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Hi,

Suppose we have a (real, separable) Banach space $V$ and a (linear) set $A\subseteq V$. I presume in general it might not be possible to write every element of the closed span of $A$ as an infinite linear combination $\sum_{i=1}^\infty\beta_i a_i$ of elements of $A$. Are there simple (non-trivial) conditions guaranteeing that the closed linear span of $A$ coincides with its infinite linear span (perhaps with unconditional/absolute convergence)?

My example of interest is the following: Let $X$ be a compact metric space and $F:X\to X$ a continuous map. My space $V$ is the space $C(X)$ of continuous (real-valued) functions on $X$, and $A$ is the subset of functions that can be written as $\varphi\circ F - \varphi$ for some $\varphi\in C(X)$.

Thank you for any help.

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Just to clarify: we can define linear subspaces $A_u$ and $A_a$ where the former consists of all unconditionally convergent sums of things in $A$, and the latter consists of all absolutely convergent sums of things in $A$, and you are asking under what "reasonable" conditions $A_u=A$ or $A_a=A$? –  Yemon Choi Aug 14 '12 at 8:39
    
The question was under what reasonable conditions $A_u$ or $A_a$ is the same as the closure of the linear span of $A$. But in case $A$ is a linear subspace itself (which was my primary interest), the answer is trivial as Wolfgang pointed out. Thanks for your time. –  Algernon Aug 14 '12 at 13:57
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1 Answer 1

up vote 4 down vote accepted

In the case of linear $A$, which seems to be your case of interest, you can simply do it as follows. For $x$ in the closure of $A$ take a sequence $(x_n)$ in $A$ converging to $x$ such that $\|x_n-x_{n+1}\|<2^{-n}$. Then set $a_i=x_i-x_{i-1}$, $\beta_i=1$ and the sum will converge absolutely, hence also unconditionally to $x$. Or did I miss something?

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Oh well... You are right. What was I thinking! I ill-posed my problem, but your answer made things clearer in my mind. Thank you very much. –  Algernon Aug 14 '12 at 13:51
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