Let $f_\xi = f\ast h_\xi$ with $\widehat{h_\xi} = 1_{[-\xi,\xi]}$($h_\xi(t) = \frac{\sin \xi t}{\pi t}$). Suppose that $f$ has a bounded variation $||f||_V < +\infty$ and that it is continuous in a neighborhood of $t_0$. Prove that in a neighborhood of $t_0$, $f_\xi(t)$ converges uniformly to $f(t)$ when $\xi$ goes to $+\infty$.
Here is my attempt: \begin{align} |f_\xi(t) - f(t)| \leq |\int_{|y|\leq \delta}(f(t-y)-f(t)) \frac{\sin \xi y}{\pi y}dy| + |\int_{|y|> \delta}(f(t-y)-f(t)) \frac{\sin \xi y}{\pi y}dy| \end{align} The first part can be controlled using the continuity near $t_0$. Here the problem is that the function $\frac{\sin \xi y}{\pi y}$ is not absolutely integrable on $\mathbb{R}$. So how can I control the second part?