For $m\in L^\infty$, define the Fourier multiplier operator $T_m$ by $T_m f=\mathscr F^{-1}[m \widehat f\ ]$ for $f\in\mathscr S$, where $\mathscr S$ is Schwarz class. ($\widehat f$ is Fourier transform of $f$, $\mathscr F^{-1}$ is Fourier inverse transform.)
I don't understand this statement.
Let $1<p<\infty$, suppose there is $C>0$ s.t. $\|T_m f\|_p\leqq C\|f\|_p$ holds for all $f\in\mathscr S.$ Then, since $\mathscr S$ is dense in $L^p$, $T_m$ has the bounded extension on $L^p.$
Maybe, this says that for $f\in L^p$, there is $\{f_j\}_{j}\subset\mathscr S$ s.t. $\|f-f_j\|_p\to 0$, and we have $\|T_m f_j\|_p\leqq C\|f_j\|_p$ and taking the limit we get $\| T_m f\|_p\leqq C\|f\|_p\ \cdots(\ast)$, thus $T_m$ is bounded on $L^p.$
But I think there is a problem : Can $T_m f$ in $(\ast)$ be defined ? When $f$ is an element of $L^p$, $T_m f(=\mathscr F^{-1}[m \widehat f \ ])$ is not always defined, so I wonder how I should regard $T_m f$ for $f\in L^p$.
One idea is regarding $f\in L^p$ as an element of $\mathscr S'$. Then $T_m f$ is defined (as tempered distribution), but then $T_m f$ is also an element of $\mathscr S'$ so I don't know what $\|T_m f\|_p$ is.(p-norm of tempered distribution?)
I think the statement should be: "[...], $T_m$ has a unique bounded extension on $L^p$."
The statement is an application of the BLT theorem.
To be precise, applied to this case, it says that there is a unique continuous linear operator $\tilde{T} : L^p \to L^p$ with $\tilde{T}|_\mathscr{S} = T_m$.