Fourier transform of $C^\infty_0$, smooth functions vanishing at infinity

Is there a proper description of the space $$\{\hat f\ | \ f\in C^\infty \ s.t. \forall \alpha\in\mathbb{N}^n,\forall \epsilon>0\exists K\subseteq \mathbb{R}^n\ K\ \text{compact};\ \sup_{x\in \mathbb{R}^n\backslash K}|\partial^\alpha_x f(x)|< \epsilon\}$$Especially I'm interested in the order of these distributions and their decay properties.

To be precise are there constants $C_{K,m}$, for every $K\subseteq\mathbb{R}^n$ compact and $m\in\mathbb{R}$, s.t. $$\left|\int \hat f(\xi)\phi(\xi-b)\ d \xi\right|\leq C_{K,m}(1+b^2)^{-m}\|\phi\|_{\infty}\quad \forall\phi\in C_c(K)$$ or anything slightly worse?

Edit: The assertion automatically holds for all $m\in\mathbb{R}$ as long as it holds for any $m_0\in\mathbb{R}$, since all derivatives are again of the same form. This maximally exploits smoothness, so to establish this starting estimate or to give a counterexample, it is no loss assuming only $f\in C_0(\mathbb{R}^n)$.
Further the condition vanishing at infinity must be exploited, since on $C^\infty_b(\mathbb{R}^n)$ the assertion does not hold! (e.g. for $\Theta$ the Heaviside step function and $\phi$ some Schwartz-function $\widehat{\phi\ast\Theta}(\xi)=\hat\phi\cdot 1/2(\delta(\xi)-\text{p.v.}\frac{i}{\pi \xi})$ is of order 1)

• I'm far from an expert here, but this space looks very close to (if not the same as) the Schwartz space of rapidly decreasing functions. en.wikipedia.org/wiki/Schwartz_space – Stefan Behrens Sep 15 '16 at 11:17
• @StefanBehrens No, functions of the Schwartz space decay (together with all derivatives) much faster. – Jochen Wengenroth Sep 15 '16 at 12:35
• The space you describe is called $\dot{\mathscr B}$ in Laurent Schwartz' Theorie des Distributions (page 199). – Jochen Wengenroth Sep 15 '16 at 12:41
• it says that for every $\alpha$, $\partial_x^\alpha f(x) \to 0$ as $|x| \to \infty$, taking the FT, It means that $\hat{f}$ is rapidly decreasing (for example $\displaystyle f(x) = \frac{1}{(1+|x|^2)^{(n+1)/2}}, \hat{f}(\xi) = e^{-2\pi |\xi|}$) – reuns Sep 15 '16 at 14:10
• @user1952009 What do you mean precisely? I think that such an $f$ need not be $L^1$ so that $\hat{f}$ a priory is only a tempered distribution. The first thing to do would be to determine the order. – Jochen Wengenroth Sep 16 '16 at 6:58