$S_{y}(x):=\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))y^{2}(k)\Big{)}^{p/2}=\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{(}\sum_{k=0}^{n-1}(x^{p}(\rho(k))y^{p}(k))^{q}\Big{)}^{1/q}.$

Applying [45, Theorem 1.2] to the right hand side of the equality above, we
obtain

$\frac{1}{5n}\Big{(}\sum_{k=0}^{n-1}\mu(k,x^{p}\otimes
y^{p})+n^{1-\frac{1}{q}}\Big{(}\sum_{k=n}^{\infty}\mu^{q}(k,x^{p}\otimes
y^{p})\Big{)}^{\frac{1}{q}}\Big{)}\leq S_{y}(x)\\\
\leq\frac{1}{n}\Big{(}\sum_{k=0}^{n-1}\mu(k,x^{p}\otimes
y^{p})+n^{1-\frac{1}{q}}\Big{(}\sum_{k=n}^{\infty}\mu^{q}(k,x^{p}\otimes
y^{p})\Big{)}^{\frac{1}{q}}\Big{)}.$ (96)

By Lemma 41 there exists the Orlicz function $M_{n}\in\mathcal{O}_{1,q}$ that
satisfies (89) for any finitely supported vector. Thus, applying (89) for the
vector $x^{p}\otimes y^{p}$ to the right and left hand sides of (96), we infer

$\frac{1}{20n}\|x^{p}\otimes y^{p}\|_{l_{M_{n}}}\leq
S_{y}(x)\leq\frac{4}{n}\|x^{p}\otimes y^{p}\|_{l_{M_{n}}}.$ (97)

Define the function $N_{n,y}$ by setting

$N_{n,y}(t):=\sum_{k\geq
0}M_{n}\Big{(}\frac{ty^{p}(k)}{\|y^{p}\|_{l_{M_{n}}}}\Big{)},\ \ \
t\in[0,\infty).$

Observe that

$N_{n,y}(1)=\sum_{k\geq
0}M_{n}\Big{(}\frac{y^{p}(k)}{\|y^{p}\|_{l_{M_{n}}}}\Big{)}=1$

and $q$-concavity of the function $N_{n,y}$ follows from the $q$-concavity of
$M_{n}$. It follows that $N_{n,y}\in\mathcal{O}_{1,q}.$ Moreover, by
definition of the Orlicz norm, we have

$\displaystyle\|x^{p}\otimes y^{p}\|_{l_{M_{n}}}$
$\displaystyle=\inf\Big{\\{}\lambda:\ \sum_{k,l\geq
0}M_{n}\Big{(}\frac{x^{p}(l)y^{p}(k)}{\lambda}\Big{)}\leq 1\Big{\\}}$ (98)
$\displaystyle=\inf\Big{\\{}\lambda:\ \sum_{l\geq
0}N_{n,y}\Big{(}\frac{\|y^{p}\|_{l_{M_{n}}}x^{p}(l)}{\lambda}\Big{)}\leq
1\Big{\\}}=\|y^{p}\|_{l_{M_{n}}}\|x^{p}\|_{l_{N_{n,y}}}.$

Combining (97) and (98), we obtain

$\frac{1}{20n}\|x^{p}\|_{l_{N_{n,y}}}\|y^{p}\|_{l_{M_{n}}}\leq
S_{y}(x)\leq\frac{4}{n}\|x^{p}\|_{l_{N_{n,y}}}\|y^{p}\|_{l_{M_{n}}}.$ (99)

Using Lemma 41 for the sequence $y\in\mathbb{R}^{n}$, we infer

$\frac{1}{4}\|y\|_{p}^{p}\leq\|y^{p}\|_{l_{M_{n}}}\leq 4\|y\|_{p}^{p}.$ (100)

Setting $M_{n,y}(t):=N_{n,y}(t^{p}),$ $t\geq 0,$ we obtain that
$\|x^{p}\|_{l_{N_{n,y}}}=\|x\|_{l_{M_{n,y}}}^{p}.$ Application of the latter
and (100) in (99) completes the proof of (95). Since
$N_{n,y}\in\mathcal{O}_{1,q},$ it follows that $M_{n,y}\in\mathcal{O}_{p,2}.$
∎

######  Lemma 43.

Let $1\leq p\leq 2$. The set $\mathcal{O}_{p,2}$ equipped with the topology of
pointwise convergence is a compact metrizable space.

###### Proof.

Using that the set of rational numbers $\mathbb{Q}$ is countable, we numerate
$\mathbb{Q}_{+}$ as $\mathbb{Q}_{+}=\\{a_{1},a_{2},\cdots\\}.$ We claim that
the metric

$d(M_{1},M_{2}):=\sum_{n\geq
1}2^{-n}\frac{|(M_{1}-M_{2})(a_{n})|}{1+|(M_{1}-M_{2})(a_{n})|},\ \ \
M_{1},M_{2}\in\mathcal{O}_{p,2},$

generates the topology of pointwise convergence on $\mathcal{O}_{p,2}$. Proof
of the completeness is routine and, therefore, omitted.

Let us show compactness. Take an arbitrary sequence $\\{M_{n}\\}_{n\geq
1}\subset\mathcal{O}_{p,2}.$ Since $M_{n}$ is $2$-concave for any $n\geq 1,$
it follows that the sequence $\\{M_{n}(a_{1})\\}_{n\geq 1}$ is bounded by
$\max\\{1,a_{1}^{2}\\}$. Let $m_{1}$ be a limit of some subsequence of
$\\{M_{n,1}(a_{1})\\}_{n\geq 1}.$ Select a subsequence $\\{M_{n,1}\\}_{n\geq
1}\subset\\{M_{n}\\}_{n\geq 1}$ such that $|M_{n,1}(a_{1})-m_{1}|\leq 1/n$ for
all $n\geq 1.$ Using the same argument, we have that the sequence
$\\{M_{n,1}(a_{2})\\}_{n\geq 1}$ is bounded. Let $m_{2}$ be a limit of some
subsequence of $\\{M_{n,1}(a_{1})\\}_{n\geq 1}.$ Select a subsequence
$\\{M_{n,2}\\}_{n\geq 1}\subset\\{M_{n,1}\\}_{n\geq 1}$ such that
$|M_{n,2}(a_{2})-m_{2}|\leq 1/n$ for all $n\geq 1.$ Also, we have
$|M_{n,2}(a_{1})-m_{1}|\leq 1/n$ for all $n\geq 1.$ Proceeding this process ad
infinitum, for every $l\geq 1,$ we obtain a sequence $\\{M_{n,l}\\}_{n\geq
1}\subset\mathcal{O}_{p,2}$ such that $|M_{n,l}(a_{k})-m_{k}|\leq 1/n$ for all
$n\geq 1$ and for all $k\leq l.$ Taking a diagonal subsequence
$\\{M_{n,n}\\}_{n\geq 1}\subset\\{M_{n,l}\\}_{n\geq 1,l\geq 1},$ we have
$|M_{n,n}(a_{k})-m_{k}|\leq 1/n$ for all $1\leq k\leq n$. Hence, the
subsequence $\\{M_{n,n}\\}_{n\geq 1}$ converges at every rational point.

Next we prove that $\\{M_{n,n}\\}_{n\geq 1}$ converges at every $t>0.$ Fix
$\varepsilon>0$ and select rational number $a\in(t,t(1+\varepsilon)).$ It
follows from convexity and $2$-concavity of $M_{n,n}$ that

$M_{n,n}(t)\leq M_{n,n}(a)\leq(1+\varepsilon)^{2}M_{n,n}(t),\ \ \ n\geq 1.$

Passing to the limit, we obtain

$\limsup_{n\to\infty}M_{n,n}(t)\leq\lim_{n\to\infty}M_{n,n}(a)\leq(1+\varepsilon)^{2}\liminf_{n\to\infty}M_{n,n}(t).$

Since $\varepsilon$ is arbitrary and the sequence $\\{M_{n,n}(a)\\}_{n\geq 1}$
converges, it follows that

$\limsup_{n\to\infty}M_{n,n}(t)=\liminf_{n\to\infty}M_{n,n}(t),$

which guarantees that the sequence $M_{n,n}(t)$ converges. Thus, compactness
is proved. ∎

The following assertion is proved in [56, Theorem 1.1] for $p=1.$ We provide a
proof (the same as in [56, Theorem 1.1]) for convenience of the reader. Due to
previous lemma, we equip $\mathcal{O}_{p,2}$ with Borel $\sigma$-algebra.

