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Let $X$ be a Banach space and $\mathcal{A}$ be an operator ideal. A sequence $(x_{n})_{n=1}^{\infty}$ in $X$ is called $\mathcal{A}$-convergent to zero if there exist an operator $S\in \mathcal{A}(Z,X)$ for some Banach space $Z$ and a sequence $(z_{n})_{n=1}^{\infty}$ in $Z$ converging to zero in norm such that $x_{n}=S(z_{n}),n=1,2,...$ We denote the set of all such sequences by $c^{\mathcal{A}}_{0}(X)$. My questions are the following:

$\bullet$ Is $c^{\mathcal{A}}_{0}(X)$ a Banach space with the sup norm?

$\bullet$ What is the dual space of $c^{\mathcal{A}}_{0}(X)$?

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Finite rank operators are in all operator ideals, so any finitely supported sequence of vectors from $X$ is always in $c^{\mathcal{A}}_{0}(X)$ regardless of the ideal $\mathcal{A}$. So if we endow $c^{\mathcal{A}}_{0}(X)$ with the sup norm, it will only be complete when it is all of $c_0(X)$.

I would guess that taking the infimum of $\|(z_n)_{n=1}^\infty\|_{\infty} \cdot \|S\|_{\mathcal{A}}$ as the definition of the norm on $c^{\mathcal{A}}_{0}(X)$ might give rise to something interesting.

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  • $\begingroup$ You are right,Alejandro. But I do not even know how to show that $c^{\mathcal{A}}_{0}(X)$ is a linear subspace of $c_{0}(X)$. Thank you very much. $\endgroup$
    – Dongyang Chen
    Commented Feb 8, 2014 at 7:54
  • $\begingroup$ Closure under $+$ is no problem, Dongyang. If $x_n = Tw_n$ and $y_n = Sz_n$, $T:W\to X$, $S:Z \to X$, consider $S\oplus T: W\oplus Z \to X$ defined by $S\oplus T(w,z) = Sw + Tz$. $\endgroup$
    – Bill Johnson
    Commented Feb 8, 2014 at 21:15
  • $\begingroup$ @BillJohnson: The problem is whether the operator $S\oplus T:W\oplus Z\rightarrow X$ is in $\mathcal{A}(W\oplus Z,X)$. $\endgroup$
    – Dongyang Chen
    Commented Feb 10, 2014 at 2:44
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    $\begingroup$ @DongyangChen: Consider the canonical projections $\pi_W$ and $\pi_Z$ defined on $W \oplus Z$. Then $S \oplus T$ is nothing but $T\circ\pi_W + S \circ \pi_Z$, which belongs to $\mathcal{A}$ because both $T$ and $S$ do. $\endgroup$
    – J. Alejandro Chávez-Domínguez
    Commented Feb 10, 2014 at 16:36
  • $\begingroup$ @Alejandro:as you suggested,we can define the new norm on $c_{0}^{\mathcal{A}}(X)$ by taking the infimum. But it seems that the new norm does not satisfy the triangle inequality. $\endgroup$
    – Dongyang Chen
    Commented Mar 16, 2014 at 8:46

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