The spectrum of the algebra of Fourier multipliers Let $\mathcal{M}_p(\mathbb{R}^n)$ denote the space of Fourier multipliers on $L^p(\mathbb{R}^n)$, i.e. the set of tempered distributions $m$ such that the operator $T_m:f\mapsto \mathcal{F}^{-1}(m \hat{f})$ extends to a bounded operator on $L^p$. It is well known that, once we equip $\mathcal{M}_p$ with the norm $\|m\|=\|T_m\|_{op}$, the space is a Banach algebra.
My question is: what is known on the spectrum of this Banach algebra? If $p=2$, $\mathcal{M}_2=L^\infty(\mathbb{R}^n)$ and so the specctrum can be determined; in general, however, it's nontrivial to determine whether $m\in \mathcal{M}_p$, so I guess the characterization of the spectrum to prove harder.
Any reference, suggestion or information is welcome
 A: This post provides a decomposition of the spectrum $\sigma(\mathcal{M}_p)$ of the commutative Banach algebra $\mathcal{M}_p$, which is nowhere near to a complete description of $\sigma(\mathcal{M}_p)$.
Let's write $A=L^1(\mathbb{R}^n)$ and $E=L^p(\mathbb{R}^n)$. By Young's inequality
$$\|f*g\|_p\leq \|f\|_1\|g\|_p $$
for all $f\in A$ and $g\in E$. Thus, $E$ is a commutative Banach $A$-module. Since $A$ has a bounded approximate identity, then $A*E = E$ by Cohen's factorization theorem.
Let $\sigma(A)=\{h_x:x\in\mathbb{R}^n\}$ denote the spectrum of the commutative Banach algebra $A$, where $h_x(f) = \widehat{f}(x)$ for each $x\in\mathbb{R}^n$ and $f\in A$.
Let $B(E)$ denote the space of all bounded linear operators on $E$. For each $f\in A$, let $L_f\in B(E)$ be the convolution operator defined by $L_fg = f*g$. Clearly $L_f$ is a bounded $A$-module homomorphism. Since $A$ has a bounded approximate identity, then $\|L_f\|_{op}=\|f\|_1$. Thus, we identify $A$ with $L_A$.
$T_m:E\to E$ is a Fourier multiplier iff it commutes with convolution. Equivalently, $T_m\in\mathcal{M}_p$ iff $T_m\in B(E)$ and it is a bounded $A$-module homomorphism. As expressed in the question, $\mathcal{M}_p$ is a commutative closed subalgebra of $B(E)$. By the previous paragraph, $A$ is a closed subalgebra of $\mathcal{M}_p$.
Every $h\in\sigma(A)$ has an extension to a $\tilde{h}\in\sigma(\mathcal{M}_p)$. This is manifested explicitly by $\tilde{h}_x(T_m) = m(x)$. This extension is, in fact, is unique. The map $\sigma(A)\to \sigma(\mathcal{M}_p)$ defined by $h_x\to\tilde{h}_x$ is a homeomorphic embedding.
Let $\textrm{hull}(A) = \{h\in\sigma(\mathcal{M}_p): h(A)=\{0\}\}$. Then, $$\sigma(\mathcal{M}_p) = \sigma(A) \cup \textrm{hull}(A)$$ as a disjoint union. It is not difficult to show that $\textrm{hull}(A)$ is a closed subset of $\sigma(\mathcal{M}_p)$. Since $\mathcal{M}_p$ is unital, then $\sigma(\mathcal{M}_p)$ is compact. Consequently, $\textrm{hull}(A)$ is a compact (and so) $\sigma(A)$ is an open subset of $\sigma(\mathcal{M}_p)$.
It is highly nontrivial to give a complete description of $\textrm{hull}(A)$. As a final note, let's write down a property of this subset: A subset $F\subseteq \sigma(\mathcal{M}_p)$ is called a set of uniqueness if $$\big(h(T_m) = 0\hspace{4mm}\forall h\notin F \big)\Rightarrow T_m = 0 $$ for every $T_m\in\mathcal{M}_p$. Clearly, $\textrm{hull}(A)$ is a set of uniqueness for $\mathcal{M}_p$.
