# derivative operator is unbounded in the sup norm

Consider ${C}^{\mathrm{\infty}}([-1,1])$ the vector space of functions with derivatives^{} or arbitrary order on the set $[-1,1]$.

This space admits a norm called the supremum norm^{} given by

$$|f|=sup\{|f(x)|:x\in [-1,1]\}$$ |

This norm makes this vector space into a metric space.

We claim that the derivative operator^{} $D:(Df)(x)={f}^{\prime}(x)$ is an unbounded operator.

All we need to prove is that there exists a succession of functions ${f}_{n}\in {C}^{\mathrm{\infty}}([-1,1])$ such that $\frac{|D({f}_{n})|}{|{f}_{n}|}$ is divergent as $n\to \mathrm{\infty}$

consider

$${f}_{n}(x)=\mathrm{exp}(-{n}^{4}{x}^{2})$$ |

$$(D{f}_{n})(x)=-2x{n}^{4}\mathrm{exp}(-{n}^{4}{x}^{2})$$ |

Clearly $|{f}_{n}|={f}_{n}(0)=1$

To find $|D{f}_{n}|$ we need to find the extrema of the derivative of ${f}_{n}$, to do that calculate the second derivative and equal it to zero. However for the task at hand a crude estimate will be enough.

$$|D{f}_{n}|\ge |(D{f}_{n})(\frac{1}{{n}^{2}})|=\frac{2{n}^{2}}{e}$$ |

So we finally get

$$\frac{|D{f}_{n}|}{|{f}_{n}|}\ge \frac{2{n}^{2}}{e}$$ |

showing that the derivative operator is indeed unbounded^{} since $\frac{2{n}^{2}}{e}$ is divergent as $n\to \mathrm{\infty}$.

Title | derivative operator is unbounded in the sup norm |
---|---|

Canonical name | DerivativeOperatorIsUnboundedInTheSupNorm |

Date of creation | 2013-07-14 20:22:55 |

Last modified on | 2013-07-14 20:22:55 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 8 |

Author | cvalente (11260) |

Entry type | Proof |

Classification | msc 47L25 |