# An inequality involving the $L^1$ norm and a Fourier transform

Can someone please tell me where this inequality comes from: $$\int_{\infty}^{\infty} |g(\xi)|\cdot | \hat{\eta}(\xi) |^2 \;d\xi \leq \frac{1}{2\pi} \| \eta \|_{L^1(\mathbb{R})}^2 \int_{-\infty}^{\infty} |g(\xi)| \; d\xi$$ where the hat denotes the Fourier transform.

I came across this when reading "On the Korteweg-de Vries-Kurumoto-Sivashinsky equation" (Biagioni, Bona, Iorio and Scialom 1996).

I'd really like the Fourier series analogue of this inequality.

## 1 Answer

Starting with Hölder, we have $$\def\norm#1{\left\|#1\right\|} \norm{g \hat \eta^2}_{L^1} \le \norm{g}_{L^1} \norm{\hat \eta^2}_{L^\infty}$$ By definition of $\hat \eta$, we have \begin{align*} \def\abs#1{\left|#1\right|}\abs{\hat \eta(\xi)} &=\abs{ \frac 1{(2\pi)^{1/2}} \int_{\mathbf R} \eta(x) \exp(-ix\xi)\, dx}\\ &\le \frac 1{(2\pi)^{1/2}} \int_{\mathbf R} \abs{\eta(x)}\, dx\\ &\le \frac 1{(2\pi)^{1/2}} \norm{\eta}_{L^1} \end{align*} and hence $$\norm{\hat \eta^2}_{L^\infty} = \norm{\hat \eta}_{L^\infty}^2 \le (2\pi)^{-1}\norm{\eta}_{L^1}^2$$ Altogether $$\norm{g \hat \eta^2}_{L^1} \le \norm g_{L^1} (2\pi)^{-1} \norm{\eta}_{L^1}^2$$