I'm reading Peters Lax's Functional Analysis and I do not understand the proof of the compactness of the closed unit ball in the weak* topology.

Let $U$ be the dual of $X$. He considers the product space $P$ of $I_x$ for $x \in X$ where $ I =[-|x|, |x| ]$ and asserts that the map $f$ which map $u$ into the array of is values $u(x)$ is an embedding from the weak* topology on $U$ to the weakest topology on $P$ such that the projections are continuous. Then he proves that the image of the unit ball is closed in $P$, saying that since a point $p$ in the closure of the image of the unite ball it lies in the weak* closure of a point $q$ of the image of unit ball.

I do not understand why the map $f$ is an embedding and why the image of the unit ball is closed. Any help is really appreciated.


1 Answer 1


The product space $\prod_{x\in X}\mathbb{F}_x$ consists of all functions $f$ from an index set $X$, with $f(x)\in\mathbb{F}_x$.

If $X$ is a Banach space over the field $\mathbb{F}$ of real or complex numbers, and if $X^*$ is its dual, then $x^* \in \prod_{x\in X}\mathbb{F}$. More specifically, if $\|x^*\| \le 1$, then $x^* \in \prod_{x\in X}B_{\|x\|}[0]$, where $B_{r}[0]$ is the closed ball in $\mathbb{F}$ centered at the origin of radius $\|x\|$. Assuming $\mathbb{F}=\mathbb{R}$, then $B_{\|x\|}[0]=[-\|x\|,\|x\|\,]$ is a closed interval. The product topology on $\prod_{x\in X}I$, where $I=[-1,1]$, is the topology of pointwise convergence, which is the same as the weak* topology on the closed unit ball of $X^*$. So $X^*$ may be viewed as a closed topological subspace of $\Pi_{x\in X}I$ with the Tychonoff product topology. Every element of $\Pi_{x\in X}I$ is a function $f: X\rightarrow I=[-1,1]$, which includes every $x^*$ in the unit ball of $X^*$, assuming $X$ is a real space.

  • $\begingroup$ Thank you very much now is clear $\endgroup$ Aug 15, 2016 at 7:25

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