######  Theorem 44.

Let $1\leq p\leq 2.$ If a symmetric sequence space $I$ isomorphically embeds
into $L_{p}(0,1),$ then there exists a Radon probability measure $\nu$ on the
space $\mathcal{O}_{p,2}$ such that

$\|x\|_{I}\sim\Big{(}\int_{M\in\mathcal{O}_{p,2}}\|x\|_{l_{M}}^{p}d\nu(M)\Big{)}^{1/p}$
(101)

for every finitely supported $x\in l_{\infty}.$

###### Proof.

Let $x=\\{x(k)\\}_{k\geq 0}\in l_{\infty}$ be finitely supported, let
$\\{e_{k}\\}_{k\geq 1}$ be the standard unit basis of $c_{0}$ and let $T$ be
the isomorphic embedding of $I$ into $L_{p}(0,1).$ Fix a constant $C$ such
that

$C^{-1}\|x\|_{I}\leq\|Tx\|_{p}\leq C\|x\|_{I},\ \ \mbox{for any}\ \ x\in I.$

Taking any $n\geq|{\rm supp}(x)|,$ we may assume that
$x=\sum_{k=0}^{n-1}x(k)e_{k}.$ Denoting $f_{k}:=Te_{k}\in L_{p}(0,1),$ $k\geq
0,$ for every $t\in(0,1)$ and for every permutation $\rho\in\mathfrak{S}_{n},$
we have

$\|x\|_{I}=\Big{\|}\sum_{k=0}^{n-1}x(\rho(k))r_{k}(t)e_{k}\Big{\|}_{I}\leq
C\Big{\|}\sum_{k=0}^{n-1}x(\rho(k))r_{k}(t)f_{k}\Big{\|}_{p}$

and, similarly,

$\|x\|_{I}\geq
C^{-1}\Big{\|}\sum_{k=0}^{n-1}x(\rho(k))r_{k}(t)f_{k}\Big{\|}_{p},$

where by $r_{k},$ $k\geq 0,$ we denote the Rademacher functions defined in
(14). Hence,

$\|x\|_{I}\leq
C\Big{(}\int_{0}^{1}\Big{\|}\sum_{k=0}^{n-1}x(\rho(k))r_{k}(t)f_{k}\Big{\|}_{p}^{p}dt\Big{)}^{\frac{1}{p}}=C\Big{\|}\Big{(}\int_{0}^{1}\Big{|}\sum_{k=0}^{n-1}x(\rho(k))r_{k}(t)f_{k}\Big{|}^{p}dt\Big{)}^{\frac{1}{p}}\Big{\|}_{p}$

and, similarly

$\|x\|_{I}\geq
C^{-1}\Big{\|}\Big{(}\int_{0}^{1}\Big{|}\sum_{k=0}^{n-1}x(\rho(k))r_{k}(t)f_{k}\Big{|}^{p}dt\Big{)}^{\frac{1}{p}}\Big{\|}_{p}$

for every permutation $\rho\in\mathfrak{S}_{n}.$

By Khinchine inequality (see [46, Theorem 2.b.3] and its proof), we have

$\frac{1}{36C}\Big{\|}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))f_{k}^{2}\Big{)}^{\frac{1}{2}}\Big{\|}_{p}\leq\|x\|_{I}\leq
C\Big{\|}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))f_{k}^{2}\Big{)}^{\frac{1}{2}}\Big{\|}_{p}$

for every permutation $\rho\in\mathfrak{S}_{n}.$ Summing by all permutations
from $\mathfrak{S}_{n}$ the inequality above, we obtain

$\displaystyle\|x\|_{I}\leq
C\Big{(}\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{\|}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))f_{k}^{2}\Big{)}^{\frac{1}{2}}\Big{\|}_{p}^{p}\Big{)}^{\frac{1}{p}}=C\Big{\|}\Big{(}\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))f_{k}^{2}\Big{)}^{\frac{p}{2}}\Big{)}^{\frac{1}{p}}\Big{\|}_{p}$
(102)

and, similarly,

$\displaystyle\|x\|_{I}\geq\frac{1}{36C}\Big{\|}\Big{(}\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))f_{k}^{2}\Big{)}^{\frac{p}{2}}\Big{)}^{\frac{1}{p}}\Big{\|}_{p}.$
(103)

For a fixed $s\in(0,1)$, by Lemma 42 applied for
$y_{s}=\\{f_{k}(s)\\}_{k=0}^{n-1}\in\mathbb{R}^{n},$ there exists the Orlicz
function $M_{n,y_{s}}\in\mathcal{O}_{p,2}$ such that

$\frac{1}{80n}\|y_{s}\|_{p}^{p}\cdot\|x\|_{l_{M_{n,y_{s}}}}^{p}\leq\frac{1}{n!}\sum_{\rho\in\mathfrak{S}_{n}}\Big{(}\sum_{k=0}^{n-1}x^{2}(\rho(k))f_{k}^{2}(s)\Big{)}^{\frac{p}{2}}\leq\frac{16}{n}\|y_{s}\|_{p}^{p}\cdot\|x\|_{l_{M_{n,y_{s}}}}^{p}.$
(104)

Combining the computation

$\Big{\|}\Big{(}\frac{1}{n}\|y_{s}\|_{p}^{p}\cdot\|x\|_{l_{M_{n,y_{s}}}}^{p}\Big{)}^{\frac{1}{p}}\Big{\|}_{p}=\Big{(}\int_{0}^{1}\frac{1}{n}\Big{(}\sum_{k=0}^{n-1}|f_{k}(s)|^{p}\Big{)}\|x\|_{l_{M_{n,y_{s}}}}^{p}ds\Big{)}^{\frac{1}{p}}$

with (102),(103) and (104), we arrive at

$\|x\|_{I}\sim\Big{(}\int_{0}^{1}\frac{1}{n}\Big{(}\sum_{k=0}^{n-1}|f_{k}(s)|^{p}\Big{)}\|x\|_{l_{M_{n,y_{s}}}}^{p}ds\Big{)}^{\frac{1}{p}}=\Big{(}\int_{0}^{1}\|x\|_{M_{n,y_{s}}}^{p}d\mu_{n}(s)\Big{)}^{\frac{1}{p}},$
(105)

where

$\mu_{n}(A):=\frac{1}{n}\sum_{k=0}^{n-1}\int_{A}|f_{k}(s)|^{p}ds,\quad
A\subseteq(0,1).$

Define a positive Radon measure $\nu_{n}$ on $\mathcal{O}_{p,2}$ equipped with
Borel $\sigma$-algebra by setting

$\nu_{n}(G):=\mu_{n}(\\{s\in(0,1):\ M_{n,y_{s}}\in G\\}),\quad
G\subseteq\mathcal{O}_{p,2}.$

Via (105), we conclude

$\|x\|_{I}\sim\Big{(}\int_{M\in\mathcal{O}_{p,2}}\|x\|_{l_{M}}^{p}d\nu_{n}(M)\Big{)}^{1/p}$

for every finitely supported $x\in l_{\infty}$ and for every $n\geq|{\rm
supp}(x)|.$

For every finitely supported $x\in l_{\infty},$ the function
$M\to\|x\|_{l_{M}}$ is continuous on $\mathcal{O}_{p,2}$ equipped with the
topology of pointwise convergence. Therefore, every Radon measure $\nu_{n},$
$n\geq|{\rm supp}(x)|,$ generates a bounded positive functional
$\varphi_{n}\in C(\mathcal{O}_{p,2})^{*},$ where $C(\mathcal{O}_{p,2})^{*}$
denotes the dual space of the space $C(\mathcal{O}_{p,2})$ of all continuous
functions on $\mathcal{O}_{p,2}.$ By Banach-Alaoglu theorem, there exists a
subnet $\varphi_{\psi(i)},$ $i\in\mathbb{I},$ and $\varphi\in
C(\mathcal{O}_{p,2})^{*}$ such that $\varphi_{\psi(i)}\to\varphi.$ By Riesz-
Markov theorem, the (positive) functional $\varphi$ is generated by a Radon
measure $\nu$ on $\mathcal{O}_{p,2}.$ Hence, we have

$\|x\|_{I}\sim\Big{(}\int_{M\in\mathcal{O}_{p,2}}\|x\|_{l_{M}}^{p}d\nu(M)\Big{)}^{1/p}$

for every finitely supported $x\in l_{\infty}.$ ∎

##  7\. Proof of Theorem 3

In this section, we prove our main result. Recall, that by Theorem 44 the
space $\mathcal{O}_{p,2}$ is equipped with a Radon probability measure $\nu$,
defined on the Borel $\sigma$-algebra.

######  Lemma 45.

There exists a Bochner measurable function
$\xi:\mathcal{O}_{p,2}\to\mathcal{L}_{p}(\mathcal{R}),1\leq p<2$ such that

$\frac{1}{12}\|A\|_{\mathcal{L}_{M}}\leq\|\xi(M)\otimes
A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}(\mathcal{R}\overline{\otimes}\mathcal{L}(H))}\leq
5\|A\|_{\mathcal{L}_{M}}$ (106)

for every finite rank operator $A\in\mathcal{L}(H)$ and for every
$M\in\mathcal{O}_{p,2}$ satisfying the condition

$\lim_{t\to 0}\frac{1}{t^{p}}M(t)=0.$ (107)

###### Proof.

Let $M\in\mathcal{O}_{p,2}$. Define an Orlicz function $N_{M}$ by setting

$N_{M}(t)=\begin{cases}M(t),\quad 0\leq t\leq 1\\\
1+\frac{1}{p}M^{\prime}(1-0)(t^{p}-1),\quad t\geq 1.\end{cases}$ (108)

It is straightforward to verify that $N_{M}\in\mathcal{O}_{p,2}.$ Furthermore,
if in addition $M$ satisfies (107), we define a quasi-concave function
$\varphi_{M}:\mathbb{R}_{+}\to\mathbb{R}_{+}$ by setting

$\varphi_{M}(t)=\frac{N_{M}(t^{\frac{1}{2-p}})}{t^{\frac{p}{2-p}}}.$ (109)

By Lemma 40 with $d=2/p$ there exists a positive decreasing function $x_{M}\in
L_{1}(0,1)$ given by formula (84).

