We consider the Fourier multiplier operator $T_0$ defined by the explicit expression $$(T_0f)(x)=\int_{\mathbb{R}^n}{e^{ix\cdot \xi}m(\xi)\hat{f}(\xi)d\xi}, \ f\in S(\mathbb{R}^n),$$ where $S(\mathbb{R}^n)$ is the Schwartz function space. Here we assume that the multiplier $m(\xi)\in L^\infty(\mathbb{R}^n)$ satisfies the conditions in the H\"ormander's multiplier theorem, which implies that $T_0$ can be extended to a bounded operator $T$ from $L^p(\mathbb{R}^n)$ to $L^p(\mathbb{R}^n)$, $1<p<\infty$. Then it is natural to ask the following question. Do we have \begin{equation}(1)\quad\quad\quad (Tf)(x)=\int_{\mathbb{R}^n}{e^{ix\cdot \xi}m(\xi)\hat{f}(\xi)d\xi},\ a.e., \ f\in L^p(\mathbb{R}^n)\cap L^1(\mathbb{R}^n),\end{equation} whenever $m(\xi)\hat{f}(\xi)\in L^1(\mathbb{R}^n)$?

$\begingroup$ Can you be more specific about how the words "makes sense" are defined? $\endgroup$ – fedja Sep 16 '16 at 6:17

$\begingroup$ @fedja It means the integral is finite a.e.. $\endgroup$ – Mr.right Sep 16 '16 at 13:36

$\begingroup$ Just for others reading: @fedja's comment was in response to an earlier version of this qyestion $\endgroup$ – Yemon Choi Sep 16 '16 at 15:28

1$\begingroup$ @ChristianRemling We define $Tf$ by the $L^p$limit of $T_0f_n$, but we don't know whether this limit equals to the right hand side of (1), which is just what we want to prove. $\endgroup$ – Mr.right Sep 17 '16 at 4:02

1$\begingroup$ @Mr.right: Why don't you try to flesh out the sketch I provided, I think that'll answer all your questions. (As for your last concern, if the RHS has a pointwise limit [as I showed it has], then this is the $L^p$ limit that we also know exists because we can pass to an a.e. convergent subsequence.) $\endgroup$ – Christian Remling Sep 18 '16 at 21:32
We construct a sequence $f_k\in S(\mathbb{R}^n)$ s.t. $f_kf_p\to 0$ and $$T_0f_k\int_{\mathbb{R}^n}e^{ix\cdot\xi}m(\xi)\hat{f}(\xi)d\xi\to 0,\ a.e..$$ By the bounded extension of the multiplier operator, there is a function $g\in L^p$ s.t. $T_0f_k g_p\to 0$. Then there is a subsequence $T_0f_{k_j}g\to 0,\ a.e.$. Hence, $g=\int_{\mathbb{R}^n}e^{ix\cdot\xi}m(\xi)\hat{f}(\xi)d\xi$, which gives the desired equality.
The construction of the sequence: Let $\phi\in C_0^\infty$ and $\varphi=\hat{\phi}$. $\phi(0)=1$. $\varphi_\epsilon(x)=\epsilon^{n}\varphi(x/\epsilon)$, $\phi_\delta=\phi(\delta x)$. Let $f_{\epsilon,\delta}=(\varphi_\epsilon\ast f)\phi_\delta$. Since $$f_{\epsilon,\delta}f_p\le(\varphi_\epsilon\ast f)\phi_\delta\varphi_\epsilon\ast f_p+\varphi_\epsilon\ast ff_p, $$ we can choose some $\epsilon_k$ and $\delta_k$ to make $f_{\epsilon_k,\delta_k}f_p$ smaller than $1/k$.
We also need to show that $$\int e^{ix\cdot \xi}m(\xi)(\hat{f_{\epsilon,\delta}}\hat{f})d\xi$$ can be small.
We see that $$\intm\hat{f_{\epsilon,\delta}}\hat{f}\le \intm(1\hat{\varphi_\epsilon})\hat{f}+\intmh_\epsilonh_\epsilon\ast\hat{\phi_\delta}=I_1+I_2,$$ where $h_\epsilon=\hat{\varphi_\epsilon}\hat{f}$. Since $$m(1\hat{\varphi_\epsilon})\hat{f}\le m\hat{f}\in L^1,$$by dominated convergence, we can choose a subsequence $\epsilon_{k_j}$ s.t. $I_1\le 1/k$.
Fix $\epsilon=\epsilon_{k_j}$. We claim that $$h_{\epsilon}\ast\hat{\phi_\delta})(\xi)=\int h_\epsilon(\xi\delta y)\hat{\phi}(y)dy\le C_{\epsilon,N}(1+\xi)^{N},$$ where $C_{\epsilon,N}$ is independent of $\delta$. To see this, we need to use the facts that $h_\epsilon_\infty\le f_1$ and ${\rm supp}\ h_\epsilon\subset B(0,\epsilon^{1})$. Indeed, when $\xi\le 10/\epsilon$, $$\int h_\epsilon(\xi\delta y)\hat{\phi}(y)dy\le C\int\hat{\phi}(y)dy\le C,$$ and when $\xi\ge 10/\epsilon$, $$\int h_\epsilon(\xi\delta y)\hat{\phi}(y)dy\le C\int_{\xi\delta y\le \epsilon^{1}}\hat{\phi}(y)dy\le C_N\int_{y\ge (\xi\epsilon^{1})/\delta}y^{N}dy \le C_N\xi^{nN}.$$ This proves the claim.
So $$mh_\epsilonh_\epsilon\ast\hat{\phi_\delta}\le m\hat{f}+C_{\epsilon,N}m(1+\xi)^{N}\in L^1.$$ Since $h_\epsilon$ is bounded and continuous and $\epsilon=\epsilon_{k_j}$ is fixed, we have $${\rm limsup}_{\delta\to 0}h_\epsilonh_\epsilon\ast\phi_\delta(\xi)\le {\rm limsup}_{\delta \to 0}\int h_\epsilon(\xi\delta y)h_\epsilon(\xi)\hat{\phi}(y)dy=0,$$ then by dominated convergence, we can choose a subsequence $\delta_{k_{j_l}}$ s.t.$I_2\le 1/k$. Now, the sequence $f_{\epsilon_{k_{j}}, \delta_{k_{j_l}}}$ is what we need.
Any comments are welcome:)

$\begingroup$ I haven't checked all the small details, but overall this looks good I think. $\endgroup$ – Christian Remling Sep 19 '16 at 16:57