Let $T:X\rightarrow Y$ be an operator satisfying that $Q\_{N}T$ is not surjective for every finite-dimensional subspace $N$ of $Y$, where $Q\_{N}:Y\rightarrow Y/N$ is the canonical quotient map. My question is: given any $\epsilon>0$, is there an infinite-codimensional subspace $M$ with a Schauder basis such that $Q\_{M}T$ is compact and the norm of $Q\_{M}T$ is less than $\epsilon$?


I have another elementary question: Let $M$ be a closed subspace of $X^{\*}$. Is the closed unit ball of $\overline{M}^{w\*}$ necessarily contained in $\overline{B\_{M}}^{w\*}$?


Thank you!



A question on strictly cosingular operators
Your first question is silly as stated. Consider $0\oplus I$ on $\ell\_2 \oplus \ell\_\infty$. 


As for the second question, Google "total non norming subspaces" to find a wealth of counterexamples.


Added 11/29/16: Oh, so Pietsch does not claim that the subspace $M$ is separable. For your first question to make sense you should assume that $Y$ is separable. Even so, the answer is negative: Consider $0\oplus I$ on $X:= \ell\_2\oplus Z$, where $Z$ is a separable space that fails the approximation property (AP). Suppose that $M$ is a subspace of $X$ that has the AP and $\|{Q\_M}\_{|Z}\| < \epsilon$ for some small enough $\epsilon$ to make the claim below true. Let $P\_Z$ be the natural projection from $X$ onto $Z$. 


Claim: $P\_Z M = Z$. 


Assume the claim. Let $W:= \ell\_2 \cap M$. Since $W$ is necessarily complemented in $X$, you can write $M$ as the direct sum of $W$ and some closed subspace $M\_0$ of $M$, and clearly $P\_Z$ is injective on $M\_0$ and $P\_Z M\_0 = Z$. Thus $M\_0$ is isomorphic to $Z$ and hence $M\_0$, whence also $M$, must fail the approximation property.


Proof of claim. There should be a book reference for this, but I don’t know one. The argument I give is basically the proof of what is often called the ``little open mapping theorem”. By the hypothesis that $\|{Q\_M}\_{|Z}\| < \epsilon$, for each non zero $z\in Z$ there is $x\_z \in M$ s.t. $\| z- x\_z\| < \epsilon \|z\|$, and necessarily $\|x\_z\| < (1+\epsilon)\|z\|$. Since $\|P\_Z\| = 1$, we also have $\|z - P\_z x\_z\| < \epsilon$. Repeat this, replacing $z$ by $z - P\_Z x\_z \in Z$ and iterate in the obvious way. This successive approximation argument produces a sequence $(x\_n)$ in $M$ s.t. $\|x\_n\| < \epsilon^{n-1} (1+\epsilon) \|z\|$ and $z = \sum P\_Z x\_n$. So you just need $\epsilon < 1$.


