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 Tui0 in , and if K is a compact set in U, and {ui}i=1 is a sequence in 𝒟K such that for any multi-index α, we have


in the supremum norm as i. For a contradictionMathworldPlanetmathPlanetmath, suppose there is a compact set K in U such that for all constants C>0 and k{0,1,2,} there exists a function u𝒟K such that


Then, for C=k=1,2, we obtain functions u1,u2, in 𝒟(K) such that |T(ui)|>i|α|i||Dαui||. Thus |T(ui)|>0 for all i, so for vi=ui/|T(ui)|, we have


It follows that ||Dαui||<1/i for any multi-index α with |α|i. Thus {vi}i=1 satisfies our assumptionPlanetmathPlanetmath, whence T(vi) should tend to 0. However, for all i, we have T(vi)=1. This contradiction completesPlanetmathPlanetmathPlanetmathPlanetmath 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