Re: Cardinality of basis of continuous functions

In article
<waderameyxiii-5F5AFB.10043205072006@xxxxxxxxxxxxxxxxxxxxxxxx>,
The World Wide Wade <waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:

In article
<1152109727.749113.288160@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Dave L. Renfro" <renfr1dl@xxxxxxxxx> wrote:

Michael Stemper wrote:

Problem 5 in Section 3.3 of _Real Analysis_ by
Haaser & Sullivan asks for proof that C[a,b]
(the continuous functions on the closed interval
[a,b]) has an infinite Hamel basis. I have a proof,
but it seems unnecessarily complicated.

[snip]

It seems to me that my families of functions are much
too complicated; there must be a simpler definition
that I can use. Any suggestions?

Off the top of my head ...

Can you show that {1, x, x^2, x^3, ...} is a linearly
independent set? [It is. But, without having given any
thought to this, I'm not sure how easy it is to prove.]

It's clear that dim {x, x^2, ..., x^n} = n for every n: If p(x) =
sum(k=1,n) a_k*x^k = 0 on [a,b], then all coefficients of the
polynomial p must vanish. Therefore C[a,b] has an infinite Hamel
basis.

Generalization: If f in C[a,b] is nonconstant, then dim {f, f^2,
...., f^n} = n for every positive integer n.

Lemma: Suppose g > 0 on (c,d) and g(x) -> 0 as x -> c+. Then for
every positive integer k, g^k is not a linear combination of 1,
g, ..., g^(k-1).

The lemma should be clear as you can never obtain vanishing of
order g^k from a finite linear combination of g^j's with j < k.

proof of Generalization: Let K be the set where the minimum value
m of f is taken on. Because f is nonconstant, K is not all of
[a,b]. K is closed in [a,b], hence the complement of K in [a,b]
is a non-empty open subset of [a,b]. This complement contains a
connected component of one of the following forms: (c,d), [a,c),
or (c,b]. Suppose it's of the first form (the others are handled
similarly, with slightly different versions of the Lemma). Then
f(c) = 0, but f > 0 on (c,d). All statements will now refer to
(c,d). We can write f = m + g, where g is positive and g(x) -> 0
as x -> c+.

Now suppose a_n*f^n + ... + a_1*f = a_n*(m+g)^n + ... + a_1*(m+g)
= 0, not all a_j = 0. Let k be the largest j for which a_j is
nonzero. Expand the terms (m+g)^j, j = 1, ..., k. Only the kth of
these involves g^k, so solving for g^k, we see g^k is a linear
combintation of 1, g, ..., g^(k-1), contradicting the Lemma.
.

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