Proving the closed unit ball of a Hilbert space is weakly sequentially compact

I bumped into this statement in Hofer-Zehnder in the middle of proving a Hamiltonian field always has a periodic orbit over a level set of the hamiltonian if that set is a regular compact and strictly convex energy surface. I looked around the internet but a proof was elusive. How do I prove this?

Solutions Collecting From Web of "Proving the closed unit ball of a Hilbert space is weakly sequentially compact"

OK, I lied :). Then again, I felt like I couldn’t really post a question saying “I found a proof scattered over a few Wiki entries and am posting just to collect the info in one place”, and that such a statement belonged in an answer, just like this one. I also found this, which proves the result for separable spaces. That proof seems similar, though longer than what I pieced together. So here goes.

Preliminary lemmata from topology

Equivalent characterizations of closedness

Definition of closed set

If $(X,\tau)$ is a topological space, we call $A\subseteq X$ closed if $X\smallsetminus A\in\tau$.

Definition of limit point

If $(X,\tau)$ is a topological space and $A\subseteq X$, we say $x\in X$ is a limit point for $A$ if for all $V\subseteq X$ open neighborhood of $x$ we have that $(V\cap A)\smallsetminus\{x\}\neq\varnothing$.

Lemma 1

$A\subseteq X$ is closed iff it contains all its limit points

Proof

$\Longleftarrow$

A set containing all its limit points has open complement since if a point is in the complement it is not a limit point so it has a neighborhood contained in the complement.

$\implies$

A set with open complement can’t have limit points outside itself since otherwise those would need a neighborhood contained in the complement which violates the definition of limit point. $\square$

Definition of poset, filtered set and net

Let $S$ be a set. A partial order on $S$ is a subset $R$ of $S\times S$ such that:

  1. $(a,a)\in R$ for all $a\in S$, otherwise known as reflexivity of $R$;
  2. $(a,b)\in R,(b,a)\in R\implies a=b$ for all $a,b\in S$, aka antisymmetry of $R$;
  3. $(a,b)\in R,(b,c)\in R\implies(a,c)\in R$, aka transitivity of $R$.

The couple $(S,\leq)$ is a poset. If $(S,\leq)$ has the property that for any two $a,b\in S$ there exists $c\in S:c\geq a,c\geq b$, then $(S,\leq)$ is a filtered set. A net with values in a set $X$ is a function from a filtered set to $X$.

Definition of net convergence

Let $(X,\tau)$ be a topological space and $x_\lambda$ be a net in $X$. We say $x_\lambda$ converges to some $x\in X$, and write $x_\lambda\to x$, if for any neighborhood $V$ of $x$ in $X$ there exists $\lambda(V)$ such that $\beta\geq\alpha(V)\implies x_\beta\in V$.

Lemma 2

$A\subseteq X$ is closed iff for any net $x_\lambda$ converging to $x$ in $X$, $x\in A$.

Proof

$\implies$

The limit of any net in $C$ is limit point of $C$ since a net is eventually in every neighborhood of its limit, implying that any neighborhood of the limit of a net in $C$ has points from the net in it, thus points in $C$, making it a limit point. So if a net in a closed set converges, the limit is a limit point of the closed set, thus by lemma 1 an element of the set.

$\Longleftarrow$

Conversely, if all nets in $C$ either converge to a limit in $C$ or don’t converge at all, there is a particular net in $C$ converging to any limit point of $C$, allowing us to conclude. Such a net is constructed by taking all the neighborhoods (open neighborhoods) of the limit point, ordering them by reversed inclusion (i.e. $U\leq V\iff U\supseteq V$) and making them into a filtered set, then choosing for each such $U$ an $x_U\in U$. That requires, of course, the axiom of choice. Such a net must have a limit in $C$ or no limit at all, but the limit is evidently the limit point, by construction. So we are done. $\square$

Continuity and equivalent characterizations

Definition of continuous function

A function $f:X\to Y$ between topological spaces is continuous if every open set $B\subseteq Y$ has open preimage $f^{-1}(B)\subseteq X$.

Remark

Evidently that is equivalent to every closed subset of $Y$ having closed preimage in $X$, since the complement of a preimage is the preimage of the complement. It is also true that continuity is equivalent to $\overline{f(A)}\subseteq f(\overline A)$ for any $A\subseteq X$.

Lemma 3

A function $f:X\to Y$ is continuous iff it preserves net convergence, i.e. iff for any $x_\lambda$ converging to $x$ in $X$, one has $f(x_\lambda)\to f(x)$ in $Y$.

