Order of distribution Let $T$ be Schwartz distribution. Assume that the following inequality holds $T(\phi) \leq \textrm{const} ~\| \tilde{\phi}\|_1$ for any $\phi \in S(\mathbb{R})$ ($\tilde{\phi}=\mathcal{F}(\phi)$ is the Fourier transform of $\phi$ and $\|\cdot \|_1$ is $L^1$ norm). What might be said about the order of this distribution?
If $\phi_a(x)=\phi(ax)$ then $\| \tilde{\phi}_a\|_1=\| \tilde{\phi}\|_1$ so $T(\phi_a)\leq \textrm{const} ~\| \tilde{\phi}\|_1$. Because of it,
I suppose that $T$ is of order 0, that means $T(\phi)\leq \textrm{const} ~\sup_{x\in\mathbb{R}} |\phi(x)|$. Am I right?
The order of $T$ is certainly not greater then 1:
Let $t=1$, $\phi \in \mathcal{D}(K)$, $K\subset \mathbb{R}$, compact,
$\| \tilde{\phi} \|_1 = \int d p \frac{1}{1+|p|^t} (1+|p|^t) |\tilde{\phi}(p)| \leq \left(\int d p \frac{1}{(1+|p|^t)^2}  \right)^{1/2} \left( \int d p (1+|p|^t)^2 |\tilde{\phi}(p)|^2\right)^{1/2} \leq \textrm{const} \left( \int d p ||p|^t\tilde{\phi}(p)|^2\right)^{1/2} \leq \textrm{const} \left( \int d p |\mathcal{F}(\phi')(p)|^2\right)^{1/2}=\textrm{const} \left( \int d p |\mathcal{F}(\phi')(p)|^2\right)^{1/2} = \textrm{const} \left( \int d x |\phi'(x)|^2\right)^{1/2} \leq \textrm{const} \sup_{x} |\phi'(x)|$.
How one can estimate $||p|^t\tilde{\phi}(p)|$ for $1/2<t<1$ (for $t=1$ we have $||p|^t\tilde{\phi}(p)|< |\mathcal{F}(\phi')(p)|$)?
 A: Let $\phi \in \mathcal{S}$ a Schwartz function such that $\text{spt} \, \phi \subseteq K$ for some compact set $K$. By Plancherel's theorem,
$$\|\tilde{\phi}\|_{L^2} = c_1 \|\phi\|_{L^2}. \tag{1}$$
(The constant $c_1$ depends on the definition of the Fourier transform you are using). $\tilde{\phi}$ is a Schwartz function, therefore
$$\begin{align*} \int_{\mathbb{R} \backslash [-1,1]} |\tilde{\phi}(\xi)| \, d\xi \leq \underbrace{\left( \int_{\mathbb{R} \backslash [-1,1]} \frac{1}{\xi^2} \, d\xi \right)^{\frac{1}{2}}}_{=:c_2} \cdot \left( \int \xi^2 \cdot |\tilde{\phi}(\xi)|^2 \, d\xi \right)^{\frac{1}{2}} =  c_2 \cdot \|\widetilde{\partial_x \phi}\|_{L^2} \stackrel{(1)}{=} c_1 \, c_2 \, \|\partial_x \phi\|_{L^2}\end{align*}$$
using that $-\imath \xi \tilde{\phi}(\xi) = \widetilde{\partial_x \phi}(\xi)$ where $\partial_x \phi(x) := \frac{d}{dx} \phi(x)$. Moreover,
$$\begin{align*} 
 \int_{-1}^1 |\tilde{\phi}(\xi)| \, d\xi &\leq 2 \cdot \|\tilde{\phi}\|_{\infty} \leq 2 \|\phi\|_{L^1} \end{align*}$$
Consequently, we obtain
$$|T\phi| \leq c \|\tilde{\phi}\|_{L^1} \leq c \cdot \left(c_1 \cdot c_2  \|\partial_x \phi\|_{L^2}+2 \|\phi\|_{L^1} \right)$$
Thus,
$$|T\phi| \leq C (\|\phi'\|_{\infty}+\|\phi\|_{\infty}) \cdot (\sqrt{\lambda(K)}+\lambda(K))$$
where $\lambda$ denotes the Lebesgue measure. This shows that $T$ is of order less than $2$.
