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.
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.
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 $c_0 (\Gamma) \cap Y = 0$.
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.
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 $X$.
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.
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.