Proof

$\implies$

Suppose $f:X\to Y$ is continuous. If $x_\alpha\to x$ in $X$, then for all open $V\ni x$ the net is eventually in $V$. In particular, for any open $U\ni f(x)$, by continuity of $f$ and convergence of the net $x_\alpha$ is eventually in $f^{-1}(U)$, i.e. $f(x_\alpha)$ is eventually in $U$, proving $f(x_\alpha)\to f(x)$ in $Y$.

$\Longleftarrow$

Conversely, let us prove that if $f$ preserves net convergence then closed sets have closed preimages, which by the Remark above is equivalent to continuity. Let $C$ be closed in $Y$ and $C’=f^{-1}(C)($. Let $x_\alpha$ be a net in $C$ converging to $x$ in $X$. Then $f(x_\alpha)\to f(x)$ in $Y$ by assumption, but $C$ is closed so $f(x)\in C$ by Lemma 2, hence $x\in C’$ which is therefore closed, completing the proof.

Compactness and continuous functions

Definition of compact set

If $(X,\tau)$ is a topological space, $A\subseteq X$ is compact if for any open cover of $A$ there is a finite subcover, i.e. whenever $A_i\in\tau$ for all $i$ in a certain set of indexes and $A\subseteq\bigcup A_i$, then there exists a finite subset $I$ of the index set for which $\bigcup_IA_i\supseteq A$.

Lemma 4

If $f:X\to Y$ is continuous and $A\subseteq X$ is compact, then $f(A)\subseteq Y$ is compact too.

Proof

Take an open cover of the image $f(A)$, take it back to an open cover of $C$ via $f$-preimages, $A$ is compact so the cover admits a finite subcover, take it back to part of the cover of $f(A)$ to find the desired subcover and prove $f(A)$ is compact. $\square$

Definition of Hausdorff space

If $(X,\tau)$ is a topological space, we say it is Hausdorff if for all $x\neq y\in X$ there exist $U\ni x,V\ni y$ open neighborhoods such that $U\cap V=\varnothing$.

Lemma 5

If $X$ is Hausdorff, then a compact $A\subseteq X$ is closed.

Proof

We will prove that if $A$ is compact in $X$ Hausdorff, then $X\smallsetminus A$ is open. Let $a\in X\smallsetminus A$. For $x\in A$, by the Hausdorff condition, there exist neighborhoods $U_x,V_x$ such that $U_x\ni a,V_x\ni x$ and $U_x\cap V_x=\varnothing$. The $V_x$ cover $A$ which is compact, so there is a finite subcover of $A$, say $V_{x_1},\dotsc,V_{x_n}$. Let $U=\bigcap_iU_{x_i}$ Then $U$ is open because it is a finite intersection of open sets, all those contain $a$ so $a\in U$, and $U$ is disjoint from all $V_{x_i}$ so it is disjoint from $A$ which is contained from their union. Hence $a$ is in the interior of the complement, which implies the complement is open. $\square$

Lemma 6

Let $X$ be a compact topological space. Then a closed $C\subseteq X$ is compact.

Proof

Take a cover of $C$, $\{A_i\}$. $X\smallsetminus C$ is open since $C$ is closed, and adding this to $\{A_i\}$ gives a cover of $X$. $X$ is compact, so that cover admits a finite subcover. Take that as the subcover of $\{A_i\}$, removing $X\smallsetminus C$ if necessary. $\square$

Lemma 7

Let $f:X\to Y$ be continuous. If $X$ is compact and $Y$ is Hausdorff, then $f$ is closed, i.e. maps closed sets to closed sets.

Proof

Let $C$ be closed in $X$. Then it is compact by Lemma 6. So $f(C)$ is compact in $Y$ by Lemma 4, which is Hausdorff, making $f(C)$ closed by Lemma 5, and proving this lemma. $\square$

Lemma 8

If $X$ is a set and $T_1,T_2$ two topologies, where $T_2\subseteq T_1$, $(X,T_2)$ is Hausdorff and $(X,T_1)$ is compact, then $T_1=T_2$.

Proof

The assumption $T_2\subseteq T_1$ is equivalent to the continuity of $id:(X,T_1)\to(X,T_2)$, but that means that $id$ is a homeomorphism, since it is a continuous bijection from a compact space to Hausdorff space, and thus $T_1=T_2$. $\square$

Alexander and Tychonoff

Definition of subbase for a topology.