We claim that the mapping $\eta:\mathcal{O}_{p,2}\to L_{1}(0,1)$ defined by
setting $\eta:M\mapsto x_{M}$ is Bochner measurable (if $M$ fails the
condition (107), then we set $x_{M}=0$).

Let $t_{k}(M),s_{k}(M)$ and $u_{k}(M),k\geq 0,$ be the numbers constructed for
the function $\varphi_{M}$ in the proof of Lemma 40. First note that
$t_{k}(M)$ and $s_{k}(M),$ $k\geq 0,$ considered as functions of $M$, are
measurable on $\mathcal{O}_{p,2}$. Indeed, since the function $\varphi_{M}$ is
quasi-concave, for every $s\in(0,1),$ by definition of numbers $t_{k}$ we have
that

$\\{M:\ t_{k}(M)\geq s\\}=\\{M:\ \varphi_{M}(s)\leq 2^{-k}\\}=\\{M:\
M(s^{\frac{1}{2-p}})\leq s^{\frac{p}{2-p}}2^{-k}\\}$

is a closed set in $\mathcal{O}_{p,2}$ (equipped with topology of poitwise
convergence) and therefore, the functions $t_{k}(\cdot)$ are measurable, and
similarly $s_{k}(\cdot)$ are measurable for all $k\geq 0$. By definition of
the numbers $u_{k}(M),$ $k\geq 0,$ we have that the functions $u_{k}(\cdot)$
are also measurable functions of $M.$ Furthermore, since
$\varphi_{M}(u_{k}(M)),k\geq 0$ takes only the values $2^{-n},$ $n\geq 0,$ and

$\\{M:\ \varphi_{M}(u_{2k}(M))=2^{-n}\\}=\\{M:\ u_{2k}(M)=t_{n}(M)\\}$

is a measurable set, we have that the function
$M\mapsto\varphi_{M}(u_{2k}(M)),$ $k\geq 0$ is also measurable. Set

$v_{k}(M)=\sum_{l\geq k}\frac{1}{2}u_{2l}^{1/(d-1)}(M)\varphi_{M}(u_{2l}(M)),$

where the convergence of the series follows from (82). We have that
$v_{k}(\cdot)$ also measurable functions of $M.$

Fix a positive element $y\in L_{\infty}(0,1)$ and denote its primitive (which
is monotone) by $z.$ It follows from (84) that

$\langle x_{M},y\rangle=\int_{0}^{1}x_{M}(s)y(s)ds=\sum_{k\geq
0}u_{2k}^{-1/(d-1)}(M)\Big{(}z(v_{k}(M))-z(v_{k+1}(M))\Big{)},$

where the series converges since

$\left|u_{2k}^{-1/(d-1)}(M)\Big{(}z(v_{k}(M))-z(v_{k+1}(M))\Big{)}\right|\leq$
$\leq
u_{2k}^{-1/(d-1)}(M)\|y\|_{\infty}|v_{k}(M)-v_{k+1}(M)|=\|y\|_{\infty}\varphi_{M}(u_{2k}(M))$

and $\varphi_{M}(u_{2k}(M))$ is a geometric sequence. Since $z$ is monotone,
it follows that $z\circ v_{k}$ is a measurable function of $M$ for every
$k\geq 0.$ Hence, the mapping $M\to\langle x_{M},y\rangle$ is also measurable.
This proves that the function $\eta:M\to x_{M}\in L_{1}(0,1),$ is weakly
measurable. Since the space $L_{1}(0,1)$ is separable, by Pettis theorem (see
e.g. [58, Proposition 2.15]), we have that $\eta$ is Bochner measurable.

Let $i:L_{1}(0,1)\to\mathcal{L}_{1}(\mathcal{R})$ be an isometric embedding.
It is clear, that the function
$\xi:\mathcal{O}_{p,2}\to\mathcal{L}_{p}(\mathcal{R})$ defined by the setting
$\xi(M)=(i(\eta(M)))^{1/p}$ is Bochner measurable.

Now, let us show that the function $\xi$ satisfies inequality (106). Fix a
function $M\in\mathcal{O}_{p,2}$ satisfying condition (107). By definition of
the function $N_{M}$ (see (108)) we have that $M(1)=N_{M}(1)=1$, and therefore
for every finitely supported $x\in l_{\infty}$ we obtain

$\displaystyle\\{\lambda:\ \sum_{k\geq 0}M(\frac{x(k)}{\lambda})\leq
1\\}=\\{\lambda\geq\|x\|_{\infty}:\ \sum_{k\geq 0}M(\frac{x(k)}{\lambda})\leq
1\\}$ $\displaystyle=\\{\lambda\geq\|x\|_{\infty}:\ \sum_{k\geq
0}N_{M}(\frac{x(k)}{\lambda})\leq 1\\}=\\{\lambda:\ \sum_{k\geq
0}N_{M}(\frac{x(k)}{\lambda})\leq 1\\}.$

Consequently

$\|x\|_{l_{N_{M}}}=\|x\|_{l_{M}}$ (110)

for every finitely supported $x\in l_{\infty}.$ By Lemma 40 we have that

$\frac{1}{4}\varphi_{M}(t)\leq\int_{0}^{1}\min\\{x_{M}(s),tx_{M}^{d}(s)\\}ds\leq\frac{5}{2}\varphi_{M}(t),\quad
t\in[0,1],$ (111)

in particular for $t=1$ we obtain $\frac{1}{4}\leq\|x_{M}\|_{1}$. The
$2$-concavity of the function $M$ implies that $M(2)\leq 4M(1)$ (see e.g.
[43]) Hence, for the function $\varphi_{M}$ given by (109) we have

$\varphi_{M}(\infty)=\frac{1}{p}M^{\prime}(1-0)\leq M(2)-M(1)\leq 3M(1)=3.$

Therefore, for $t\geq 1$ we infer that

$\int_{0}^{1}\min\\{x_{M}(s),tx_{M}^{d}(s)\\}ds=\int_{0}^{1}x_{M}(s)ds\geq\frac{1}{4}\geq\frac{1}{12}\varphi_{M}(\infty)\geq\frac{1}{12}\varphi_{M}(t).$

On the other hand, definition of the function $x_{M}$ (see (84)) implies that
$x_{M}\geq\mathrm{supp}x_{M}$. In addition by Lemma 40 $\|x_{M}\|_{1}\leq 1$.
Consequently, we have

$\int_{0}^{1}\min\\{x_{M}(s),tx_{M}^{d}(s)\\}ds=\int_{0}^{1}x_{M}(s)ds\leq
1=\varphi_{M}(1)\leq\frac{5}{2}\varphi_{M}(t),t\geq 1.$

Thus, combining the last two inequalities with (111) we obtain that

$\frac{1}{12}\varphi_{M}(t)\leq\int_{0}^{1}\min\\{x_{M}(s),tx_{M}^{d}(s)\\}ds\leq\frac{5}{2}\varphi_{M}(t)$
(112)

for all $t\geq 0$.

Next, for the function $M_{p}(t)$ defined by (79) we have

$\min\\{t^{2},t^{p}\\}\leq M_{p}(t)\leq\frac{2}{p}\min\\{t^{2},t^{p}\\},t\geq
0.$

Consequently, from the one hand for all $t\geq 0$ we have

$\displaystyle\begin{split}\int_{0}^{1}M_{p}(tx_{M}^{1/p}(s))ds&\leq\frac{2}{p}\int_{0}^{1}\min\\{t^{2}x_{M}^{2/p}(s),t^{p}x_{M}(s)\\}ds\\\
&\leq
2t^{p}\int_{0}^{1}\min\\{t^{2-p}x_{M}^{d}(s),x_{M}(s)\\}ds\stackrel{{\scriptstyle\eqref{lem_all_t}}}{{\leq}}5t^{p}\varphi_{M}(t^{2-p})\\\
&=5N_{M}(t)\leq N_{M}(5t).\end{split}$

And similarly, from another hand

$\displaystyle\begin{split}\int_{0}^{1}M_{p}(tx_{M}^{1/p}(s))ds&\geq\int_{0}^{1}\min\\{t^{2-p}x_{M}^{d}(s),x_{M}(s)\\}ds\stackrel{{\scriptstyle\eqref{lem_all_t}}}{{\geq}}\frac{1}{12}t^{p}\varphi_{M}(t^{2-p})\\\
&=\frac{1}{12}N_{M}(t)\geq N_{M}(\frac{1}{12}t),\end{split}$

That is for every finitely supported $x\in l_{\infty}$

$\frac{1}{12}\|x\|_{L_{N_{M}}}\leq\|x\|_{L_{M^{\prime}}}\leq
5\|x\|_{L_{N_{M}}},$

where the Orlicz function $M^{\prime}$ is defined by the equality
$M^{\prime}(t)=\int_{0}^{1}M_{p}(tx_{M}^{1/p}(s))ds$. By Lemma 39 we have that
$\|x_{M}^{1/p}\otimes x\|_{L_{p}+L_{2}}=\|x\|_{L_{M^{\prime}}}.$ Consequently,
due to (110) we obtain that

$\frac{1}{12}\|x\|_{l_{M}}\leq\|x_{M}^{1/p}\otimes x\|_{L_{p}+L_{2}}\leq
5\|x\|_{l_{M}}$

for every finitely supported $x\in l_{\infty}.$ The last inequality
immediately implies (106) for $\xi(M)$ defined by
$\xi(M)=(i(\eta(M)))^{1/p}=(i(x_{M}))^{1/p}$. ∎

We are now ready to prove the main result of the paper.

