5
$\begingroup$

I am finding serious difficulties in understanding some things about relative countable compactness and the use of sequences in proving it by my functional analysis text, Kolmogorov-Fomin's.

For example, here in corollary 2, it says that a subset of the space $E^{\ast}$ conjugate to a separable Banach space $E$ is bounded if it is relatively countably compact (as in definition 5 here) in the weak$^{\ast}$ topology as a consequence of theorem 2' here. I would like to understand why, if $M$ is not bounded, it cannot be relatively countably compact, but from theorem 2' I only get that, if $\{f_n\}_n$ is not bounded, then it is not weakly$^{\ast}$ convergent: how can that prove that any infinite subset of $M$ has an accumulation point?

My book, without giving a proof and treating it as trivial, might appear to treat countable compactness in the weak$^{\ast}$ topology as equivalent to the fact that any sequence has a a weak$^{\ast}$-convergent subsequence, but, if it is true, I am too stupid to see it as trivial, though I am not sure that it is true...

If $M$ were a subset of a metric space I would know that $x$ would be an accumulation point of $M$ if and only if it were the limit of an eventually non-constant sequence of points belonging to $M$, but $E^{\ast}$ with the weak$^{\ast}$ topology is not a metric space (though a sphere centred in 0 is metrisable, but, if we do not a priori know that a subset $M$ is bounded...).

I also think, thanks to what, and to who has written what, I read here, that if $E$ is a separable Banach space then every relatively weak$^{\ast}$-compact subset of $E^{\ast}$ is relatively sequentially weak$^{\ast}$-compact (definition as here), but here we only have countable weak$^{\ast}$-compactness... Does relative countable weak$^{\ast}$-compactness implies relative sequential weak$^{\ast}$-compactness? If it does, I see that if $M$ is not bounded, and infinite, we can chose from it a sequence $\{f_n\}$ (even such that $\forall i\ne j\quad f_i\ne f_j$, if we desire so) such that $\|f_n\|\to+\infty$, which, by theorem 2', will not have any convergent subsequence. Then, if relative countable weak$^{\ast}$-compactness implied relative sequential weak$^{\ast}$-compactness, $M$ would not be countabye weak$^{\ast}$-compact, so that the "only if part" of corollary 2, which is what causes some problems to me, would be proven. But I am far from being conviced that such implication is true...

I uncountably thank you for any help! ;-)

$\endgroup$
3
$\begingroup$

I agree that the book should have explained Corollary 2 a bit better, just as you explained.

Nevertheless, we can prove Corollary 2 by contradiction (as far as I noticed, this book does not use Banach-Steinhauss/Uniform Boundedness Principle, which I will use): Suppose $M$ is an unbounded rcc (relatively countably compact). Let $(f_n)_{n=1}^\infty\subseteq M$ with $\Vert f_n\Vert\to\infty$. Since $(f_n)$ is not norm-bounded in $E^*$, the Uniform Boundedness Principle implies that it is not pointwise-bounded, so there exists $x\in E$ with $\sup_{n\in\mathbb{Z}_{>0}}|f_n(x)|=\infty$. Taking a subsequence if necessary, suppose $|f_n(x)|\geq n$.

But $M$ is rcc, so $(f_n)$ has a limit point $F$. The set $U=\left\{g\in E^*:|g(x)-F(x)|<1\right\}$ is a weak*-neighbourhood of $F$, so $U$ contains infinitely many $f_n$'s, so, for infinitely many $n\in\mathbb{Z}_{>0}$, $|f_n(x)-F(x)|<1$, hence $n-1\leq |f_n(x)|-1\leq|f_n(x)-F(x)|+|F(x)|-1<|F(x)|$, a contradiction.

Alternatively, I believe you can proceed with a proof similar to that of Theorem 2.


Your comment on rcc implying sequential weak*-compactness is true. Moreover, we can prove that the following are equivalent, given $M\subseteq E^*$, with $E$ separable:

  1. $M$ is a closed rcc.
  2. $M$ is sequentially weak*-compact.
  3. $M$ is bounded and weak*-closed.
  4. $M$ is weak*-compact.

(1.$\Rightarrow$2.) Suppose $M$ is a closed rcc. Let $(f_n)\subseteq M$ be an arbitrary sequence. By Corollary 2, $(f_n)$ is bounded, so Theorem 4 implies that $(f_n)$ has a weak*-convergent subsequence. Therefore $M$ is sequentially weak*-compact.

(2.$\Rightarrow$3.) Suppose $M$ is sequentially weak*-compact. First let's show that $M$ is bounded: Suppose not. Then, by Uniform Boundedness, $M$ is not pointwise bounded, so there is $y\in E$ and $(f_n)\subseteq M$ such that $\Vert f_n(y)\Vert \to\infty$. By sequentially weak*-compactness, we can assume, going to a subsequence if necessary, that $f_n$ converges weakly* to some $F\in E^*$. But then $|F(y)|=\lim |f_n(y)|=\infty$, a contradiction. Hence $M$ is bounded.

Now we show that $M$ is closed. Let $(x_i)\subseteq E$ be a dense sequence. Suppose $F\in\overline{M}$ (the closure in the weak*-topology). For every $n$, choose $f_n\in M$ such that $|f_n(x_i)-F(x_i)|<1/n$ for every $i=1,\ldots,n$. In particular $f_n(x_i)\to F(x_i)$ as $n\to\infty$ for every $i$. Going to a subsequence, assume $f_n$ converges weakly* to some $G\in M$. Then $G(x_i)=\lim f_n(x_i)=F(x_i)$ for every $i$. Since $(x_i)$ is dense, $F=G\in M$, so $M$ is closed.

(3.$\Rightarrow$4.) If $M$ is bounded and closed, then it is a closed subset of some closed ball in $E^*$, which is weak*-compact by Banach-Alaoglu, so $M$ is weak*-compact itself.

(4.$\Rightarrow$1.) If $M$ is weak*-compact (in particular it is closed) and $S$ is an infinite subset of $M$, let $(f_n)\subseteq S$ be an infinite subsequence. By compactness, $(f_n)$ has a subnet which converges to some $F\in E^*$, and $F$ is clearly a limit point of $S$, so $M$ is rcc.

$\endgroup$
  • $\begingroup$ It's a pity that the maximum score is +1. You would deserve $n$ where $n\to\infty$ :-) I admit that I don't perceive the proof "immediate" as the book says. Banach-Steinhaus theorem hasn't even been proven nor stated in the book at this point. You saved me from being blocked in a situation where my studies couldn't go further. $\aleph_1$ graças!!! $\endgroup$ – Self-teaching worker Sep 15 '14 at 16:04

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.