Let $f$ be a bounded lipschitz continuous function. It is not to hard to show that $$\|D^{-h}_i f\|_\infty \le C,$$ for $C$ the Lipschitz constant and $$D^{-h}_i f (x)= \frac{f(x ) - f(x - he_i)}{h}, \quad h \in \mathbb R.$$ In my functional analysis course, we were given the following exercise:
If $(h_k)_k \subset \mathbb R$ with $h_k \to 0$, then there exists $f_i \in L^\infty_{loc}(\mathbb R^n)$ such that, for all compact set $K \subset \mathbb R^n$, $(D^{-h_k}_i f)_k$ converges $\star$-weakly to $f_i$ in $L^\infty(K)$ (up to a subsequence). Moreover, show that $f_i \in L^\infty(\mathbb R^n)$.
I don't really understand the condition "then there exists $f_i \in L^\infty_{loc}(\mathbb R^n)$". Indeed, from Banach-Alaoglu theorem we have that there is a subsequence of $(D^{-h_k}_i f)_k$ that weakly-$\star$ converges to some $f_i \in L^\infty(\mathbb R^n)$. Why do we have to go through $L^\infty_{loc}(\mathbb R^n)$ and compact sets ?
What is the dual space of $L^\infty(\mathbb{R}^n)$? It's the space of Radon measures that need not be bounded. Hence, using Banach-Alaoglu is not justified because you might be working with an unbounded sequence.
However, if you're working with a compact set $K$, then these measures are in fact bounded and one can use Banach-Alaoglu as you say.