For $\epsilon > 0$ and $f \in L^p, p \geq 1$ then it is not difficult to show that the function $$ H_\epsilon f(x) = \frac{1}{\pi} \int_{|y| > \epsilon} \frac{f(x-y)}{y}dy $$ is well defined. In fact, much like the Hilbert transform, $H_\epsilon$ is weak $(1,1)$ and strong $(p,p)$ for $1< p < \infty$.
What I'm working to understand is why if $f \in L^p, 1 \leq p < \infty$ then the sequence $\{H_\epsilon f\}$ converges to $Hf$ in $L^p$. For now let's assume $p > 1$ (the $p=1$ case boils down to replacing norm convergence with convergence in measure in the following statements). Letting $\{f_n\}$ be a sequence of Schwartz functions converging to $f$ in $L^p$ then I have $$ \|Hf - H_\epsilon f\|_p \leq \|Hf - Hf_n\|_p + \|Hf_n - H_\epsilon f_n\|_p + \|H_\epsilon f_n - H_\epsilon f\|_p . $$
Given the strong $(p,p)$ inequalities of $H_\epsilon$ and $H$, then I am good with the first and third terms on the right-hand side of the inequality. My struggle is with the middle term: I can't seem to understand why $H_\epsilon f_n$ would converge to $Hf_n$ in $L^p$. I tried using the Minkowski inequality for integrals on said term, but to no avail.
Thank you for the help!
For $f$ a Schwartz function, we have that $$ (H_\epsilon f-Hf)(x)=p.v. \int_{-\epsilon}^\epsilon \frac{f(x-y)}{y}dy=\int_{-\epsilon}^\epsilon \frac{f(x-y)-f(x)}{y}dy $$ where we consider the integrand of the latter function to be continued to be a smooth function at $y=0$. We can then apply the mean value theorem to get pointwise the integrand is bounded by $\max_{y\in (x-\epsilon,x+\epsilon)}|f'(y)|$, so $$ |(H_\epsilon f-Hf)(x)|\leq 2\epsilon \max_{y\in (x-\epsilon,x+\epsilon)}|f'(y)|. $$ If we let $g(x)=\max_{y\in (x-1,x+1)}|f'(y)|$, then $g(x)$ will have rapid decay and for all $\epsilon<1$, $$ |(H_\epsilon f-Hf)(x)|\leq 2 \epsilon \cdot g(x)\implies ||H_\epsilon f-Hf||_p\leq 2\epsilon ||g||_p. $$ Taking $\epsilon$ to zero we are done.