Is it true that Hamel basis of $C[a,b]$ can be dense in $C[a,b]$?

Question

Can a Hamel basis of $C[a,b]$ be dense in $C[a,b]$? Also, can ALL the elements in C[a,b] be expressed as linear combination of the Hamel basis? by Hamel Basis $\{e_\alpha\}$ I just mean a maximal linear independent set of elements in $X$. Therefore, in general, It's not necessary to have $\text{span}\{e_\alpha\} = X$.

Answer 1

Yes; in fact

Theorem 1. If $X$ is an infinite-dimensional separable Banach space then there exists a Hamel basis which is dense in $X$.

Note The assumption that $X$ is separable is not needed; see below. We include this version just because the proof is more elementary.

Lemma If $X$ is a Banach space and $Y$ is a proper subspace then $Y$ has empty interior.

Proof: If $B(x,r)\subset Y$ then $B(0,r) = B(x,r) - x\subset Y$, hence $Y=X$.

Proof of the theorem: Say $B_1,B_2,\dots$ is a base for the topology of $X$. The lemma shows we can recursively choose $x_n\in B_n$ so that $x_{n+1}$ is not in the span of $x_1,\dots, x_n$. So $\{x_1,\dots\}$ is independent, hence it is a subset of some basis. QED.

Theorem 2. Same as Theorem 1 except $X$ need not be separable.

Proof. Just a transfinite version of the previous proof.

Say the dimension is $\kappa$. Say $B$ is a Hamel basis for $X$ of cardinality $\kappa$. Then there is a base for the topology of cardinality $\kappa$, consisting of all $B(x,r)$ where $x$ is a rational linear combination of (finitely many) elements of $B$ and $r>0$ is rational. Say $(B_\alpha)_{\alpha<\kappa}$ is such a base (the index $\alpha$ running over all ordinals less than $\kappa$.)

Now suppose $\alpha<\kappa$ and $(x_\beta)_{\beta<\alpha}$ is an independent set with $x_\beta\in B_\beta$ for all $\beta<\alpha$. Since $\alpha<\kappa$ the span of $(x_\beta)_{\beta<\alpha}$ cannot be $X$, so there exists $x_\alpha\in B_\alpha$ which is not in this span. So by induction we obtain an independent dense set $(x_\alpha)_{\alpha<\kappa}$.

Answer 2

To construct a clear example, take a Hamel basis and renormalize it so that all elements have norm equal to one. Then the zero function is at distance 1 from the basis (and any element with norm other than 1 will have positive distance to the basis).

As for your second question: yes, that is in the definition of basis.

Answer 3

A separable Banach space $X$ with $\dim X\ge\aleph_0$ (and thus $\dim X=\beth_1$) always admits a dense Hamel basis.

Since $X$ is separable, it has a dense subspace $V$ such that $\dim V=\aleph_0$. This subspace has a basis $\mathcal B$ and $\overline{\operatorname{span}_{\Bbb Q} \mathcal B}=X$. $\mathcal B$ may be extended to a Hamel basis $\mathcal B\cup \mathcal W$ by some set such that $\#\mathcal W=\beth_1$. The idea is to use the elements of $\mathcal W$ to approximate the elements of the countable set $\operatorname{span}_{\Bbb Q}\mathcal B$. Since $\aleph_0\cdot\aleph_0=\aleph_0<\beth_1$, there is an injective function $\Psi:\Bbb N\times\operatorname{span}_{\Bbb Q}\mathcal B\hookrightarrow \mathcal W$. Let's call $\Phi(n,v)=v+\frac{1}{2^n\lVert \Psi(n,v)\rVert}\Psi(n,v)$ for $n\in\Bbb N$ and $v\in\operatorname{span}_{\Bbb Q}\mathcal B$.

Now, I claim that $\mathcal X=\mathcal B\cup\{\Phi(n,v)\,:\,n\in\Bbb N,\ v\in\operatorname{span}_{\Bbb Q}\mathcal B\}$ is a linearly independent set/sequence. In fact, if $b_i\in\mathcal B$ and $$\alpha_1 b_1+\cdots+\alpha_m b_m+\beta_1\Phi(n_1,v_1)+\cdots+\beta_k\Phi(n_k,v_k)=0$$ then $$\underbrace{\sum_i\alpha_ib_i+\sum_j\beta_j v_j}_{\in\ \operatorname{span}\mathcal B}+\underbrace{\sum_j\frac{\beta_j}{2^{n_j}\lVert\Psi(n_j,v_j)\rVert}\Psi(n_j,v_j)}_{\in\ \operatorname{span}\mathcal W}=0$$

By $\operatorname{span}\mathcal W$ and $\operatorname{span}\mathcal B$ being in direct sum, the former implies $$\begin{cases}\sum_j\frac{\beta_j}{2^{n_j}\lVert\Psi(n_j,v_j)\rVert}\Psi(n_j,v_j)=0\\\sum_i\alpha_ib_i+\sum_j\beta_j v_j=0\end{cases}\iff\begin{cases}\forall j,\ \beta_j=0\\\sum_i\alpha_ib_i+\underbrace{\sum_j\beta_j v_j}_{=0}=0\end{cases}\iff\\\iff\begin{cases}\forall j,\ \beta_j=0\\\forall i,\ \alpha_i=0\end{cases}$$

It is clear that $\lim_{n\to\infty}\Phi(n,v)=v$, and thus $\overline{\mathcal X}\supseteq\operatorname{span}_{\Bbb Q}\mathcal B$, making it a dense subset. Finally, if you so desire, $\mathcal X$ may be completed to a Hamel basis.