqref{lov6} and \eqref{lov5}, $$ \sum^{s_\alpha(x_1)}_{j=s_{\alpha-1}(x_1) + 1} w_{j+M} = \sum^{s_\alpha(x_1)}_{j= s_{\alpha-1}(x_1)+1} w_j. $$ Since $\{w_j\}$ is a decreasing sequence of numbers we immediately conclude that, for all $ \alpha$ with $2 \leq \alpha \leq n $, $$ w_{s_{\alpha-1}(x_1)+1} = w_{s_\alpha (x_1) + M}. $$ %page 61 Since $M \geq 1 $ we get \begin{equation} \label{lov7} w_{s_1(x_1)+1} = w_{s_2(x_1) +M} = w_{s_2(x_1)+1}=w_{s_3(x_1)+M} = \dots = w_{s_n(x_1)+M}. \end{equation} Finally, choose $ \{ b_i \}^{\mu}_{i=1} \subset \R_+ $ in such a way that, for all $ i$ with $2 \leq i \leq \mu $, $$ b_i (\widetilde{{x}_{i}})_{j_i} > b_1 (\widetilde{{x}_{1}})_{1}. $$ %page 62 Now set $ y = \sum^{\mu}_{i=1} b_i x_i $. Then there exists $ t \in \N $, with $1 \leq t \leq 1 + \sum^{\mu}_{i=2} j_i$, such that for all $\alpha$ with $1 \leq \alpha \leq n $ we have \begin{equation} \label{lov8} b_1 (\widetilde{{x}_{1}})_{ \alpha} = \tilde{y}_{t + (\alpha -1 )}\ \ ; A_\alpha(x_1) = A_{t+(\al -1)}(y). \end{equation} Similarly, as before, $$ s_{t+(\alpha -1)}(y) = s_\alpha(x_1)+M $$ and \begin{equation} \label{lov9} L_{t+(\alpha-1)}(y) = M + L_\alpha (x_1) = \{ M + s_{\alpha-1} (x_1)+1,\, \dots,\, M + s_\alpha(x_1) \}. \end{equation} Again, by Proposition \ref{conj} there exist norming functionals $y^N$ for $y$ and $x_i^N$ for $x_i$, % where $i=1, \dots , \mu $, and constants $ K_i'$, where $i=1, \dots, \mu $, such that $$ y^N = \sum^\mu_{i=1} K_i' x_i^N. $$ %page 63 Thus, by \eqref{l1n} and \eqref{lov8} we get, for all $ \alpha$ with $1 \leq \alpha \leq n $, $$ \left ( \frac{\tilde{y}_{t+(\alpha-1)}}{\| y \| }\right )^{p-1} \cdot \sum_{j \in L_{t+(\alpha-1)}(y)} w_j = K_1' \! \left ( \frac{(\widetilde{{x}_{1}})_{ \alpha}}{\|x\|}\right )^{p-1} \cdot \sum_{j \in L_\alpha(x_1)} w_j. $$ Hence, by \eqref{lov9}, \begin{equation} \label{lov10} \left (\frac{1}{\| y \| }\right)^{p-1} \cdot \sum^{s_\alpha(x_1)}_{j=s_{\alpha-1}(x_1) + 1} w_{j + M} = K_1' \cdot \left (\frac{1}{\|x\|}\right )^{p-1} \sum^{s_\alpha(x_1)}_{j=s_{\alpha-1}(x_1)+1} w_j. \end{equation} If $ \alpha = 2 $, by \eqref{lov7} we conclude that $$ \left (\frac{1}{\| y \|}\right )^{p-1} = K_1' \cdot \left (\frac{1}{\|x\|}\right )^{p-1}. $$ %page 64 Thus, when $ \alpha = 1 $, \eqref{lov10} becomes $$ \sum^{s_1(x_1)}_{j=1} w_{j+M} = \sum^{s_1(x_1)}_{j=1}w_j. $$ Thus $$ w_1 = w_{s_1(x_1) + M} $$ and since $ M \geq 1 $, by \eqref{lov7} we get $$ w_1 = w_{s_n(x_1) + M}.$$ Since $ \nu \leq s_n(x_1) + M $ we conclude that $$ 1 = w_1 = w_\nu,$$ which ends the proof. \end{proof} \bibliographystyle{alpha} \bibliography{tref} \begin{thebibliography}{ACL} \bibitem[ACL]{ACL} Z.~Altshuler, P.G. 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