N. V. Volosova, “The $p$-convexity functor for $L_p(X)$-spaces”, Sb. Math., 213:6 (2022), 734

In this paper we consider the so-called $\mathbf L$-spaces, which occupy an intermediate position between the classical and quantum (operator) normed spaces.

Objects of this kind were first considered by Lambert in his Ph.D. thesis [1]. Whereas in the theory of quantum spaces (see [2]) one considers sequences of spaces of matrices with entries that lie in some normed spaces and satisfy certain conditions (the Ruan axioms), Lambert’s proposal was to equip with norms columns of elements of a normed space, rather than matrices. The requirements imposed by Lambert on these norms (the contractiveness and convexity axioms), which were formulated in terms of $\ell_2^n$-spaces, made it possible to regard the resulting objects as amplifications of linear spaces by means of $\ell_2^n$-spaces. Subsequently, several authors considered more general $p$-multi-normed spaces defined in terms of $\ell_p^n$-spaces, $1\leqslant p\leqslant\infty$ (see [3]–[6]).

A further generalization of this construction was due to Helemskii (see [7] and [8]), who considered $L_p(X,\mu)$-spaces for arbitrary measurable $X$ as base spaces. This advance was made by using the ‘noncoordinate’ (‘index-free’) approach to the definition of the amplification of a normed space (for a systematic treatment of this approach for operator spaces, see [9]). In the noncoordinate approach quantum spaces are not described by defining a sequence of norms on the spaces $M_n(E)$ of matrices with entries from a normed space $E$, but rather by defining a single norm (of course, satisfying certain conditions) on the tensor product $\mathcal F\otimes E$ of this space and the space $\mathcal F$ of bounded finite-dimensional operators on a separable infinite-dimensional Hilbert space. Correspondingly, $p$-multi-normed spaces can similarly be defined by introducing a norm on the tensor product of a linear space and the space $\ell_p$. The next step, which is the transition from $\ell_p$ to an arbitrary $L_p(X,\mu)$-space, leads to the concept of an $\mathbf L$-space.

The most interesting among such spaces are $p$-convex spaces, which satisfy an analogue of the convexity axiom (the spaces considered by Lambert in [1] have this property for $p=2$). For example, in [8] a tensor product of $p$-convex $\mathbf L$-spaces was constructed — this product has a convenient form for a certain class of factors, for example, the ‘$p$-convex’ tensor product of the spaces $L_q(Y_1,\nu_1)$ and $L_q(Y_2,\nu_2)$ (where $1/p+1/q=1$) which are equipped with the so-called minimal $\mathbf L$-norms, is known to be $L_q(Y_1\times Y_2,\nu_1\times \nu_2)$.

In our paper we propose a method for transforming an arbitrary $\mathbf L$-space into a $p$-convex space. In passing, we construct an $\mathbf L$-space with the maximal $p$-convex norm — such spaces can be regarded as a generalization of maximal $\ell_2$-spaces, considered by Lambert in [1].

Acknowledgements

The author is grateful to A. Ya. Helemskii for his generous advice and encouragement.

§ 2. Basic concepts

In this section we mainly follow [8], where the structures described below were introduced for the first time with the degree of generality and in the form required in our paper.

Let $X$ be a measurable separable space satisfying the following property: $X$ can be represented as a disjoint union of subspaces $X_1$ and $X_2$ which are isomorphic to $X$ as measurable spaces. This is equivalent to saying that either $X$ is atomless or it contains an infinite number of atoms. Such measurable spaces are called convenient. We denote by $\mathbf L$ the $L_p(X)$-space of $p$-integrable functions on $X$ (${1\leqslant p\leqslant\infty}$). Let $E$ be a normed space. Consider the tensor product $\mathbf LE=\mathbf L\otimes E$; in the notation for elementary tensors in it, the tensor product sign will also be omitted: $fx=f\otimes x$. The product $\mathbf LE$ is a left module over the algebra $\mathcal B(\mathbf L)$ of bounded operators on $\mathbf L$ with respect to outer multiplication which is defined on elementary tensors by $A\cdot fx=(Af)x$.