###### Proof of Theorem 3.

Let us use means provided by Theorem 44 to represent the norm in $I$. That is,
we write

$\|x\|_{I}\sim(\int_{M\in\mathcal{O}_{p,2}}\|x\|_{l_{M}}^{p}d\nu(M))^{1/p}$
(113)

for every finitely supported $x\in l_{\infty}.$

Let us show firstly that condition (107) holds in fact for almost every
$M\in\mathcal{O}_{p,2}$. For every $M\in\mathcal{O}_{p,2},$ it follows from
$p-$convexity of $M$ that $M(t)\leq t^{p}$ for $t\in[0,1].$ Similarly,
$2-$concavity of $M$ implies that $M(t)\leq t^{2}$ for $t\geq 1.$ Denote the
Orlicz function $t\to\max\\{t^{2},t^{p}\\}$ by $F.$ We have $M\leq F$ and,
therefore, $\|x\|_{l_{M}}\leq\|x\|_{l_{F}}$ for every finitely supported $x\in
l_{\infty}.$ However, for Orlicz spaces $l_{F}$ and $l_{p}$ we have
$l_{F}=l_{p}$ and $\|\cdot\|_{F}\sim\|\cdot\|_{p}$. Hence, we have

$\|x\|_{I}\stackrel{{\scriptstyle\eqref{th_I-
sim}}}{{\leq}}\mathrm{const}\|x\|_{p}$

for every finitely supported $x\in l_{\infty}$.

Define measurable functions $\alpha_{t}:\mathcal{O}_{p,2}\to\mathbb{R}$ by
setting

$\alpha_{t}(M):=\frac{1}{t^{p}}M(t),t\in(0,1).$

Since $M(t)\leq t^{p}$, the limit $\alpha(M):=\lim_{t\rightarrow
0}\alpha_{t}(M)$ exists. Observe that $\alpha(M)$ is a measurable function as
a pointwise limit of measurable functions. Again using $p-$convexity of
$M\in\mathcal{O}_{p,2}$ we infer that $M(t)\geq\alpha(M)t^{p}$ for all $t\geq
0.$ Therefore, we have $\|x\|_{l_{M}}\geq\alpha(M)^{1/p}\|x\|_{p}.$ If
$\nu(\\{M:\alpha(M)>0\\})>0,$ then again using (113) we have
$\|x\|_{I}\geq\mathrm{const}\|x\|_{p}$ for every finitely supported $x\in
l_{\infty}.$ Since we already established that
$\|x\|_{I}\leq\mathrm{const}\|x\|_{p},$ it follows that $l_{p}=I,$ that
contradicts to the assumption $\mathcal{I}\neq\mathcal{L}_{p}(H)$. Hence,
$\nu(\\{M:\alpha(M)>0\\})=0,$ and therefore equality (107) holds for almost
every $M.$

Let $\xi:\mathcal{O}_{p,2}\to\mathcal{L}_{p}(\mathcal{R}),$ be the Bochner
measurable function constructed in Lemma 45. By the preceding argument
inequality (106) holds for almost every $M$ in $\mathcal{O}_{p,2}$. Therefore,
combining (113) and (106) we obtain that

$\|A\|_{\mathcal{I}}\sim\left(\int_{M\in\mathcal{O}_{p,2}}\|\xi(M)\otimes
A\|_{\mathcal{L}_{p}+\mathcal{L}_{2}}^{p}d\nu(M)\right)^{1/p}$

for every finite rank operator $A\in\mathcal{L}(H).$ Furthermore, Lemma 38
implies that

$\|A\|_{\mathcal{I}}\sim\left(\int_{M\in\mathcal{O}_{p,2}}\|\mathscr{K}(\xi(M)\otimes
A\otimes r)\|_{p}^{p}d\nu(M)\right)^{1/p}$ (114)

for every finite rank operator $A\in\mathcal{L}(H).$

Since every finite rank operator is contained in $\mathcal{L}_{p}(H)$ we have
that $B\otimes
A\in(\mathcal{L}_{p}+\mathcal{L}_{2})(\mathcal{R}\overline{\otimes}\mathcal{L}(H))$
for every $B\in\mathcal{L}_{p}(\mathcal{R})$. Hence by Lemma 38, for a fixed
finite rank operator $A,$ the mapping $B\to\mathscr{K}(B\otimes A\otimes r)$
is continuous on $\mathcal{L}_{p}(\mathcal{R}).$ Since
$\xi:\mathcal{O}_{p,2}\to\mathcal{L}_{p}(\mathcal{R})$ is Bochner measurable,
we infer that the function $M\to\mathscr{K}(\xi(M)\otimes A\otimes r)$ from
$\mathcal{O}_{p,2}$ into $\mathcal{L}_{p}(\mathcal{R}),$ is also Bochner
measurable. Define an operator $\mathscr{J}$ from the ideal of finite rank
operators into the set of Bochner measurable
$\mathcal{L}_{p}(\mathcal{R})-$valued functions on $\mathcal{O}_{p,2}$ by
setting

$(\mathscr{J}(A))(M)=\mathscr{K}(\xi(M)\otimes A\otimes r),\quad
M\in\mathcal{O}_{p,2}.$

It is proved in [31] (see Theorem 2 there and the preceding paragraph) that
every metrizable compact equipped with atomless Radon probability measure is
isomorphic to a unit interval equipped with Lebesgue measure. Hence, every
metrizable compact equipped with (not necessarily atomless) Radon probability
measure is a standard probability space in the sense of Rokhlin. That is, (up
to an isomorphism) it is an interval equipped with Lebesgue measure plus at
most countably many atoms. Therefore, by Theorem 44 $(\mathcal{O}_{p,2},\nu)$
is (isomorphic to) an interval equipped with Lebesgue measure plus at most
countably many atoms.

It follows from [10, Lemma 6.2] that Bochner space
$L_{p}(\mathcal{O}_{p,2},\nu,\mathcal{L}_{p}(\mathcal{R}))$ is naturally
isomorphic to the space
$\mathcal{L}_{p}(\mathcal{R}\overline{\otimes}L_{\infty}(\mathcal{O}_{p,2},\nu),\tau\otimes
d\nu)$ and

$\left(\int_{M\in\mathcal{O}_{p,2}}\|\mathscr{K}(\xi(M)\otimes A\otimes
r)\|_{p}^{p}d\nu(M)\right)^{1/p}=\|\mathscr{J}(A)\|_{\mathcal{L}_{p}(\mathcal{R}\otimes
L_{\infty}(\mathcal{O}_{p,2},\nu))}.$ (115)

Combining (114) and (115), we infer that
$\|A\|_{\mathcal{I}}\sim\|\mathscr{J}(A)\|_{p}$ for every finite rank operator
$A.$ Since $\mathcal{I}$ is separable, the set of all finite rank operators is
dense in $\mathcal{I}$, and therefore, $\mathcal{I}$ isomorphically embeds
into $\mathcal{L}_{p}(\mathcal{R}\otimes L_{\infty}(\mathcal{O}_{p,2},\nu))$.

