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Recall that an operator $T:X\rightarrow Y$ is called absolutely summing if there exists a constant $C>0$ such that $$\sum_{i=1}^{n}\|Tx_{i}\|\leq C \sup_{x^{*}\in B_{X^{*}}}\sum_{i=1}^{n}|\langle x^{*},x_{i}\rangle|,$$ for all finite families $(x_{i})_{i=1}^{n}$ in $X$. The least $C$ for which the above inequality always holds is denoted by $\pi(T)$. The set of all absolutely summing operators from $X$ to $Y$ is denoted by $\Pi(X,Y)$.

An operator $T:X\rightarrow Y$ is said to be nuclear provided that there are sequences $(x^{*}_{n})_{n}$ in $X^{*}$ and $(y_{n})_{n}$ in $Y$ such that $\sum\limits_{n}\|x^{*}_{n}\|\|y_{n}\|<\infty$ and $Tx=\sum\limits_{n}x^{*}_{n}(x)y_{n}$ for $x\in X$. The nuclear norm $\nu(T)$ is defined to be the infimum of $\sum\limits_{n}\|x^{*}_{n}\|\|y_{n}\|<\infty$ over all possible such representations. The space of all nuclear operators from $X$ to $Y$ is denoted by $\mathcal{N}(X,Y)$. It is known that every nuclear operator is absolutely summing.

Question 1. For $1<p,q<\infty$, when is $\Pi(l_{p},l_{q})$ reflexive ?

Question 2. For $1<p,q<\infty$, $\Pi(l_{p},l_{q})=\mathcal{N}(l_{p},l_{q})$ ?

Question 3. For $1<p<q<\infty$, is the inclusion map $i_{p,q}$ from $l_{p}$ to $l_{q}$ absolutely summing ?

Thank you !

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3 Answers 3

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As for Question 3, note that an absolutely summing operator is completely continuous, i.e., maps weakly null sequences to null sequences, a property not shared by $i_{p,q}$ if $p>1$. Therefore this question has a negative answer.

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  • $\begingroup$ You are right, Dirk. What about Question 3 if $p=1$ and $q<2$. Obviously, Question 3 is true if $p=1$ and $q\geq 2$. $\endgroup$
    – Dongyang Chen
    Jan 15 at 0:38
  • $\begingroup$ For $1<p,q<\infty$, is there a compact but not absolutely continuous operator from $l_{p}$ to $l_{q}$ ? $\endgroup$
    – Dongyang Chen
    Jan 15 at 0:50
  • $\begingroup$ Absolutely summing operators factor through $\ell_2$ and operators from $\ell_2$ to $\ell_q$ are compact when $q<2$, so question 3 has a negative answer for $p=1$. $\endgroup$
    – Bill Johnson
    Jan 15 at 1:33
  • $\begingroup$ @Bill: For the Pisier space $P$, all tensor norms on $P\otimes P$ are equivalent. So every bounded operator from $P$ to $P^*$ is integral, in particular 1-summing, with equivalent norms. [So there are infinite dimensional spaces between which the operator norm is equivalent to the 1-summing norm.] $\endgroup$
    – Dirk Werner
    Jan 15 at 14:27
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    $\begingroup$ @Dirk: Thanks. I deleted and made a correct comment. $\endgroup$
    – Bill Johnson
    Jan 15 at 17:04
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Question 2 has a negative answer.

Hint: Show that if question 2 has a positive answer, then every Hilbert-Schmidt operator must be of trace class.

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  • $\begingroup$ Thanks, Bill. Do you know something about the reflexivity of spaces of absolutely summing operators, even $p$-summing operators? $\endgroup$
    – Dongyang Chen
    Jan 15 at 1:12
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Question 1 has a negative answer.

In 1977, D. R. Lewis proved that $\Pi_{p}(X,Y)$($1\leq p<\infty$) is reflexive if both $X$ and $Y$ are reflexive and either $X$ or $Y$ has the approximation property.

But what about $\Pi(l_{1},l_{p})$ for $1<p<\infty$ ?

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  • $\begingroup$ $\Pi(X,Y)$ contains an isometric copy of $X^*$ whenever $Y$ is infinite dimensional. $\endgroup$
    – Bill Johnson
    Jan 15 at 17:06

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