# example of construction of a Schauder basis

Consider an uniformly continuous function $f:[0,1]\to \mathbb{R}$. A Schauder basis ${\{{f}_{n}(x)\}}_{0}^{\mathrm{\infty}}\in C[0,1]$ is constructed. For this purpose we set ${f}_{0}(x)=1$, ${f}_{1}(x)=x$. Let us consider the sequence of semi-open intervals in $[0,1]$

$${I}_{n}=[{2}^{-k}(2n-2),{2}^{-k}(2n-1)),{J}_{n}=[{2}^{-k}(2n-1),{2}^{-k}2n),$$ |

where $$, $k\ge 1$. Define now

${f}_{n}(x)$ | $=$ | $\{\begin{array}{cc}{2}^{k}[x-({2}^{-k}(2n-2)-1)]\hfill & \text{if}x\in {I}_{n},\hfill \\ 1-{2}^{k}[x-({2}^{-k}(2n-1)-1)]\hfill & \text{if}x\in {J}_{n},\hfill \\ 0\hfill & \text{otherwise.}\hfill \end{array}$ |

Geometrically these functions form a sequence of triangular functions of height one and width ${2}^{-(k-1)}$, sweeping $[0,1]$. So that if $f\in C([0,1])$, it is expressible in Fourier series $f(x)\sim {\sum}_{n=0}^{\mathrm{\infty}}{c}_{n}{f}_{n}(x)$ and computing the coefficients ${c}_{n}$ by equating the values of $f(x)$ and the series at the points $x={2}^{-k}m$, $m=0,1,\mathrm{\dots},{2}^{k}$. The resulting series converges uniformly to $f(x)$ by the imposed premise.

Title | example of construction of a Schauder basis |
---|---|

Canonical name | ExampleOfConstructionOfASchauderBasis |

Date of creation | 2013-03-22 17:49:18 |

Last modified on | 2013-03-22 17:49:18 |

Owner | perucho (2192) |

Last modified by | perucho (2192) |

Numerical id | 5 |

Author | perucho (2192) |

Entry type | Example |

Classification | msc 15A03 |

Classification | msc 42-00 |