Since there exists a trace preserving $*-$homomorphism from
$L_{\infty}(\mathcal{O}_{p,2},\nu)$ into $L_{\infty}(0,1)$ and a trace
preserving $*-$homomorphism from $L_{\infty}(0,1)$ into $\mathcal{R}$, we have
that there exists a trace preserving $*-$homomorphism from $\mathcal{R}\otimes
L_{\infty}(\mathcal{O}_{p,2},\nu)$ into $\mathcal{R}\otimes\mathcal{R}.$ The
latter von Neumann algebra is $*-$isomorphic to $\mathcal{R}.$ Thus,
$\mathcal{L}_{p}(\mathcal{R}\otimes L_{\infty}(\mathcal{O}_{p,2},\nu))$
isomorphically embeds into $\mathcal{L}_{p}(\mathcal{R}).$ Consequently, since
$\mathcal{I}$ is isomorphically embeds into
$\mathcal{L}_{p}(\mathcal{R}\otimes L_{\infty}(\mathcal{O}_{p,2},\nu))$ we
infer that $\mathcal{I}$ isomorphically embeds into
$\mathcal{L}_{p}(\mathcal{R}).$ ∎

##  8\. Proof of Theorem 6

Throughout this section, let $\mathcal{N}$ be a finite von Neumann algebra
equipped with a faithful normal tracial state $\tau.$

###  8.1. Definitions of auxiliary algebras and homomorphisms

Here we define finite von Neumann algebras $(\mathcal{N}_{m},\tau_{m})$ by
means of the GNS construction of the algebra
$l_{\infty}(\mathbb{Z}_{+}^{m})\otimes\mathcal{N}$ with respect to a certain
tracial state $\tau_{m}.$ After that, we discuss certain properties of trace
preserving $*-$isomorphisms $\varrho_{k},$ $0\leq k<m.$

Fix a free ultrafilter $\omega$ on $\mathbb{Z}_{+}$ and denote by
$\mathcal{N}_{m}$ the von Neumann algebra obtained via the GNS construction
from the algebra $l_{\infty}(\mathbb{Z}_{+}^{m})\otimes\mathcal{N}$
(identified with the algebra of all bounded $\mathcal{N}-$valued functions on
$\mathbb{Z}_{+}^{m}$) and the tracial state 333Here, $\lim_{n\to\omega}a(n)$
denotes the limit of a sequence $a=\\{a(n)\\}_{n\geq 0}$ along the ultrafilter
$\omega.$

$\tau_{m}:x\to\lim\limits_{n_{0}\to\omega}\cdots\lim\limits_{n_{m-1}\to\omega}\tau(x(n_{0},\cdots,n_{m-1})),\quad
x\in l_{\infty}(\mathbb{Z}_{+}^{m})\otimes\mathcal{N}.$

That is, $\mathcal{N}_{m}$ is the quotient of the algebra
$l_{\infty}(\mathbb{Z}_{+}^{m})\otimes\mathcal{N}$ by the ideal

$\Big{\\{}x\in l_{\infty}(\mathbb{Z}_{+}^{m})\otimes\mathcal{N}:\
\lim\limits_{n_{0}\to\omega}\cdots\lim\limits_{n_{m-1}\to\omega}\tau(|x^{2}|(n_{0},\cdots,n_{m-1}))=0\Big{\\}}.$

The state $\tau_{m}$ generates a faithful normal tracial state on
$\mathcal{N}_{m}$ (also denoted by $\tau_{m}$).

Let $\varrho_{k}:{\mathcal{N}}_{1}\to{\mathcal{N}}_{m},$ $0\leq k<m,$ be the
unital $*-$homomorphism defined by the evaluation at the $k-$th coordinate.
That is,

$(\varrho_{k}(x))(n_{0},\cdots,n_{m-1})=x(n_{k}),\quad(n_{0},\cdots,n_{m-1})\in\mathbb{Z}_{+}^{m}.$

We have

$\tau_{m}(\varrho_{k}(x))=\lim\limits_{n_{0}\to\omega}\cdots\lim\limits_{n_{m-1}\to\omega}\tau(x(n_{k}))=\lim\limits_{n_{k}\to\omega}\tau(x(n_{k}))=\lim\limits_{n_{0}\to\omega}\tau(x(n_{0}))=\tau_{1}(x).$

That is, $\varrho_{k}$ is trace preserving for every $0\leq k<m.$

In what follows, $M_{m}(\mathbb{C})$, $m\geq 1$ is equipped with the
normalised trace $\frac{1}{m}{\rm Tr}.$ We identify the tensor product algebra
$(M_{m}(\mathbb{C}),\frac{1}{m}{\rm Tr})^{\otimes m}$ with the algebra
$M_{m^{m}}(\mathbb{C})$ equipped with a normalised trace. Let
$\theta_{k}:M_{m}(\mathbb{C})\to M_{m^{m}}(\mathbb{C}),$ $0\leq k<m,$ be a
unital $*-$homomorphism defined by the setting

$\theta_{k}(A)=1^{\otimes k}\otimes x\otimes 1^{\otimes(m-1-k)},\quad x\in
M_{m}(\mathbb{C}).$

In what follows, $\Omega_{m}=\\{0,1,\cdots,m-1\\}$ (respectively,
$\Omega_{m^{m}}=\\{0,\cdots,m^{m}-1\\}$) equipped with the normalised measure
and the algebra $L_{\infty}(\Omega_{m})$ (respectively,
$L_{\infty}(\Omega_{m^{m}})$) is identified with the diagonal subalgebra of
$M_{m}(\mathbb{C})$ (respectively, of $M_{m^{m}}(\mathbb{C})$). Each
$\theta_{k},$ $0\leq k<m,$ maps a diagonal subalgebra $L_{\infty}(\Omega_{m})$
of $M_{m}(\mathbb{C})$ into the diagonal subalgebra
$L_{\infty}(\Omega_{m^{m}})$ of $M_{m^{m}}(\mathbb{C}).$

Finally, let us define the probability space

$(\mathcal{M}_{m},\sigma_{m}):=(M_{m^{m}}(\mathbb{C}),\frac{1}{m^{m}}{\rm
Tr})\otimes(\mathcal{N}_{m},\tau_{m}).$

###  8.2. Equivalent representation for the norm $\|\cdot\|_{\mathcal{I}}$

In this subsection, we assume that the commutative core $I$ of a Banach ideal
$\mathcal{I}$ admits an isomorphic embedding
$T:I\to\mathcal{L}_{p}(\mathcal{N},\tau),$ $1\leq p<2$. Our main objective in
this subsection is Lemma 48 providing an equivalent representation of the norm
$\|\cdot\|_{\mathcal{I}}$.

######  Lemma 46.

Let $I\neq l_{p}$ be a symmetric sequence space and let
$T:I\to\mathcal{L}_{p}(\mathcal{N},\tau),$ $1\leq p<2,$ be an isomorphic
embedding. If $\\{e_{n}\\}_{n\geq 0}$ is a standard basis of $I,$ then the
sequence $\\{|T(e_{n})|^{p}\\}_{n\geq 0}$ is uniformly integrable.

###### Proof.

Assume the contrary. By [29, Corollary 2.11], there exists a strictly
increasing sequence $n_{k},$ $k\geq 0,$ such that the sequence
$\\{T(e_{n_{k}})\\}_{k\geq 0}$ is equivalent to the standard basis of $l_{p}.$
Thus, the sequence $\\{e_{n_{k}}\\}_{k\geq 0}$ is equivalent to the standard
basis of $l_{p}.$ Since $\\{e_{n}\\}_{n\geq 0}$ is a symmetric basis of $I,$
it follows that $\\{e_{n}\\}_{n\geq 0}$ is equivalent to the standard basis of
$l_{p}.$ In particular, we have $I=l_{p}$, which is a contradiction with the
assumption. ∎

The following lemma provides an equivalent representation for the norm
$\|\cdot\|_{{I}}$.

######  Lemma 47.

Let $I$ be a symmetric sequence space. For every $\alpha\in\mathbb{C}^{m},$ we
have444Here, the notation $f\otimes e_{k},$ $f\in L_{p}(\Omega_{m^{m}})$
stands for the element of $L_{p}(\Omega_{m^{m}},I)$ acting by $u\to f(u)e_{k}$
for every $u\in\Omega_{m^{m}}.$

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\alpha)\otimes
e_{k}\Big{\|}_{L_{p}(\Omega_{m^{m}},I)}\stackrel{{\scriptstyle
C_{abs}}}{{\sim}}\|\alpha\|_{I},\quad 1\leq p\leq 2,$

Here, $C_{abs}$ is the absolute constant.

###### Proof.

For convenience of the reader, we restate here the main result of [36]. If
$\\{f_{k}\\}_{k=0}^{m-1}$ is a sequence of independent random variables on a
probability space $\Omega_{m^{m}}$, then

$\displaystyle\Big{\|}\sum_{k=0}^{m-1}f_{k}\otimes
e_{k}\Big{\|}_{L_{p}(\Omega_{m^{m}},I)}\stackrel{{\scriptstyle
C_{abs}}}{{\sim}}\Big{(}m\int_{0}^{1/m}\mu^{p}(s,h)ds\Big{)}^{\frac{1}{p}}+\Big{\|}\sum_{i=1}^{m}\Big{(}m\int_{(i-1)/m}^{i/m}\mu^{p}(s,h)ds\Big{)}^{\frac{1}{p}}e_{i}\Big{\|}_{I},$

where the function $h\in L_{\infty}(\Omega_{m^{m}}\times(0,1))$ is given by
the formula

$h(u,t)=\sum_{i=1}^{m}\chi_{(\frac{i-1}{m},\frac{i}{m})}(t)f_{i}(u),\quad
u\in\Omega_{m^{m}},t\in[0,1].$

Here, $\chi_{(\frac{i-1}{m},\frac{i}{m})}$ is the indicator function of the
interval $(\frac{i-1}{m},\frac{i}{m}).$ Setting $f_{k}=\theta_{k}(\alpha),$ we
see that $h$ is equimeasurable with $\alpha$ (considered as an element of
$L_{\infty}(\Omega_{m})$ equipped with a normalised measure). Therefore,

$m\int_{(i-1)/m}^{i/m}\mu^{p}(s,h)ds=\mu^{p}(i-1,\alpha),\quad 1\leq i\leq m.$

In the latter formula, $\mu(\alpha)$ is computed with respect to the counting
measure. In this setting, the above result reads as follows

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\alpha)\otimes
e_{k}\Big{\|}_{L_{p}(\Omega_{m^{m}},I)}\stackrel{{\scriptstyle
C_{abs}}}{{\sim}}\|\alpha\|_{\infty}+\|\alpha\|_{I}.$

Since $\|\alpha\|_{\infty}\leq\|\alpha\|_{I}$, the assertion follows. ∎

In the next lemma, we provide an equivalent representation for the norm
$\|\cdot\|_{\mathcal{I}}$ of the Banach ideal $\mathcal{I}$ under the
assumption that the commutative core $I$ admits an isomorphic embedding into
$\mathcal{L}_{p}(\mathcal{N},\tau).$ We recall that trace preserving
$*-$homomorphisms $\\{\varrho_{k}\\}_{k=0}^{m-1}$ from $\mathcal{N}_{1}$ to
$\mathcal{N}_{m}$ were introduced in the preceding subsection.

