I am trying to understand this particular step in a solution, $f \in C_c^2(\mathbb{R}^n)$: $||\frac{\partial f}{\partial_j\partial_k}||_2^2 = C|| |\xi_j||\xi_k||\mathcal{F}f||_2^2$.
I believe they are using Plancharel and that $g(x) = -ixf(x) \implies \triangledown(\mathcal{F}f) = \mathcal{F}g$, but I do not see how this implies we can in some sense say the derivative is just some constant.
I couldn't find anything online, except a formula on Wikipedia that in general $\frac{d^nf(x)}{dx^n} = (\xi^n)\mathcal{F}f(\xi)$ without an explanation of why, nor where I could find why. Any help would be appreciated!
The fact that is being used here is that $\mathcal F\left [ \frac{\partial f}{\partial x_j} \right ](\xi) = 2\pi i\xi_j \mathcal F[f](\xi) $. To see where this comes from we can just use the definition of the Fourier transform (there are a few possible conventions on how to define the Fourier transform):
$$\mathcal F\left [ \frac{\partial f}{\partial x_j} \right ](\xi) = \int e^{-2\pi i x \cdot \xi}\frac{\partial f}{\partial x_j}(x) dx =-\int f(x) \frac{\partial}{\partial x_j}e^{-2\pi ix\cdot \xi} dx= 2\pi i \xi_j \int f(x) e^{-2\pi i x \cdot \xi}dx = 2\pi i \xi_j \mathcal F[f](\xi) $$
(if you use another convention on how to define the Fourier transform you might even have the formula $\mathcal F\left [ \frac{\partial f}{\partial x_j} \right ](\xi) = \xi_j \mathcal F[f](\xi) $ without the factor $2 \pi i$).
Now we are using first Plancherel and then this fact twice : $$\left \lVert \frac{\partial^2f}{\partial x_j \partial x_k} \right \rVert_2^2 = \left \lVert \mathcal F \left [ \frac{\partial^2f}{\partial x_j \partial x_k} \right ] \right \rVert_2^2 = \left \lVert 2\pi i \xi_j\mathcal F \left [ \frac{\partial f}{ \partial x_k} \right ] \right \rVert_2^2 =\left \lVert 2\pi i \xi_j \cdot 2 \pi i \xi_k \mathcal F[f] \right \rVert_2^2 = C \lVert \xi_j \xi_k \mathcal F[f]\rVert_2^2$$
as claimed with $C= 4 \pi^2$.