We will show that the space contains isomorphically the space $l_1$ therefore the space is not reflexive. We start with the following that I posed as a question in a previous comment.

**Fact 1:** For every $0<\delta < 1 $ $lim_n \frac{\int_1^{1+\delta} n^{1/p} dp } {\int_1^2 n^{1/p} dp } = 1$.

**Proof:** Since the function $n^{1/p}$, $1\leq p \leq 2$ is decreasing we have that $\int_1^{1+\delta} n^{1/p} dp >\delta n^{1/1+\delta}$ and

$n^ {1/1+2\delta} > \int_{1+2\delta}^2 n^{1/p} dp$

Now for $n\in N$ we have that $ \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2 n^{1/p} dp } < \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^{1+\delta} n^{1/p} dp } < \frac{n^{1/1+2\delta }} {\delta n^{1/1+ \delta }} = \frac{1} { \delta} \frac {1}{n^{\delta/ (1+\delta)(1+2\delta)}}$.

Hence for every $0<\delta <1/2$

$lim_n \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2 n^{1/p} dp } =0$ which finishes the proof of Fact 1.

**Fact 2:** We start with the following classical result.

If $(f_n)_n$ is a normalized sequence in $L^1 [1,2]$ which is not uniformly integrable (i.e. there exists $\epsilon>0$ such that for every $\delta>0$
there exists a Borel set $A$ with $\lambda (A)<\delta$ and $\int_A |f_n| > \epsilon$ for infinite $n\in N$ ) then $(f_n)_n$ has a subsequnce equivalent to $l_1$ basis.

This result is due to Kadec and Pelczynski ( see **J. Diestel: Sequences and Series in Banach Spaces (Graduate Texts in Mathematics, 92) p. 93**).

Next in the space we denote $(e_i)_i\in N$ the basis of $l_1$ which is a symmetric basis for the space.For $n\in N $ we set $z_n = \sum_ {i=1} ^{n} e_i$ and $x_n = \frac{1} {\int_1^2 n^{1/p} dp } z_n$ which has norm 1.

Consider the function $f_n(p) = |x_n|_p $ $1\leq p \leq 2 $ and Fact 1 yields that the sequence $(f_n)$ it is not uniformly integrable.Therefore $(f_n)$ has a subsequence equivalent to $l_1$ basis which implies that $(x_n)$ satisfies the same property in the norm of the space. We will adapt Kadec - Pelczynski's proof in the setting of $(x_n)$.

**Step 1 :** There exists a decreasing sequence $(\delta_k )$ and a subsequence $(x_{n_k})$ such that for all $k$ we have that $\int_{1+\delta_{k+1}}^{1+\delta_k} |{x_{n_k}}|_p dp > \frac{1} { 4}$.

The proof uses induction and the following :

From Fact 1 for $ 0< \delta < 1$ there exists $n\in N$ such that $\int_1^{1+\delta}| {x_n}|_p dp > \frac{1} {2}$. For this $n$ there exists $\delta_1 < \delta $ such that $\int_1^{1+\delta_1}|{x_n}|_p dp < \frac{1} {4}$. Hence $\int_{1+\delta_1}^{1+\delta} |{x_n}|_p dp > \frac{1} {4}$.

**Step 2:** We set $A_k = [\delta_{k+1}, \delta_{k }]$. There exists an infinite $I \subset N $ such that for every $k\in I$ setting $ B_k = \cup \{ A_j : j\in I, j\neq k \} $ we have that $\int_{B_k} |{x_{n_k}|_p}dp < \frac {1} {8}$.

This is a classical result due to H. P. Rosenthal and an elegant and short proof was given by J. Kupka ( see Page 82 in the aforementioned reference). We assume that $ I = N $

**Claim:** The sequence $(x_{n_k} )_{k\in N}$ is equivalent to $l_1$ basis.

Indeed

$\int_1^2 |\sum_{j=1}^k \lambda_{j} x_{n_j} |_p dp > \int_{\cup_{i\in N} A_i} |\sum_{j=1}^k \lambda_{j} x_{n_j} |_pdp \geq \sum_{j=1} ^{k} (\int_{A_j} |\lambda_{j} x_{n_j}|_pdp - \int_{B_j} |\lambda_{j} x_{n_j}|_p dp)\geq
\frac {1}{8} \sum_{j=1}^{k} |\lambda_j|$.

I have two questions related to this result.

**Question 1 :** Does the space contain a complemented subspace isomorphic to $l_1$ ?

**Edit:** The answer to Question 1 is affirmative hence the dual of the space contains isomorphically the space $l_\infty $.

**The functionals $f_A^x $ where A is a Borel subset of [1,2] and $x=\sum_{i=1}^n \lambda_i e_i (\lambda_i \geq 0). $**

For $x$ as above and $ p\in (1,2] $ we set $f_p^x = \frac{1} {(\sum_{i=1}^n \lambda_i^p)^1/q } \sum_{i=1}^n \lambda_i^{1/q-1} e_i$ where $ 1/p + 1/q = 1 $.

The functional $f^x_p$ is the unique normalized element of $l_q$ that norms $x$ as an element of $l_p $.

