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.
Best Answer
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: