Why is the space of $L^p$ (Fourier) multipliers a Banach space? (Stein) I am reading Stein's book 'Singular Integrals and Differentiability Properties of Functions', and he introduces the class $\mathcal{M}_p$ of $L^{\infty}$ multipliers $m$ such that the operator $T_m$ defined for $f\in{L^2}{\cap}{L^p}$ by $\widehat{(T_mf)}(x)=m(x)\hat{f}(x)$ is bounded on $L^p$, endowing the class with the operator norm of $T_m$.
He then states that this class is a Banach algebra, but I can't prove it. If $(m_n)_n$ is a sequence in $\mathcal{M}_p$, we can deduce that $T_{m_n}f\rightarrow{Tf}$ for some $T\in{\mathcal{B}}(L^p,L^p)$ as ${\mathcal{B}}(L^p,L^p)$ is a Banach space. But how can we show that $T$ corresponds to a Fourier multiplier? I think it would be enough to show that $T_{m_n}f\rightarrow{Tf}$ in $L^2$ also, to try and show that $(m_n)_n$ is also Cauchy in $L^{\infty}$ and so has a limit there, but I can't see how to prove this.
Any ideas greatly appreciated!
 A: Here's an argument that at least works for $1 < p < \infty$. We take for granted at the outset that $\mathcal{M}_2 = L^{\infty}$ and so is complete.
Step 1
First off, by a duality argument we can show that if $m \in \mathcal{M}_p$, then also $m \in \mathcal{M}_{p'}$, and that the norms are identical. (So these spaces are isomorphic.)
To see this, let $f$ and $g$ be Schwartz functions. Then using Plancherel's theorem, we obtain
$$
\int (T_m f) \overline{g} \, dx
= \int m \hat{f} \overline{\hat{g}} \, d\xi
= \int (T_m \tilde{g}) \overline{\tilde{f}} \, dx,
$$
where $\tilde{f}(x) = \overline{f(-x)}$. So
$$
| (T_m f, g) |
= | (T_m \tilde{g}, \tilde{f}) |
\leq \| T_m \tilde{g} \|_p \| \tilde{f} \|_{p'}
\leq \| m \|_{\mathcal{M}_p} \| g \|_p \| f \|_{p'}.
$$
This shows that $T_m$ is bounded on $L^{p'}$, with an operator norm not more than the operator norm on $L^p$. Inverting the roles of $p$ and $p'$ gives the equality of norms.
Step 2
Based on the above, we can extend $T_m$ to an operator on $L^p + L^{p'}$. Complex interpolation then gives us that $\log \| T_m \|_{L^q \to L^q}$ is a convex function of $1/q$, for $q$ between $p$ and $p'$. In particular, the equality of operator norms at the endpoints implies that $\| T_m \|_{L^q \to L^q} \leq \| m \|_{\mathcal{M}_p}$ for $q$ in this range.
Step 3
With these preliminaries set, suppose that $(m_n)_n$ is Cauchy in $\mathcal{M}_p \subseteq \mathcal{B}(L^p, L^p)$, with $p \leq 2$. By the norm comparison property in Step 2, it follows that the sequence is also Cauchy in $\mathcal{M}_q$ for all $q$ between $p$ and $p'$, and in particular for $\mathcal{M}_2$. So for each $q$ we can extract a limiting operator $T^{(q)} \in \mathcal{B}(L^q, L^q)$. Since $\mathcal{M}_2 = L^{\infty}$ is complete, we know that $T^{(2)}$ corresponds to a multiplier $m$.
Step 4
We now have a candidate multiplier, $m$, and it just remains to show that $T^{(p)}$ corresponds to $T_m$. For this, fix a (Schwartz, or $L^p \cap L^2$) function $f$. Then the sequence $(T_{m_n} f)_n$ converges in $L^p$ to $T^{(p)} f$. The corresponding convergence in $L^2$ to $T^{(2)} f = T_m f$ implies, via Hölder's inequaltiy, that if we restrict the $T_{m_n} f$ to a set $E$ of finite measure, then $T_{m_n} f |_E \to T_m f |_E$ in $L^p$. Piecing together the finite-measure sets, we see that the overall $L^p$ limit $T^{(p)} f$ must be identical to $T_m f$.
Step 5
Steps 3 and 4 show that $\mathcal{M}_p$ is complete for $1 < p < 2$. By the isomorphism indicated in Step 1, we also obtain completeness for $2 < p < \infty$.
