Strong convergence does not implies weak star convergence

Question

Let $E$ be a normed vector space and $E^*$ its dual space. Denote by $\mathcal{T}_{E^*}$ the norm topology in $E^{*}$, by $\sigma(E^*,E^{**})$ the weak topology in $E^{*}$ and by $\sigma(E^*,E)$ the weak star topology in $E^*$. We have the fallowing:

a) $f_n \rightarrow f$ in $\mathcal{T}_{E^*}$ $\implies f_n \rightharpoonup f$ in $\sigma(E^*,E^{**})$;

b) $f_n \rightharpoonup f$ in $\sigma(E^*,E^{**})$ $\implies f_n \overset{\ast}{\rightharpoonup} f$ in $\sigma(E^*,E)$;

c) $f_n \rightarrow f$ in $\mathcal{T}_{E^*} \implies f_n \overset{\ast}{\rightharpoonup} f$ in $\sigma(E^*,E)$

These cames from the inclusions $$ \sigma(E^*,E) \subset \sigma(E^*,E^{**}) \subset \mathcal{T}_{E^*}. $$ I would like some examples which show that the reciprocal of the convergences in itens a), b) and c) are not always true.

Answer 1

Counterexamples for a) and c).
Let's take $E = c_0, E^* = l^1, E^{**} = l^\infty$.
Let $e_n$ be the unit vectors in $l^1$, that is $e_n$ has coordinate $1$ in the $n$th position, zero elsewhere. Then:
$e_n \to 0$ in $\sigma(E^*,E)$,
$e_n \not\to 0$ in $\sigma(E^*,E^{**})$,
$e_n \not\to 0$ in norm.

Counterexample for b). Let's take $E = l^2, E^* = l^2, E^{**} = l^2$.
Let $e_n$ be the unit vectors in $l^2$, that is $e_n$ has coordinate $1$ in the $n$th position, zero elsewhere. Then:
$e_n \to 0$ in $\sigma(E^*,E)$,
$e_n \not\to 0$ in norm.