Dual space of the function $f$ in Fourier Transform

Question

Let $f\in L^1{(\mathbb{R})}$. Why the Fourier Transform $\hat{f}\in L^{\infty}{(\mathbb{R})}$.

Is it because $(L^1{(\mathbb{R})})^*=L^{\infty}{(\mathbb{R})}$?

Answer 1

This has nothing to do with the dual of $L^1$.

This is just an immediate consequence of the definition : for every $\xi$, $\hat{f}(\xi) = \int e^{-it \xi} f(t) \, dt$, and so $\left| \hat{f}(\xi) \right| \le \int \left|f(t) \right|dt$ which is a finite constant by definition of $L^1$.