I wondered if anyone could help me prove these inequalities for $f\in \mathcal{S}(\mathbb{R})$ and $\lambda>0$:
$(1) \int_{|\xi|\geq\lambda}|\hat{f}(\xi)|^2\mathrm{d}\xi\leq\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2.$
$(2)\|f\|_2^2\leq 2\lambda\|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2.$
For the first one,
$$\int_{|\xi|\ge \lambda} |\widehat{f}(\xi)|^2 d\xi = \frac1{\lambda^2} \int_{|\xi|\ge \lambda} |\lambda \widehat{f}(\xi)|^2 d\xi\le \frac1{\lambda^2} \int_{\mathbb{R}} |\xi \widehat{f}(\xi)|^2 d\xi = \frac1{c\lambda^2} \|f'\|_2^2,$$
where the last equality follows by Plancherel's theorem.
For the second one, first use Plancherel's theorem and then decompose the integral as
$$\int_{\mathbb{R}} |\widehat{f}(\xi)|^2 d\xi=\int_{|\xi|\le \lambda} |\widehat{f}(\xi)|^2 d\xi + \int_{|\xi|\ge \lambda} |\widehat{f}(\xi)|^2 d\xi$$
We already estimated the second integral on the right; the first integral is bounded as follows: $$\int_{|\xi|\le \lambda} |\widehat{f}(\xi)|^2 d\xi \le 2\lambda\|\widehat{f}\|_\infty ^2\le 2\lambda \|f\|_1^2$$
