nLab
basis in functional analysis

Idea

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:

1. Is complete: every point in the space can be described in this fashion.
2. Has no redundancies: the description of a point is unique.

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.

Definitions

Let $V$ be a topological vector space and $B\subseteq V$ a subset.

1. We say that $B$ is a Hamel basis if:

   1. Every element of $v$ is a finite linear combination of elements of $B$,
   2. If $v={\sum }_{b\in B}{\alpha }_{b}b$ then the ${\alpha }_b$ are unique.

   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.

2. We say that $B$ is a topological basis if:

   1. Every element $v\in V$ is a limit of a sequence or (more generally) a net of finite linear combinations of elements of $B$,
   2. No element of $B$ is a limit of a sequence or net of finite linear combinations of the other elements of $B$.

   Alternatively, $B$ is total? (meaning that its span is dense) but no proper subset of $B$ is total.

3. We say that $B$ is a Schauder basis if:

   1. Every element of $v$ is a (possibly infinite) sum of scales of elements of $B$,
   2. If $v={\sum }_{b\in B}{\alpha }_{b}b$ then the ${\alpha }_b$ are unique.

Properties

1. In the presence of the axiom of choice, Hamel bases always exist.

2. 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 $B\setminus \{b\}\subseteq \ker f$, $f(b\prime )=0$ for all $b\prime \in B$, $b\prime \ne b$.

3. 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 $v$: $v=\sum f_b(v)b$.

Examples

1. In $C([0,1],ℂ)$ with the norm $\parallel f\parallel =\max \{\mid f(t)\mid \}$:

   1. 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 $t$.

   2. 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.

   3. 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 $k<n$, and $f_n$ has the least “breaks”. Then $f_n$ forms a Schauder basis for $C([0,1],ℂ)$. This is the classical Faber-Schader basis.

References

• 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.

category: functional analysis