Just a follow up to the following two questions:

Rudin's functional analysis 3.18, every originally bounded subset of a locally convex space is weakly bounded.

Theorem 3.18, Rudin's functional analysis

The second question has almost all the proof, I'm trying to understand the second part.

Since $E$ is weakly bounded, there corresponds to each $\Lambda \in X^*$ a number $\gamma(\Lambda) < \infty$ such that $$ |\Lambda x | \leq \gamma(\Lambda) \;\;\;\; (x \in E) $$

It's not entirely clear to me how we get such bound, if $E$ is weakly bounded than for any arbitrary union of finite intersections of counter images of linear functionals (call such set $\Omega$) there's a $t > 0$ such that $$ E \subset t \Omega $$ I suppose the bound $| \Lambda x | \leq \gamma(\Lambda)$ comes from choosing $$ \gamma(\Lambda) = \sup_{x \in E} |\Lambda x|, $$ why does such sup exists though? how do I use the weak-bounded condition to prove it exists.

My attempt to elaborate since $E$ is weakly bounded for every weakly neighborhood of $0$ there's a $t>0$ such that

$$ E \subset t \bigcup_{j \in J} \bigcap_{i=1}^{n_j} \Lambda^{-1}(\Phi_{i,j}) $$

I guess from here I can (somehow) get to

$$ \Lambda(E) \subset t \Omega_\Lambda $$

where $\Omega_{\Lambda}$ is the image of the set $$ \Omega_{\Lambda} = \Lambda\left(\bigcup_{j \in J} \bigcap_{i=1}^{n_j} \Lambda^{-1}(\Phi_{i,j})\right) $$

but I'm still missing bits, can you please help?

The rest of the proof seems clear to me.


Just had one more thought, I suppose without loss of generality I can assume that $0 \in E$. I also guess that in the context of weak-topology boundness translates as

$E$ is weakly bounded if for each weak neighborhood $U$ there's a $t>0$ such that $E \subset t U$

And more specifically since this should hold for each weak neighborhood I guess we can say

If $E$ is weakly bounded then for each $\Lambda \in X^*$ there are both a neighborhood $U_{\Lambda}$ in the scalar field and a constant $\gamma(\Lambda) > 0$ such that $$ E \subset \gamma(\Lambda) \Lambda^{-1}(U_{\Lambda}) $$

I suppose the above can be simply proven by contradiction. Now since the above hold if $E$ is weakly bounded we have

$$ \Lambda(E) \subset \gamma(\Lambda) U_{\Lambda} $$

which implies

$$ |\Lambda x| \leq \gamma(\Lambda) \sup \; U_{\Lambda} \;\;, x \in E $$

  • $\begingroup$ Can anyone help? $\endgroup$ – user8469759 Feb 6 at 14:46

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