A space $E$ is called an $\mathbf L$-space if $\mathbf LE$ is equipped with a norm $\|\,{\cdot}\,\|$ such that this module is contractive, that is, $\|A\cdot u\|\leqslant\|A\|\,\|u\|$ for all $A\in\mathcal B(\mathbf L)$, $u\in\mathbf LE$, and $\|fx\|=\|f\|\,\|x\|$ for all $f\in\mathbf L$, $x\in E$ (here $\|f\|$ is the norm on the space $\mathbf L$, and $\|x\|$ is the norm on $E$). This norm on $\mathbf LE$ will be referred to as the quantization of the norm of the space $E$ or simply the $\mathbf L$-norm on $E$; the corresponding $\mathbf L$-space is called the quantization of the space $E$.

On $\mathcal B(\mathbf L)$ we consider the family of projections $\mathcal P=\{P_Y|Y\subset X\}$ onto spaces of functions with support $Y\subset X$ ($P_Yf=\chi_Yf$, where $\chi_Y$ is the characteristic function of the set $Y$). Projections $P_{Y_1},P_{Y_2}\in\mathcal P$ are called orthogonal if $P_{Y_1}P_{Y_2}=P_{Y_2}P_{Y_1}=0$ (this condition is satisfied if $Y_1\cap Y_2$ is a nullset). A projection $P\in\mathcal P$ is called a support of an element $u\in\mathbf LE$ if $P\cdot u=u$.

We say that an $\mathbf L$-space $E$ is $p$-convex if, for any $u,v\in\mathbf LE$ with orthogonal supports,

Consider the problem of the minimal and maximal $p$-norms on $E$. The injective $\|\cdot\|^{\mathrm{inj}}$ and projective $\|\cdot\|^{\mathrm{pr}}$ tensor norms on $\mathbf LE$ are defined by

Now let us define an appropriate class of linear operators for $\mathbf L$-spaces. Let $E$ and $F$ be $\mathbf L$-spaces, and $T\colon E\to F$ be a linear operator. The amplification of $T$ is defined by

We denote by $\mathfrak{L}$ the category of $\mathbf L$-spaces for which morphisms are completely bounded operators; we denote by $\mathfrak{L_p}$ the subcategory of $\mathfrak{L}$ consisting of the $p$-convex $\mathbf L$-spaces for which morphisms are completely bounded operators. We will construct a functor from $\mathfrak{L}$ into $\mathfrak{L}_p$ that acts identically on $\mathfrak{L}_p$.

§ 3. The results

Let $E$ be an $\mathbf L$-space with $\mathbf L$-norm $\|\cdot\|$. We describe a construction for ‘transforming’ this norm $\|\cdot\|$ into a $p$-convex norm. Given $u\in\mathbf LE$, we set

Proof. First we show that $\|\cdot\|_p$ is a $p$-convex semi-norm on $\mathbf LE$. The equality $\|Cu\|_p=|C|\|u\|_p$ with an arbitrary constant $C\in\mathbb C$ is clear. The following lemma is required for the proof of the triangle inequality and $p$-convexity.

Lemma 1. 1. Let $A,B\in\mathcal B(\mathbf L)$ satisfy $AP=A$ and $BQ=B$ for some orthogonal projections $P$ and $Q$. Then

$$ \begin{equation*} \|A+B\|\leqslant(\|A\|^q+\|B\|^q)^{1/q},\quad\textit{where } \frac1{p}+\frac1{q}=1. \end{equation*} \notag $$

2. Let $A,B\in\mathcal B(\mathbf L)$ and let $P_1AP_2=A$, $Q_1BQ_2=B$ for some projections $P_1$, $P_2$, $Q_1$ and $Q_2$ such that $P_1\perp Q_1$ and $P_2\perp Q_2$. Then

$$ \begin{equation*} \|A+B\|=\max\{\|A\|,\|B\|\}. \end{equation*} \notag $$

Proof. 1. Consider a function $f\in\mathbf L$. It can be assumed that $P\cup Q=\mathbf 1_{\mathbf L}$ (otherwise we can take, for example, the projection $P\,{\cup}\,(\mathbf 1_{\mathbf L}\,{-}\,Q$) for $P$). Then $\|f\|^p=\|Pf\|^p+ \|Qf\|^p$.

