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I was wondering if $C_{c}(\mathbb{R})=\text { continuous functions with compact support }$ have countable Hamel basis. My intuition tells me try to use Baire Category Theorem and I found one corollary saying that for an infinite-dimensional Banach space every Hamel basis is uncountable. However $C_c(\mathbb{R})$ s not even complete. Any help or hint? Thanks in advance.

Maskoff
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  • It is known that $C_c(\Bbb R)$ does not have a countable Hamel basis, see here. – Dietrich Burde Apr 23 '20 at 12:39
  • I only see the author claims at an exercise that $C_{c}(\mathbb{R}) \text { is a meager, dense subspace of } C_{0}(\mathbb{R})$, is there a proof? – Maskoff Apr 23 '20 at 12:44
  • Yes I was trying to search articles and Math StackExchange, haven't found any proofs so far. Maybe it's well known but I'm a beginner at functional analysis. – Maskoff Apr 23 '20 at 12:49
  • I think it is in several lecture notes on functional analysis. It takes a bit of searching. I also found this post. – Dietrich Burde Apr 23 '20 at 12:52
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    Any linear image of a space with countable Hamel basis also has a countable (or finite) Hamel basis. Now note that $C_c \to C([0,1]), f \mapsto f|_{[0,1]}$ is surjective, where $C([0,1])$ is an infinite dimensional Banach space and thus does not have a countable Hamel basis, by a standard Baire category theorem application. – PhoemueX Apr 23 '20 at 12:53
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    And the latter statement has been proved already here on this site. – Dietrich Burde Apr 23 '20 at 12:56
  • "Any linear image of a space with countable Hamel basis also has a countable (or finite) Hamel basis. " Why this implies "$C_c$ has no Hamel basis because $C[0,1]$ has no Hamel basis?" – Maskoff Apr 23 '20 at 13:11
  • @Maskoff: Because $C[0,1]$ is the linear image of $C_c$. So if $C_c$ had a countable Hamel basis, so would $C[0,1]$. – PhoemueX Apr 23 '20 at 19:54
  • @PhoemueX Thanks for your reply! One last question: Does "Any linear image of a space with countable Hamel basis also has a countable (or finite) Hamel basis. " this lemma has a name? Or how to show this? – Maskoff Apr 24 '20 at 00:34
  • @Maskoff You say that $C_c(\mathbb{R})$ is not complete. It is not complete when given the supremum norm from $C_0(\mathbb{R})$, but it is complete when given its (finer) locally convex topology as a colimit of the sequence $C([-1,1]) \rightarrow C([-2,2]) \rightarrow \cdots$. This fact is important if you want a version of the Riesz representation theorem general enough to include Lebesgue measure on $\mathbb{R}$. – Robert Furber Apr 24 '20 at 03:42
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    @Maskoff: Any spanning set contains a basis (using Zorn's lemma, take a maximal linearly independent subset of the spanning set). Now if X has a countable Hamel basis and $\varphi : X \to Y$ is surjective and linear, then the image of the Hamel basis under $\varphi$ spans Y, so you can select a subset of it which is a basis for Y. – PhoemueX Apr 24 '20 at 05:34

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You can also provide an explicit uncountable linearly independent set. For $a > 0$ consider the function $f_a \in C_c(\Bbb{R})$ given by $$f_a(x) = (a-|x-a|)\chi_{[0,2a]}.$$ Then $S = \{f_a : a > 0\}$ is uncountable and linearly independent. Indeed, let $0 < a_1 < a_2 < \cdots < a_n$ and let $\alpha_1, \ldots, \alpha_n \in \Bbb{R}$ be such that $$0 = \sum_{j=1}^n \alpha_jf_{a_j}.$$ If $\alpha_1 \ne 0$, then for every $x \in \langle 0,\min\{2a_1, a_2\}\rangle$ we have $$\sum_{j=1}^n \alpha_jf_{a_j}(x) = \alpha_1(a_1-|x-a_1|) + \sum_{j=2}^n \alpha_jx_j$$ so $\sum_{j=1}^n \alpha_jf_{a_j}$ is not differentiable at $a_1$, which is a contradiction since it is equal to $0$. Therefore $\alpha_1 = 0$ and inductively we conclude that $\alpha_1= \cdots = \alpha_n = 0$. Hence, $S$ is linearly independent.

Therefore, $C_c(\Bbb{R})$ does not have a countable Hamel basis.

mechanodroid
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