# proof of convergence theorem

Let us show the equivalence of (2) and (3). First, the proof that (3) implies (2) is a direct calculation. Next, let us show that (2) implies (3): Suppose $T{u}_{i}\to 0$ in $\u2102$, and if $K$ is a compact set in $U$, and ${\{{u}_{i}\}}_{i=1}^{\mathrm{\infty}}$ is a sequence in ${\mathcal{D}}_{K}$ such that for any multi-index $\alpha $, we have

$${D}^{\alpha}{u}_{i}\to 0$$ |

in the supremum norm ${\parallel \cdot \parallel}_{\mathrm{\infty}}$ as $i\to \mathrm{\infty}$.
For a contradiction^{}, suppose there is a compact set $K$ in $U$
such that for all constants $C>0$ and $k\in \{0,1,2,\mathrm{\dots}\}$ there exists
a function $u\in {\mathcal{D}}_{K}$ such that

$$|T(u)|>C\sum _{|\alpha |\le k}{||{D}^{\alpha}u||}_{\mathrm{\infty}}.$$ |

Then, for $C=k=1,2,\mathrm{\dots}$ we obtain functions ${u}_{1},{u}_{2},\mathrm{\dots}$ in $\mathcal{D}(K)$ such that $|T({u}_{i})|>i{\sum}_{|\alpha |\le i}{||{D}^{\alpha}{u}_{i}||}_{\mathrm{\infty}}.$ Thus $|T({u}_{i})|>0$ for all $i$, so for ${v}_{i}={u}_{i}/|T({u}_{i})|$, we have

$$1>i\sum _{|\alpha |\le i}{||{D}^{\alpha}{v}_{i}||}_{\mathrm{\infty}}.$$ |

It follows that $$
for any multi-index $\alpha $ with $|\alpha |\le i$.
Thus ${\{{v}_{i}\}}_{i=1}^{\mathrm{\infty}}$ satisfies our assumption^{}, whence $T({v}_{i})$ should tend to $0$.
However, for all $i$, we have $T({v}_{i})=1$. This contradiction
completes^{} the proof.

Title | proof of convergence theorem |
---|---|

Canonical name | ProofOfConvergenceTheorem |

Date of creation | 2013-03-22 13:44:36 |

Last modified on | 2013-03-22 13:44:36 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 10 |

Author | matte (1858) |

Entry type | Proof |

Classification | msc 46-00 |

Classification | msc 46F05 |