finite rank approximation on separable Hilbert spaces


Theorem Let ℋ be a separable Hilbert space and let T∈L⁢(ℋ). Then T is a compact operatorMathworldPlanetmath iff there is a sequence {Fn} of finite rank operators with ∥T-Fn∥→0.

Proof.

(⇒): Assume T is compactPlanetmathPlanetmath on ℋ and {en} is an orthonormal basis of ℋ. Define:

Pn⁢f =∑k=0n⟨f,ek⟩⁢ek

It is clear that the Pn have finite rank and that we have ∥Pn⁢f∥≤∥f∥ for all n∈ℕ, f∈ℋ.

Let ℬ be the unit ball in ℋ. We have that Pn→I pointwise. Since the Pn are contractive they are equicontinuous, hence Pn converges uniformly to I on compact sets, and in particular on T⁢(ℬ)¯, which is compact by assumptionPlanetmathPlanetmath. Therefore Pn⁢T→T uniformly on ℬ, hence ∥Pn⁢T-T∥→0. Since Pn⁢T is boundedPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and of finite rank the first direction follows.

(⇐): Now let {Fn} be a sequence of bounded operatorsMathworldPlanetmathPlanetmath of finite rank with ∥T-Fn∥→0. We have to show that T⁢(ℬ) is relatively compact in ℋ. This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to T⁢(ℬ) being totally boundedPlanetmathPlanetmath in ℋ. So we are left to show that for all ϵ>0 there is an ϵ-net x1,⋯,xn∈ℋ so that:

T⁢(ℬ) ⊆⋃k=1nBϵ⁢(xk)

So choose ϵ>0 and n∈ℕ fixed so that:

∥Fn-T∥<ϵ2

Choose x1,⋯,xm∈ℋ with:

Fn⁢(ℬ) ⊆⋃k=1mBϵ2⁢(xk)

Hence (by the triangle inequality):

T⁢(ℬ) ⊆⋃k=1mBϵ⁢(xk)

and we are done. ∎

Title finite rank approximation on separable Hilbert spaces
Canonical name FiniteRankApproximationOnSeparableHilbertSpaces
Date of creation 2013-03-22 18:23:18
Last modified on 2013-03-22 18:23:18
Owner karstenb (16623)
Last modified by karstenb (16623)
Numerical id 10
Author karstenb (16623)
Entry type Theorem
Classification msc 46B99