A subbase for the topology $\tau$ of a topological space $(X,\tau)$ is a collection of open sets such that the topology generated by it is the topology of our topological space and the finite intersections of this collection form a basis, i.e. every open set is a union of finite intersections of the subbase.

Alexander subbase theorem

If $X$ is a topological space with a subbase such that every subbasic cover has a finite subcover, then $X$ is compact.

Proof comes from here

Find a special cover without subcover

By contradiction, suppose $X$ is not compact, but every subbasic cover admits a finite subcover. By Zorn’s lemma, we can find a cover $C$ which has no finite subcover, but any $C\cup\{U\}$ with $U$ open does. This is obtained by considering the poset of finite subcover-free covers with the partial order that $C\leq C’$ if $C\subseteq C’$. Taking a chain $\{C_i\}$, obviously the union is an upper bound, because if that admitted a finite subcover, there would be a cover in the chain containing all the elements of the subcover, but no member of the chain admits a finite subcover. By Zorn’s lemma, there is a maximal element in this poset, and that cover $C$ has the required property, for if $C$ is maximal and yet there is $U$ open with $C\cup\{U\}$ has no finite subcovers, then $C\leq C\cap\{U\}$ which is in the poset, contradicting maximality.

Extract and analyze the subbasic family from that cover

Let us call the subbase $B$. Take $C\cap B$, the subbasic subfamily of $C$. Of course, there are no finite subcovers of $C\cap B$, but if it were a cover there would be one, so it is not a cover.

Reach a contradiction

Now take $x\in X$ uncovered by $C\cap B$. There exists $U\in C$ with $x\in U$. $B$ is a subbase, so there exist a finite number of $S_1,\dotsc,S_n$ such that $x\in S_1\cap\dotso\cap S_n\subseteq U$. By our assumption $C\cup\{S_i\}$ has a finite subcover for all $i$. Take that subcover, remove $S_i$ (which is necessarily in it or the subcover comes from $C$, contradicting our hypothesis), and call the result $C_i$. This means $\bigcup_iC_i\cup\{U\}$ cover all of $X$, but then $C$ had a finite subcover after all. $\square$

Tychonoff’s theorem

A product of compact spaces is compact in the product topology.

Proof

Indeed, the product topology on the cartesian product of a family of sets has by definition a subbase made by “cylinder sets”, the inverse projections of open sets in one factor. Just prove this subbase satisfies the above theorem’s hypothesis, and we are done. So suppose a subbasic cover $C$ has no finite subcover. Partition it into subfamilies $\bigcup_iC_i$ where $C_i$ is made by inverse projections of open sets in $X_i$. Naturally, none of the $C_i$ can have a finite subcover. Projecting them onto $X_i$ gives projections that have no finite subcovers, and thus cannot be covers since $X_i$ is compact. This means for all $i$ there is $x_i\in X_i\smallsetminus\bigcup_{U\in C_i}\pi_i(U)$, $\pi_i$ being the projection onto $X_i$. But then $(x_i)_{i\in\mathbb{N}}\in X\smallsetminus\bigcup_{U\in C}U$, so $C$ is not a subbasic cover, contradiction. $\square$

Weak-* topology

The canonical injection of a Banach space into its bidual

If $X$ is a normed space, then the dual $X^\ast$ also is, with the norm operator, i.e. if $f\in X^\ast$, $\|f\|=\sup_{\|x\|=1}|f(x)|$. It is easy to see that, defining $J(x)$, for $x\in X$, as the element of the bidual $X^{\ast\ast}$ which acts in the following way:

$$J(x)(f)=f(x),$$

one obtains an isometry $J:X\to X^{\ast\ast}$. This map is therefore continuous and injective, but not necessarily surjective.

Weak-* topology on the dual

The weak-* topology on $X^\ast$ is the initial topology w.r.t. $J(E)$, i.e. the coarsest topology making each element of $J(E)$ continuous. This is coarser than the weak topology on $X^\ast$, in general, and naturally coincides with it iff $X$ is reflexive, which means iff $J$ is surjective.

Characterisation of open sets in an initial topology

$U\subseteq X$ is open in the initial topology w.r.t. a family $\{f_i\}$ of functions to topological spaces $Y_i$ iff it is an arbitrary union of finite intersections of sets of the form $f_i^{-1}(U_i)$, with $U_i$ open in $Y_i$.