######  Lemma 48.

Let $I\neq l_{p}$ be a symmetric sequence space and let
$T:I\to\mathcal{L}_{p}(\mathcal{N},\tau)$ be an isomorphic embedding. There
exists555Note that $F$ does not depend on the choice of $m.$
$F\in\mathcal{L}_{p}(\mathcal{N}_{1},\tau_{1})$ such that

$\|A\|_{\mathcal{I}}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes\varrho_{k}(F)\Big{\|}_{\mathcal{L}_{p}(\mathcal{M}_{m},\sigma_{m})},\quad
A=A^{*}\in M_{m}(\mathbb{C}).$

Here, $C(I)$ is the constant which only depends on $I$ (but not on the integer
$m\geq 1$) and whose value may change from line to line.

###### Proof.

Set $F=\\{T(e_{k})\\}_{k\geq 0}\in\mathcal{L}_{p}(\mathcal{N},\tau)^{\omega}.$
It follows from Lemma 46 that $|F|^{p}$ is a uniformly integrable sequence.
Using Lemma 18, we infer that

$F\in\mathcal{L}_{p}(\mathcal{N}^{\omega},\tau^{\omega})\stackrel{{\scriptstyle
def}}{{=}}\mathcal{L}_{p}(\mathcal{N}_{1},\tau_{1}).$

Fix a self-adjoint operator $A\in M_{m}(\mathbb{C})$ and select a unitary
operator $U\in M_{m}(\mathbb{C})$ such that666Here, $\lambda(A)\in
L_{\infty}(\Omega_{m})$ denotes the eigenvalue sequence of the operator $A.$
$A=U^{-1}\lambda(A)U.$ Define a unitary operator $V\in
M_{m^{m}}(\mathbb{C})\otimes\mathcal{N}$ by setting

$V=\Big{(}\prod_{k=0}^{m-1}\theta_{k}(U)\Big{)}\otimes 1.$

We have

$V^{-1}\Big{(}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
T(e_{n_{k}})\Big{)}V=\sum_{k=0}^{m-1}\theta_{k}(A)\otimes T(e_{n_{k}}).$

Here, $(n_{0},\cdots,n_{m-1})\in\mathbb{Z}_{+}^{m}$ is arbitrary. Therefore,
for the algebra $M_{m^{m}}(\mathbb{C})\otimes\mathcal{N}$ equipped with a
tensor product trace, we have

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes
T(e_{n_{k}})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}=\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
T(e_{n_{k}})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}.$

Since each $\theta_{k}$ maps the diagonal subalgebra $L_{\infty}(\Omega_{m})$
of $M_{m}(\mathbb{C})$ into the diagonal subalgebra
$L_{\infty}(\Omega_{m^{m}})$ of $M_{m^{m}}(\mathbb{C}),$ it follows that

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes
T(e_{n_{k}})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}=\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
T(e_{n_{k}})\Big{\|}_{L_{p}(\Omega_{m^{m}},L_{p}({\mathcal{N}},\tau))},$

where on the right hand side we use the Bochner space norm. Taking
$C(I)\geq\|T\|,\|T^{-1}\|,$ we obtain

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
T(e_{n_{k}})\Big{\|}_{L_{p}(\Omega_{m^{m}},L_{p}({\mathcal{N}},\tau))}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
e_{n_{k}}\Big{\|}_{L_{p}(\Omega_{m^{m}},I)}.$

Therefore, we have

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes
T(e_{n_{k}})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
e_{n_{k}}\Big{\|}_{L_{p}(\Omega_{m^{m}},I)}.$

Since $\\{e_{k}\\}_{k\geq 0}$ is a symmetric basis in $I,$ it follows that

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes
T(e_{n_{k}})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(\lambda(A))\otimes
e_{k}\Big{\|}_{L_{p}(\Omega_{m^{m}},I)}$

whenever $n_{0}<n_{1}<\cdots<n_{m-1}.$ It follows now from Lemma 47 that

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes
T(e_{n_{k}})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\|\lambda(A)\|_{I}=\|A\|_{\mathcal{I}}$ (116)

whenever $n_{0}<n_{1}<\cdots<n_{m-1}.$ By definitions of $F$ and
$\varrho_{k},$ $0\leq k<m,$ we have that

$(\varrho_{k}(F))(n_{0},\cdots,n_{m-1})=T(e_{n_{k}}),\quad(n_{0},\cdots,n_{m-1})\in\mathbb{Z}_{+}^{m}.$

Therefore, we can restate (116) as follows:

$\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes(\varrho_{k}(F))(n_{0},\cdots,n_{m-1})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\|A\|_{\mathcal{I}}$

whenever $n_{0}<n_{1}<\cdots<n_{m-1}.$ Since
$F\in\mathcal{L}_{p}(\mathcal{N}_{1},\tau_{1}),$ it follows that
$\varrho_{k}(F)\in\mathcal{L}_{p}(\mathcal{N}_{m},\tau_{m}).$ Hence,
$\sum_{k=0}^{m-1}\theta_{k}(A)\otimes\varrho_{k}(F)\in\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N}_{m}).$
Clearly, $M_{m^{m}}(\mathbb{C})\otimes\mathcal{N}_{m}$ is an ultrapower of
$M_{m^{m}}(\mathbb{C})\otimes\mathcal{N}.$ Thus, we have

$\displaystyle\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes\varrho_{k}(F)\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N}_{m})}$
$\displaystyle=\lim\limits_{n_{0}\to\omega}\cdots\lim\limits_{n_{m-1}\to\omega}\Big{\|}\sum_{k=0}^{m-1}\theta_{k}(A)\otimes(\varrho_{k}(F))(n_{0},\cdots,n_{m-1})\Big{\|}_{\mathcal{L}_{p}(M_{m^{m}}(\mathbb{C})\otimes\mathcal{N})}\stackrel{{\scriptstyle
C(I)}}{{\sim}}\|A\|_{\mathcal{I}}.$

This proves the assertion. ∎

###  8.3. Proof of uniform integrability

In this subsection, we show that, for every rank one projection $p,$ there
exists an element $z\in L_{p}(0,1)$ such that

$\mu(\sum_{k=0}^{m-1}\theta_{k}(p)\otimes\varrho_{k}(F))\leq\mu(z),\quad m\geq
1,$

where $F$ is defined in Lemma 48 above. The following lemma shows that
$\\{\varrho_{k}\\}_{k\geq 0}$ are subsymmetric copies in the sense of
Koestler.

######  Lemma 49.

For every $\mathcal{A}\subset\\{0,\cdots,m-1\\},$ we have

$\mu\Big{(}\sum_{k\in\mathcal{A}}\varrho_{k}(x)\Big{)}=\mu\Big{(}\sum_{k=0}^{|\mathcal{A}|-1}\varrho_{k}(x)\Big{)},\quad
x\in\mathcal{S}(\mathcal{N}_{1},\tau_{1}).$

###### Proof.

Without loss of generality, $x$ is bounded. It suffices777Indeed, if all
moments of bounded random variables $x$ and $y$ coincide, then so do their
characteristic functions. Hence, their distributions also coincide. to show
that

$\tau_{m}\Big{(}\Big{(}\sum_{k\in\mathcal{A}}\varrho_{k}(x)\Big{)}^{N}\Big{)}=\tau_{m}\Big{(}\Big{(}\sum_{k=0}^{M-1}\varrho_{k}(x)\Big{)}^{N}\Big{)},\quad
N\in\mathbb{N},$

where we denoted for brevity, $M=|\mathcal{A}|.$ Let
$\mathcal{A}=\\{k_{0},\cdots,k_{M-1}\\}.$ It is clear that

$\tau_{m}\Big{(}\Big{(}\sum_{k\in\mathcal{A}}\varrho_{k}(x)\Big{)}^{N}\Big{)}=\lim\limits_{n_{0}\to\omega}\cdots\lim\limits_{n_{m-1}\to\omega}\tau\Big{(}\Big{(}\sum_{i=0}^{M-1}x(n_{k_{i}})\Big{)}^{N}\Big{)}=$
$=\lim\limits_{n_{k_{0}}\to\omega}\cdots\lim\limits_{n_{k_{M-1}}\to\omega}\tau\Big{(}\Big{(}\sum_{i=0}^{M-1}x(n_{k_{i}})\Big{)}^{N}\Big{)}.$

