Finding closed subspaces whose sum isn't closed Let $V_0$ be a closed infinite-dimensional subspace of a Banach space $V$ such that the quotient $V/V_0$ is also infinite-dimensional. Is it always possible to find a closed subspace of $V$ whose sum with $V_0$ isn't closed?
A positive solution would let me answer this question.
 A: Probably Spyros has in mind something like the following.
Suppose you have a semi-normalized basic sequence $(x_n)$ in $V$ with biorthogonal functionals $(f_n)_n$
in $V_0^\perp \subset V^*$. Take any normalized basic sequence $(y_n)_n$ in $V_0$. If $\epsilon_n \to 0$ sufficiently quickly with all $\epsilon_n >0$, then $(y_n + \epsilon_n x_n)_n$ is a basic sequence that is even equivalent to $(y_n)_n$ and $(\epsilon_n^{-1} f_n)_n$ are biorthogonal to $(y_n + \epsilon_n x_n)_n$. Since $(y_n + \epsilon_n x_n)_n$ is basic, $(f_n)_n$ separates the points in the closed linear span $W$ of $(y_n + \epsilon_n x_n)_n$, and hence $W \cap V_0 =\{0\}$. By construction, $W+V_0$ is not closed. (If it were closed, then its image under the projection from $V$ to $V/V_0$ would be closed and thus the projection would take $W$ isomorphically onto its image, but $\|y_n + \epsilon_n x_n\| \to 1$ and $\|\epsilon_n x_n\| \to 0$.)
To get such $(x_n)$ and $(f_n)$, pull back to a semi-normalized sequence $V$ any normalized basic sequence in $V/V_0$. That sequence in $V$ might not be basic, but you can pass to a subsequence of differences that is basic; see e.g. the early part of the chapter on basic sequences in the book of Albiac and Kalton.
A: 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  $c_0 (\Gamma) \cap Y = 0$.
A: We will discuss an extension of Haskell's Theorem stated in the previous answer. The motivation is to show that if $l_1$ is a subspace of $Y^\bot$ and $Y^*$ is $w^*$ separable then $Y$ is quasi complemented in $X$.
The extension of Rosenthal's Theorem states the following.
$Theorem 2$ Let $X$ be a Banach space. If $Y$ is a closed subspace of $X$ such that $Y^*$ is $w^*$ separable and there exists a separable $W$ with a shrinking basis such that $W^*$ is $w^*$ isomorphic to a subspace of $Y^\bot$
then $Y$ is quasi complemented in $X$.
The proof is identical with Haskell's proof with one critical difference. For this we need the following concept and result.I do not know if these( all or some) are known.
$Definition$ Let $X^*$ be a dual Banach space. A $w^*$ closed subspace $Z$ is $w^*$ quasi complemented if there exists a $w^*$ closed subspace $V$ such that $Z \cap V =0$ and $Z+V$ is $w^*$ dense in $X^*$.
$Proposition$ Let  $X$  be a Banach space and $Z$ a $w^*$ closed subspace of $X^*$ such that $Z_\bot$ is quasi complemented in $X$. Then $Z$ is $w^*$ quasi complented in $X^*$.In particular if $X$ is separable then every $w^*$ closed subspace of $X^*$ is $w^*$ quasi complemented.
$Proof$:Let $W$ be the quasi complement of $Z_\bot$.Then it is easy to check that  $W^\bot$ is the $w^*$ quasi complement of $Z$.For the second part we use  that every subspace of a separable space is quasi complemented. This an old result due to G. Mackey and F. J. Murray.
$Proof of Theorem 2$:We proceed as in Haskell's proof.
$Step 1$: We select a sequence $(g_n)$ of norm one vectors such that $(g_n)_\bot \cap Y = 0$. We can do it since $Y^*$ is $w^*$ separable.
$Step 2$:Let $(w_n)$ be the basis of $W^*$ which is a $w^*$ closed subspace of $X^*$. We set $R_1$ the subspace of $X^*$ generated by the sequence $ (w_n + 2^{-n}g_n )$ which remains $w^*$ closed with a boundedly  complete basis.
Consider the $ Y^\bot \cap  R_1 $ which is a $w^*$ closed subspace of $R_1$ and set R its $w^*$  quasi complement in $R_1$. Then $R_\bot$ is the quasi complement of $Y$. (We need the subspace $R$ be $w^*$ closed. If $R_1$ is reflexive then we take this from the norm closure  of a quasi complemt of  $ Y^\bot \cap  R_1 $ in $R_1$. But in the general case we need the $w^*$ quasi complementation.)
$Remark$: Bill, with a commend, pointed out that he- Rosenthal - Lindenstrauss, indepentedly, have proved a stronger version of what I call "extension of Rosenthal's Theorem"(PACIFIC JOURNAL OF MATHEMATICS. Vol. 48, No. 1, 1973. ON QUASI-COMPLEMENTS. WILLIAM B. JOHNSON ) In particular the following holds.
