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

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 $$$$|f_n(x)g(x)| \le|g(x)| \quad \forall g \in L^1(-1,1)$$$$

So the result follows easily from the dominated convergence theorem.

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

• 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 $L^\infty$. Dec 26 '18 at 15:18
• 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.. Dec 26 '18 at 15:21
• Correct, it should say $\phi$ is in $(L^\infty)^*$. Dec 26 '18 at 15:25
• you mean the definition I've written is wrong? Just to understand Dec 26 '18 at 15:26
• Your definition is correct, where $X^*$ is $L^\infty$. My correction is now corrected. Dec 26 '18 at 15:28

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 $$C[-1,1]$$: $$$$\delta_0(f) = f(0) \quad \forall f \in C[-1,1]$$$$
Then if we call $$T$$ this extension it follows $$T(f_n) = f_n(0)=1$$.