0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $W_1\cap P=W_2\cap P$.

$\endgroup$
1
  • $\begingroup$ What about the second bit? $\endgroup$ Jan 10 '19 at 17:39
0
$\begingroup$

Let me give an indirect answer to your second question by providing a alternative proof for why the topologies should coincide on K. This hopefully helps with understanding why $W_1\cap K=W_2\cap K$.

The product topology on P (this is the topology $\tau$ in your notation) is defined as the smallest topology which makes the set of all projection maps continuous.

The weak* topology is defined as the smallest topology which makes the set of all the evaluation maps continuous.

Now we note that the projection map to the x-th coordinate coincides on $K$ with the evaluation map in x. Indeed, for any $f\in K$ and $x\in X$ we have $\pi_x(f)=f(x)=\iota_x(f)$. Here $\pi_x(f)$ is the projection map to the x-th coordinate and $\iota_x(f)$ the evaluation map in x.

Now we can come to the conclusion for why the topologies coincide on $K$ using only logic. On $K$, both topologies are defined to be the smallest topologies such that they make some set of maps continuous (this statement requires a proof, but it is not hard). However, the sets of maps that they make continous are identical on $K$. This means that the on $K$ topolgies must coincide.

If we now look back at the definition for $W_1$ and $W_2$, we see that they are infact identical sets on $K$. Because they define the same basis in each of their topologies, and they coincide on $K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.