$Theorem 3$ Let $X$ be a Banach space and $Y$ a closed subspace of $X$ with $w^*$ separable dual. Assume that $W$ is a separable space such that $W^*$ is $w^*$ isomorphic to a subspace of $Y^\bot$. Then $Y$ is quasi complemented in $X$.
I would like to note that this result can be also proved  following the proof of the extension, presented before, if we consider a subspace Z of $W^*$ which is $w^*$  closed and  $w^*$  generated by a $w^*$ basic sequence $(z_n)$.Working with $Z$ instead of $W^*$ we derive the result.
The $l_1$ case.
Assume that $l_1$ is isomorphic to a subspace of $X^*$. then either $l_2$ is isomorphic to a subspace of $X^*$ or $c_0$ is a quotient of $X$. The later yields that $l_1$ as the dual of $c_0$ is $w^*$ isomorphic to a subspace of $X^*$. This result is a consequence of the well known J. Hagler- W.B. Johnson theorem  https://link.springer.com/article/10.1007/BF02760638. For details see https://link.springer.com/article/10.1007/s00208-007-0179-y (prop. 16).
For the case where $l_1$ is isomorphic to a subspace of $Y^\bot $
we use the previous result for the dual pair $X/Y$ and $Y^\bot$ to conclude that either $l_2$ is isomorphic to a subspace of $Y^\bot$ or $l_1$ is $w^*$ isomorphic to a subspace of $Y^\bot$. In either case the extension of Rosenthal's Theorem yields the result.
A: The motivation for the following result was Bill's key observation that the Separable Quotient Problem (SQP) is  equivalent to Separable Quasi-Complemented Problem (S Q-C P).In particular we know that SQP has an affirmative answer for the dual Banach spaces (https://link.springer.com/article/10.1007/s00208-007-0179-y ) and naturally we could ask if the same is true for the S Q-C P.The answer to this is positive, it is a consequence of the existing results and we will describe it next. The proof will be based on J-L-R  Theorem stated in my previous answer (Th. 3) and for this we need  something stronger than the existence of a separable quotient of $X^*$.
$ Proposition $: Let $A$ be a separable subspace of $X^*$. Then the space $X^*/A$ has a separable quotient.
This and  Theorem 3 (previous answer ) yield that:
$Corollary$: Every separable subspace of $X^*$ is quasi complemented.
Haskell has shown that separable subspaces of $l^\infty $ are quasi complemented.(https://www.sciencedirect.com/science/article/pii/0022123669900111 )
$Step1$: We assume that $X$ is separable and A a separable subspace of $X^*$.
$Lemma  $: Let  X, A as before. If $X^*$ is non separable then there exists an uncountable unconditional family $\{ x_\gamma ^{**} \} \subset A^\bot $.
$Proof$: (a)If  $l^1 $  is isomorphic to a subspace of $X$ then $(l^1)^{**} $ is isomorphic to a subspace of $X^{**}$ and hence $ L^1 [0,1]^\Gamma$ is also a subspace of $X^{**}$ which yields that $l^2 (\Gamma)$ is a subspace of $X^{**}$. Let $\{ e_\gamma : \gamma \in \Gamma\}$ be the basis of $l^2 (\Gamma)$ as a subspace of $X^{**}$. Observe that for any $x^*\in X^*$  the set $ \{ \gamma : e_\gamma (x^*) \neq 0 \}$ is at most countable, hence for $(x_n^*)$  norm dense subset of $A$ there exists a countable $\Delta$ such that for every $\gamma \in \Gamma \backslash \Delta$ we have that   $e_\gamma (x_n^*) =0$ for all $n\in N$ which yields that $e_\gamma$ belongs to $A^\bot$.
(b) The space $X$ does not contain $l^1$.
Then by https://link.springer.com/article/10.1007/s00208-007-0179-y there exists an uncountable unconditional family $\{ x_\gamma ^{**} \} \subset  X^{**} $ which is $w^*$ discrete and has one point compactification 0. As in the previous case we conclude that there exists a countable $\Delta $ such that $\{ x_\gamma ^{**}: \gamma \in \Gamma \backslash \Delta \} \subset A^\bot$.
$Step 2$: Assume that X is arbitrary.
$ Lemma $ Let $A$ be a separable subspace of $X^*$. Then $A^\bot$ either contains an uncountable unconditional family or there exists a separable space $W$ such that $W^*$ is $w^*$ isomorphic to a subspace of $A^\bot$.
Proof:  Choose a separable subspace $Y$ of $X$  which 1-norms $A$. Then $A$ is naturally embedded in $Y^*$ and assume that it is of infinite codimension. Denote $A_Y^\bot $ the annihilator of $A$ in $Y^{**}$.Observe that $A_Y^\bot $ is  $w^*$ isomorphic to a subspace of $A^\bot$.
