Is it possible to find a function $f$ which is smooth, compactly supported, identically $1$ in a neighborhood of $0$, and such that $\|\hat{f}\|_{L^1} = 1$?
This would be very useful in cutting off a convolution, and still obtain strict estimates on the Fourier side. My apologies if this question is naive.
Assuming you are defining $\hat{f}(\xi) = \int\limits_{\mathbb{R}}{f(x)e^{-2\pi i \xi x}\,dx}$ so that $f(x) = \int\limits_{\mathbb{R}}{\hat{f}(\xi)e^{2\pi i \xi x}\,d\xi}$ for $f$ nice enough (or some definition where you can obtain $\|f\|_{L^{\infty}}\le\|\hat{f}\|_{L^1}$):
The answer is no. Indeed, if we assume $\int\limits_{\mathbb{R}}{|\hat{f}(\xi)|\,d\xi} = 1$, and that for some $x$ we have $1 = f(x) = \int\limits_{\mathbb{R}}{\hat{f}(\xi)e^{2\pi i \xi x}\,d\xi}$, then we would have $$ 0 = \int\limits_{\mathbb{R}}{|\hat{f}(\xi)|-\hat{f}(\xi)e^{2\pi i \xi x}\,d\xi}. $$ In particular, the real part must vanish. But $$\text{Re}(\hat{f}(\xi)e^{2\pi i \xi x})\le |\hat{f}(\xi)|,$$ with equality if and only if $\hat{f}(\xi)e^{2\pi i \xi x} = |\hat{f}(\xi)|$. It follows that $|\hat{f}(\xi)|-\text{Re}(\hat{f}(\xi)e^{2\pi i \xi x})\ge 0$ for all $\xi$, and hence $$ 0 = \int\limits_{\mathbb{R}}{|\hat{f}(\xi)|-\text{Re}(\hat{f}(\xi)e^{2\pi i \xi x})\,d\xi}\implies |\hat{f}(\xi)|-\text{Re}(\hat{f}(\xi)e^{2\pi i \xi x}) = 0\text{ for almost every }\xi.$$ Thus $\hat{f}(\xi) = |\hat{f}(\xi)|e^{-2\pi i \xi x}$ for almost every $\xi$. This cannot hold for two different values of $x$ since $\hat{f}$ is not zero almost everywhere, so in particular this cannot hold on a neighborhood of $0$.