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 $|m(\xi)\hat{f}(\xi)|\in L^1(\mathbb{R}^n)$?



Does the bounded extension of the Fourier multiplier operator agrees with its original explicit definition?
We construct a sequence $f\_k\in S(\mathbb{R}^n)$ s.t. $||f\_k-f||\_p\to 0$ 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
$$||f\_{\epsilon,\delta}-f||\_p\le||(\varphi\_\epsilon\ast f)\phi\_\delta-\varphi\_\epsilon\ast f||\_p+||\varphi\_\epsilon\ast f-f||\_p, $$
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 $$\int|m||\hat{f\_{\epsilon,\delta}}-\hat{f}|\le \int|m||(1-\hat{\varphi\_\epsilon})\hat{f}|+\int|m||h\_\epsilon-h\_\epsilon\ast\hat{\phi\_\delta}|=I\_1+I\_2,$$ 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 $|\xi|\ge 10/\epsilon$, $$\int |h\_\epsilon(\xi-\delta y)\hat{\phi}(y)|dy\le C\int\_{|\xi-\delta y|\le \epsilon^{-1}}|\hat{\phi}(y)|dy\le C\_N\int\_{|y|\ge (|\xi|-\epsilon^{-1})/\delta}|y|^{-N}dy \le C\_N|\xi|^{n-N}.$$
This proves the claim.


So $$|m||h\_\epsilon-h\_\epsilon\ast\hat{\phi\_\delta}|\le |m\hat{f}|+C\_{\epsilon,N}|m|(1+|\xi|)^{-N}\in L^1.$$
Since $h\_\epsilon$ is bounded and continuous and $\epsilon=\epsilon\_{k\_j}$ is fixed, we have
 $${\rm limsup}\_{\delta\to 0}|h\_\epsilon-h\_\epsilon\ast\phi\_\delta|(\xi)\le {\rm limsup}\_{\delta \to 0}\int |h\_\epsilon(\xi-\delta y)-h\_\epsilon(\xi)||\hat{\phi}(y)|dy=0,$$
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:)


