Estimate the $L^1$-norm of the Fourier transform Let $f\in C^\infty_0$ (smooth function with compact support). Is there any way to estimate the $L^1$-norm of $\hat f$ (Fourier transform of $f$) in terms of the $L^\infty$-norm of $f$?
Context: I'm trying to understand the following estimate
$$|S_N(u,\lambda)|\le\frac{C_Q}{N!}(2\pi)^{-n/2}\lambda^{-n/2-N}\int\left|\left(\frac12\langle\xi,Q^{-1}\xi\rangle\right)^N\hat u(\xi)\right|\,d\xi$$
$$\le\tilde C_Q(N!)^{-1}\lambda^{-N-n/2}\sum_{|\alpha|<n+1}\left\Vert D^\alpha\left(\frac12\langle D,Q^{-1}D\rangle\right)^Nu\right\Vert_{L^1(\mathbb R^n)}$$
from Grigis & Sjöstrand's Microlocal Analysis for differential operators page 21, where $u\in C^\infty_0$ and $Q$ is a symmetric matrix.
I thought this question together with a Sobolev embedding could have explained the second inequality, but apparently this is not the case.
If someone could explain why the inequality holds I would be very grateful.
 A: You have the inequality 
$$
\|f\|_\infty\le \|\hat f\|_1,$$
that you can immediately establish via the formula $f(x)=\int_{\mathbb R^d} \hat{f}(\xi)e^{i2\pi x\cdot \xi}\, d\xi.$ 
The right hand side can be infinite with the left hand side being finite, though. Take for example 
$$f(x)=\begin{cases} 1, & -1\le x \le 1 \\ 0, & \text{otherwise},\end{cases}$$ 
which is such that $\hat{f}(\xi)=C \frac{\sin \xi}{\xi}$ (here $C>0$ depends on the chosen normalization for the Fourier transform). 
This example is not smooth, and indeed if $f\in C^\infty_c$ then $\hat{f}\in L^1$; see Netchaiev's answer. However, you cannot hope for an inequality such as 
$$\tag{FALSE}
\|\hat{f}\|_1\le C \|f\|_\infty, \quad \forall f\in C^\infty_c$$ 
with $C>0$ independent of $f$. If this were true, you could take a sequence $f_n$ of smooth compactly supported functions approximating the previous example $f$ and the inequality $\|\hat{f}_n\|_1\le C\|f_n\|_\infty$ would extend to $\infty=\|\hat{f}\|_1\le C\|f\|_\infty=1.$
P. S. Please note that the last approximation process has to be done with some care, because $C^\infty_c$ is not dense in $L^\infty $. It suffices to take a sequence of smooth compactly supported functions approximating $f$ with the property $\|f_n\|_\infty \le 1$, like the ones in this picture: 

A: You can have an upper-bound for the $L^2$ norm of $\hat{f}$ : 
$$ \int |\hat{f}|^2 = \int|f|^2\leq \mathrm{supp}(f)\times\sup(1,\|f\|_{L^{\infty}}^2) $$
And since $\hat{f}$ is in the Schwartz space (fast decreasing), you can cut the tails of $\int |\hat{f}|$ : 
$$ \int |\hat{f}| = \int_{-n,n} |\hat{f}| + \varepsilon_n \leq 2n\int|\hat{f}|^2 + \varepsilon_n$$
So 
$$ \|\hat{f}\|_{L^1} \leq 2n \times \mathrm{supp}(f)\times\sup(1,\|f\|_{L^{\infty}}^2) + \varepsilon_n$$ 
with $ \varepsilon_n\rightarrow 0$.
A: Note that $(1+\Delta_x)^d e^{-ix\xi} = (1+\vert \xi\vert^2)^d e^{-ix\xi} $, thus 
$$\Vert \hat f \Vert_{L^1}= \int  \vert \int e^{- i x \xi}f(x) dx \vert d\xi = \int (1+\vert \xi\vert^2)^{-d} \vert \int (1+ \Delta_x)^{d} e^{- i x \xi}f(x) dx \vert d\xi. $$
Integrate partially w.r.t. $x$ and take the absolute value inside, then
$$
\Vert \hat f \Vert_{L^1} \le \int\int (1+\vert \xi\vert^2)^{-d} \vert (1+\Delta_x)^d f(x) \vert dxd\xi <\infty,
$$
because on $\mathbb{R}^d$, the function $(1+\vert \xi\vert^2)^{-d}$ is integrable. This shows that $\hat f\in L^1$ and by the usual continuity of the Fouriertransform  $F\colon L^1 \rightarrow C_0$ you get
$$
\Vert f \Vert_\infty = (2\pi)^{-d}\Vert F \hat f \Vert_\infty \le (2\pi)^{-d}\Vert \hat f \Vert_{L^1},
$$
because $F \hat f(x) = (2 \pi)^d f(-x)$.
