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

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

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 $c_j \in \mathbb{R}$, $x_j&lt;x_{j+1}$, $j=1,\ldots,n$. 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.
Correct answer is:

$f(x)\in L^1$ does not imply $\lim\limits_{\omega\to \infty}\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}dx=0$ even if $f$ is R-integrable.

Sample: $f(x)=\arcsin(x)$, where $f\colon [-1,1] \to \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$.

Please pay attention that notation $\int f(x) dx$ determines using Riemann integral, if you want to use Lebesgue integral, you can just skip $dx$ or write $\mu(dx)$. 



Below is a simple option of proving Riemann-Lebesgue Lemma:

Let $f(x)$ be numeric $L^1(\mathbb{R})$, measurable function determined for any $x\in \mathbb{R}$.

$$\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}=-\int\limits_{\mathbb{R}}f(x) e^{-i(x\omega+\pi)}=-\int\limits_{\mathbb{R}}f(x) e^{-i\omega\left(x+\frac{\pi}{\omega}\right)}=-\int\limits_{\mathbb{R}}f\left(x-\frac{\pi}{\omega}\right) e^{-i\omega x}$$

Note: I am using substitution $v=x+\frac{\pi}{\omega}$ with renaming $v$ into $x$ what is not changing a value of the integral.

So:

$$2\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}=-\int\limits_{\mathbb{R}}\left[f\left(x-\frac{\pi}{\omega}\right)-f(x) \right]e^{-i\omega x}
$$
$$2\left|\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}\right|= \left|\int\limits_{\mathbb{R}}\left[f\left(x-\frac{\pi}{\omega}\right)-f(x) \right]e^{-i\omega x}\right|\leqslant \left\|f\left(x-\frac{\pi}{\omega}\right)-f(x)\right\|_{L^1}$$
what implies:
$$\lim\limits_{\omega\to \infty}\left|\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}\right|\leqslant \frac{1}{2}\lim\limits_{\omega\to \infty}\left\|f\left(x-\frac{\pi}{\omega}\right)-f(x)\right\|_{L^1}=\frac{1}{2}\lim\limits_{h \to 0}\|f(x-h)-f(x)\|_{L^1}$$

Functions from $L^1(\mathbb{R})$ space keep continuity of translation in meaning of $L^1$ norm what is described here and here, hence:

$$\lim\limits_{\omega\to \infty}\left|\int\limits_{\mathbb{R}}f(x) e^{-ix\omega}\right|\leqslant \frac{1}{2}\lim\limits_{h \to 0}\|f(x-h)-f(x)\|_{L^1}=\frac{1}{2}\cdot 0$$

Note: If I am asking a question I am not looking for something what is described in popular sources like wikipedia or wikiproof. You can assume that aforementioned places of the Internet are checked before asking the question and it is not bringing a suitable answer.
After analyzing of some users I feel happy that I am a disabled person who is treating mathematics as a hobby quite similar as an old man solitaire. 

Points here means nothing in the real life.