Goldstine theorem

Question

Given the embedding $j:X\to X''$ defined by,

$$j=(x\mapsto(\phi\mapsto\phi(x)))\,,$$

according to my interpretation of the wikipedia page, Goldstine theorem says the following:

$$\overline{jB_X}^{weak\ast}=B_{X''}$$

with $B_Y$ the closed unit ball of $Y$. But I weak $\ast$ topology is defined on $X'$ right? So how can we take the closure with respect to this topology of elements of $X''$?

Answer 1

The weak $\ast$ topology used in the Goldstine theorem is not the normal weak $\ast$ topology of $X$ defined on $X'$. It is the weak $\ast$ topology of $X'$ defined on the dual of $X'$, i.e. $X''$.

Goldstine theorem can thus be formulated as: $$\overline{jB_X}^{\sigma(E'',E')}=B_{X''}$$