Bound $L^1$-norm of Schwartz function using fourier transform I have a function $f \in \mathcal S({\mathbb R^d})$, where $\mathcal S(\cdot)$ denote the Schwartz Space. In this case, the fourier transform of $f$:
$$
\hat f (x) = \int_{\mathbb {R}^d} e^{i \langle t,x\rangle } f(t)\mathrm dt
$$
can uniquely decide $f$.
My question is, can we bound $\| f\|_1$ using some term of $\| \hat f\|_1$, like $$\| f\|_1 \leq C \|\hat f\|_1$$Here $\| \cdot \|_1$ is the L1 norm, i.e. $\|f\|_1 = \int_{\mathbb R^d} |f|$.
If not, can we have a weaker result: there exists a small positive $\epsilon$, s.t. if $\| \hat f\|_1\leq \epsilon$, then $$\|f\|_1  \leq C \epsilon.$$
I think it might be true since when $\epsilon = 0$, the statement holds.
 A: No, you need to look at the partial derivatives of $\hat{f}$
$$\|f\|_1 \le \| \frac1{1+|x|^{2d}}\|_1 \| (1+|x|^{2d})f\|_\infty$$  where $$\| (1+|x|^{2d})f\|_\infty \le 2^d\sum_{|\alpha|=2d} \|x^\alpha f\|_\infty\le (1/\pi)^d\sum_{|\alpha|=2d} \| \partial^{\alpha}\hat{f}\|_1$$
A: The map $\hat{f} \mapsto f$ is linear so the two things you asked for are equivalent.  (A linear map is bounded if and only if it's continuous at zero.)
The bound $\|f\|_{1} \leq C \|\hat{f}\|_{1}$ can't be true.  First, notice that this is true if and only if we have $\|\hat{g}\|_{1} \leq C \|g\|_{1}$ for Schwartz $g$.  The reason is $\check{f}$ equals $\hat{f}$ up to a reflection (which leaves the $L^{1}$ norm unchanged), and $f \mapsto \hat{f}$ is a bijection on the Schwartz space.  ($\check{f}$ is the inverse Fourier transform.)
Now the problem is we can't have a bound like $\|\hat{g}\|_{1} \leq C \|g\|_{1}$ --- even assuming it only holds in the Schwartz space.  To see this, observe that if we take $g_{\epsilon}(x) = (2 \pi \epsilon)^{-\frac{d}{2}} \exp(-\|x\|^{2}/2\epsilon)$, which is a Schwartz function, then $\|g_{\epsilon}\|_{1} = 1$ while $\hat{g}_{\epsilon}(\xi) = \exp(- 2 \epsilon \|\xi\|^{2})$ and $\|\hat{g}_{\epsilon}\|_{L^{1}(\mathbb{R}^{d})} \to \infty$ as $\epsilon \to 0^{+}$.
