From what I understand, \begin{align} \int \left(1+\left|\xi\right|^2\right)^s\left|\hat{f}(\xi)\right|^2\,d\xi &= \sum_{k=0}^s {s\choose k} \int \left|\xi\right|^{2k}\left|\hat{f}(\xi)\right|^2\,d\xi\\ &\overset{\color{red}{(?)}}{=} \sum_{k=0}^s {s\choose k} \int \left|D^k f(x)\right|^2\,dx \end{align} where $\hat{f}(\xi)$ is the Fourier transform of $f(x)$.
I understand that Plancherel theorem was applied in $\overset{\color{red}{(?)}}{=}$ and that $$ \mathcal{F}\left\{D^\alpha f(x)\right\} = \xi^\alpha \hat{f}(\xi) \tag{1} $$
But my confusion lies primarily in the fact that $\alpha$ is a multi-index notation but $k$ is an integer, so I am not sure how/why $(1)$ can be applied in $\overset{\color{red}{(?)}}{=}$.
Any guidance regarding this is much appreciated; thank you!
This is just some integration by parts. For example, with $2$ derivatives: $$ \int |\partial_{i} \partial_{j} f|^2 = \int (\partial_{i} \partial_{j} f) (\partial_{i} \partial_{j} f) \\= \int (\partial_{i}^2 f) (\partial_{j}^2 f)$$ and the result follows by summing on $i$ and $j$ and taking the Fourier transform of the right-hand side. I let you generalize to $k$ derivatives ;)