Renaming the index $n_{k_{0}}$ with $n_{0},$ we obtain

$\tau_{m}\Big{(}\Big{(}\sum_{k\in\mathcal{A}}\varrho_{k}(x)\Big{)}^{N}\Big{)}=\lim\limits_{n_{0}\to\omega}\lim\limits_{n_{k_{1}}\to\omega}\cdots\lim\limits_{n_{k_{M-1}}\to\omega}\tau\Big{(}\Big{(}x(n_{0})+\sum_{i=1}^{M-1}x(n_{k_{i}})\Big{)}^{N}\Big{)}.$

Renaming the index $n_{k_{1}}$ with $n_{1},$ we obtain

$\tau_{m}\Big{(}\Big{(}\sum_{k\in\mathcal{A}}\varrho_{k}(x)\Big{)}^{N}\Big{)}=\lim\limits_{n_{0}\to\omega}\lim\limits_{n_{1}\to\omega}\lim\limits_{n_{k_{2}}\to\omega}\cdots\lim\limits_{n_{k_{M-1}}\to\omega}\tau\Big{(}\Big{(}x(n_{0})+x(n_{1})+\sum_{i=2}^{M-1}x(n_{k_{i}})\Big{)}^{N}\Big{)}.$

Repeating the process, we conclude the proof. ∎

Consider the probability space $(\mathcal{N}_{{\rm ext}},\tau_{{\rm ext}})$

$\mathcal{N}_{{\rm ext}}=\mathbb{C}\bigoplus\Big{(}\bigoplus_{m\geq
1}\mathcal{N}_{m}\Big{)},\quad\tau_{{\rm
ext}}=\frac{1}{e}\bigoplus\Big{(}\bigoplus_{k=0}^{\infty}\frac{1}{e\cdot
m!}\tau_{m}\Big{)}.$

Define the mapping
$\mathscr{L}:\mathcal{L}_{1}(\mathcal{N}_{1},\tau_{1})\to\mathcal{L}_{1}(\mathcal{N}_{{\rm
ext}},\tau_{{\rm ext}})$ defined by the setting

$\mathscr{L}(x)=0\bigoplus\Big{(}\bigoplus_{m\geq
1}\Big{(}\sum_{k=0}^{m-1}\varrho_{k}(x)\Big{)}\Big{)},\quad
x\in\mathcal{L}_{1}(\mathcal{N}_{1},\tau_{1}).$

The expression for the mapping $\mathscr{L}$ could be understood as a
subsymmetric version of the Kruglov operator replacing tensor copies by the
copies given by the $\\{\rho_{k}\\}_{k\geq 0}.$ However, we do not need these
properties and, hence, do not state and prove them.

######  Lemma 50.

$\mathscr{L}:\mathcal{L}_{p}(\mathcal{N}_{1},\tau_{1})\to\mathcal{L}_{p}(\mathcal{N}_{{\rm
ext}},\tau_{{\rm ext}})$ for all $1\leq p\leq 2.$

###### Proof.

Let $0\leq x\in\mathcal{L}_{2}(\mathcal{N}_{1},\tau_{1}).$ We have

$\tau_{{\rm ext}}\big{(}(\mathscr{L}x)^{2}\big{)}=\sum_{m\geq
1}\frac{1}{e\cdot
m!}\tau_{m}\Big{(}\Big{(}\sum_{k=0}^{m-1}\varrho_{k}(x)\Big{)}^{2}\Big{)}=\sum_{m\geq
1}\frac{1}{e\cdot
m!}\sum_{k_{1},k_{2}=0}^{m-1}\tau_{m}(\varrho_{k_{1}}(x)\varrho_{k_{2}}(x)).$

Since the $*-$homomorphisms $\varrho_{k},$ $0\leq k<m,$ are trace preserving,
it follows that

$0\leq\tau_{m}(\varrho_{k_{1}}(x)\varrho_{k_{2}}(x))\leq\left(\tau_{m}((\varrho_{k_{1}}(x))^{2})\tau_{m}((\varrho_{k_{2}}(x))^{2})\right)^{1/2}=\tau_{1}(x^{2}).$

Therefore, we have

$\tau_{{\rm ext}}((\mathscr{L}x)^{2})\leq\sum_{m\geq 1}\frac{m^{2}}{e\cdot
m!}\tau(x^{2})=2\tau(x^{2}).$

It is immediate that
$\|\mathscr{L}\|_{\mathcal{L}_{2}\to\mathcal{L}_{2}}\leq\sqrt{2}.$ Since
$\tau_{{\rm ext}}(\mathscr{L}x)=\tau_{1}(x)$ for every
$x\in\mathcal{L}_{1}(\mathcal{N}_{1},\tau_{1}),$ it follows that
$\|\mathscr{L}\|_{\mathcal{L}_{1}\to\mathcal{L}_{1}}\leq 1.$ The assertion
follows by interpolation. ∎

######  Lemma 51.

For every rank one projection $p\in M_{m}(\mathbb{C}),$ we have

$\mu\Big{(}\sum_{k=0}^{m-1}\theta_{k}(p)\otimes\varrho_{k}(x)\Big{)}\leq
D_{e}\mu(\mathscr{L}(x)),\quad x\in\mathcal{S}(\mathcal{N}_{1},\tau_{1}).$

Here, the left hand side is computed in $(\mathcal{M}_{m},\sigma_{m})$ and the
right hand side is computed in $(\mathcal{N}_{{\rm ext}},\tau_{{\rm ext}}).$

###### Proof.

Consider the family $\\{\theta_{k}(p)\\}_{k=0}^{m-1}$ of commuting projections
with trace $\frac{1}{m}$. Define the family of $2^{m}$ pairwise orthogonal
projections

$p_{\mathcal{A}}=\Big{(}\prod_{k\in\mathcal{A}}\theta_{k}(p)\Big{)}\cdot\Big{(}\prod_{k\notin\mathcal{A}}(1-\theta_{k}(p))\Big{)},\quad\mathcal{A}\subset\\{0,1,\cdots,m-1\\}.$

A moment’s reflection shows that $\sum_{\mathcal{A}}p_{\mathcal{A}}=1.$ We
have

$p_{\mathcal{A}}\theta_{k}(p)=\begin{cases}p_{\mathcal{A}},\quad
k\in\mathcal{A}\\\ 0,\quad k\notin\mathcal{A}.\end{cases}$

Therefore,

$\sum_{k=0}^{m-1}\theta_{k}(p)\otimes\varrho_{k}(x)=\sum_{\mathcal{A}}\sum_{k=0}^{m-1}p_{\mathcal{A}}\theta_{k}(p)\otimes\varrho_{k}(x)=\sum_{\mathcal{A}}p_{\mathcal{A}}\otimes\Big{(}\sum_{k\in\mathcal{A}}\varrho_{k}(x)\Big{)}.$

Since the projections $p_{\mathcal{A}},$
$\mathcal{A}\subset\\{0,1,\cdots,m-1\\},$ are pairwise orthogonal, it follows
from Lemma 49 that

$\mu\Big{(}\sum_{k=0}^{m-1}\theta_{k}(p)\otimes\varrho_{k}(x)\Big{)}=\mu\Big{(}\sum_{\mathcal{A}}p_{\mathcal{A}}\otimes\Big{(}\sum_{k=0}^{|\mathcal{A}|-1}\varrho_{k}(x)\Big{)}\Big{)}.$

Set

$p_{l}=\sum_{|\mathcal{A}|=l}p_{\mathcal{A}},\quad 0\leq l\leq m.$

These projections are pairwise orthogonal and $\sum_{l=0}^{m}p_{l}=1.$ We have

$\mu\Big{(}\sum_{k=0}^{m-1}\theta_{k}(p)\otimes\varrho_{k}(x)\Big{)}=\mu\Big{(}\sum_{l=0}^{m}p_{l}\otimes\Big{(}\sum_{k=0}^{l-1}\varrho_{k}(x)\Big{)}\Big{)}.$
(117)

Since the projections $\theta_{k}(p),$ $0\leq k<m,$ are independent (see
Subsection 2.8), it is immediate from the definition of $p_{\mathcal{A}}$ that

$\frac{1}{m^{m}}{\rm
Tr}(p_{\mathcal{A}})=\Big{(}\frac{1}{m}\Big{)}^{|\mathcal{A}|}\Big{(}1-\frac{1}{m}\Big{)}^{m-|\mathcal{A}|},\quad\mathcal{A}\subset\\{0,1,\cdots,m-1\\}.$

Therefore, we have

$\frac{1}{m^{m}}{\rm
Tr}(p_{l})=\binom{m}{l}\Big{(}\frac{1}{m}\Big{)}^{l}\Big{(}1-\frac{1}{m}\Big{)}^{m-l}\leq\frac{1}{l!}=e\cdot\frac{1}{e\cdot
l!},\quad 0\leq l\leq m.$

The assertion follows by combining (117) with the definition of
$\mathscr{L}(x)$ and the definition of the dilation operator. ∎

###  8.4. Proof of Theorem 6

We are now ready to prove our second main result.

