prove that $\mathcal{F} (\partial_{x_j}f)(\xi) = i \xi_j \mathcal{F} f(\xi) $ Let $f \in L^{1}(\mathbb{R}^n) \cap C^{1}(\mathbb{R}^n)$, and $\partial_{x_j}f \in L^{1}(\mathbb{R}^n)$. Then, prove that $\mathcal{F} (\partial_{x_j}f)(\xi) = i \xi_j \mathcal{F} f(\xi) $ where $\mathcal{F}f$ is the Fourier transform of $f$.
My attempt: $\mathcal{F} (\partial_{x_j}f)(\xi) = \int_{\mathbb{R}^n} e^{-i x \cdot \xi} (\partial_{x_j}f) dx$. I don't know what to do with this partial derivative.
 A: I had rewritten completely the answer for a cleanest and shorter one. For fixed $\xi\in \mathbb{R}^n$ set
$$
G(y):=\int_{\mathbb{R}^n}e^{i\langle y-x,\xi \rangle}f(x)\mathop{}\!d x=\int_{\mathbb{R}^n}e^{i\langle x,\xi \rangle}f(y-x)\mathop{}\!d x
$$
and note that the second identity follow by the substitution rule. Now as the integrand of both integrals belongs to $L^1 \cap C^1(\mathbb{R}^n)$ then by the measure-theoretic version of the Leibniz integral rule we can differentiate $G$ and we get
$$
\partial _jG(y)=\int_{\mathbb{R}^n}\partial _je^{i\langle y-x,\xi \rangle}f(x)\mathop{}\!d x=\int_{\mathbb{R}^n}i\xi_je^{i\langle y-x,\xi \rangle}f(x)\mathop{}\!d x
$$
and
$$
\partial _jG(y)=\int_{\mathbb{R}^n}e^{i\langle x,\xi \rangle}\partial _jf(y-x)\mathop{}\!d x=\int_{\mathbb{R}^n}e^{i\langle y-x,\xi \rangle}\partial _jf(x)\mathop{}\!d x
$$
where $\partial_j f(v)$ is the partial derivative of $f$ at $v$ in direction $e_j$ and the last identity follows again from the substitution rule. Then setting $y=0$ we find the desired identity.∎
