Convergence in $L^∞(\mathbb{R})$

Question

Let $(x_n)\to x$ in $\mathbb{R}$. Is it true that $e_{x_n}\to e_x$ in $L^∞(\mathbb{R})$, where $e_x:\mathbb{R}\to \mathbb{C}$ defined by $e_x(y)=e^{ixy}$?

What I can show is that $e_{x_n}\to e_x$ pointwise.

Answer 1

No (unless $x_n$ eventually is constant with value $x$). For any distinct real numbers $x$ and $t$, there exists $y\in\Bbb R$ such that $e^{ixy} = e^{-ity}$ (take $y=\pi/(x-t)$ for example). Therefore $|e_x-e_t|_\infty = 2$ for all $x\ne t$, and in particular no such sequence can converge in $L^\infty(\Bbb R)$.