We have $(A+B)f=A(Pf)+B(Qf)$, and therefore

$$ \begin{equation*} \begin{aligned} \, \|(A+B)f\| &\leqslant\|A\|\,\|Pf\|+\|B\|\,\|Qf\| \\ &\leqslant(\|A\|^q+\|B\|^q)^{1/q}(\|Pf\|^p+\|Qf\|^p)^{1/p} =(\|A\|^q+\|B\|^q)^{1/q}\|f\|. \end{aligned} \end{equation*} \notag $$

The corresponding inequality $\|A+B\|\leqslant\|A\|+\|B\|$ for $p=\infty$ is clear.

2. Consider a function $f\in\mathbf L$. It can be assumed that $P_2\cup Q_2=\mathbf 1_{\mathbf L}$. Then $\|f\|^p=\|P_2f\|^p+\|Q_2f\|^p$ (for $p=\infty$ we have $\|f\|=\max\{\|P_2f\|,\|Q_2f\|\}$).

We have $(A+B)f=A(P_2f)+B(Q_2f)$, and moreover, $A(P_2f)$ and $B(Q_2f)$ have pairwise orthogonal supports. Hence

$$ \begin{equation*} \begin{aligned} \, \|(A+B)f\| &=(\|A(P_2f)\|^p+\|B(Q_2f)\|^p)^{1/p}\leqslant(\|A\|^p\|P_2f\|^p+\|B\|^p\|Q_2f\|^p)^{1/p} \\ &\leqslant\max\{\|A\|,\|B\|\}(\|P_2f\|^p+\|Q_2f\|^p)^{1/p}=\max\{\|A\|,\|B\|\}\|f\|; \end{aligned} \end{equation*} \notag $$

for $p=\infty$ we have

$$ \begin{equation*} \begin{aligned} \, \|(A+B)f\| &=\max\{\|A(P_2f)\|,\|B(Q_2f)\|\}\leqslant\max\{\|A\|\,\|P_2f\|,\|B\|\,\|Q_2f\|\} \\ &\leqslant\max\{\|A\|,\|B\|\}\max\{\|P_2f\|,\|Q_2f\|\}=\max\{\|A\|,\|B\|\}\|f\|, \end{aligned} \end{equation*} \notag $$

which proves the lemma.

Proposition 1. Let $u,v\in\mathbf LE$. Then $\|u+v\|_p\leqslant\|u\|_p+\|v\|_p$. If, in addition, $u$ and $v$ have pairwise orthogonal supports, then

$$ \begin{equation*} \|u+v\|_p\leqslant(\|u\|^p_p+\|v\|^p_p)^{1/p}, \end{equation*} \notag $$

or, for $p=\infty$,

$$ \begin{equation*} \|u+v\|_\infty\leqslant\max\{\|u\|_\infty,\|v\|_\infty\}. \end{equation*} \notag $$

Proof. Consider an arbitrary $\varepsilon>0$. Let $u=A\cdot\sum u_i$, $v=B\cdot\sum v_j$ be representations such that

$$ \begin{equation*} a=\|A\|\Bigl(\sum\|u_i\|^p\Bigr)^{1/p}\leqslant\|u\|_p+\varepsilon\quad\text{and} \quad b=\|B\|\Bigl(\sum\|v_j\|^p\Bigr)^{1/p}\leqslant\|v\|_p+\varepsilon, \end{equation*} \notag $$

or, for $p=\infty$,

$$ \begin{equation*} a=\|A\|\max\{\|u_i\|\}\leqslant\|u\|_\infty+\varepsilon\quad\text{and} \quad b=\|B\|\max\{\|v_i\|\}\leqslant\|v\|_\infty+\varepsilon. \end{equation*} \notag $$

