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

Let $X, Y$ be Banach spaces and let $T\colon X\to Y$ be a bounded linear operator. Assume that $X$ admits a Markushevic basis. Does $\overline{T(X)}\subseteq Y$ admit a Markushevic basis as well?

What if $(x_\alpha, x^*_\alpha)$ is a Markushevic of $X$. Let $M$ be a maximal linearly independent subset of $\{Tx_\alpha\colon \alpha\in A\}$. Is there any chance that there is a Markushevic basis for $\overline{T(X)}$ of this form $\{ (Tx_{\alpha}, ?)\colon\; Tx_\alpha\in M \} $?

share|cite|improve this question

The answer is no. To make a counterexample, we will use the fact that every space Banach space of the form $\ell_1(\Gamma)$ has an $M$-basis (easily checked) and that every Banach space $Z$ is the continuous linear image of a space of the form $\ell_1(\Gamma)$; for this latter fact, just note that the map $$\ell_1(B_{Z})\longrightarrow Z: (a_z\delta_z)_{z\in B_Z}\mapsto \sum_{z\in B_Z}a_zz$$ is is continuous, linear and surjective, where $B_Z$ denotes the closed unit ball of $Z$.

Let $\Gamma$ be an infinite set and let $Y=\ell_\infty(\Gamma)$. By a theorem of Johnson (No infinite-dimensional $P$-space admits a Markushevich basis, Proc. Amer. Math. Soc. 26 (1970), 467-468; see also Section 5.1 of the book Biorthogonal systems in Banach spaces by Hajek et al), $Y$ does not have an $M$-basis. Thus, from the previous paragraph, we have a Banach space without an $M$-basis that is the continuous linear image of a Banach space that does have an $M$-basis.

Epilogue: The moral of the story is that if there is a nice property that is enjoyed by spaces of the form $\ell_1(\Gamma)$, but not by some other Banach space, then the property is not preserved by linear maps. Note also that in the first paragraph of this answer I could have used in place of $B_Z$ a norm-dense subset of $B_Z$; the proof that the map is still surjective in this case is slightly more work, but it is classical and not difficult. Of course, to answer the OP's question we only need our operator into a space without an $M$-basis to have dense range, and this is easily checked to be so if $B_Z$ is replaced by any of its norm-dense subsets. The reason I mention the fact that one can use a norm dense subset of $B_Z$ in place of $B_Z$ is that the resulting $\ell_1(\Gamma)$ space is then of the same density character with respect to the norm topology as $Z$; in some situations, this is certainly desirable.

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.