$L^p(\mathbb{R})$ convergence of Fourier truncated integral to $f$ for $f\in L^p(\mathbb{R})$ and $p\in(2,+\infty)$. From the Hilbert transform theory it follows that $$\forall p\in(1,2], \forall f\in L^p(\mathbb{R}), \left\|f-\left(x\mapsto\int_{-M}^M\hat{f}(\xi)e^{2\pi i\xi x}\operatorname{d}\xi \right)\right\|_{L^p(\mathbb{R})}\rightarrow0, M\rightarrow+\infty,\ \ \ \ (*)$$
where $\hat f$ is the Fourier transform of $f$.

What about if $p\in(2,+\infty)$?

Obviously, there's a problem in defining what should be the meaning of $\int_{-M}^M\hat{f}(\xi)e^{2\pi i\xi x}\operatorname{d}\xi$, but using the relation between convolution and Fourier transform valid at least in $L^2(\mathbb{R})$, we can easily see that we can replace it with $\varphi_M*f$ where:
$$\varphi_M(x):= \frac{\sin (2\pi Mx)}{\pi x}.$$
Thanks to Young inequality for convolutions, we have that $\varphi_M*f \in L^\infty(\mathbb{R})$ if $f\in L^p(\mathbb{R})$  with $2<p<\infty$, but I don't see a way to guarantee that $\varphi_M*f \in L^p(\mathbb{R})$ in order to obtain something meaningful in $(*)$.

Is it true in general that $\varphi_M*f\in L^p(\mathbb{R})$? If it is, does $(*)$ hold? If not, is there a way to recover something analogous?

 A: Fix $2<p<\infty$.
It is known from the Hilbert transform theory that the characteristic function of an interval $I$ is a Fourier multiplier of $L^p(\mathbb{R})$, i.e. the operator initially defined on the dense subspace $L^2(\mathbb{R})\cap L^p(\mathbb{R})$ by 
$$T_I(f)=\mathcal{F}^{-1}\left(\chi_I\mathcal{F}(f)\right),$$
where $\mathcal{F}$ is the Fourier transform, extends uniquely by continuity to a continuous operator from $L^p(\mathbb{R})$ into itself.
Now, if $f\in L^p(\mathbb{R})$ choose a sequence of simple functions $(f_n)_{n\in\mathbb{N}}$ that converges in $L^p(\mathbb{R})$ to $f$ and get $M>0$. Then $\varphi_M*f_n\in L^p(\mathbb{R})\cap L^2(\mathbb{R})$ and so:
$$\|\varphi_M*f_n-T_{[-M,M]}(f)\|_p =\|T_{[-M,M]}(f_n)-T_{[-M,M]}(f)\|_p\rightarrow0, n\rightarrow\infty.$$
On the other hand, for Young inequality we have that $$\|\varphi_M*f_n -\varphi_M*f\|_\infty = \|\varphi_M*(f_n-f)\|_\infty\le \|\varphi_M\|_{p'}\|f_n-f\|_p\rightarrow0, n\rightarrow\infty$$
where $p'\in(1,2)$ is such that $\frac{1}{p}+\frac{1}{p'}=1$.
Then, taking if necessary a subsequence, we can guarantee that $\varphi_M*f_n\rightarrow T_{[-M,M]}(f), n\rightarrow\infty$ pointwise a.e., so we get for a.e. $x\in\mathbb{R}$ that: $$(\varphi_M*f)(x)=T_{[-M,M]}(f)(x)$$
and so $\varphi_M*f$ is a.e. equal to a member of $L^p(\mathbb{R})$ and so it is in $L^p(\mathbb{R})$.
This argument also shows that $$\varphi_M* = T_{[-M,M]}.$$
Now, from the fact that the set of Fourier transform of simple functions are dense in $L^p(\mathbb{R})$ and from the fact that if $g$ is the Fourier transform of a simple function then:
$$\|\varphi_M*g-g\|_p\rightarrow0, M\rightarrow\infty$$
we get that:
$$\forall f\in L^p(\mathbb{R}), \|\varphi_M*f-f\|_p\rightarrow0, M\rightarrow\infty$$
is equivalent to:
$$\sup_{M>0}\|\varphi_M*\|_{p\rightarrow p}<\infty.$$
But we have shown that:
$$\varphi_M*=T_{[-M,M]}$$
and from the Hilbert transform theory (see e.g. Javier Duoandikoetxea - Fourier Analysis, chapter 3 on Hilbert transform, paragraph 5 on multipliers, proposition 3.6) we know that:
$$\sup_{M>0}\|T_{[-M,M]}\|_{p\rightarrow p}<\infty$$
and so we have obtained that:
$$\forall f\in L^p(\mathbb{R}), \|\varphi_M*f-f\|_p\rightarrow0, M\rightarrow\infty.$$