###### Proof of Theorem 6.

The proof of the implication $\eqref{ji}\to\eqref{jii}$ is immediate. We now
proceed with the proof of the implication $\eqref{jii}\to\eqref{ji}.$ Suppose
that there exists an isomorphic embedding from the commutative core $I$ of the
Banach ideal $\mathcal{I}$ to $\mathcal{L}_{p}(\mathcal{N},\tau)$ for some
probability space $(\mathcal{N},\tau).$ Let $\\{p_{m}\\}_{m\geq 0}$ be an
increasing sequence such that ${\rm Tr}(p_{m})=m.$ We identify
$p_{m}\mathcal{L}(H)p_{m}=M_{m}(\mathbb{C}).$ Take an element
$F\in\mathcal{L}_{p}(\mathcal{N}_{1},\tau_{1})$ constructed in Lemma 48.
Define the mapping
$T_{m}:\mathcal{C}_{00}\to\mathcal{L}_{p}(\mathcal{M}_{m},\sigma_{m})$ by
setting

$T_{m}(A)\stackrel{{\scriptstyle
def}}{{=}}\sum_{k=0}^{m-1}\theta_{k}(p_{m}Ap_{m})\otimes\varrho_{k}(F),\quad
m\geq 0,$

where $\mathcal{C}_{00}$ denotes the ideal of finite rank operators in
$\mathcal{L}(H).$ Set

$T(A)\stackrel{{\scriptstyle def}}{{=}}\\{T_{m}(A)\\}_{m\geq
0}\in\Big{(}\mathcal{L}_{p}(\mathcal{M}_{m},\sigma_{m})\Big{)}_{m\geq
0}^{\omega}.$

If $A=\sum_{k=0}^{N-1}\lambda(k,A)q_{k},$ where each $q_{k},$ $0\leq k<N,$ is
a rank one projection, then (see (6))

$\mu(T_{m}(A))=\mu\Big{(}\sum_{k=0}^{N-1}\lambda(k,A)T_{m}(q_{k})\Big{)}\leq\|A\|_{\infty}ND_{N}\mu(T_{m}(q_{0})).$

Using Lemma 51, we infer that

$\mu(T_{m}(A))\leq N\|A\|_{\infty}D_{eN}\mu(\mathscr{L}(F)).$

Since $\mathscr{L}(F)\in\mathcal{L}_{p}(\mathcal{N}_{{\rm ext}},\tau_{{\rm
ext}}),$ it follows that the sequence $\\{|T_{m}(A)|^{p}\\}_{m\geq 0}$ is
uniformly integrable. By Lemma 18, we have

$T(A)\in\mathcal{L}_{p}(\Big{(}\mathcal{M}_{m}\Big{)}_{m\geq
0}^{\omega},\Big{(}\sigma_{m}\Big{)}_{m\geq 0}^{\omega}),\quad
A\in\mathcal{C}_{00}.$

It follows immediately from Lemma 48 that

$\|T(A)\|_{p}=\lim_{m\to\omega}\|T_{m}(A)\|_{p}\sim\lim_{m\to\omega}\|p_{m}Ap_{m}\|_{\mathcal{I}}=\|A\|_{\mathcal{I}},\quad
A\in\mathcal{C}_{00}.$

Thus, $T$ extends to an isomorphic embedding

$T:\mathcal{I}\to\mathcal{L}_{p}(\Big{(}\mathcal{M}_{m}\Big{)}_{m\geq
0}^{\omega},\Big{(}\sigma_{m}\Big{)}_{m\geq 0}^{\omega}).$

∎

## Appendix A

The following result shows that for the Orlicz ideal the notion of
$p$-convexity (resp., $q$-concavity) coincides with upper $p$-estimate (resp.,
lower $q$-estimate).

######  Theorem 52.

Let $1\leq p\leq 2$ and $2\leq q<\infty.$ The following conditions are
equivalent

  1. (i)

The Orlicz ideal $\mathcal{L}_{M}(H)$ is $p$-convex and $q$-concave.

  2. (ii)

The Orlicz ideal $\mathcal{L}_{M}(H)$ satisfies an upper $p$-estimate and a
lower $q$-estimate.

  3. (iii)

The Orlicz space $l_{M}$ is $p$-convex and $q$-concave.

  4. (iv)

The Orlicz space $l_{M}$ satisfies an upper $p$-estimate and a lower
$q$-estimate.

  5. (v)

The Orlicz function $M$ on $(0,\infty)$ is equivalent on $[0,1]$ to some
$p$-convex and $q$-concave Orlicz function.

###### Proof.

The implications (i)$\Rightarrow$(iii) and (ii)$\Rightarrow$(iv) are obvious.
The equivalence of (iii) and (iv) is proved in [41, p. 121,124]. The
implications (iii)$\Rightarrow$(i) and (iii)$\Rightarrow$(ii) are shown in
[22, Theorems 3.8 and 4.7] and [22, Proposition 5.1], respectively.

(iv)$\Rightarrow$(v). Let $l_{M}$ satisfies upper $p$-estimate and lower
$q$-estimate. Without loss of generality, we may assume that $M(1)=1.$
Applying (12) and (13) to the sequence

$\sum_{j=0}^{nm-1}e_{j}=\sum_{k=0}^{n-1}\sum_{j=(k-1)m}^{km-1}e_{j}\in l_{M},$

where $\\{e_{j}\\}_{j=0}^{\infty}$ is the standard vector basis in $l_{M}$ and
observing

$\Big{\|}\sum_{j=(k-1)m}^{km-1}e_{j}\Big{\|}_{l_{M}}=\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}\
\ \ \mbox{for any}\ \ \ 1\leq k\leq n,$

we obtain that

${\rm const}\cdot
n^{1/q}\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}\leq\Big{\|}\sum_{j=0}^{nm-1}e_{j}\Big{\|}_{l_{M}}\leq{\rm
const}\cdot n^{1/p}\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}$ (118)

for all $n,m\in\mathbb{N}.$

Let $0<s\leq t\leq 1.$ Set $u:=\frac{1}{M(s)}$ and $v:=\frac{1}{M(t)}.$ Then
$u\geq v\geq 1.$ Let $m:=[v]$ and $n:=\Big{[}\frac{u}{m}\Big{]}+1$. Observe
that since $v<2m$ and $v\leq u$, we have

$n\leq\frac{u}{m}+1<\frac{2u}{v}+\frac{u}{v}=\frac{3u}{v}\ \ \mbox{and}\ \
n=\Big{[}\frac{u}{m}\Big{]}+1\geq\frac{u}{m}.$ (119)

Observe also that (see e.g. [54, I, 1.2, Example 9])

$\Big{\|}\sum_{j=0}^{m-1}e_{j}\Big{\|}_{l_{M}}=\frac{1}{M^{-1}\Big{(}\frac{1}{m}\Big{)}}\
\ \ \mbox{for any}\ \ \ m\in\mathbb{N}.$ (120)

Now using the fact that $M^{-1}$ is increasing and applying the right hand
side inequality of (118) together with (120), we obtain

$t=M^{-1}\Big{(}\frac{1}{v}\Big{)}\leq M^{-1}\Big{(}\frac{1}{m}\Big{)}\leq$
$\stackrel{{\scriptstyle\eqref{eq_(ii)to(iii)1}}}{{\leq}}{\rm const}\cdot
n^{\frac{1}{p}}M^{-1}\Big{(}\frac{1}{mn}\Big{)}\stackrel{{\scriptstyle\eqref{eq_(ii)to(iii)2}}}{{\leq}}{\rm
const}\cdot\Big{(}\frac{u}{v}\Big{)}^{\frac{1}{p}}M^{-1}\Big{(}\frac{1}{u}\Big{)}={\rm
const}\cdot\Big{(}\frac{M(t)}{M(s)}\Big{)}^{\frac{1}{p}}\cdot s.$

Raising to the power $p$ both sides of the inequality above, we obtain (8).
The left hand side inequality of (118) similarly implies (9). By Lemma 11,
there exists a $p$-convex and $q$-concave Orlicz function equivalent to $M.$

(v)$\Rightarrow$(iii). Suppose that an Orlicz function $M$ satisfies (v).
Without loss of generality we may assume that $M$ is $p$-convex. Define a
convex Orlicz function $M_{1}$ by setting

$M_{1}(t):=M(t^{1/p}),\ \ \ t>0.$

Since for any finite sequence $\\{x_{k}\\}_{k=1}^{n}\subseteq l_{M},$

$\Big{\|}\Big{(}\sum_{k=1}^{n}|x_{k}|^{p}\Big{)}^{1/p}\Big{\|}_{l_{M}}=\Big{\|}\sum_{k=1}^{n}|x_{k}|^{p}\Big{\|}_{l_{M_{1}}}^{1/p}\leq\Big{(}\sum_{k=1}^{n}\|x_{k}\|_{l_{M_{1}}}^{p}\Big{)}^{1/p}\leq\Big{(}\sum_{k=1}^{n}\|x_{k}\|_{l_{M}}^{p}\Big{)}^{1/p},$

it follows that $l_{M}$ is $p$-convex. Similarly, one can deduce that $l_{M}$
is $q$-concave from $q$-concavity of $M.$

∎

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