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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?

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    $\begingroup$ You do not need to identify the dual of $L^\infty$ (which is very complicated). You only need to construct one element $\phi$ of $L^\infty$ such that $\phi(f_n)\rightarrow \phi(f)$ fails. A hint for constructing continuous linear functionals: choose a Banach subspace $E \subset L^\infty$ that contains all the $f_n$, use a continuous functional on $E$, then cite the Hahn-Banach theorem to extend it to all of $L^\infty$. $\endgroup$
    – GEdgar
    Dec 26 '18 at 15:18
  • $\begingroup$ but following your reasonment, shouldn't $\phi$ be in $(L^{\infty})^{*}$? Why you say that $\phi$ is in $L^\infty$? Assuming the definition I've written is right.. $\endgroup$ Dec 26 '18 at 15:21
  • $\begingroup$ Correct, it should say $\phi$ is in $(L^\infty)^*$. $\endgroup$
    – GEdgar
    Dec 26 '18 at 15:25
  • $\begingroup$ you mean the definition I've written is wrong? Just to understand $\endgroup$ Dec 26 '18 at 15:26
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    $\begingroup$ Your definition is correct, where $X^*$ is $L^\infty$. My correction is now corrected. $\endgroup$
    – GEdgar
    Dec 26 '18 at 15:28
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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).

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