About weak convergence in $L^{\infty}$

Question

doing my homework I'm dealing with this:

Let, for all $n\in \mathbb{N} \quad f_n(t) := e^{-nt^2}, \quad t \in [-1,1]$ Show that

1)$f_n \overset{\ast}{\rightharpoonup} 0$ in $L^\infty(-1,1)$

2)$f_n$ does not converge weakly to $0$ in $L^\infty (-1,1)$

So I did 1) simply considering \begin{equation} |f_n(x)g(x)| \le|g(x)| \quad \forall g \in L^1(-1,1) \end{equation}

So the result follows easily from the dominated convergence theorem.

But for 2) I know that given $f_n, f \in X^*$ \begin{equation} f_n \rightharpoonup f \quad \Leftrightarrow \quad \phi(f_n)\rightarrow \phi(f) \quad \forall \phi \in X^{**} \end{equation} But how can I identify the dual of $L^\infty$ to solve this?

Answer 1

Ok I think I solved this.

In this case, we don't have weak convergence of $f_n$ to $0$ because we can consider the extension of Dirac $\delta_0$ as linear functional of $(L^\infty (-1,1))^*$, which is a linear bounded functional on $C[-1,1]$: \begin{equation} \delta_0(f) = f(0) \quad \forall f \in C[-1,1] \end{equation}

Then if we call $T$ this extension it follows $T(f_n) = f_n(0)=1$.

(we have the extension because of Hahn Banach Theorem, thanks for your suggestions).