Proof

First of all, since the $f_i$ must be continuous, $f_i^{-1}(U_i)$ has to be open whenever $U_i$ is open in $Y_i$, and since we must have a topology, arbitrary unions of finite intersections of such sets must be open. Take the collection of arbitrary unions of finite intersections. This is a topology. Therefore, the initial topology must be contained in it, yet must contain it by the above. Hence they coincide. That the collection above is a topology is easily verified:

  • The empty set is the preimage of the empty set under any $f_i$, and $X=f_i^{-1}(Y_i)$ for any $i$;
  • The collection is one of arbitrary unions, hence it is evidently closed under arbitrary unions;
  • Take $A=\bigcup_I\bigcap_1^kA_{ij}$ and $B=\bigcup_{I’}\bigcap_1^{k’}B_{ij}$. If we prove $A\cap B$ is in the collection, we are done. Let:

$$C_{ij}:=\left\{\begin{array}{cc}
A_{ij} & i\in I, j\leq k \\
X & i\in I,k<j\leq k+k’ \\
B_{ij} & i\in I’,j\leq k’ \\
X & i\in I’,k'<j\leq k+k’
\end{array}\right.$$

Then $A\cap B=\bigcup_{I\cup I’}\bigcap_1^{k+k’}C_{ij}$, and $C_{ij}$ is a preimage of an open set under some $f_i$, so $A\cap B$ is again in the collection. By induction, this implies our collection is closed under finite intersections, hence by the two points above a topology. $\square$

Characterization of weak-* convergence

$f_n\to f$ weakly-* iff for all $x\in X$ we have $f_n(x)\to f(x)$ in $\mathbb{R}$.

Proof

$\implies$

$f\mapsto f(x)$ is made continuous by the weak-* topology, so if $f_n\to f$ in that topology, naturally $f_n(x)\to f(x)$.

$\Leftarrow$

The weak-* open sets are arbitrary unions of finite intersections of sets of the form $J(x)^{-1}(\omega)$ with $x\in X,\omega\subseteq\mathbb{R}$ open. So if $U$ is a weak-* open set containing $f$, it must contain a finite intersection $\bigcap_1^kJ(x_i)^{-1}(\omega_i)$ which contains $f$. Naturally, since $f_n(x)\to f(x)$ for all $x$, we have $f_n(x_i)\in\omega_i$ from some $n_i$ on, by definition of convergence. Take $N=\max n_i$. Then from $N$ on, $f_n(x_i)\in\omega_i$ for all $i$, or in other words, if $n\geq N$, $f_n$ belongs to that intersection of preimages contained in $U$, and therefore belongs to $U$. This proves that $f_n$ is eventually in any weak-* open set containing $f$, and thus $f_n\to f$ in the weak-* topology. $\square$

Characterization of continuity of functions to spaces endowed with initial topologies

Let $X$ be endowed with the initial topology w.r.t. a family $\{f_i\}$ of functions $f_i:X\to Y_i$, where $Y_i$ are any topological spaces. Let $f:Z\to X$ be a function with $Z$ a topological space. Then $f$ is continuous iff $f_i\circ f$ is continuous for any $i$.

Proof

$\implies$

This is easy: the initial topology makes the $f_i$ continuous, so if $f$ is continuous, $f_i\circ f$ is continuous as it is a composition of continuous functions.

$\Leftarrow$

Take an open set $U$ in $X$. It is an arbitrary union of finite intersections of $f_i^{-1}(U_i)$ with $U_i\subseteq Y_i$ open, by the above characterization of initial open sets. Take its preimage under $f$, you get the arbitrary union of the finite intersections of the preimages. So the preimage of your set is an arbitrary union of finite intersections of sets of the form $f^{-1}(f_i^{-1}(U_i))$. This can be rewritten as $(f_i\circ f)^{-1}(U_i)$, but $f_i\circ f$ is continuous by hypothesis, hence the preimages in question are open, and so are the arbitrary unions of finite intersections of those, making $f$ continuous as claimed. $\square$

Banach-Alaoglu-Bourbaki theorem

The closed unit ball of the dual of a Banach space is weakly-* compact.

Proof

Define a compact space

For $x\in X$, define $D_x=\{z\in\mathbb{C}:|z|\leq\|x\|\}$. This is compact in the complex plane. By Tychonoff’s theorem above, $D=\prod_{x\in X}D_x$ is compact in the product topology.

Map the unit ball into it

Let us now map $B_1(X^\ast)$ into $D$. Define $\pi(f)=(f(x))_{x\in X}$.