Multiplying and dividing the operator $A$ and the sum $\sum u_i$ by the same scalar, one can ensure that

$$ \begin{equation*} \|A\|=a^{p/(p+q)}\quad\text{and} \quad \Bigl(\sum\|u_i\|^p\Bigr)^{1/p}=a^{q/(p+q)},\quad\text{where } \frac1{p}+\frac1{q}=1. \end{equation*} \notag $$

A similar argument shows that

$$ \begin{equation*} \|B\|=b^{p/(p+q)}\quad\text{and} \quad \Bigl(\sum\|v_j\|^p\Bigr)^{1/p}=b^{q/(p+q)}. \end{equation*} \notag $$

For $p=\infty$, we have

$$ \begin{equation*} \|A\|=a, \qquad \max\{\|u_i\|\}=1, \qquad \|B\|=b \quad\text{and} \quad \max\{\|v_i\|\}=1. \end{equation*} \notag $$

Recall that the above measurable space $X$ is convenient, that is, $X$ can be represented as a disjoint union of $X_1$ and $X_2$ such that $X_1$, $X_2$ and $X$ are isomorphic as measurable spaces. Then $L_p(X_1)$ and $L_p(X_2)$ are isometrically isomorphic to $\mathbf L$, where $\mathbf L=L_p(X_1)\oplus_p L_p(X_2)$. We denote the corresponding isomorphisms by $i_1\colon \mathbf L\to L_p(X_1)$ and $i_2\colon \mathbf L\to L_p(X_2)$, respectively. Given a function $f\in\mathbf L$, we denote by $f^{(1)}$ the function equal to $i_1(f)$ on $X_1$ and vanishing on $X_2$, and we denote by $f^{(2)}$ the function agreeing with $i_2(f)$ on $X_2$ and vanishing on $X_1$. Next, for $w=\sum f_ix_i\in\mathbf LE$ we define $w^{(1)}=\sum f^{(1)}_ix_i$ and $w^{(2)}=\sum f^{(2)}_ix_i$. So, for $k=1,2$ we have $w^{(k)}=(j_k\circ i_k)\cdot w$, where $j_k$ is an embedding of $L_p(X_k)$ in $\mathbf L$. Hence

$$ \begin{equation*} \|w^{(k)}\|\leqslant\|j_k\circ i_k\|\,\|w\|=\|w\|. \end{equation*} \notag $$

On the other hand, $w=i_k^{-1}P_k\cdot w^{(k)}$, where $P_k$ is the projection of $\mathbf L$ onto $L_p(X_k)$. Hence

$$ \begin{equation*} \|w\|\leqslant\|i_k^{-1}P_k\|\,\|w^{(k)}\|=\|w^{(k)}\|, \end{equation*} \notag $$

and therefore $\|w\|=\|w^{(1)}\|=\|w^{(2)}\|$.

Given an operator $M\in\mathcal B(\mathbf L)$, we denote by $M^{(1)}$ the operator that acts on the space $L_p(X_1)$ as $Mi_1^{-1}$ and acts on $L_p(X_2)$ as the zero operator, that is, $M^{(1)}= Mi_1^{-1}\oplus\mathbf 0$ under the identification $\mathbf L=L_p(X_1)\oplus_p L_p(X_2)$; similarly, $M^{(2)}=\mathbf 0\oplus Mi_2^{-1}$.

