I'd like help with the following problem:
Let $f\in L^p(\Bbb R)$ for $1 < p\le 2$. Show that $f(t)\ast \frac{\sin \pi t}{\pi t}$ is defined, and $f(t)\ast \frac{\sin \pi t}{\pi t} \in L^p(\Bbb R)$.
My work: I started by observing that $g(t):= \frac{\sin \pi t}{\pi t} = \widehat{\mathbf{1}_{[-\frac12,\frac12]}}(t)$. Hence, $\widehat g(t) = \widehat{\widehat{\mathbf{1}_{[-\frac12,\frac12]}}}(t) = \mathbf{1}_{[-\frac12,\frac12]}(-t) = \mathbf{1}_{[-\frac12,\frac12]}(t)$. We have $$\begin{align*} (f\ast g)(x) &= \int_{\Bbb R} \widehat{\mathbf{1}_{[-\frac12,\frac12]}}(t) f(x-t)\, dt\\ &= \int_{\Bbb R}\int_{-1/2}^{1/2} e^{-2\pi i ut} f(x-t)\, du dt \\ &= \int_{-1/2}^{1/2}\int_{\Bbb R} e^{-2\pi i ut} f(x-t)\, dt du \\ &=\int_{-1/2}^{1/2} (h_u \ast f)(x)\, du \end{align*} $$ where $h_u(t) := e^{-2\pi i ut}$ for $t\in \Bbb R$. I don't know where to go from here. Also, I have used Fubini's theorem to obtain the third equality - but I'm not sure if it is applicable here.
Thank you for your time!
Note: We use the definition $$\widehat{f}(\omega) := \int_{\Bbb R} f(t)\, e^{-2\pi i \omega t}\, dt.$$
I ignore if there is an elementary proof that involves only basic properties of the Fourier transform (extended to $L_p$ with $1<p<2$ and applications of Fubini-Tonelli's).
Using some classic results from singular operators (see Stein, E., Singular integrals, Princeton University Press, 1970, Chapter 2; Duoandikoetxea, J. Fourier Analysis, AMS 2001, chapter 3, for example) the operator $S:f\mapsto f *\operatorname{sinc}$ can be expressed in terms of the Hilbert operator in L$_2$, and then use properties of the Hilbert operator to show that $S$ defines a bounded operator from $L_p$ to itself (for $1<p<\infty$). Here is a sketch of how this works:
Denote by $\mathcal{F}$ the Fourier transform operator. Consider the operator $S$ on $L_2(\mathbb{R})$ given by $$ Sf(x)=f*\operatorname{sinc}(y)=\int_{\mathbb{R}}f(x-y)\frac{\sin \pi x}{\pi x}\,dx$$
Since $\mathcal{F}\mathbb{1}_{[-1/2,1/2]}(\xi)=\frac{\sin pi\xi}{\pi\xi}=\operatorname{sinc}(\pi\xi)$
$$\mathcal{F}(Sf)(\xi)=\mathbf{1}_{[-1/2,1/2]}(\xi)\mathcal{F}f(\xi)$$ where $\mathcal{F}f$ is the Fourier transform of $f$.
The operator $S$ can be expressed in terms of the Hilbert transform $H$:
$$Hf(x)=\frac{1}{\pi}\lim_{\varepsilon\rightarrow0+}\int_{|y|>\varepsilon}\frac{f(x-y)}{y}\,dx$$ (convergence exists in $L_2$; some heavy lifting is required to properly defined this operator). In terms of multipliers, $H$ is the operator on $L_2$ such that $$\mathcal{F}Hf(\xi)=-i\operatorname{sign}(\xi)\mathcal{F}f(\xi)$$
$$ Sf=\frac{i}{2}\big(M_{-1/2}HM_{1/2}-M_{1/2}HM_{-1/2}\big)f$$ where and
$$M_bf(x)=e^{2\pi ibx}f(x)$$ for every $b\in\mathbb{R}$. To check this, notice that \begin{align} \mathcal{F}(M_aHM_-a - M_bHM_{-b})f&=(\mathcal{F}HM_{-a}f)(\xi-a)-(\mathcal{F}HM_{-a}f)(\xi-b)\\ &=-i\big(\operatorname{sign}(\xi-a)\mathcal{F}H_{-a}f(\xi-a)-\operatorname{sign}(\xi-b)\mathcal{F}H_{-b}f(\xi-b)\big)\\ &=-i\big(\operatorname{sign}(\xi-a)\mathcal{F}f(\xi)-\operatorname{sign}(\xi-b)\mathcal{F}f(\xi)\big)\\ &=-2i\mathbf{1}_{[a,b]}(\xi)\mathcal{F}f(\xi) \end{align}
The Hilbert transform is known to have a $(p, p)$-strong extension to $L_p$ for all $1<p<\infty$; that means that $H$ can be defined on $L_p(\mathbb{R})$, as a bounded operator for $1<p<\infty$: there is constant $C_p>0$ depending on $p$ only, such that $$\|Hf\|_p\leq C_p\|f\|_p,\qquad f\in L_p$$ The operators $M_b$ are unitary in any $L_p$ and so $$\|Sf\|_p\leq C_p\|f\|,\qquad f\in L_p$$