If $f \in L^1(\mathbb{R})$, then $\hat{f}(\omega)‎\to0$ as $‎\omega‎ \to+‎\infty‎$?

Question

The Fourier transform of $f$ is as follows: $$\hat{f}(\omega)=\frac{1}{‎‎‎\sqrt{2‎\pi‎}}\int_{-\infty}^{+\infty} e^{-i\omega t} f(t) \, dt.$$ I need to know that if $f \in L^1(R)$, then can we conclude that $\hat{f}$ is bounded and continuous and $\hat{f}(\omega)‎\to 0$ as $‎\omega‎ \to+‎\infty‎$? It would be appreciated if someone could help me.

Answer 1

Suppose $f\in L^1(\mathbb{R})$. For $0 < R < \infty$ and $f \in L^1(\mathbb{R})$, the function $$ \hat{f_R}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt $$ is infinitely differentiable with $$ \hat{f_{R}}^{(n)}(s)=\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)(-it)^{n}e^{-ist}dt. $$ Furthermore, one has the uniform convergence to $0$ of the function: $$ |\hat{f_{R}}(\omega)-\hat{f}(\omega)| \le \frac{1}{\sqrt{2\pi}}\int_{|t| \ge R}|f(t)|dt \rightarrow 0,\mbox{ as $R\rightarrow\infty$}. $$ The uniform limit of continuous functions is continuous. So $\hat{f}$ is continuous on $\mathbb{R}$. The Riemann-Lebesgue lemma gives $$ \lim_{s\rightarrow\infty}\hat{f}(s)= \lim_{s\rightarrow\pm\infty}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}dt =0. $$

Answer 2

Note that the $\mathcal{C}^\infty_c$ functions are dense in $\mathcal{L}^1$. This theorem is true for a smooth function; you can see this by integrating by parts. Now invoke the density to see it holds for any $f\in\mathcal{L}^1$.

Answer 3

Yes. This is usaully called the Riemann-Lebesgue Lemma. See for instance the section "The Fourier transform on $L^1$" in Hunter's notes: