I have to proof that $$ \|\mathcal{F}((1+|\cdot|^2)^d\partial^\alpha g(\cdot))\|_{L^\infty} \lesssim \|(\text{Id} - \Delta)^d((\cdot)^\alpha \Phi)\|_{L^1} $$ where Id is the identity matrix, $\Delta$ the Laplace operator and $g = \mathcal{F}^{-1}\Phi$.
I know I can use the inverse Fourier theorem to use the properties of the Fourier transform for the inverse Fourier transform as well. So first of all we can use the inequality $\|\mathcal{F}^{-1} f\|_{L^\infty} \lesssim \|f\|_{L^1}$. Second, I know the equalities of the $1$-dimensional Fourier transform
$$ \frac{\text{d}^n f(x)}{\text{d}x^n} = (2\pi i \xi)^n \widehat{f}(\xi) \quad \text{and} \quad x^n f(x) = \left(\frac{i}{2\pi} \right)^n \frac{\text{d}^n \widehat{f}(\xi)}{\text{d} \xi^n}$$
although I'm pretty sure, without a precise proof, that I can also apply this to the $n$-dimensional case.
Now to my questions. Is there a theorem that I can ignore the exponent $d$? And how does the $1$ become the identity matrix? I know that in the $1$-dimensional case $1$ becomes the Dirac delta. But I don't see how that helps me. Can anyone give me a tip on the best way to show this? I don't see it as making too much sense to put it in the definition, since then I can't use the sentences without further ado.
I think I've come to a proof with the help of LL 3.14.
\begin{align*} \mathcal{F}((1+ |\, \cdot \,|^2)^d \partial^{\alpha}g) &= \mathcal{F}((1+|\cdot|^2)(1+ |\, \cdot \,|^2)^{d-1} \partial^{\alpha}g) \\&= \mathcal{F}((1+ |\, \cdot \,|^2)^{d-1} \partial^{\alpha}g) + \mathcal{F}(|\cdot|^2(1+ |\, \cdot \,|^2)^{d-1} \partial^{\alpha}g) \\&= \mathcal{F}((1+ |\, \cdot \,|^2)^{d-1} \partial^{\alpha}g) + (i^2\Delta)(\mathcal{F}((1+ |\, \cdot \,|^2)^{d-1} \partial^{\alpha}g)) \\ &= (\text{Id} - \Delta)(\mathcal{F}((1+ |\, \cdot \,|^2)^{d-1} \partial^{\alpha}g)) \\ &= (\text{Id}-\Delta)^d(\mathcal{F}(\partial^{\alpha}g)) \\ &= (\text{Id}-\Delta)^d (i^{\alpha} (\cdot)^\alpha \mathcal{F}(g)) \end{align*}
If you now use the properties of the norm, you get the desired result.