The question
The question is 2.5.10 from Grafakos' Classical Fourier Analysis. It starts by defining a multiplier
$$m_{\text{per}}(\omega)=\sum_{k\in\mathbb Z^d}m(\omega -k),$$
where $m$ is a Fourier multiplier in the $\mathcal M_p$ space. It also holds that $m$ has support in $[0,1]^d$.
It is to be proven that $m_{\text{per}}$ belongs in $\mathcal M_p$ as well.
Work so far
The multiplier $m_{\text{per}}$ is defined by the relation
$$T_{m_{\text{per}}}f(t)=\int\limits_{\mathbb R^d}m_{\text{per}}(\omega)\hat f(\omega)e^{2\pi i \langle\omega\mid t\rangle} d\omega.$$ We now go to the definition of $m_{\text{per}}$ to express the last integral as
$$\int\limits_{\mathbb R^d}\sum_{k\in\mathbb Z^d}m(\omega -k)\hat f(\omega)e^{2\pi i \langle\omega\mid t\rangle} d\omega\underbrace{=}_{(??)}\sum_{k\in\mathbb Z^d}\int\limits_{\mathbb R^d}m(\omega -k)\hat f(\omega)e^{2\pi i \langle\omega\mid t\rangle} d\omega.$$
Here comes my first question: Is it possible to exchange that sum with the integral? I would hazard a guess that it is indeed possible so I continued while assuming it.
I think that a possible explanation would be Lebesgue's DCT or the fact that $m$ is compactly supported. Is that the case, or is it completely different?
Assuming the last equality is true we can make a change of variables $\omega\mapsto\omega+k$, and the differential doesn't change so we get
\begin{align*} \sum_{k\in\mathbb Z^d}\int\limits_{\mathbb R^d}m(\omega)\hat f(\omega +k)e^{2\pi i \langle\omega+k\mid t\rangle} d\omega&=\sum_{k\in\mathbb Z^d}e^{2\pi i \langle k\mid t\rangle}\int\limits_{\mathbb R^d}m(\omega)(e^{-2\pi i \langle\cdot\mid k\rangle}f)^{\land}e^{2\pi i \langle\omega\mid t\rangle} d\omega\\ &=\sum_{k\in\mathbb Z^d}e^{2\pi i \langle k\mid t\rangle}T_m(e^{-2\pi i \langle\cdot\mid k\rangle}f)(t) \end{align*}
where the first equality is true after separating the exponential and also recalling the formula for a translation of the Fourier transform. The second one is the definition of the Fourier multiplier. We now take $L_p$-norm and use Minkowski's inequality as follows:
\begin{align*} \|T_{m_{\text{per}}}f\|_{L^p}&\leq\sum_{k\in\mathbb Z^d}\|e^{2\pi i \langle k\mid \cdot\rangle}T_m(e^{-2\pi i \langle\cdot\mid k\rangle}f)\|_{L^p}\\ &\leq \sum_{k\in\mathbb Z^d}\|T_m(e^{-2\pi i \langle\cdot\mid k\rangle}f)\|_{L^p}\\ &\leq \sum_{k\in\mathbb Z^d}\|T_m\|_{L^p\to L^p}\|e^{-2\pi i \langle\cdot\mid k\rangle}f\|_{L^p}\\ &=\|T_m\|_{L^p\to L^p}\|f\|_{L^p}|\mathbb Z^d|.\ (??) \end{align*}
I feel like I've taken many liberties, but talking about the inequalities, I've used the fact that the exponential function $e^{i\dots}$ has complex modulus 1. That is not a problem, but when we get to the last line, I think that it is wrong. That is because no term depends on $k$ and the sum is infinite and I also haven't used the fact that $m$ is compactly supported. At least not that I have noticed.
So, in concrete, my second question would be: Is my idea for this part correct? Except for the part where I have $\mathbb Z^d$'s measure. From here, how can I finish?
The only thing I need to prove is that $m_{\text{per}}$'s $\mathcal M_p$-norm is finite and that norm is the $L^p$ operator norm of $T_{m_{\text{per}}}$, so my strategy has been to find a bound with the $\mathcal M_p$ norm of $m$. I haven't been able to finish this up, so any hint, suggestion or answer is most welcome.