# Absolutely summing operators from $l_{p}$ to $l_{q}$

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 $$1, when is $$\Pi(l_{p},l_{q})$$ reflexive ?

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

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

Thank you !

## 3 Answers

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.

• You are right, Dirk. What about Question 3 if $p=1$ and $q<2$. Obviously, Question 3 is true if $p=1$ and $q\geq 2$. Jan 15 at 0:38
• For $1<p,q<\infty$, is there a compact but not absolutely continuous operator from $l_{p}$ to $l_{q}$ ? Jan 15 at 0:50
• 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 $p=1$. Jan 15 at 1:33
• @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.] Jan 15 at 14:27
• @Dirk: Thanks. I deleted and made a correct comment. Jan 15 at 17:04

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.

• Thanks, Bill. Do you know something about the reflexivity of spaces of absolutely summing operators, even $p$-summing operators? Jan 15 at 1:12

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 ?

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