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 $Tu_{i}\to 0$ in $\mathbb{C}$, and if $K$ is a compact set in $U$, and $\{u_{i}\}_{i=1}^{\infty}$ 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 $\lVert\cdot\rVert_{\infty}$ as $i\to\infty$. For a contradiction   , suppose there is a compact set $K$ in $U$ such that for all constants $C>0$ and $k\in\{0,1,2,\ldots\}$ there exists a function $u\in\mathcal{D}_{K}$ such that

 $|T(u)|>C\sum_{|\alpha|\leq k}||D^{\alpha}u||_{\infty}.$

Then, for $C=k=1,2,\ldots$ we obtain functions $u_{1},u_{2},\ldots$ in $\mathcal{D}(K)$ such that $|T(u_{i})|>i\sum_{|\alpha|\leq i}||D^{\alpha}u_{i}||_{\infty}.$ Thus $|T(u_{i})|>0$ for all $i$, so for $v_{i}=u_{i}/|T(u_{i})|$, we have

 $1>i\sum_{|\alpha|\leq i}||D^{\alpha}v_{i}||_{\infty}.$

It follows that $||D^{\alpha}u_{i}||_{\infty}<1/i$ for any multi-index $\alpha$ with $|\alpha|\leq i$. Thus $\{v_{i}\}_{i=1}^{\infty}$ 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 ProofOfConvergenceTheorem 2013-03-22 13:44:36 2013-03-22 13:44:36 matte (1858) matte (1858) 10 matte (1858) Proof msc 46-00 msc 46F05