proof of Banach-Alaoglu theorem

For any xX, let Dx={z:|z|x} and D=ΠxXDx. Since Dx is a compact subset of , D is compact in product topology by Tychonoff theoremMathworldPlanetmath.

We prove the theoremMathworldPlanetmath by finding a homeomorphism that maps the closed unit ball BX* of X* onto a closed subset of D. Define Φx:BX*Dx by Φx(f)=f(x) and Φ:BX*D by Φ=ΠxXΦx, so that Φ(f)=(f(x))xX. Obviously, Φ is one-to-one, and a net (fα) in BX* convergesPlanetmathPlanetmath to f in weak-* topologyMathworldPlanetmathPlanetmath of X* iff Φ(fα) converges to Φ(f) in product topology, therefore Φ is continuousPlanetmathPlanetmath and so is its inversePlanetmathPlanetmathPlanetmathPlanetmath Φ-1:Φ(BX*)BX*.

It remains to show that Φ(BX*) is closed. If (Φ(fα)) is a net in Φ(BX*), converging to a point d=(dx)xXD, we can define a function f:X by f(x)=dx. As limαΦ(fα(x))=dx for all xX by definition of weak-* convergence, one can easily see that f is a linear functionalMathworldPlanetmath in BX* and that Φ(f)=d. This shows that d is actually in Φ(BX*) and finishes the proof.

Title proof of Banach-Alaoglu theorem
Canonical name ProofOfBanachAlaogluTheorem
Date of creation 2013-03-22 15:10:03
Last modified on 2013-03-22 15:10:03
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 12
Author Mathprof (13753)
Entry type Proof
Classification msc 46B10