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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!

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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 $\epsilon < 1$.

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  • $\begingroup$ If the infinite-codimensional subspace $M$ in my first question is not required to have a basis, my first question is Lemma 1.10. 1 in the book of A. Pietsch, Operator Ideals. But the proof of Lemma 1. 10. 1 is omitted. Could you give me a sketch of the proof? $\endgroup$
    – Dongyang Chen
    Nov 23, 2016 at 13:35
  • $\begingroup$ You must be misquoting Pietsch; see my example. I don't have Pietsch's book to check what he said. $\endgroup$
    – Bill Johnson
    Nov 23, 2016 at 16:12
  • $\begingroup$ I think that the difference between the statement attributed to Pietsch and Question 1 of Dongyang is substantial, the counterexample of Bill will not work if we allow $M$ to be $\ell_\infty$. As for proof of the lemma of Pietsch, this is a kind of a dual form of Proposition 2.c.4 in Lindenstrauss-Tzafriri, volume I. I needed something similar in my paper Studia Math. 105 (1993), no. 1, 37–49, see Lemma 3.4 and Proposition 3.5; you can try to adjust that argument. $\endgroup$
    – Mikhail Ostrovskii
    Nov 23, 2016 at 17:04
  • $\begingroup$ I have Pietsch's book and check Lemma 1. 10.1 that says: if an operator $T:X\rightarrow Y$ satisfies $Q_{N}T$ is not surjective for every finite-dimensional subspace $N$ of $Y$, then, for every $\epsilon>0$, there exists an infinite-codimensional subspace $M$ of $Y$ such that $Q_{M}T$ is compact and the norm of $Q_{M}T$ is less than $\epsilon$. $\endgroup$
    – Dongyang Chen
    Nov 24, 2016 at 1:09
  • $\begingroup$ To prove this lemma observe the following: To satisfy the assumption of the Pietsch lemma the operator $T$ should satisfy at least one of the following conditions: (1) The closure of the image has infinite codimension; (2) The image is non-closed. In the first case let $M$ be the closure of the image, and you are done. In the second case use the argument of my 1993 paper to find a perturbation of $R$ of $T$ with a nuclear operator, such that the closure of the image of $R$ has infinite codimension and let $M$ be the closure of the image of $R$. $\endgroup$
    – Mikhail Ostrovskii
    Nov 24, 2016 at 4:09
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On your second question: by now the answer is well known, but this question was first asked by S. Banach in terms of iterated weak* sequential closures and answered by S. Mazurkiewicz [Studia Math. 2 (1930), 68–71]. Later this result of Mazurkiewicz was developed in many different directions, mostly using the term "total non-norming subspace" mentioned by Bill Johnson. You can find some account on these developments (as of 2000) in my survey https://arxiv.org/abs/math/0203139.

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