1

I'm reading the proof of below Theorem 3.26 in Brezis's book of Functional Analysis. A similar proof can be found here. However, the idea behind the construction of such a countable dense subset is not very clear. After a while, I come up with my intuition. Please feel free to give comments on my attempt or add your thoughts on how we arrive at the proof.

Let $E$ be a normed space and $E'$ its topological dual. If $E'$ is separable, then so is $E$.

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, then I will happily accept theirs.

Analyst
  • 6,351
  • I find that intuition is a dangerous thing to have in the context of functional analysis. – J. De Ro Feb 12 '22 at 09:06
  • @QuantumSpace You meant that there are spaces that behave strangely and quite unintuitively? I guess I should replace "intuition" by "idea". – Analyst Feb 12 '22 at 09:08
  • @QuantumSpace: you can get rid of most of intuition if you change the vocabulary. If chicken means norm, and soup means vector space, you can use the expression "chicken soup"! If $E$ is a chicken soup and $E'$ is its extra spicy version... if $E'$ has mushrooms, does it imply that $E$ had mushrooms in the first place? Wait!!! I did not get rid of intuition!!! Actually, the proof got much easier!!! If all I did was to add some spice and now I have mushrooms, it does CERTAINLY mean that the mushrooms were already there!!! – André Caldas Feb 12 '22 at 10:44
  • @QuantumSpace: of course, you would need the definition of a "complete spice". A spice is called complete when it has no trace of mushrooms. It is widely known that $E'$ is complete, and therefore, you can conclude that the added spice also never contains any trace of mushroom. – André Caldas Feb 12 '22 at 10:52

1 Answers1

0

Let $F$ be a countable dense subset of $E'$. By Hahn-Banach theorem, we need to find a countable subset $D$ of $E$ such that $$\forall f \in E' \big [ f \restriction \operatorname{span}_\mathbb R D \equiv 0 \implies f \equiv 0 \big ].$$

Clearly, $\operatorname{span}_\mathbb R D$ is a vector subspace of $E$. If $D$ satisfies above condition, then $\overline{\operatorname{span}_\mathbb R D} = E$. Clearly, $\overline{\operatorname{span}_\mathbb Q D} = \operatorname{span}_\mathbb R D$ and thus $\overline{\operatorname{span}_\mathbb Q D} = E$. Moreover, $\overline{\operatorname{span}_\mathbb Q D}$ is also countable.

Let's characterize $D$. Fix $f\in E'$ such that $f \restriction \operatorname{span}_\mathbb R D \equiv 0$. There is a sequence $(f_n)$ in $F$ such that $f_n \to f$. We have $$ \sup_{x \text{ s.t. } |x|=1} \langle f_n-f, x \rangle = \|f_n-f\| \to 0. $$

The hypothesis $f \restriction \operatorname{span}_\mathbb R D \equiv 0$ implies $f \restriction D \equiv 0$. It follows that $$ \sup_{x \in D \text{ s.t. } |x|=1} \langle f_n, x \rangle \to 0. $$

It suffices that $f_n \to 0$ or equivalently $\|f_n\| \to 0$. This suggests us a "bound" $$ \|f_n\| \le\sup_{x \in D \text{ s.t. } |x|=1} \langle f_n, x \rangle. $$

This "bound" maybe too strong to be true, but a weaker bound (by a positive constant $\alpha <1$) is enough for our goal, i.e., $$ \alpha\|f_n\| \le\sup_{x \in D \text{ s.t. } |x|=1} \langle f_n, x \rangle. $$

For this inequality to hold, it suffices that for each $n$, there is $x_n \in D$ such that $|x_n|=1$ and $$ \alpha\|f_n\| \le \langle f_n, x_n \rangle. $$

Such pick of $x_n$ is possible due to $\alpha \in (0, 1)$. This completes the proof.

Analyst
  • 6,351