# continuity of composition of functions

All functions in this entry are functions from $\mathbb{R}$ to $\mathbb{R}$.

Example 1
Let $f(x)=1$ for $x\le 0$ and $f(x)=0$ for $x>0$,
let $h(x)=0$ when $x\in \u2102$ and $1$ when $x$ is irrational,
and let $g(x)=h(f(x))$. Then $g(x)=0$ for all $x\in \mathbb{R}$, so
the composition^{} of two discontinuous functions can be
continuous^{}.

Example 2 If $g(x)=h(f(x))$ is continuous for all functions $f$, then $h$ is continuous. Simply put $f(x)=x$. Same thing for $h$ and $f$. If $g(x)=h(f(x))$ is continuous for all functions $h$, then $f$ is continuous. Simply put $h(x)=x$.

Example 3 Suppose $g(x)=h(f(x))$ is continuous and $f$ is continuous. Then $h$ does not need to be continuous. For a conterexample, put $h(x)=0$ for all $x\ne 0$, and $h(0)=1$, and $f(x)=1+|x|$. Now $h(f(x))=0$ is continuous, but $h$ is not.

Example 4 Suppose $g(x)=h(f(x))$ is continuous and $h$ is continuous. Then $f$ does not need to be continuous. For a counterexample, put $f(x)=0$ for all $x\ne 0$, and $f(0)=1$, and $h(x)=0$ for all $x$. Now $h(f(x))=0$ is continuous, but $f$ is not.

Title | continuity of composition of functions |
---|---|

Canonical name | ContinuityOfCompositionOfFunctions |

Date of creation | 2013-03-22 14:04:55 |

Last modified on | 2013-03-22 14:04:55 |

Owner | bbukh (348) |

Last modified by | bbukh (348) |

Numerical id | 7 |

Author | bbukh (348) |

Entry type | Result |

Classification | msc 54C05 |

Classification | msc 26A15 |