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.

Note that your argument about boundaries on limit does not prove the existence of the limit: it just says that the limit doesn't diverge to infinity; it might very well be oscillating. To prove the existence of the limit, you need to use some additional argument. — TZakrevskiy, Mar 04 '14 at 08:22
The inequality you write simply states that $\forall \omega$ $|\hat f(\omega)|\le |f,L^1(\Bbb R)|$. The link you so patronisingly provided has nothing to do with the behaviour of $\hat f(\omega)$ at infinity, only with its existence for each $\omega$. — TZakrevskiy, Mar 04 '14 at 12:33
I canno tmake sense of the last sentence. What possible relevace does that function have to this queestion? — Mariano Suárez-Álvarez, Mar 10 '14 at 05:59
Well, I am somewhat fluent in questions and answers here, and to me it is not at all clear what relation that function has here and —much less, really— what is the purpose of «all math tricks acceptable» announcement. — Mariano Suárez-Álvarez, Mar 10 '14 at 13:16
@MarianoSuárez-Alvarez I just wanted to encourage open-minded people. — Darius, Mar 10 '14 at 13:53
Citing a web page with the proof? The best "math trick" yet! — GEdgar, Mar 11 '14 at 21:47

saz · Answer 1 · 2014-11-10T16:07:05.423

10

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.

