# Basis for L_infty(R)

Let $V$ be the Banach space of bounded sequences of reals with the sup norm. Does there exists a subset $B$ of $V$ such that

• Linear Independence: For all functions $c$ in $\mathbb{R}^B$, if $\sum_{b \in B} c(b) \cdot b = 0$, then $c$ is identically zero.
• Spanning Set: For all vectors $v$ in $V$, there exists a function $c$ in $\mathbb{R}^B$ such that $\sum_{b \in B} c(b) \cdot b = v$.

If so, is an explicit such $B$ known?

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he Banach space V with which you start would seem to be $\ell^\infty_{\mathbb R}$ and not $L^\infty({\mathbb R})$ -- as it happens, the two are (non-isometrically) isomorphic as Banach spaces, but this is not trivial and the isomorphism is slightly mysterious. Could you clarify whether you had one or the other in mind when you asked this question? –  Yemon Choi Aug 9 '10 at 4:50
Why the set of $e_j=(0,\ldots,0,1,0\ldots)$ does not work ? –  Leandro Aug 9 '10 at 4:51
Yemon, I had the sequence space in mind. Leandro, that's not a spanning set. –  Ricky Demer Aug 9 '10 at 4:55
Are you requiring these functions to be zero for all but finitely many $b$? If so, this is just the statement that every vector space has a basis. I think you can't find an explicit basis. –  Kevin Ventullo Aug 9 '10 at 5:24
No, and that's why I made my question more explicit than the topic. –  Ricky Demer Aug 9 '10 at 5:38

## 2 Answers

The space $\ell^\infty_R$ does not have even an M-basis; i.e., a biorthogonal set $(x_t,x_t^*)$ such that the span of the $x_t$ is dense and the $x_t^*$ are total (Lindenstrauss, late 1960s IIRC), so it has nothing like a Schauder basis. Later I proved [PAMS 26. no. 3 467-468 (1970)] that $\ell^\infty$ also does not have an M-basis. However, each of these spaces does have a biorthogonal set $(x_t,x_t^*)$ such that the span of the $x_t$ is dense. This is in my paper with W.J. Davis [Studia Math. 45 173-179 (1973)].

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What does "the $x_t^*$ are total" mean? –  Ricky Demer Aug 9 '10 at 14:53
Separate points. –  Bill Johnson Aug 9 '10 at 15:03

It exists for any linear space (Banach structure is not essential here), is called Hamel base. No explicit construction (without use of Axiom of Choice) exists (and, I guess, it may proved in a sense - that existence of Hamel base in any linear space implies the axiom of choice or smth similar).

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No, a Hamel base is what you would get if you required the cs to be zero for all but finitely many b. –  Ricky Demer Aug 9 '10 at 5:59
If $c$ is not assumed to be finitely supported, then how do you understand this sum? Should it be absolutely convergent series in almost every point? –  Fedor Petrov Aug 9 '10 at 6:09
–  Ricky Demer Aug 9 '10 at 6:12
So, $B$ is not countable, but $c$ is countably supported and the convergence is unconditional in $L_{\infty}$ norm. Right? –  Fedor Petrov Aug 9 '10 at 6:20
Ricky: regarding the Wikipedia page, my mistake. I still think the question might be a little clearer if you said explicitly that you were talking about unconditional convergemce of the "series" - especially to indicate that you are talking about something a little different from a Schauder basis –  Yemon Choi Aug 9 '10 at 8:49