# Rudin's functional analysis Theorem 3.18, second part.

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.

Update

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

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

• Can anyone help? – user8469759 Feb 6 at 14:46