Is the Fourier transform a tame linear operator? $\mathcal{F}:C^{\infty}_{0}(B^d)\to L_{1}^{\infty}(\mathbb{R}^{d},\mu,w)$
$\mathcal{F}(f)=\hat{f}$
I'd like show that $\left\|\mathcal{F}(f)\right\|_{n}\leq\left\| f \right\|_{C^{n+r}(\mathbb{R}^d)}$ for someone $r\geq 0$.
And 
$\mathcal{F}^{-1}:L_{1}^{\infty}(\mathbb{R}^{d},\mu,w)\to C^{\infty}_{0}(\mathbb{R}^d) $
$\mathcal{F}^{-1}=\check{f}$
I'd like show that $\left\|\mathcal{F}^{-1}(f)\right\|_{C^{n}(\mathbb{R}^d)}\leq\left\| f \right\|_{n+s}$ for someone $s\geq 0$.
The objects are defined down:
$L_{1}^{\infty}(\mathbb{R}^{d},\mu,w):=\{f\in L^1(\mathbb{R}^{d},\mu):\int_{\mathbb{R}^d} e^{nw}|f|d_{\mu}<\infty \quad\forall n\}$, with seminorms $\left\| f \right\|_{n}=\int_{\mathbb{R}^d} e^{nw}|f|d_{\mu}$, $\mu$ is Lebesgue mesure and $w(x)=\log(1+|x|)$.
$C^{\infty}_{0}(\mathbb{R}^d):=\{f\in C^{\infty}(\mathbb{R}^{d}):\lim_{|x|\to\infty}D^{n}f(x)=0  \quad\forall n\} $, with seminorms $\left\| f \right\|_{C^n(\mathbb{R}^d)}=\max_{|j|\leq n} \sup_{x\in\mathbb{R}^d} |D^{j}f(x)|$.
$C^{\infty}_{0}(B^d):=\{f\in C^{\infty}(\mathbb{R}^{d}): supp(f)\subset B^{d}\}$ is a subspace of $C^{\infty}_{0}(\mathbb{R}^d)$, than its seminorms is the same.
 A: We can observe that
\begin{eqnarray*}
\left|| f \right||_{n}&=&\int_{\mathbb{R}^{d}} (1+|x|)^{n}|f(x)|dx
\\ &=& \sum_{j=0}^{n}\binom{n}{j} \int_{\mathbb{R}^{d}} |x|^{j}|f(x)|dx.
\end{eqnarray*}
Doing the calculus for ${\mathcal{F}}^{-1}$,
\begin{eqnarray*}
\sup_{x}|D^{j}\mathcal{F}[f](x)|&=&\sup_{x}\left|D^{j}\int_{\mathbb{R}^{d}} e^{ix\cdot \xi}f(\xi)d\xi\right|
\\ &=& \sup_{x}\left|\int_{\mathbb{R}^{d}} e^{ix\cdot \xi}\xi^{j}f(\zeta)d_{\xi}\right|
\\ &\leq& \int_{\mathbb{R}^{d}}|\xi|^{j}|f(\xi)|d\xi
\\&\leq & \left|| f \right||_{n}
\end{eqnarray*}
implies
$$\left||\mathcal{F}^{-1}[f] \right||_{C^{n}}\leq \left|| f \right||_{n}.$$
for $\mathcal{F}$ we have,
\begin{eqnarray*}
\left|| \mathcal{F}[g] \right||_{n} & = &\int_{\mathbb{R}^{d}} (1+|x|)^{n}|\mathcal{F}[g](x)|dx
\\ &=& \int_{\mathbb{R}^{d}} (1+|x|)^{n+r}|\mathcal{F}[g](x)|\frac{dx}{(1+|x|)^{r}}
\\ &=& \sum_{j=0}^{n+r} \binom{n+r}{j}\int_{\mathbb{R}^{d}} |x|^{j}|\mathcal{F}[g](x)|\frac{dx}{(1+|x|)^{r}}
\\&\leq & \sum_{j=0}^{n+r} \binom{n+r}{j}\int_{\mathbb{R}^{d}} |\mathcal{F}[D^{j}g](x)|\frac{dx}{(1+|x|)^{r}}
\\&\leq & \sum_{j=0}^{n+r} \binom{n+r}{j}\sup_{x\in\mathbb{R}^{d}}|\mathcal{F}[D^{j}g](x)|\int_{\mathbb{R}^{d}}\frac{dx}{(1+|x|)^{r}}
\\&\leq & C \left||g \right||_{C^{n+r}} \int_{\mathbb{R}^{d}}\frac{dx}{(1+|x|)^{r}}
\end{eqnarray*}
and make a analysis about convergence of this integral 
\begin{eqnarray*}
\int_{\mathbb{R}^{d}}\frac{dx}{(1+|x|)^{r}} &=& \mbox{área}(S^{d-1})\int_{0}^{\infty} \frac{\rho^{d-1}}{(1+\rho)^{r}}d\rho
\\ &\leq & \mbox{área}(S^{d-1})\int_{0}^{\infty} \frac{(1+\rho)^{d-1}}{(1+\rho)^{r}}d\rho
\\ &\leq & \mbox{área}(S^{d-1})\int_{1}^{\infty} \frac{1}{(\rho)^{r+1-d}}d\rho
\end{eqnarray*}
the integral is convergent if $r+1-d>1$, so $r>d$. Then,
$$\left||\mathcal{F}[g] \right||_{n}\leq \left|| g \right||_{C^{n+r}}.$$
Sorry for my mistakes in writing english.
