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.
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 $\epsilon \to 0^{+}$.