I am trying to understand the solution to the following problem:
Let $u \in \mathcal{D}'(\mathbb{R}^{n})$ such that $u(x) = c \log(|x|)$ when $|x|>1$, where $c \in \mathbb{C}$. Show that $u \in \mathcal{S}'(\mathbb{R}^{n})$ and that the Fourier transform $\hat{u}(\xi) \in C^{\infty}$ when $\xi \neq 0$.
So what I have on my lecture notes is the following:
Let $\mathcal{X}(x) = 1$ when $|x| \leq 2$, I don't know if this is smooth function or the indicator function.
Then $u = \mathcal{X}(x)u + (1-\mathcal{X}(x))u = u_{1} + u_2.$
$u_1 = \mathcal{X}u \in \mathcal{E}' \subset \mathcal{S}' \implies \hat{u}_1 \in C^{\infty}$.
Now, why does it follow immediately that $\hat{u}_{1} \in C^\infty$, and what is actually meant with that a Fourier transform of a distribution in general is $C^\infty$?
Does it mean that $\langle \xi^\alpha \hat{u}, \phi\rangle = \langle u, (-i)^{|\alpha|}\partial_{\xi}^{\alpha}\phi\rangle$ is a well-defined distribution?
If so then $\int u(\xi)(-i)^{|\alpha|}\partial_{\xi}^{\alpha}\hat{\phi}(\xi) d\xi$ is finite for all $u\in \mathcal{E}'$, so that's why all $u \in \mathcal{E}'$ have a Fourier transform in $C^\infty$?
Now $u_2 \in C^\infty$ and $|u_2| \leq c \log(|x|) \leq c_{1}|x| \implies u_2 \in \mathcal{S}' \implies u \in \mathcal{S}'$.
This is clear since the $\int c_1 |x|\phi(x) dx < \infty $ $\forall \phi \in \mathcal{S}$.
Now it shall be shown thath $\hat{u}$ is $C^\infty$, which is equivalent to $u_2 \in C^{\infty}$ since $u_1 \in \mathcal{E}'$.
The approach is to consider
$\xi^{\alpha}\hat{u_2}(\xi) = \int e^{ix\xi}(i\partial_x)^\alpha u_2(x)dx$
$\partial_{x_j}u_2 = \partial_{x_j}(1-\mathcal{X})u = -\partial_{x_j}(\mathcal{Xu})-(1-\mathcal{X})\partial_{x_j}u$
The derivatives seems a bit strange to me.
Now $\partial_{x}^{\alpha}u_2 = v_\alpha + w_\alpha$
$v_\alpha \in \mathcal{E}'$ $ (|x| \leq 2)$
$w_\alpha = (1-\mathcal{X})\partial^{\alpha}u = |w_\alpha(x)| \leq \frac{C}{|x|^{\alpha}}$ $(|x| > 2)$
This shall then imply that $\xi^\alpha \hat{u} = \hat{v}_{\alpha} + \hat{w}_\alpha$ and that $\hat{v}_\alpha \in C^\infty$ and $\hat{w}_{\alpha} \in C^k$
With $|\alpha| = N:$
$|\int e^{ix\xi}w_\alpha dx | \leq C\int_{|x|>2} \frac{1}{|x|^N}dx < \infty$ (If N > n I guess?)
$\partial_{\xi} \int e^{ix\xi}w_\alpha dx = \int e^{ix\xi}(ix_j)^\alpha \partial_j^\alpha w_\alpha dx \leq C$
So exactly what is the idea in the proof, how do we conclude that $\hat{w}_ {\alpha} \in C^{\infty}$?
That $\hat{u}_1$ is smooth follows from one of the Paley Wiener theorems, right? I think 7.2.1 in Strichartz.