Understanding Fourier transform of derivative of distribution $\mathcal S$ is the Schwartz space over $\mathbb R^n$ and $\mathcal S'$ is its continuous dual, that is the space of tempered distributions. The Fourier transform $\mathscr F$ on $\mathcal S'$ is defined by 
\begin{align}
\mathscr F\colon \mathcal S' &\to \mathcal S' \\
f &\mapsto \bigl(g \mapsto f(\mathcal Fg)\bigr),
\end{align}
where $\mathcal F$ is the Fourier transform for the integrable functions from $\mathbb R^n$ to $\mathbb C$. In my textbook, it is stated that, for any partial derivative $\partial_j$ and any distribution $u \in \mathcal S'$, 
$$ \mathscr F (\partial_j u) = i\xi_j \mathscr F(u). $$
What I don't understand is the meaning of that  $\xi_j$. The left-hand side is a function from $\mathcal S$ to $\mathbb R$ and so must be the right-hand side. Since $\mathscr F(u) \in \mathcal S'$, my guess would be that $\xi_j$ is in $\mathcal S'$, but I don't see what it could be. 
My question is, if $f\in \mathcal S'$ and $g\in \mathcal S$ and $\mathscr F$ and $\partial_j$ are as above, what is $\mathscr F (\partial_j f)(g)$?
 A: The Fourier transform of a tempered distribution is defined by duality, if $u \in \mathcal{S}'(\mathbb{R}^n)$ we put
\begin{align*} 
\displaystyle \langle \varphi, \widehat{u} \rangle:= \langle \widehat{\varphi}, u \rangle
\end{align*}
Now if $\varphi \in \mathcal{S}(\mathbb{R}^n)$ then $D^{\alpha}_x \widehat{\varphi}(x)=(-1)^{|\alpha|} \widehat{\xi^\alpha \varphi}$, and with notation $D_x^{\alpha}= 1/i D_x^{\alpha}$
\begin{align*}
\displaystyle \langle \varphi, \widehat{D^{\alpha}_x u} \rangle &= \langle \widehat{\varphi} , D^{\alpha}_x u \rangle \\
&= (-1)^{|\alpha|} \langle D^{\alpha}_x \widehat{\varphi}, u \rangle \\ 
&= (-1)^{|\alpha|+|\alpha|} \langle \widehat{\xi^\alpha \varphi} , u \rangle \\
&= \langle \xi^\alpha \varphi , \widehat{u} \rangle \\
&= \langle \varphi , \xi^\alpha \widehat{u} \rangle
\end{align*}
therefore
$$\widehat{D^\alpha u}=\xi^{\alpha} \widehat{u}$$
This is the general case. Without notation $D_x^{\alpha}= 1/i D_x^{\alpha}$, we have $D^{\alpha}_x \widehat{\varphi}(x)=(-i)^{|\alpha|} \widehat{\xi^\alpha \varphi}$. If $|\alpha|=1$
$$\displaystyle D_x \widehat{\varphi}(x) = D_x \int_{\mathbb{R}^n} e^{- i(x,\xi)} \varphi(\xi) d\xi = \int_{\mathbb{R}^n} -i\xi e^{-i(x,\xi)} \varphi(\xi) d\xi = -i\widehat{\xi \varphi}$$
