Estimate for $a_m$ in the Fourier multiplier $\sum_{m\in \mathbb{Z}^n}a_me^{-2\pi i x\cdot m}$

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I'm dealing with some type of transference problem about $L^p$ multiplier. It is stated as follows:

Let $1<p<\infty$ and $0<A<\infty$. Prove that the following are equivalent:

(a) The operator $f \mapsto \sum_{m \in \mathbf{Z}^n} a_m f(x-m)$ is bounded on $L^p\left(\mathbf{R}^n\right)$ with norm $A$.

(b) The $\mathscr{M}_p$ norm of the function $\sum_{m \in \mathbf{Z}^n} a_m e^{-2 \pi i m \cdot x}$ is exactly $A$.

(c) The operator given by convolution with the sequence $\left\{a_m\right\}$ is bounded on $\ell^p\left(\mathbf{Z}^n\right)$ with norm $A$.

I can deduce that $a\Longleftrightarrow c$. And I deduced that $a\Longrightarrow b$. The final problem for me is whether $b\Longrightarrow c$ or $b\Longrightarrow a$.

When I deduce $b$ from $a$, by constructing a Schwartz funcion $f$ whose Fourier transform $\hat{f}$ is supportted in $(-\frac 12,\frac 12)^n$, we can easily see that $$ |a_m|<\infty. $$ And we can then use the same way as proving the Poisson summation formula to change the integral and the summation.

But I cannot find any estimate for $a_m$ when I deduce $a$ from $b$. The only trivial estimate is $$ |\Sigma_m a_me^{-2\pi im\cdot x}|<\infty. $$ And this is far away from $|a_m|<\infty$. So I cannot use the same way as proving Poisson summation formula.

I'm wondering are there any more concrete estimates for $a_m$?
Or can we deduce $c$ from $b$? (I tried this, but it seems that we also need more information about $a_m$)

Do you have any advices? Everything helps will be appreciate!