In this notation we have the representation

$$ \begin{equation*} u+v=(A^{(1)}+B^{(2)})\cdot\Bigl(\sum u_i^{(1)}+\sum v_j^{(2)}\Bigr). \end{equation*} \notag $$

Note that $u_i^{(1)}$ and $v_j^{(2)}$ have pairwise orthogonal supports. Hence

$$ \begin{equation*} \begin{aligned} \, \|u+v\|_p &\leqslant\|A^{(1)}+B^{(2)}\|\Bigl(\sum\|u_i^{(1)}\|^p+\sum\|v_j^{(2)}\|^p\Bigr)^{1/p} \\ &=\|A^{(1)}+B^{(2)}\|\Bigl(\sum\|u_i\|^p+\sum\|v_j\|^p\Bigr)^{1/p}; \end{aligned} \end{equation*} \notag $$

or, for $p=\infty$,

$$ \begin{equation*} \|u+v\|_\infty\leqslant\|A^{(1)}+B^{(2)}\|\max\{\|u_i^{(1)}\|,\|v_j^{(2)}\|\} =\|A^{(1)}+B^{(2)}\|\max\{\|u_i\|,\|v_j\|\}. \end{equation*} \notag $$

Moreover,

$$ \begin{equation*} \sum\|u_i\|^p=a^{pq/(p+q)}=a \quad\text{and}\quad \sum\|v_j\|^p=b^{pq/(p+q)}=b. \end{equation*} \notag $$

For the operators $A^{(1)}$ and $B^{(2)}$ we have $A^{(1)}=A^{(1)}P_{X_1}$ and $B^{(2)}=B^{(2)}P_{X_2}$, where $P_{X_1}$ and $P_{X_2}$ are the projections onto $L_p(X_1)$ and $L_p(X_2)$, respectively. These operators are orthogonal, hence from assertion 1 of Lemma 1 we get that

$$ \begin{equation*} \|A^{(1)}+B^{(2)}\|\leqslant(\|A\|^q+\|B\|^q)^{1/q} =(a^{pq/(p+q)}+b^{pq/(p+q)})^{1/q}=(a+b)^{1/q}. \end{equation*} \notag $$

Thus, $\|u+v\|_p\leqslant(a+b)^{1/q}(a+b)^{1/p}=a+b\leqslant\|u\|+\|v\|+2\varepsilon$.

For $p=\infty$, setting $\max\{\|u_i\|\}=\max\{\|v_j\|\}=1$, we have

$$ \begin{equation*} \begin{aligned} \, \|u+v\|_\infty &\leqslant\|A^{(1)}+B^{(2)}\|\max\{\|u_i^{(1)}\|,\|v_j^{(2)}\|\} =\|A^{(1)}+B^{(2)}\|\max\{\|u_i\|,\|v_j\|\} \\ &\leqslant\|A\|+\|B\|=a+b\leqslant\|u\|+\|v\|+2\varepsilon. \end{aligned} \end{equation*} \notag $$

Now let $u,v\!\in\!\mathbf LE$ have pairwise orthogonal supports $P$ and $Q$, and let ${u\!=\!A\cdot\sum u_i}$ and $v=B\cdot\sum v_j$ be representations such that

or, for $p=\infty$,

$$ \begin{equation*} a=\|A\|\max\{\|u_i\|\}\leqslant\|u\|_\infty+\varepsilon\quad\text{and} \quad b=\|B\|\max\{\|v_j\|\}\leqslant\|v\|_\infty+\varepsilon. \end{equation*} \notag $$

It can be assumed that $PA=A$, $QB=B$. By dividing and multiplying the operators $A$ and $B$ and the elements $\sum u_i$ and $\sum v_j$ by appropriate scalars, one can ensure that $\|A\|=\|B\|=1$. Again, we use the representation

$$ \begin{equation*} u+v=(A^{(1)}+B^{(2)})\cdot\Bigl(\sum u_i^{(1)}+\sum v_j^{(2)}\Bigr). \end{equation*} \notag $$

