Triangular frequency cutoff

The cutoffs ${\theta }_n(\xi )=(1-|\xi |/n{)}_{+}$ increase pointwise to one, while their Fourier transforms are nonnegative and have integral $2\pi$ under the angular-frequency convention.

If $f\in L^1\cap L^{\infty }$ and $\hat{f}\ge 0$, testing against triangular frequency cutoffs and using monotone convergence gives $\Vert \hat{f}{\Vert }_1\le 2\pi \Vert f{\Vert }_{\infty }$.