Due to the Hausdorff-Young inequality the Fourier transform \(\mathcal{F}\colon L^p(\mathbb{R}^n)\to L^q(\mathbb{R}^n)\) is a bounded operator, where \(1\le p\le 2\) and \(\frac{1}{p}+\frac{1}{q}=1\).
Due to the Hausdorff-Young inequality the Fourier transform \(\mathcal{F}\colon L^p(\mathbb{R}^n)\to L^q(\mathbb{R}^n)\) is a bounded operator, where \(1\le p\le 2\) and \(\frac{1}{p}+\frac{1}{q}=1\).