Does $f(x)\in L^1$ imply that $\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx=0$?

Question

Suppose that $f(x)$ is $L^1$ and R- integrable function, problem is to resolve if it is possible existence of such a $f(x)$ that: $$\lim_{\omega \to \infty } \, \int_{-\infty }^{+\infty } f(x) e^{-i x \omega } \, dx\neq 0$$

All math tricks acceptable.

Answer 1

For any $f \in L^1$ we have

$$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}$$

this result is known as Riemann-Lebesgue lemma.

As $f \in L^1$ we can choose a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions such that $f_n \stackrel{L^1}{\to} f$. Since $\|\hat{g}\|_{\infty} \leq \|g\|_{L^1}$ for any $g \in L^1$, we get by the triangle inequality

$$\begin{align*} |\hat{f}(\omega)| &\leq |\hat{f}(\omega)-\hat{f}_n(\omega)|+|\hat{f}_n(\omega)| \leq \|f_n-f\|_{L^1} + |\hat{f}_n(\omega)|\end{align*}$$

for all $\omega \in \mathbb{R}$. By virtue of our choice, $\|f_n-f\|_{L^1} \to 0$ as $n \to \infty$. Therefore, we see that it suffices to prove $(1)$ for simple functions. So let

$$f(x) = \sum_{j=1}^n c_j \cdot 1_{[x_j,x_{j+1}]}$$

for some constants $c_j \in \mathbb{R}$, $x_j<x_{j+1}$, $j=1,\ldots,n$. Then,

$$\hat{f}(\omega) = \sum_{j=1}^n c_j \int_{x_j}^{x_{j+1}} e^{-\imath \, x \omega} \, dx = \sum_{j=1}^n \frac{c_j}{-\imath \omega} \bigg( e^{-\imath \, x_{j+1} \omega}-e^{-\imath \, x_j \omega} \bigg) \stackrel{\omega \to \infty}{\to} 0,$$

i.e. $(1)$ holds. This finishes the proof.

Answer 2

If $f$ and $g$ are in $L^{1}$, then $$ \left|\int_{-\infty}^{\infty}fe^{i\omega t}\,dt-\int_{-\infty}^{\infty}ge^{i\omega t}\,dt\right| \le \|f-g\|_{1}. $$ So, if you can prove the limit property $$ \lim_{|\omega|\rightarrow\infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}\,dt=0 $$ for a dense subspace of $L^{1}$, then you'll have it for all $f\in L^{1}$.

The limit property $$ \lim_{|\omega|\rightarrow\infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}\,dt=0 $$ definitely holds for characteristic functions $f$ of finite intervals. And it holds for all sets of Lebesgue measure 0. Let $\mathcal{F}$ be the collection of subsets $E$ of $[-R,R]$ for which the characteristic functions $f=\chi_{E}$ satisfy the above. Can you show this is a $\sigma$-algebra of subsets of $[-R,R]$? If so, then you can get the limit property for all compactly-supported simple functions $f$, which is a dense subspace of $L^{1}$.