In this case the operators $A^{(1)}$ and $B^{(2)}$ satisfy the hypotheses of assertion 2 of Lemma 1 (for $P_1=P$, $Q_1=Q$, $P_2=P_{X_1}$ and $Q_2=P_{X_2}$). Hence $\|A^{(1)}+B^{(2)}\|=\max\{\|A^{(1)}\|,\|B^{(2)}\|\}=1$. Therefore,

$$ \begin{equation*} \begin{aligned} \, \|u+v\|_p &\leqslant\Bigl(\sum\|u_i^{(1)}\|^p+\sum\|v_j^{(2)}\|^p\Bigr)^{1/p} \\ &=\Bigl(\sum\|u_i\|^p+\sum\|v_j\|^p\Bigr)^{1/p}=(a^p+b^p)^{1/p}, \end{aligned} \end{equation*} \notag $$

or, for $p=\infty$,

$$ \begin{equation*} \begin{aligned} \, \|u+v\|_\infty &\leqslant\max\{\|u_i^{(1)}\|,\|v_j^{(2)}\|\}=\max\{\|u_i\|,\|v_j\|\} \\ &=\max\bigl\{\max\{\|u_i\|\},\max\{\|v_j\|\}\bigr\}=\max\{a,b\}. \end{aligned} \end{equation*} \notag $$

Now we claim that $\|\cdot\|_p$ is an $\mathbf L$-norm on $E$.

Let us show that $\|B\cdot u\|_p\leqslant\|B\|\,\|u\|_p$ for any $B\in\mathcal B(\mathbf L)$ and $u\in\mathbf LE$. For any representation $u=A\cdot\sum u_i$ consider the corresponding representation $B\cdot u=BA\cdot\sum u_i$. Then we have $\|BA\|\leqslant\|B\|\,\|A\|$, and so

$$ \begin{equation*} \|BA\|\Bigl(\sum\|u_i\|^p\Bigr)^{1/p}\leqslant\|B\|\,\|A\|\Bigl(\sum\|u_i\|^p\Bigr)^{1/p}. \end{equation*} \notag $$

Now the required result follows by taking the infimum over all such representations.

The proof for $p=\infty$ is similar.

Proposition 1 is proved.

To complete the proof of Theorem 1 it remains to show that $\|fx\|_p=\|f\|\,\|x\|$. In the proof of this result we also show that $p$-convex $\mathbf L$-norms are invariant under the transformation under consideration.

Proof. For any $\mathbf L$-norm $\|\cdot\|$ on $E$ and all $u\in\mathbf LE$ we have $\|u\|_p\leqslant\|u\|$, because $u$ can be written as $u=\mathbf 1_{\mathbf L}\cdot u$. If $\|\cdot\|$ is $p$-convex, then, for any representation $u=A\cdot\sum u_i$, where $u_i$ have pairwise orthogonal supports, we have

$$ \begin{equation*} \|u\|\leqslant\|A\|\Bigl\|\sum u_i\Bigr\|\leqslant\|A\|\Bigl(\sum\|u_i\|^p\Bigr)^{1/p}, \end{equation*} \notag $$

and so $\|u\|\leqslant\|u\|_p$. The proof for $p=\infty$ is similar.

Proposition 2 is proved.

To complete the proof of Theorem 1 we require the second auxiliary result, which has importance in its own right. It states that the above transformation of $\mathbf L$-norms preserves the complete boundedness of linear operators between $\mathbf L$-spaces.

