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The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.

$Theorem$: Let $X$ be a Banach space with its dual $w^*$ separable.If $Y$ is a closed subspace of $X$ such that $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.

I think that the result remains valid if $Y^\bot$ contains $l_1$.

It holds that $X^{**}$ and $JT^{**}$ are $w^*$ separable and also $JT_*^\bot$ is isometric to $l^{2} (2^N)$ while $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.

Following Rosenthal's proof we can describe the quasi-complement of $JT_*$ as follows.

First $JT^{**}= JT \oplus l^{2} (2^N)$. Let ($e_n)$ be the basis of $JT$ and let $(e_{\gamma_n})$ be a sequence of the basis of $l^{2} (2^N)$.We set $W$ the closed space generated by $(e_{\gamma_n} + 1/2^n e_n)$. Then $W$ is $w^*$ closed and the subspace $W_\bot$ of $JT^*$ is the quasi-complement of $JT_*$.

The proof is an easy consequence of the following observation due also to Rosenthal.

If $V$, $W$ are $w^*$ closed subspaces of $X^*$ such that $V\cap W=0$ and $V_\bot \cap W_\bot =0$ then $V_\bot$ , $W_\bot$ are quasi -complementary.

Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.

The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.

$Theorem 1$: Let $X$ be a Banach space. If $Y$ is a closed subspace of $X$ with its dual $w^*$ separable and $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.

I think that the result remains valid if $Y^\bot$ contains $l_1$.

It holds that $X^{**}$ and $JT^{**}$ are $w^*$ separable and also $JT_*^\bot$ is isometric to $l^{2} (2^N)$ while $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.

Following Rosenthal's proof we can describe the quasi-complement of $JT_*$ as follows.

First $JT^{**}= JT \oplus l^{2} (2^N)$. Let ($e_n)$ be the basis of $JT$ and let $(e_{\gamma_n})$ be a sequence of the basis of $l^{2} (2^N)$.We set $W$ the closed space generated by $(e_{\gamma_n} + 1/2^n e_n)$. Then $W$ is $w^*$ closed and the subspace $W_\bot$ of $JT^*$ is the quasi-complement of $JT_*$.

The proof is an easy consequence of the following observation due also to Rosenthal.

If $V$, $W$ are $w^*$ closed subspaces of $X^*$ such that $V\cap W=0$ and $V_\bot \cap W_\bot =0$ then $V_\bot$ , $W_\bot$ are quasi -complementary.

Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.

The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.

$Theorem$: Let $X$ be a Banach space with its dual $w^*$ separable.If $Y$ is a closed subspace of $X$ such that $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.

I think that the result remains valid if $Y^\bot$ contains $l_1$.

It holds that $X^{**}$ and $JT^{**}$ are $w^*$ separable and also $JT_*^\bot$ is isometric to $l^{2} (2^N)$ while $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.

Following Rosenthal's proof we can describe the quasi-complement of $JT_*$ as follows.

First $JT^{**}= JT \oplus l^{2} (2^N)$. Let ($e_n)$ be the basis of $JT$ and let $(e_{\gamma_n})$ be a sequence of the basis of $l^{2} (2^N)$.We set $W$ the closed space generated by $(e_{\gamma_n} + 1/2^n e_n)$. Then $W$ is $w^*$ closed and the subspace $W_\bot$ of $JT^*$ is the quasi-complement of $JT_*$.

Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.

The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.

$Theorem$: Let $X$ be a Banach space with its dual $w^*$ separable.If $Y$ is a closed subspace of $X$ such that $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.

I think that the result remains valid if $Y^\bot$ contains $l_1$.

It holds that $X^{**}$ and $JT^{**}$ are $w^*$ separable and also $JT_*^\bot$ is isometric to $l^{2} (2^N)$ while $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.

Following Rosenthal's proof we can describe the quasi-complement of $JT_*$ as follows.

First $JT^{**}= JT \oplus l^{2} (2^N)$. Let ($e_n)$ be the basis of $JT$ and let $(e_{\gamma_n})$ be a sequence of the basis of $l^{2} (2^N)$.We set $W$ the closed space generated by $(e_{\gamma_n} + 1/2^n e_n)$. Then $W$ is $w^*$ closed and the subspace $W_\bot$ of $JT^*$ is the quasi-complement of $JT_*$.

The proof is an easy consequence of the following observation due also to Rosenthal.

If $V$, $W$ are $w^*$ closed subspaces of $X^*$ such that $V\cap W=0$ and $V_\bot \cap W_\bot =0$ then $V_\bot$ , $W_\bot$ are quasi -complementary.

Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.

The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.

$Theorem$: Let $X$ be a Banach space with its dual $w^*$ separable.If $Y$ is a closed subspace of $X$ such that $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.

I think that the result remains valid if $Y^\bot$ contains $l_1$.

It holds that $X^{**}$ and $JT^{**}$ are $w^*$ separable and also $JT_*^\bot$ is isometric to $l^{2} (2^N)$ while $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.

Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.

The space $X$, defined above, as well as the space $JT$ could not serve as counterexamples to the question I asked.More precisely the subspaces $X_*$, $JT_*$ are quasi-complemented in the corresponding space.We recall that a closed subspace $Y$ of $X$ is quasi-complemented if there exists a closed subspace $W$ of $X$ such that $Y \cap W = 0$ and $Y+W$ dense in $X$.Since the subspaces $X_*$, $JT_*$ are separable their quasi-complement is non separable thus for each one of them there is a non separable closed subspace with trivial intersection.This is a consequence of an old theorem due to H.P. Rosenthal https://www.sciencedirect.com/science/article/pii/0022123669900111 (Th.2.1)that states the following.

$Theorem$: Let $X$ be a Banach space with its dual $w^*$ separable.If $Y$ is a closed subspace of $X$ such that $Y^\bot$ contains a reflexive subspace then $Y$ is quasi-complemented in $X$.

I think that the result remains valid if $Y^\bot$ contains $l_1$.

It holds that $X^{**}$ and $JT^{**}$ are $w^*$ separable and also $JT_*^\bot$ is isometric to $l^{2} (2^N)$ while $X_*^\bot$ is isomorphic to $l^{1} (2^N)$ and this yields the quasi-complementation of both subspaces.

Following Rosenthal's proof we can describe the quasi-complement of $JT_*$ as follows.

First $JT^{**}= JT \oplus l^{2} (2^N)$. Let ($e_n)$ be the basis of $JT$ and let $(e_{\gamma_n})$ be a sequence of the basis of $l^{2} (2^N)$.We set $W$ the closed space generated by $(e_{\gamma_n} + 1/2^n e_n)$. Then $W$ is $w^*$ closed and the subspace $W_\bot$ of $JT^*$ is the quasi-complement of $JT_*$.

Hence my question remains open.The only known example of a subspace which is not quasi-complement is $c_0 (\Gamma)$ as a subspace of $l^\infty (\Gamma)$ with $\Gamma$ an uncountable set. This is a result of J. Lindenstrauss. However it is easy to see that there exists a non separable closed subspace $Y$ of $l^\infty (\Gamma)$ with $c_0 (\Gamma) \cap Y = 0$.