weak$^*$ topology is strictly coarser than weak topology ; a remark in Brezis

Question

Let $E$ be a banach space. $\sigma(E^*,E)$ be the weak $^*$ topology, $\sigma(E^*, E^{**})$ the weak topology on $E^*$ induced by the functionals on $E^{**}$.

It is proven in Brezis, pg65,

Corollary 3.15: If $H$ is a hyperplane in $E^*$ closed in $\sigma(E^*,E)$ then $H$ has the form $$H= \{ f \in E^* \, :\, f(x_0)= \alpha \}$$for some $x_0 \in E$, $\alpha \in \Bbb R$.

He stated, if canonical injection $j:E \rightarrow E^{**}$ is not surjective then

Let $\xi \notin J(E)$, the set $$ H= \{ f \in E^* \, : \, \xi(f) = 0 \} $$ is not closed in $\sigma(E^*,E)$.

I do not see how this follows from the corollary.

Answer 1

Suppose $ \ker (\xi) = \{f\in E^* : \xi (f) = 0\}$ is closed in the weak star topology, then by corollary 3.15, $\ker(\xi) = \ker (J(x_0))$ for some $x_0\in E$.

The two elements $\xi$ and $J(x_0)$ in $E^{**}$ have the same kernel, then they must be constant multiple of each other. This is lemma 3.2 on page 64 from Brezis.

Answer 2

If $H$ were closed in $\sigma(E^*,E)$, then by Corollary 3.15 you would have that there is $x_0\in E$ with $$\tag1 H=\{f\in E^*:\ f(x_0)=\alpha\}. $$ Since $H$ is a subspace, it follows that $\alpha=0$. We can rewrite $(1)$ as $$\tag2 H=\{f\in E^*:\ j(x_0)f=0\}. $$ It follows that $j(x_0)$ and $\xi$ are linearly dependent, because otherwise there would be $f\in E^*$ with $j(x_0)f=0$, $\xi(f)=1$. But this is a contradiction, because no multiple of $\xi$ is in $j(E)$.

Thus $H$ is not closed in $\sigma(E^*,E)$, while it is closed in $\sigma(E^*,E^{**})$.