Conservation of the weak topology by homeomorphism

Question

I have some questions about Brezis book. We know that M is reflexive, so there exists an homeomorphism $J:(M,\|\|_M) \to (M'',\|\|_{L(M',R)})$ between the "strong" topology of M and M''. (Moreover J is isometric).

1. So i would like to know why $B_M=B_{M''}=\{\xi\in M'':\|\xi\|_{L(E',R)}\leq 1\}$ and why $(M,\sigma(M,M')=(M,\sigma(M'',M')$
2. when the ball $B_M$ is metrizable in the weak topology $(M,\sigma(M,M')$ by consequence $B_{M''}$ is metrizable for the weak-star topology $(M,\sigma(M'',M')$ ?

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Translation :

Theorem III.27 : Let $E$ be a reflexive Banach space, $(x_n)$ a bounded sequence in $E$. Then there exists a sub-sequence extracted from $(x_n)$ which converges for the $\sigma(E,E')$ topology

Demonstration : Let $M_0$ be the vector space generated by the $(x_n)$, et $M=\overline{M_0}$. M is a separable space (see III.23). Furthermore, M is reflexive (see III.17). Therefore $B_M$ is a compact, metrizable space for the topology $\sigma(M,M')$. Indeed, $M'$ is separable (see III.24) hence $B_{M''}(=B_M)$ is metrizable for $\sigma(M'',M')(=\sigma(M,M'))$ (see III.25). We can then extract a sub-sequence $(x_{n_k})$ which converges for $\sigma(M,M')$. We conclude that $(x_n)$ converges too for $\sigma(E,E')$ (by restricting $M$ to linear forms on, $E$).

Answer 1

We don't actually have equality, but we have a canonical identification between the two spaces. This canonical identification is habitually left implicit for notational simplicity (at the cost of temporarily confusing beginners).

With the identification made explicit, the assertions are

1. $J(B_M) = B_{M''}$, which immediately follows from the fact that $J$ is an isometric isomorphism, and

2. $J\colon (M,\sigma(M,M')) \to (M'',\sigma(M'',M'))$ is a topological isomorphism. In particular, the restriction of $J$ to $B_M$ is a homeomorphism between $B_M$ and $B_{M''}$, where both are endowed with the subspace topology induced by $\sigma(M,M')$ and $\sigma(M'',M')$ respectively.

It is clear that if two spaces are homeomorphic, each is compact resp. metrisable if and only if the other is.

To see that $J\colon (M,\sigma(M,M'))\to (M'',\sigma(M'',M'))$ is a topological isomorphism, consider the standard neighbourhood bases of $0$ in these topologies: Given $\mu_1,\dotsc,\mu_k\in M'$, we have

$$\begin{aligned} J\left(\{ x \in M : \lvert \mu_\kappa(x)\rvert < 1 \text{ for } 1 \leqslant \kappa\leqslant k\}\right) &= J\left(\{x \in M : \lvert J(x)(\mu_\kappa)\rvert < 1 \text{ for } 1 \leqslant \kappa\leqslant k\}\right)\\ &= \{ J(x)\in M'' : \lvert J(x)(\mu_\kappa)\rvert < 1 \text{ for } 1 \leqslant \kappa\leqslant k\}\\ &= \{\varphi\in M'' : \lvert \varphi(\mu_\kappa)\rvert < 1 \text{ for } 1 \leqslant \kappa\leqslant k\}, \end{aligned}$$

so $J$ induces a bijection between the two neighbourhood bases, and that implies that $J$ is a homeomorphism, since $J$ is linear.