Let $f\in L^2(\Bbb{R})$, and for each $0<r<\infty$, define $\phi_r:\Bbb{R}\to\Bbb{R}$ by \begin{align} \phi_r(y):= \begin{cases} \frac{1}{y} &\text{if $r<|y|<\infty$}\\ 0& \text{otherwise} \end{cases} \end{align} In this case, we have $\phi_r\in L^2\setminus L^1$. My question is whether it is even true that $f*\phi_r\in L^2$, and if it is, whether it is true that the Fourier transform satisfies $\widehat{f*\phi_r}= \widehat{f}\cdot \widehat{\phi_r}$.
Of course, if $g\in L^1$, then Young's inequality tells us $\lVert f*g\rVert_2\leq \lVert f\rVert_2\cdot \lVert g\rVert_1$, and we also have $\widehat{f*g}=\hat{f}\cdot \hat{g}$. But I'm struggling to determine what happens under the weaker hypotheses above where $\phi_r\in L^2\setminus L^1$.
Context: this question arose as part of a larger calculation related to the Hilbert transform (namely that if we define for $f\in L^2$, $Hf:= \left[\text{sgn}(\cdot)\cdot \hat{f}\right]^{\vee}$, then I'm trying to show that $\lVert Hf-\frac{-1}{i\pi}f*\phi_r\rVert_2\to 0$ as $r\to 0^+$; my main idea is to use Parseval's identity and then explicitly calculate $\widehat{\phi_r}$).