edited Nov 10 '14 at 16:07

answered Mar 04 '14 at 08:14

saz

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@Darius No, I don't assume this. The inequality follows from the fact that $$|\hat{f}(\omega)-\hat{f}n(\omega)| \leq |\hat{f}-\hat{f}_n|{\infty} \leq |f-f_n|_{L^1}.$$ – saz Mar 04 '14 at 12:30
@Darius that line should read as $ |\hat{f}(\omega) - \hat{f}n(\omega)| \leq |f_n-f|{L^1} $. – TZakrevskiy Mar 04 '14 at 12:31
@Darius Of course $n \neq \infty$. $n$ is a natural number! – saz Mar 04 '14 at 16:19
Would the downvoter care to comment? – saz Mar 06 '14 at 07:18
@Darius Yes, this function is integrable. In which way does this affect my proof? – saz Mar 08 '14 at 07:41
@Darius No, your counterexample doesn't work, but it is rather tedious to discuss this here in the comments. So please consider asking a new question. Moreover, as already mentioned, this result is known as Riemann-Lebesgue lemma (see here), so there is no doubt that the result holds. – saz Mar 09 '14 at 07:34
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@Darius No, not at all. My argument is simply that the step functions are dense in $L^1$ and that the statement holds for the step functions, hence for all $L^1$-functions. Actually, the idea of the proof is rather similar to the other answer to your question. And please, please, note that $$\lim_{\omega \to \infty} \int f(x) e^{\imath \omega x} , dx =0$$ holds for any $L^1$-function; independent of the fact whether you like my proof or not. – saz Mar 09 '14 at 20:24
@Darius, you should probably read carefully what saz has written before continuing to deduce strange things from it. – Mariano Suárez-Álvarez Mar 10 '14 at 06:00
That function is equal to $1/(1+x^2)$ in $L^1$. As I said above, it is not at all clear what exactly you are trying to do with it. – Mariano Suárez-Álvarez Mar 10 '14 at 13:16
@MarianoSuárez-Alvarez Do you find $\frac{\cos(\pi I_Q(x))}{1+x^2}$ as $L^1$ function? – Darius Mar 10 '14 at 13:59
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That function and $1/(1+x^2)$ have the same values, except on a countable set. They are therefore almost equal in the sense of measure theory, with respect to Lebesgue measure. It follows that one is in $L^1$ exactly when the other is, and $1/(1+x^2)$ is very much in $L^1$. – Mariano Suárez-Álvarez Mar 10 '14 at 14:04
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Ok, I will simply stop now. – Mariano Suárez-Álvarez Mar 10 '14 at 14:32
@MarianoSuárez-Alvarez Only one question: can you find a Fourier transform of $\cos(\pi I_Q(x)) e^{-|x|}$ ? – Darius Mar 10 '14 at 15:11
That function is equal in $L^1$ to $e^{-|x|}$ (for exactly the same reason as before) and its transform is (a scalar multiple of) $1/(1+x^2)$. – Mariano Suárez-Álvarez Mar 10 '14 at 15:16
@MarianoSuárez-Alvarez So according to you $\int_{-\infty}^{+\infty}e^{-|x|}e^{-i 0 x} , dx=1$ ? – Darius Mar 10 '14 at 15:50
@MarianoSuárez-Alvarez I think you should consider limitations of Lebesgue integral. Function $sinc(x)$ has Fourier transform , but is not Lebesgue integrable. – Darius Mar 10 '14 at 16:42
$\int_{-\infty}^\infty e^{-|x|}e^{-i0x}=2\int_0^\infty e^{-x}$ because the integrand in the first integral is an even function. You can surely compute this last integral. – Mariano Suárez-Álvarez Mar 10 '14 at 17:11
@Darius That's exactly what I meant: It is tedious to discuss this question here in the comments. Even if I already mentioned this several times: It is well-known that this statement holds. So there is some mistake in your counterexample; it might be a start to accept this. However, if you are interested in a further disussion of your example, it is more convenient to ask a new question. – saz Mar 10 '14 at 17:37
@MarianoSuárez-Alvarez You claimed that $\mathcal{F}_t\left\cos(\pi I_Q(t))e^{-|t|}\right=\mathcal{F}_t\lefte^{-|t|}\right=\frac{1}{1+x^2}$ what is not true. – Darius Mar 10 '14 at 17:45
@saz $\lim_{\omega \to \infty } , \int_{-\infty }^{+\infty } \frac{\cos(\pi I_Q(x))}{1+x^2} e^{-i x \omega } , dx\neq 0$ when the Riemann integral is used. Lebesgue integration has limitations , consider function $sinc(x)$ which has Fourier transform equal to $\pi\Pi \left(\frac{\omega}{2} \right)$. – Darius Mar 10 '14 at 18:04
@Darius Thanks, I know about the differences of Lebesgue and Riemann integration. No reason to downvote my questions. – saz Mar 10 '14 at 18:49
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@Darius $\dfrac{\cos (\pi I_\mathbb{Q}(x))}{1+x^2}$ isn't Riemann-integrable on any interval. – Daniel Fischer Mar 10 '14 at 23:23
@DanielFischer I know , but $\left|\frac{cos(\pi I_Q(x))}{1+x^2}\right|$ is.You can read my answer. – Darius Mar 10 '14 at 23:51
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@Darius, please do not delete comments to which others have replied and which form part of an exchange, for that renders other people's participation incomprehensible. – Mariano Suárez-Álvarez Mar 11 '14 at 21:17

score 2 · Answer 2 · answered Mar 04 '14 at 08:26

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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 $L^{1}$.

answered Mar 04 '14 at 08:26

Disintegrating By Parts

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$f(x) \in L^1$ means that it is compactly-supported? – Darius Mar 04 '14 at 11:11
@Darius No, lots of $L^1$ functions are not compactly supported. Nevertheless, as stated in the answer, the compactly supported ones (even just the compactly supported simple ones) are dense in $L^1$. – Andreas Blass Mar 04 '14 at 12:38
@AndreasBlass Do you know if $\frac{\cos(I_Q(x)\pi)}{1+x^2}$ belongs to $L^1$ space and can be called absolutely integrable? ($I_Q(x)$ is the famous Dirichlet function.) I am just an amateur of mathematics and sometimes I feel like I am lost in all that definitions. – Darius Mar 06 '14 at 12:38
@Darius Yes. Any bounded measurable function divided by $1+x^2$ will be in $L^1$. – Andreas Blass Mar 06 '14 at 12:54

score -2 · Accepted Answer · edited Apr 13 '17 at 12:21

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.

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

3 Answers3