Simple proof of Krein-Smulian

Question

I have a question concerning a simple and short proof of Krein-Shmulian, the proof can be found here https://people.math.ethz.ch/~jteichma/slides_ftap.pdf on page 35/36. Here the proof:

Let $X$ be a Banach space. The Krein-Smulian theorem tells that a convex subset $C\subset X^∗$ is weak-∗-closed if and only if its intersections with balls in $X^∗$ are weak-∗-closed.

We can conclude this theorem from a separation theorem (see Conways’ book on functional analysis): assume that for a convex set $C \subset X^*$ all its intersections with balls in $X^*$ are weak-∗-closed, and assume that the intersection of $C$ with the unit ball (centered at $0$) is empty, then there is $x \in X$ such that $(x,x∗)\geq 1$ for all $x^* \in C$.

From this we can conclude immediately: let $x^* \in X^*$ be in the weak-∗=closure of $C$ but not in $C$, then – due to the fact that $C$ is norm closed (prove it!) – there is a ball of radius $r$ around $x^∗$ which does not intersect $C$. Whence $r^{−1}(C−x^*)$ does not intersect the unit ball centered at $0$. By the previous separation statement this however means that $x^*$ cannot lie in the weak-∗-closure of $C$.

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While the statements on the first page are clear to me, I have some problems with page 36:

First, it is a well known fact that if $S \subset X$ and convex ($X$ being a vector space carrying a norm) is a weak closed convex set if and only if $S$ is also strongly closed.

But how one get's that that if $C$ is a convex set where its intersection with balls in $X^*$ are weak*-closed, then $C$ is norm closed ?

And further, having this fact, why we can find a ball of radius $r$ around $x^*$ (where $x^*$ is in the weak*-closure but not in $C$) which does not intersect $C$ ? And why this finally implies the statement ?

Answer 1

Hello and welcome to MSE!

Let's denote by $B(\cdot,r)$ the ball centered at $\cdot$ of radius $r>0$.

1. Let $X$ be a normed space, $C\subset X^*$ a convex set so that $C\cap B(f,r)$ is weak-star closed for all $f\in X^*$ and all $r>0$. Then $C$ is norm-closed.

Proof: Let $f\in\overline{C}$, the norm-closure and find a sequence $(f_n)\subset C$ so that $f_n\to f$ in norm. Note that $\{\|f_n-f\|\}$ is bounded (since it converges to 0) so $(f_n)\subset B(f,r)$ for some $r>0$. Therefore $(f_n)\subset C\cap B(f,r)$ and $f_n\to f$ in norm, we also have that $f_n\to f$ weak-star, since the norm topology is stronger than the weak-star topology. But since $C\cap B(f,r)$ is weak-star closed by assumption, we have that $f\in C\cap B(f,r)$, so $f\in C$.

Note that we didn't really use convexity for this part.

2. In the same setting as above, let $f\in\overline{C}^{wk^*}\setminus C$. Since $C$ is norm-closed and $f\not\in C$, we have that $f$ does not belong to the norm closure of $C$ (because $C=\overline{C}^{\|\cdot\|}$!), i.e. there exists $r>0$ such that $B(f,r)\cap C=\emptyset$.

As the slides suggest, this means that $\frac{1}{r}(C-f)$ does not intersect the closed unit ball. The separation theorem displayed in the first part (which is essentially Hahn-Banach) shows that there exists a point $x\in X$ so that $g(x)\geq1$ for all $g\in\frac{1}{r}(C-f)$.

Take a net $\{c_i\}_{i\in I}\subset C$ converging to $f$ weak-star. Then $\frac{1}{r}(c_i-f)\in\frac{1}{r}(C-f)$, so $\frac{1}{r}(c_i-f)(x)\geq1$, so $c_i(x)-f(x)\geq r$ and this is true for all $i\in I$. Taking limits as $i\in I$, since $c_i\to f$ weak-star, hence $c_i(x)\to f(x)$, this shows that $0\geq r$, which is a contradiction.