Observe that for a given $x$ as above and $z\in l_1$ the function $f_p ^x (z) $ with variable $ p \in (1,2]$ is continuous hence for a Borel $A \subset (1,2]$ the integral

$f_A ^x(z)= \int_A f_p^x (z) dp$

is well defined for all $z \in l_1$ and $f_A ^x$ is linear.

**Properties of $f_A ^x$**.

We denote by $|.|$ the norm of the space and by $|.|_*$ the norm of its dual.

For all $x$ , $A$ $| f_A ^x |_* \leq 1$ moreover if $ |x|\leq 1$ and $\int_A |x|_p dp \geq c > 0 $ then $| f_A ^x |_* \geq c$.

For $(A_k)_{k=1}^m $ disjoint Borel sets , $(x_k)_{k=1}^m$ in $l_1$ and $(\alpha_k)_{k=1}^m$ reals we have that.

$ | \sum_{k=1}^{m} \alpha_{k}f_{A_k}^{x_k} |_* \leq max \{|\alpha_k| : k=1,...,m \}$

In particular every sequence $(f_{A_k}^{x_k} )_k$ with $(A_k))_k$ disjoint Borel sets is weakly null since every n-average of them has norm less or equal to $1/n$.

Furthermore if for every $k \in N$ $|f_{A_k}^{x_k}|_*\geq c >0 $ then $(f_{A_k}^{x_k} )_k$ has a subsequence which is Schauder basic which yields that this subsequence is equivalent to $c_o$ basis.

**The dual of the space contains isomorphically $c_0$**.

We set $A_k= (\delta_{k+1}, \delta_k)$ and $x_k = x_{n_k}$ as they appeared in Step 1 above. Then the sequence $(f_{A_k}^{x_k} )_k$ satisfies $ |f_{A_k}^{x_k}|_* \geq \frac{1} {4}$ and $(A_k)_k$ are disjoint. Hence it has a subsequence equivalent to $c_o$ basis.

**The space has a complemented subspace isomorphic to $l_1$**.

This is an immediate consequence of the previous result. A classical Theorem states that if $c_0$ is isomorphic to a subspace of $X^*$ then $l^1$ is isomorphic to a complemented subspace of $X$.

**Question 2:** Is the space $l_1$ saturated?

**Edit 1** The answer to this question is also affirmative. In particular the following holds:

**Fact 3:** Every closed infinite dimensional subspace $Z$ of the space has a further subspace $Y$ which is complemented in the space and isomorphic to $l_1$.

To prove this we first observe that it is enough to consider block subspaces namely subspaces generated by a normalized block sequence $(x_n )_n$. For such a sequence we will show the following:

**Fact 4:** For every normalized block sequence $( x_n)_n$ there exists a sequence $(F_k )_k $ with $F_k \subset N$, $\#F_k = n_k$ such setting $z_k = \sum_{n\in F_k}x_n$ the following holds.

For every $q > 1 $ $ lim_k \frac {\int_1^{q} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} = 1 $.

With this result and repeating the proof of Fact 2 above for the sequence $(z_k)_k $ we conclude that it has a subsequence equivalent to $l_1$ basis. moreover the subspace generated by a further subsequence is complemented in the space. This follows from the answer to Question 1 above with a small modification at the last part.

**Proof of Fact 4:** We set $q_k = 1 + \frac{1}{2^k} $. Observe that for all $k \in N$

$sup \{ |x_n |_{q_k} : n\in N \} \leq 2^k $.

We assume that $ lim_n |x_n |_{q_k} = C_k $ for all $k\in N $ (otherwise we pass to a subsequence).

Observe that $(C_k)_k$ is increasing.

**Claim:** For every $k\in N$ there exists $F_k \subset N$ such that setting $z_k = \sum_ {n\in F_k} x_n $ we have that

$ \frac {\int_1^{q_k} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} > 1- \frac{1} {k} $

**Proof of the Claim:** Since $ |x_n|_{q_k} \rightarrow C_k$ and $|x_n|_{q_{k+1}} \rightarrow C_{k+1}$ for every $l\in N$ we may select $F_l \subset N$ such that $\#F_l =l$ and

$ \frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^{-k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$

where $z_l = \sum _{n\in F_{l}} x_n$.

As in Fact 1 we have that:

$ \frac {\int_{q_k}^2 |z_l|_p dp } {\int_1^{2} |z_l|_p dp} < \frac {\int_{q_k}^2 |z_k|_p dp } {\int_1^{q_{k+1}} |z_k|_p dp} <
\frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } \leq
(\frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^ {-k+1}C_{k+1} l^{1/q_{k+1}}}) + \frac { C_{k} l^{1/{q_k}}} { 2^{-k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$.

We choose $l_k$ such that $\frac {\int_{q_k}^2 |z_{l_k}|_p dp } {\int_1^{2} |z_{l_k}|_p dp} <\frac {1} { k}$ and we set $ F_k = F_{l_k} $ and $ z_k = z_{l_k} $.
Clearly $F_k , z_k $ satisfy the conclusion of the claim.

**Conclusion:** The sequence $(z_k )_k $ satisfies:

For every $q > 1 $ $ lim_k \frac {\int_1^{q} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} = 1 $.Hence as in Fact 2 there is a subsequence $(z_{k_m})_m$ equivalent to $l_1$ basis. This subsequence has a further subsequence with the property that the space that generates is complemented in the whole space.

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