Bases in linear algebra are extremely useful tools for analysing problems. Using a basis, one can often rephrase a complicated abstract problem in concrete terms, perhaps even suitable for a computer to work with. A basis provides a way of describing a vector space in a way that:
When translated into the language of linear algebra, we recover the key properties of a basis: that it be a spanning set and linearly independent.
In infinite dimensions, having a basis becomes more valuable as the spaces get more complicated. However, the notion of a basis also becomes complex because the question of what makes a description admits different answers depending on whether we want only finite sums, we allow sequences, or we want infinite sums.
Let $V$ be a topological vector space and $B \subseteq V$ a subset.
We say that $B$ is a Hamel basis if:
Alternatively, $B$ is linearly independent and $\Span(B) = V$; in other words, the span of $B$ is $V$ but no proper subset of $B$ has this property.
We say that $B$ is a topological basis if:
Alternatively, $B$ is total? (meaning that its span is dense) but no proper subset of $B$ is total.
We say that $B$ is a Schauder basis if:
In the presence of the axiom of choice, Hamel bases always exist.
If $B$ is a topological basis, then $B$ has a dual basis. Since $B \setminus \{b\}$ is not total but $B$ is total, the closure of the span of $B \setminus \{b\}$ must be a codimension $1$ subspace, whence the kernel of a non-trivial continuous linear functional on $V$, say $f_b$. By scaling, this functional can be assumed to satisfy $f_b(b) = 1$. Since $B \setminus \{b\} \subseteq \ker f$, $f(b') = 0$ for all $b' \in B$, $b' \ne b$.
If $B$ is a Schauder basis then it is a topological basis and so, as mentioned, has a dual basis. Then the coefficients in the sum $v = \sum \alpha_b b$ must be given by evaluating the dual basis on $v$: $v = \sum f_b(v) b$.
In $C([0,1],\mathbb{C})$ with the norm ${\|f\|} = \max\{{|f(t)|}\}$:
The monomials are linearly independent and have dense span, but do not form a topological basis as there is a sequence of polynomials with no linear term converging to $t$.
The trigonometric polynomials do form a topological basis. The dual basis is given by taking the Fourier coefficients of a function. However, it is not a Schauder basis as there are continuous functions which are not the uniform limit of their Fourier series.
The following is a Schauder basis. Let $(d_n)$ be the sequence $\{0, 1, \frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \dots\}$. Define $f_n$ to be the piecewise-linear function with the property that: $f_n(d_n) = 1$ and $f_n(d_k) = 0$ for $k \lt n$, and $f_n$ has the least “breaks”. Then $f_n$ forms a Schauder basis for $C([0,1],\mathbb{C})$. This is the classical Faber-Schader basis.
Enflo, P. (1973). A counterexample to the approximation problem in Banach spaces. Acta Math., 130, 309–317.
Semadeni, Z. (1982). Schauder bases in Banach spaces of continuous functions (Vol. 918). Lecture Notes in Mathematics. Berlin: Springer-Verlag.