Let $f\in L^1{(\mathbb{R})}$. Why the Fourier Transform $\hat{f}\in L^{\infty}{(\mathbb{R})}$.
Is it because $(L^1{(\mathbb{R})})^*=L^{\infty}{(\mathbb{R})}$?
Let $f\in L^1{(\mathbb{R})}$. Why the Fourier Transform $\hat{f}\in L^{\infty}{(\mathbb{R})}$.
Is it because $(L^1{(\mathbb{R})})^*=L^{\infty}{(\mathbb{R})}$?