Inequality of Fourier Transform

Question

We all know that on $\mathbb R$, $$ \|\hat f\|_\infty\leq \|f\|_1. $$ In particular, equality holds if $f\geq 0$, but in most of the cases we have strict inequality. But is the following true: for every $\epsilon>0$, there is an $0\neq f\in L^1$ with $$ \|\hat f\|_\infty\leq \epsilon\|f\|_1? $$

Answer 1

Let's define $f_n(x)$ piecewise by $f_n(x) = e^{2\pi i k x}$ if $0\leq k < x < k+1 \leq n$ and $0$ otherwise. Then $\lVert f\rVert_1 = 2\pi n$, but $$|\widehat {f_n}(\xi)| = \left| \int_{\mathbb R} e^{-2\pi i \xi x}f(x) dx\right| = \left|\sum_{k=0}^{n-1} \int_0^{1}e^{2\pi i (k-\xi) (x+k)}dx\right|\leq \\\sum_{k=0}^{n-1} |e^{2\pi i k(k-\xi)}|\frac{\left|e^{2\pi i (k-\xi)}-1\right|}{2\pi |k-\xi|} = \mathcal O(\ln(n)).$$ Thus, $\lVert \widehat{f_n} \rVert_\infty \leq \frac{C\ln n}{n}\lVert f_n \rVert_1$ for some $C$ independent of $n$. The idea here was to construct a function which "resonates" with no particular frequency, such that it's fourier transform takes no large values.