Now if $Y^*$ is separable then $A_Y^\bot $  is the dual of a separable and this covers the second alternative in the statement of the Lemma. If it is non separable then   $A_Y^\bot $  contains an uncountable unconditional family a property that passes to $A^\bot$.
Proof of Proposition: It well known that if $X^*$ contains an unconditional basic sequence then $X$ has a separable quotient ( see  https://link.springer.com/article/10.1007/s00208-007-0179-y ).
If $X^*$ contains $w^*$ isomorphically  the dual of a separable space then the space is a quotient of $X$.
A: Today I noticed that the result remains valid with the same proof in the dual of the James Tree space. In fact the arguments are easier and more clear. So I will try to describe the proof for the spaces $JT^*$ and its subspace $JT_*$.
Let $ ( D, \preceq ) $ denote the dyadic tree, $\alpha , \beta $ its nodes, $s,t$ its segments and $\sigma, \tau $ its branches. Also we denote by $W_\alpha = \{\beta : \alpha \preceq \beta \}$ the wadges of $D$.
The only information we need, from the norm structure of $JT^*$,  is the following one.
Lemma 0 : Let $(\alpha_i)_{i=1}^n$ be pairwise incomparable nodes of $D$ and $(f_i)_{i=1}^n$be elements of $JT^*$ such that $supp(f_i) \subset W_{\alpha_i}$. Then $\| \sum_{i=1}^{n} f_i \| = (\sum_{i=1}^{n} \| f_i \|^2)^{1/2}$
The corresponding formula for $JT$ is immediate from the definition of the norm and this passes  to $JT^*$ with the use of the $CS$ inequality. As consequence we get the following.
$dist ( \sum_{i=1}^{n}\lambda_i \sigma_i , JT_*) = (\sum_{i=1} ^{n} \lambda_i ^2)^{1/2} $ where $(\sigma_i)$ are pairwise diferent branches.Actually the space $JT^*/JT_*$ is isometric to $ l_{2} (2^N)$.
Next we will prove the lemma stated in the previous answer. The only difference is that $\|y\|\leq 2/\sqrt{n} $ instead of $4/n$.
Proof of  the lemma : First we choose an $m$ so that $2/\sqrt{nm} < \varepsilon $ and for every $i $  a node $ \alpha_i \in \sigma_i $ such that $(\alpha_i)_{i=1}^n$ are pairwise incomparable.Next for every $i$  we select  a set of branches $F_i \subset A$ such that $|F_i|=m $ and each $\tau \in F_i $ is separated from $\sigma_i$ after the node $\alpha_i$.The choice of $F_i$ is possible since no $\sigma \in A $ is an isolated point.
We set $y_i = 1/m  \sum _{\tau \in F_i}( \tau - \sigma_i)$. Clearly the support of $y_i$ is a subset of $W_{\alpha_i }$ and $\|y_i\| \leq 2$.
We set $ y = 1/n \sum _{i=1} ^ {n} y_i $. Then $ \| y \| \leq 2/ \sqrt {n} $  and  $x+y$ is an average of mn distinct branches in $A$.
To finish the proof we follow the arguments after the statement of the lemma in the previous answer.
Notice that the proof works for every set of branches $A$ which is dense in itself.
A: I will attempt to explain why a closed subspace generated by uncountable many branches intersects the subspace $X_*$.
First let's observe that for every $x= \sum_{i=1}^n \alpha_i\sigma_i $, with $ \{\sigma_i \}$ distinct branches, we have that $dist(x,X_*) \leq 2 max\{ |\alpha_i| : i=1...n\} $.
Let $A$ be an uncountable set of branches and assume that no element of $A$ is an isolated point.
Lemma: Let $ x = 1/n\sum_{i = 1}^{n} \sigma_i $ be an average of distinct branches in $A$.Then for every $\varepsilon > 0$ there exists $y$ in the linear span of $A$ such that:
(i) $\|y\| \leq 4/n $ , (ii)  $x+y$ is an average of branches and (iii) $dist ( x+y, X_* ) < \varepsilon$.
It is clear that if we have the above Lemma then by induction we could produce a sequence $(x_n)$ such that their norms are summable and $dist ( \sum_{j=1}^{n} x_j , X_* )\rightarrow 0$. Hence $\sum_{n=1}^ {\infty} x_n $ belongs to $X_*$.
The proof of the Lemma is a multiple application of the following observation.
Given $\sigma \in A$ and $k$ natural number. then for every  $n$ there are $\{ \sigma_i \}_{i=1} ^ {n} $ distinct branches each one coinciding to $\sigma$ in the first $k$ of its elements. Set $ y = 1/n \sum _{i = 1} ^ {n}  \sigma_i  - \sigma $. Then $ \|y \| \leq 2$ , $\sigma + y $ is an average and $dist (\sigma + y , X_* ) \leq 2/n$.
I understand that my explanation is rather incomplete but  I hope that gives an idea of the approach.