Analyze the map

  • This is clearly an injection.
  • It is also continuous if $B_1(X^\ast)$ is endowed with the weak-* topology, by definition of that topology. Indeed, the product topology is the initial topology w.r.t. the projections $\pi_x:(\omega_x)_{x\in E}\mapsto\omega_x$. Using the characterization above, let us see what happens if we compose our map with the projections. The result is precisely the family of maps $\{J(x)\}_{x\in E}$, which are continuous by definition of weak-* topology.
  • So we need to prove the range is closed, for $D$ is compact, and a closed subspace of a compact space is also compact by Lemma 6. So let $\lambda_\alpha=(f_\alpha(x))_{x\in X}$ be a net in the range of $\pi$. If it converges to $\lambda$, let $f$ be defined as $f(x)=\pi_x(\lambda)$, where $\pi_x$ projects $D$ onto $D_x$. This is a well-defined element of the closed unit ball to which $f_\alpha$ evidently converges, by the characterization of weak-* convergence, whose proof evidently extends to nets. Let us prove it is indeed in the dual, i.e. it is linear and continuous from $X$ to $\mathbb{R}$. Linearity is evident since $f(ax+by)=\lim_\alpha f_\alpha(ax+by)$, use linearity of $f_\alpha$ for all $\alpha$ and that of the limit to conclude. Continuity is because $f_\alpha$ has norm at most 1, so $|f_\alpha(x)|\leq\|x\|$, and a limit preserves an inequality, concluding $\|f\|\leq1$. This shows the range is closed.
  • The last thing is to prove the inverse of this map is also continuous, for then we will have a homeomorphism. This is done by the characterization of continuity to an initial topological space given above. Compose our map with any $J(x)$. You get a projection, which is continuous by definition of product topology. Indeed, the product topology is also known as topology of pointwise convergence, because convergence in the product topology is precisely pointwise convergence, by the characterization of initial convergence… did I give it? Well, maybe I gave it for weak-* convergence, but of course the proof does not depend on the particular instance of initial topology, but only on the fact it is an initial topology

Conclusion

Summing up, we have a compact space $D$, a map $\pi:B_{X^\ast}\to D$ constructed above, and we have shown it is a continuous and open injection with closed range. Then the range is a closed subset of the compact space $D$, hence compact, and $\pi$ is a homeomorphism onto its image, making $B_{X^\ast}$ compact, as claimed. $\square$

Separability and metrizability of the weak-* topology

Definition of separable space

A topological space $(X,\tau)$ is separable if it has a countable dense subset. This is very different from a separated space, which simply means a Hausdorff space.

Restricting ourselves to separable spaces $X$, we now need to show $X^\ast$ has a metrizable topology in the closed unit ball. First, we show that a countable family of seminorms $\{\rho_n\}_{n\in\mathbb{N}}$ always induces a metrizable topology, metrized by the metric:

$$d(x,y)=\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x-y)}{1+\rho_n(x-y)}.$$

Now my original plan was to take this for granted, but then I tried proving this is a metric and got stuck on the triangle inequality. I did some inequalities, but only proved $d(x,z)+d(z,y)$ is at least $d(x,y)$ with the denominators changed to $1+\rho_n(x-z)+\rho_n(z-y)+\rho_n(x-z)\rho_n(z-y)$, which is evidently at least the correct denominator, but most probably strictly greater, which messes up the inequality. I asked about this metric here.

Proposition

This metric induces precisely the topology of the seminorm family.

Proof

Assuming this is a metric, it clearly makes the space Hausdorff. We must prove the metrizability in the (obviously necessary) assumption that the seminorm topology makes the space Hausdorff. Now the topology the seminorm family induces is generated by the sets where $\rho_n(x-y)<\epsilon$, for $n,x,\epsilon$ arbitrary. But if we take a ball for $d$, all $\rho_n$ will be less than the radius by immediate inequalities, so that ball is contained in infinitely many such sets. Such sets are evidently open in the metric topology since each of their points has a metric ball contained in the set. To prove the reverse inclusion, we use Lemma 3. Proving the metric topology is contained in the seminorm topology is equivalent to proving the identity is continuous from the space with the seminorm topology to the space with the metric topology, which in turn, by the mentioned result, is equivalent to proving that every converging net in the seminorm topology converges to the same limit in the metric topology, or equivalently that:

$$d(x_\alpha,x)\to0\Leftarrow\rho_n(x_\alpha-x)\to0\,\,\,\,\forall n\in\mathbb{N}.$$

