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Few questions about the theorem

If $V$ is a neighborhood of $0$ in a topological vector space $X$ and if $$ K = \left\{\lambda \in X^* : |\Lambda x | \leq 1 \; \text{for every} \; x \in V \right\} $$ then $K$ is weak* compact.

I'll comment on the proof

Proof: Since neighborhoods of $0$ are abosrbing, there corresponds to each $x \in X$ a number $\gamma(x) < \infty$ such that $x \in \gamma(x)V$. Hence $$ |\Lambda x|\leq \gamma(x) \; (x \in X, \Lambda \in K) \;\;\;\;(1) $$ Let $D_x$ be the set of all scalars $\alpha$ such that $|\alpha| \leq \gamma(x)$. Let $\tau$ be the product topology on $P$, the cartesian product of all $D_x$, one for each $x \in X$. Since each $D_x$ is compact, so is $P$, by Tychinoff's theorem. The elements of $P$ are the functions $f$ on $X$ (linear or not) that satisfy $$ |f(x)| \leq \gamma(x) \;\;\; (x \in X) \;\;\;\;\; (2). $$

Why are such $f$ the elements of $P = \prod_{x \in X} D_x$?

Thus $K \subset X^* \cap P$. It follows that $K$ inherits two topologies, one from $X^*$ (it's weak* topology, to which the conclusion of the theorem refer) and the other, $\tau$, from $P$. We will see that

(a) these two topologies coincide on $K$, and

(b) $K$ is a closed subset of $P$

Since $P$ is compact, (b) implies that $K$ is $\tau$ compact, and then (a) implies that $K$ is weak*-compact. Fix $\Lambda_0 \in K$. Choose $x_i \in X, 1 \leq i \leq n$; choose $\delta > 0$. Put $$ W_1 = \left\{\Lambda \in X^* : |\Lambda x_i - \Lambda_0 x_i | < \delta \; for \; 1 \leq i \leq n \right\} \;\;\;\; (3) $$ and $$ W_2 = \left\{f \in P : |f(x_i) - \Lambda_0 x_i | < \delta \; for \; 1 \leq i \leq n \right\} \;\;\;\; (4) $$ Let $n, x_i$ and $\delta$ range over all admissible values. The resulting sets $W_1$ form a local base for the weak* topology of $X^*$ at $\Lambda_0$ and the sets $W_2$ form a local base for the product topology of $P$ at $\Lambda_0$. Since $K \subset P \cap X^*$, we have $$ W_1 \cap K = W_2 \cap K. $$ This proves (a).

Why do we have both $$ K \subset P \cap X^* $$ and $$ W_1 \cap K = W_2 \cap K. $$ ?

The rest of the theorem is clear to me.

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Let $F=\{f:X\rightarrow \mathbb{R}: |f(x)|\leq \gamma(x)\}$. $P=\prod_{x\in X} D_x$ this implies that an element of $P$ is a family $(a_x)_{x\in X}$ where $a_x\in D_x$, which is equivalent to saying that $|a_x|\leq \gamma(x)$. Write $f(x)=a_x$ then $f(x)\in F$.

Conversely, if you have $f\in F$, $|f(x)|\leq \gamma(x)$ implies that $f(x)\in D_x$, so the map $F\rightarrow \prod_{x\in X}D_x$ which assigns $(f(x))_{x\in X}$ is well defined and is a bijection.

$K=\{\lambda\in X^*:|\lambda(x)|\leq 1, x\in V\}$. Let $y\in X$, there exists $\gamma(y)$ and $z\in V$ with $y=\gamma(y)z$, we have $|f(y)|=|\gamma(y)||f(z)|\leq |\gamma(y)|$ since $|f(z)|\leq 1$, we deduce that $(f(y))_{y\in X}\in P$, this implies that $f\in P$.

$W_1=\{\Lambda\in X^*:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$, this implies that $W_1\cap K=\{\Lambda\in X^*\cap K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}=\{\Lambda\in K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$. since $K\subset X^*\cap P\subset X^*$.

$W_2=\{\Lambda\in P^*:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$, this implies that $W_2\cap K=\{\Lambda\in P\cap K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}=\{\Lambda\in K:|\Lambda x_i-\Lambda x_0|\leq \delta, 1\leq i\leq n\}$ since $K\subset X^*\cap P\subset P$.

We deduce that $W_1\cap P=W_2\cap P$.

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  • $\begingroup$ What about the second bit? $\endgroup$ – user8469759 Jan 10 at 17:39

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