Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the following assertion true/known?

Let $V$ be a Banach space and let $T\colon \ell_1\to V$ be a bounded linear operator. Is it true that $T$ is not weakly compact if and only if there is a complemented subspace $X$ of $\ell_1$ (thus, isomorphic to $\ell_1$) such that $T|_X\colon X\to T(X)$ is an isomorphism?

Of course, the part 'only if' is trivial.

I've got some evidences that it might be true, yet I am not sure one thing in my proof.

share|cite|improve this question
up vote 4 down vote accepted

It is not true. Consider a continuous linear surjection of $\ell_1$ onto $c_0$; such a map exists and is strictly singular (every operator from $\ell_1$ to $c_0$ is strictly singular).

What is true is the following result Pelczynski:

Theorem. Let $\mu$ be a non-trivial measure on the field of all Borel subsets of some topological space and $T: X\longrightarrow L_1(\mu)$ a bounded linear operator. The following are equivalent:

  1. $T$ is not weakly compact.
  2. $T$ factors the identity operator of $\ell_1$.
  3. There exists a complemented subspace $Y$ of $X$ such that $T|_Y$ is an isomorphism, $Y$ (and hence $T(Y)$ also) is isomorphic to $\ell_1$ and $T(Y)$ is complemented in $L_1(\mu)$.
  4. $T$ is strictly cosingular.

Moreover, if $X$ has the Dunford-Pettis property, then 1.-4. above are equivalent to:

$5.$ $T$ is strictly singular.

This result is proved in Pelczynski's paper On strictly singular and strictly cosingular operators. II. Strictly singular and strictly cosingular operators in $L_1(\nu)$ spaces, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 13 (1965) 37-41. It generalises earlier joint work of Kadets and Pelczynski that showed that a nonreflexive subspace of an $L_1(\mu)$ space contains a subspace that is isomorphic to $\ell_1$ and complemented in the ambient space $L_1(\mu)$.

Pelczynski's theorem above for non-weakly compact operators into $L_1(\mu)$ spaces is, in a sense, dual to his theorem asserting that an operator from a $C(K)$ space is non-weakly compact if and only if it fixes a copy of $c_0$. As Pelczynski points out in the beginning to part I. of the above paper, "these results are closely connected with criteria of weak compactness of linear operators in $C(S)$ and $L_1(\nu)$ spaces due to Grothendieck".

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.