0
$\begingroup$

Let $X$ be a Banach space having a (unconditional, normalized) Schauder basis $(e_n)_n$. Suppose that $Y$ and $Z$ are (closed) block subspaces of $X$ having normalized block bases (with respect to $(e_n)_n$) $(y_n)_n$ and $(z_n)_n$, respectively. Moreover, suppose that the two sets are linearly independent, i.e., their respective linear spans (not their closures) $Y'$ and $Z'$ are such that $Y'\cap Z'=\{0\}$. When can we conclude that $Y\cap Z=\{0\}$ as well?

I'm interested in sufficient criteria, but an exact criterion (even just for the classical sequence spaces, $c_0$, $\ell^p$) would be great.

$\endgroup$
3
  • 1
    $\begingroup$ Let $(y^∗_n)$ be the functionals biorthogonal to $(y_n)$. Extend these to $Y' + Z'$ to be zero on $Z'$. I guess the most natural sufficient condition is that for each $n$ these extensions are all continuous (with a similar condition on the functionals biorthogonal to $(z_n)$). This is the same as saying that $(x_n)\cup (y_n)$ is a minimal sequence, which is weaker than saying that $(x_n)\cup (y_n)$ is an $M$-basis for its closed linear span. – Bill Johnson $\endgroup$ Aug 25, 2015 at 2:12
  • $\begingroup$ Bill: I'm a bit unfamiliar with the terminology (Banach space theory is a outside my usual area), what do you mean by "minimal sequence" and "M-basis"? $\endgroup$ Aug 27, 2015 at 16:58
  • $\begingroup$ An $M$ basis is a biorthogonal system s.t. the vectors have dense linear span and the functionals separate points. A minimal sequence is a sequence of vectors s.t. there exist biorthogonal functionals for the sequence (but the biorthogonal functionals need not separate points of the closed linear span of the sequence of vectors). $\endgroup$ Aug 28, 2015 at 19:17

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.