# proof of Banach-Alaoglu theorem

For any $x\in X$, let ${D}_{x}=\{z\in \u2102:|z|\le \parallel x\parallel \}$ and $D={\mathrm{\Pi}}_{x\in X}{D}_{x}$. Since ${D}_{x}$ is a compact subset of $\u2102$, $D$ is compact in product topology by Tychonoff theorem^{}.

We prove the theorem^{} by finding a homeomorphism that maps the closed unit ball
${B}_{{X}^{*}}$ of ${X}^{*}$ onto a closed subset of $D$. Define ${\mathrm{\Phi}}_{x}:{B}_{{X}^{*}}\to {D}_{x}$ by
${\mathrm{\Phi}}_{x}(f)=f(x)$ and $\mathrm{\Phi}:{B}_{{X}^{*}}\to D$ by $\mathrm{\Phi}={\mathrm{\Pi}}_{x\in X}{\mathrm{\Phi}}_{x}$, so that
$\mathrm{\Phi}(f)={(f(x))}_{x\in X}$. Obviously, $\mathrm{\Phi}$ is one-to-one, and a net $({f}_{\alpha})$ in ${B}_{{X}^{*}}$ converges^{} to $f$ in weak-* topology^{} of ${X}^{*}$ iff $\mathrm{\Phi}({f}_{\alpha})$ converges to $\mathrm{\Phi}(f)$ in product topology, therefore $\mathrm{\Phi}$ is continuous^{} and so is its inverse^{} ${\mathrm{\Phi}}^{-1}:\mathrm{\Phi}({B}_{{X}^{*}})\to {B}_{{X}^{*}}$.

It remains to show that $\mathrm{\Phi}({B}_{{X}^{*}})$ is closed. If $(\mathrm{\Phi}({f}_{\alpha}))$ is a net
in $\mathrm{\Phi}({B}_{{X}^{*}})$, converging to a point $d={({d}_{x})}_{x\in X}\in D$, we can define a function
$f:X\to \u2102$ by $f(x)={d}_{x}$. As ${lim}_{\alpha}\mathrm{\Phi}({f}_{\alpha}(x))={d}_{x}$ for all $x\in X$ by definition of weak-* convergence, one can easily see that $f$ is a linear functional^{} in ${B}_{{X}^{*}}$ and that $\mathrm{\Phi}(f)=d$. This shows that $d$ is actually in $\mathrm{\Phi}({B}_{{X}^{*}})$ 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 |