Uniform convergence for bounded variational function

Question

Let $f_{\xi} = f*h_{\xi}$, where $h_{\xi}(t) = \frac{1}{\pi}\frac{sin(\xi x)}{x}$. Suppose $f$ 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 $\xi \rightarrow +\infty$.

Here $||f||_V = lim_{h\rightarrow 0} \int_{\mathbb{R}} \frac{|f(t)-f(t-h)|}{|h|}dt$.

I tried to write $f(t)-f_{\xi}(t) = \frac{1}{\pi}\int_{\mathbb{R}} (f(t)-f(t-y))\frac{sin(\xi y)}{y}dy$, and then divide the integral to two parts $\{|y|\leq \delta\}$ and $\{|y|>\delta\}$. I know how to control the first part of the integral but I have some trouble dealing with the second part because I don't know how to use the condition that f is bounded variation.

Can anyone give some ideas? Thanks in advance!

Answer 1

Hint: This is just the analog for the Fourier transform of a standard fact about Fourier series. (The function $h_\xi$ looks a lot like the Dirichlet kernel, right?)

Standard Fact. If $f$ is continuous, periodic and has bounded variation then the Fourier series for $f$ converges uniformly to $f$.

You can find a proof of that in many places; try to adapt the proof to the present context.