There is a remark that I don't understand in Brezis' functional analysis.
Here is the text:
Lemma 3.4 (Goldstine). Let $E$ be any Banach space. Then $J(B_E)$ is dense in $B_{E^{**}}$ with respect to the topology $\sigma(E^{**}, E^*)$, and consequently $J(E)$ is dense in $E^{**}$ in the topology $\sigma(E^{**}.E^*)$.
Remark 15. Note that $J(B_E)$ is closed in $B_{E^{**}}$ in the strong toplogy. Indeed, if $\xi_n = J(x_n) \to \xi$ we see that $(x_n)$ is a Cauchy sequence in $B_E$ (since $J$ is an isometry) and therefore $x_n \to x$, so that $\xi = Jx$. It follows that $J(B_E)$ is not dense in $B_{E^{**}}$ in the strong topology, unless $J(B_E) = B_{E^{**}}$, i.e., $E$ is reflexive.
$J$ is the canonical mapping from $E$ to $E^{**}$ where $J_x (f) = f(x)$ for $f \in E^*$.
I have two questions.
In the Lemma, why do we only need to prove density of the unit ball? Why does it follow that $J(E)$ is dense in $E^{**}$? My immediate thought was that, if there were a neighborhood in $E^{**}$ containing no elements of $J(E)$, then scaling by 1 over the norm might lead to a neighborhood in the unit ball that has no elements in $J(B_E)$, but I don't know how to rigorize this, and it doesn't seem like the neighborhood will even be in the unit ball always.
The second question is in the remark. Why does it follow that $J(B_E)$ is not dense in $B_{E^{**}}$ in the strong topology unless $E$ is reflexive? And does the "so that $\xi = Jx$" bit follow from the fact that $J$ is continuous / is $J$ continuous? I'm a bit confused about the argument of $J$. For any fixed $f \in E^*$, $J$ would be continuous over $x \in E$, but since $Jx \in E^**$ so its argument should be in $E^*$, right?