So we are assuming the right side, and we must prove the left side, i.e. that:

$$\sum_{n=1}^\infty2^{-n}\frac{\rho_n(x_\alpha-x)}{1+\rho_n(x_\alpha-x)}\to0.$$

We can interpret this as an integral w.r.t. the counting measure on $\mathbb{N}$, and turn the above into:

$$\lim_\alpha\int_{\mathbb{N}}f_\alpha(n)\mathrm{d}C(n),$$

where $C$ is the counting measure $C(A)=|A|$ and $f_\alpha(n)=\frac{2^{-n}\rho_n(x_\alpha-x)}{1+\rho_n(x_\alpha-x)}$. Now, the proof of the Dominated Convergence Theoerm seems to readily extend to nets of functions as is our case, and in our case the net evidently is dominated by $2^{-n}$ which is integrable (i.e. has a convergent series), so we permute sum and limit and we must show those nets of $f_\alpha(n)$ tend to 0 for all $n$, but the assumption, plus some elementary properties of limits usually proved for sequences but that are valid for nets with the same proofs (limit of ration and of sum and of product), gives precisely that. So the topologies coincide. $\square$

Theorem

The closed unit ball of the dual of a separable normed space is metrizable under the weak-* topology.

Proof

We have the weak-* topology, clearly induced by the family of seminorms $\rho_x(y)=|y(x)|$ ($y$ is in the dual so $y:X\to\mathbb{R}$). Consider a countable dense subset, which exists by the separability assumption on $X$. Let that subset be $\{x_n\}$, and consider $\{\rho_{x_n}\}$. That is a countable family, so it induces a topology $T_2$ with respect to which the closed unit ball is certainly Hausdorff as the topology is metrizable by the above. We know the ball is compact w.r.t. $T_1$ the weak-* topology by Banach-Alaoglu. If we prove $T_2\subseteq T_1$, then by Lemma 8 we have they coincide. But $T_2$ is generated by sets where $\rho_{x_n}$ for some $n$ is less than $\epsilon$ for some $\epsilon$, and those sets are amongst the generators of $T_1$. $\square$

With that, by metrizability the compactness given by Banach-Alaoglu is equivalent to weak-* sequential compactness, so we finally get that the closed unit ball of $X^\ast$, the dual of a normed space $X$, is weakly-* sequentially compact.

Moving to Hilbert spaces

Separable case

Recalling the (Fréchet-)Riesz representation theorem (see also here), we have an isomorphism from a Hilbert space into its dual; the canonical injection of $H$ into $H^{\ast\ast}$ is just the composition of the Riesz isomorphism of $H^\ast$ with that of $H$, as can be easily proved; hence, a Hilbert space is reflexive, and all the below Brezis results hold for Hilbert spaces. The weak topology on a Hilbert space thus corresponds to the weak-* topology, so Banach-Alaoglu gives weak compactness for the closed unit ball of a Hilbert space, and sequential weak compactness for separable ones.

General case

But what about non-separable Hilbert spaces then? Well, take the closed unit ball and a sequence in it. Consider the closed span of that sequence. The induced inner product makes it a Hilbert space, so by the above our sequence will have a subsequence converging in its span, and so also in the “big” ambient Hilbert space, generalizing the result with little to no effort.

Final remarks

  1. Obviously, the above only requires $X$ to be the dual of something, a Banach space actually. So this works for all reflexive Banach spaces, i.e. Banach spaces coinciding with their double dual. Wikipedia has a few more consequences, in particular the Hahn-Banach theorem;
  2. I asked about the metric
    here;
  3. Finally, I note that Brezis, Functional Analysis, Sobolev spaces and
    PDEs, has a slightly different proof of Banach-Alaoglu, and proves
    weak compactness of the unit ball is equivalent to reflexiveness of
    the space, in Chapter 3. I haven’t yet looked if it also proves
    sequential compactness, which without metrizability is not
    equivalent to compactness, though;
  4. What we proved is in fact much stronger than just the property for closed balls: Corollary 3.22 on p. 71 of Brezis proves any closed, bounded, and convex set in a reflexive space is weakly compact; in particular, any closed ball is weakly compact;
  5. Theorem 3.29 on p. 74 of Brezis states that not only does separability imply metrizability for the weak topology of the closed unit ball of a reflexive banach space, but that the converse also holds;