finite rank approximation on separable Hilbert spaces


Theorem Let be a separable Hilbert space and let TL(). Then T is a compact operatorMathworldPlanetmath iff there is a sequence {Fn} of finite rank operators with T-Fn0.

Proof.

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

Pnf =k=0nf,ekek

It is clear that the Pn have finite rank and that we have Pnff for all n, f.

Let be the unit ball in . We have that PnI pointwise. Since the Pn are contractive they are equicontinuous, hence Pn converges uniformly to I on compact sets, and in particular on T()¯, which is compact by assumptionPlanetmathPlanetmath. Therefore PnTT uniformly on , hence PnT-T0. Since PnT is boundedPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and of finite rank the first direction follows.

(): Now let {Fn} be a sequence of bounded operatorsMathworldPlanetmathPlanetmath of finite rank with T-Fn0. We have to show that T() is relatively compact in . This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to T() being totally boundedPlanetmathPlanetmath in . So we are left to show that for all ϵ>0 there is an ϵ-net x1,,xn so that:

T() k=1nBϵ(xk)

So choose ϵ>0 and n fixed so that:

Fn-T<ϵ2

Choose x1,,xm with:

Fn() k=1mBϵ2(xk)

Hence (by the triangle inequality):

T() 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