Let $f_k, f \in L^{\infty}(R)$ and $f_k \overset * \to f$ in $L^{\infty}(R)$. Is $f_k$ a bounded sequence in $L^{\infty}(R^n)$?
(Definition: if $(v_n)$ is a sequence in $V = X^*$, we say that $v_n \overset * \to v \in V$ if and only if $\lim \limits _{n \to \infty} v_n x=v x \ \forall x \in X$.)
I guess that if a sequence has a finite limit then the sequence is bounded. But the problem is that it is not an absolute convergence.
By definition weak* convergence of $f_k$ to $f$ in $X=L^{\infty}(\mathbb R)$ means:
$$lim_{k \to \infty}\langle f^*,f_k\rangle=\langle f^*,f \rangle \space (\forall f^*\in X^*)$$
This is a convergent sequence of reals, hence we have $$sup_{k \in \mathbb N}\Vert \langle f^*,f_k \rangle \Vert = sup_{k \in \mathbb N} \Vert \iota(f_k)(f^*)\Vert < \infty \space (\forall f^* \in X^*)$$ where the $\iota$ denotes the isometric embedding of $X$ into its bidual $X^{**}$. Since $\{\iota(f_k)\}_{k \in \mathbb N}$ is a family of bounded linear operator with domain $X^*$ (which is a Banach space!) we can apply Banach-Steinhaus and deduce from the pointwise boundedness the uniform boundedness $$sup_{k \in \mathbb N} \Vert \iota(f_k)\Vert < \infty$$
Therefore we obtain by the isometry property: $$sup_{k \in \mathbb N} \Vert f_k\Vert=sup_{k \in \mathbb N} \Vert \iota( f_k)\Vert < \infty $$ and the claim follows.