Lemma 2. Let $T\colon E\to F$ be a completely bounded operator between $\mathbf L$-spaces $E$ and $F$. Then

$$ \begin{equation*} \|T_\infty u\|_p\leqslant\|T_\infty\|\,\|u\|_p. \end{equation*} \notag $$

Proof. If $u\in\mathbf LE$,

$$ \begin{equation*} u=A\cdot\sum_i u_i, \quad u_i=\sum_j f_{ij}x_{ij}, \qquad \|A\|\Bigl(\sum\|u_i\|^p\Bigr)^{1/p}\leqslant\|u\|_p+\varepsilon, \end{equation*} \notag $$

then $T_\infty u=A\cdot\sum T_\infty u_i$; moreover, the $T_\infty u_i$ have pairwise orthogonal supports. Hence

$$ \begin{equation*} \begin{aligned} \, \|T_\infty u\|_p &\leqslant\|A\|\Bigl(\sum\|T_\infty u_i\|^p\Bigr)^{1/p}\leqslant\|A\|\Bigl(\sum(\|T_\infty\|\,\|u_i\|)^p\Bigr)^{1/p} \\ &\leqslant\|T_\infty\|\,\|A\|\Bigl(\sum\|u_i\|^p\Bigr)^{1/p}\leqslant\|T_\infty\|(\|u\|+\varepsilon). \end{aligned} \end{equation*} \notag $$

The proof for $p=\infty$ is similar.

Lemma 2 is proved.

Consider the amplification of the identity operator on $E$, that is, the identity operator $I_\infty\colon\mathbf LE\to\mathbf L\otimes_i E$ from the space $\mathbf LE$ with the norm $\|\cdot\|$ into the same space, but equipped with the injective tensor norm $\|\cdot\|^{\mathrm{inj}}$. The norm ${\|\cdot\|^{\mathrm{inj}}}$ is minimal among all $\mathbf L$-norms of the space $E$ (and, in general, among all cross-norms on $\mathbf LE$). Hence $\|I_\infty\|\leqslant1$, and so $\|u\|_p\geqslant\|u\|^{\mathrm{inj}}_p$ by Lemma 2. But the norm ${\|\cdot\|^{\mathrm{inj}}}$ is $p$-convex (see [8]), hence $\|\cdot\|^{\mathrm{inj}}=\|\cdot\|^{\mathrm{inj}}_p$ by Proposition 2. Therefore, $\|u\|_p\geqslant\|u\|^{\mathrm{inj}}$ and, in particular, $\|fx\|_p\geqslant\|fx\|^{\mathrm{inj}}=\|f\|\,\|x\|$. The inequality ${\|fx\|_p\leqslant\|f\|\,\|x\|}$ follows from the fact that $fx$ can be represented in the form $\mathbf 1_{\mathbf L}\cdot fx$, and ${\|fx\|=\|f\|\,\|x\|}$. So we have $\|fx\|_p=\|f\|\,\|x\|$.

This completes the proof of Theorem 1.

Thus, with each $\mathbf L$-norm on a normed space $E$ we have associated a $p$-convex $\mathbf L$-norm on $E$ (note that by Proposition 2 this transformation does not change $p$-convex $\mathbf L$-norms). We have also shown that completely bounded operators between $\mathbf L$-spaces remain completely bounded when considered as operators between the same spaces endowed with the corresponding $p$-convex $\mathbf L$-norms (Lemma 2). Consequently, the mapping

Let us show that an application of the above transformation to the projective tensor norm gives a maximal (among all $p$-convex norms) $p$-convex norm on $E$ (the resulting construction is a kind of generalization of maximal Lambert spaces, which were defined in [1], § 2.1.2). Namely, let $\|\cdot\|^{\mathrm{pr}}$ be the projective tensor norm on $\mathbf LE$ (according to the above, this is an $\mathbf L$-norm which is not necessarily $p$-convex; see [8]). Then $\|\cdot\|^{\mathrm{pr}}_p$ is a $p$-convex $\mathbf L$-norm. Recall that

We call an $\mathbf L$-space $E$ with the above maximal $p$-convex $\mathbf L$-norm the maximal $p$-convex quantization of the normed space $E$. The maximal $p$-convex quantization has the following property (cf. the analogous property of automatic complete boundedness for operators acting between quantum spaces, where for the first space we consider the maximal quantization; see [9], Proposition 2.2.8).

§ 1. Introduction

Acknowledgements

§ 2. Basic concepts

§ 3. The results

Bibliography