What are examples of a basis of $C[0,1]$? (Hamel basis or Schauder basis... related: What is the difference between a Hamel basis and a Schauder basis?)

  • $\begingroup$ Do you mean as a vector space over $\mathbb{R}$ ? $\endgroup$
    – WLOG
    Feb 7, 2014 at 11:27
  • 1
    $\begingroup$ The usual one is described here. Schauder discovered this basis, I believe; it is called the Schauder Basis of $C[0,1]$. $\endgroup$ Feb 7, 2014 at 11:28
  • $\begingroup$ Related: math.stackexchange.com/questions/198575/… $\endgroup$
    – Seirios
    Feb 7, 2014 at 11:58
  • $\begingroup$ Do you mean a Hamel basis? $\endgroup$ Feb 7, 2014 at 12:24

2 Answers 2


The Faber-Schauder basis of $C[0,1]$, mentioned by David Mitra, consists of the following functions: $f_0\equiv 1$, and after that $$f_{j,k} = (1-2^j|x-k/2^j|)^+,\quad j\ge 0, \quad 1\le k\le 2^j, \ \text{ $k$ is odd}$$ Apart from $f_0$, these are essentially antiderivatives of Rademacher functions, which hints at the reason for their good basis properties. The form a monotone Schauder basis for $C[0,1]$. In contrast, polynomials and trigonometric polynomials fail at this task.

A couple of pictures: functions up to $j=2$


and up to $j=3$



I assume you want Hamel basis. If you mean Schauder basis (or some other type of basis, like a base for the topology), you should say so.

I cannot write down a Hamel basis for $C[0,1]$. However, I can write down a linearly independent set of cardinal $\mathfrak{c} = 2^{\aleph_0}$. For $s \in (0,1)$, let $g_s \in C[0,1]$ be piecewise linear with $g_s(0)=0$, $g_s(1)=0$, $g_s(s) = 1$. The uncountable set $\{g_s : s \in (0,1) \}$ is linearly independent.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .