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 $ℂ$, and if $K$ is a compact set in $U$, and ${\{u_i\}}_{i=1}^{\mathrm{\infty }}$ is a sequence in ${𝒟}_K$ 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 $u\in {𝒟}_K$ 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 $𝒟(K)$ 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