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Let $f$ be a function satisfying $$\int_{-\infty}^{\infty}|f(x)|dx<\infty.$$

Is it true that for almost every $x\in\mathbb{R}$, $$f(x)=\lim_{R\rightarrow+\infty}\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw?$$

Here $\hat{f}$ denotes the Fourier transform of $f$.

Thanks for any comment or any suggestions.

John Hana
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    No, because there is no reason to have $\hat f$ integrable. But if $\hat f$ is integrable, then indeed the equality hold a.e. – Surb Aug 03 '19 at 15:05
  • We are not assumed $\hat{f}$ to be integrable here. You can see that the integral on the RHS is well-defined since $\hat{f}\in L^{\infty}$. – John Hana Aug 03 '19 at 15:08
  • The function $g=x$ is not integrable but the limit exists. – John Hana Aug 03 '19 at 15:13
  • @JohnHana but $e^{ikx}=\cos(kx)+i\sin(kx)$, and the integral of $\sin(kx)x$ won't be zero. – Botond Aug 03 '19 at 15:24
  • @Surb Could you give me a hint why must have $\hat{f}\in L^{1}$? I know that if $\hat{f}\in L^{1}$ then it is true. But I don't see the converse. – John Hana Aug 03 '19 at 15:30
  • So that's why it's interesting and it is the main part of my question. – John Hana Aug 03 '19 at 15:35
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    Unfortunately, the statement is not true. Kolmogorov gave an example of $f\in L^1(\mathbb T)$ whose Fourier series diverges almost everywhere, and by the transference principle, there exists a counterexample of your statement. However, if $f\in L^p(\mathbb R)$ for $1<p<\infty$, then $$\lim_{N\to\infty} \int_{-N}^N \hat f(\xi) e^{2\pi i\xi x}d\xi =f(x)$$ holds almost everywhere by the Carleson-Hunt theorem. – Myunghyun Song Aug 03 '19 at 17:19
  • @Song Thanks for your answer. I am following the book: The fast Fourier transform and its applications by E. Oran Brigham. On page 13 the abobe statement was confirmed to be true (without a given proof). – John Hana Aug 03 '19 at 17:31
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    @JohnHana I've seen it, but I should say that the condition I (that $f\in L^1$) on page 15 is not sufficient for the inversion formula to be vaild. It should be either $\mathscr{F}[f] \in L^1$ (so that $f(x) = \int_{\Bbb R} \mathscr{F}fe^{2\pi i \xi x}d\xi$) or $f\in L^p$ for some $p\in (1,\infty)$ (so that $f(x) = \lim_{N\to\infty}\int_{[-N,N]} \mathscr{F}fe^{2\pi i \xi x}d\xi$). Sadly, every existing counterexample to your statement is too monstrous to present here... – Myunghyun Song Aug 03 '19 at 17:54
  • @Surb If $f(t)=\sin(t)/t$, or for that matter $f(t)=t$, then $\lim_{R\to\infty}\int_{-R}^Rf(t),dt$ exists. – David C. Ullrich Aug 03 '19 at 20:02
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    @Song Yes for $1<p\le 2$, but not for $p>2$; in that case $\hat f$ need not even exist (except as a tempered distribution). ($\Bbb T\ne\Bbb R$. Proof: $L^4(\Bbb T)\subset L^2(\Bbb TT)$, $L^4(\Bbb R)\not\subset L^2(\Bbb R)$.) – David C. Ullrich Aug 03 '19 at 20:04
  • @DavidC.Ullrich Thank you for correcting me, sir. It seems that this wikipedia page on Carleson-Hunt theorem needs to be corrected. – Myunghyun Song Aug 09 '19 at 17:06
  • @Song No, what wikipedia says is fine. The difference is that they're talking about $L^p(\Bbb T)$, not $L^p(\Bbb R)$. As I pointed out a week ago: If $p>2$ then $L^p(\Bbb T)\subset L^2(\Bbb T)$, but $L^p(\Bbb R)\not\subset L^2(\Bbb R)$.. – David C. Ullrich Aug 09 '19 at 19:48
  • @DavidC.Ullrich Yes, I understood that. Actually I was mentioning the 'Statement of the theorem' section where it says analogous result holds for $f \in L^p(\mathbb R)$, $1<p<\infty$. – Myunghyun Song Aug 10 '19 at 20:20
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    @Song I see your point. I jst realized I could be missing something; before editing Wikipedia we should try to find an answer to the question here: https://math.stackexchange.com/questions/3320182/carleson-hunt-theorem-on-bbb-r – David C. Ullrich Aug 11 '19 at 16:01

2 Answers2

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A simpler case is when $f$ is $L^1$ and locally $a$-Hölder continuous, from $$\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw = \int_{-\infty}^\infty f(x-y)\frac{\sin(2 \pi R y)}{\pi y}dy$$

we obtain that it converges to $f$ locally uniformly.

Note that $\int_{|y|> b}|\frac{f(x-y)}{\pi y} |dy< \infty$ implies $\lim_{R \to \infty}\int_{|y|> b}\frac{f(x-y)}{\pi y} \sin(2\pi R y)dy = 0$ thus we care only of the behavior on a neighborhood of $x$.

reuns
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  • good explanation. Do you have any idea to prove the case $f\in L^{1}\cap L^{p}$, as mentioned above? – Binjiu Aug 04 '19 at 03:58
  • If $f \in L^2$ it converges to $f$ in $L^2$ so there is a subsequence converging a.e. to $f$ – reuns Aug 04 '19 at 04:03
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Since the FT of $f(x)$ as $\hat f(\omega)$ exists, we can write:$$\hat f(\omega)=\int_{-\infty}^\infty f(x)e^{-i2\pi\omega x}dx$$and by substituting we obtain$$\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw{=\int_{-R}^{R}\int_{-\infty}^\infty f(x_1)e^{-i2\pi\omega x_1}e^{2i\pi x w}dx_1dw\\=\int_{-\infty}^\infty f(x_1)\int_{-R}^{R} e^{-i2\pi\omega x_1}e^{2i\pi x w}dwdx_1\\= \int_{-\infty}^\infty f(x_1){\sin 2\pi R(x-x_1)\over R(x-x_1)}dx_1 }$$since the latter integral is absolutely convergent we can write: $$\lim_{R\rightarrow+\infty}\int_{-R}^{R}\hat{f}(w)e^{2i\pi x w}dw=\int_{-\infty}^\infty f(x_1)\lim_{R\rightarrow+\infty}{\sin 2\pi R(x-x_1)\over R(x-x_1)}dx_1\\=\int_{-\infty}^\infty f(x_1)\delta(x-x_1)dx_1=f(x)$$where $\delta(x)$ denotes the Dirac delta function.

Mostafa Ayaz
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    Well no it is not a great answer because $\frac{\sin 2\pi R x}{\pi Rx} \to \delta(x)$ is true in the dual of the Schwartz space, how do you know it is true in the dual of $L^1$ and in the almost everywhere sense – reuns Aug 03 '19 at 16:33
  • @reuns You are right. Let me think more. – John Hana Aug 03 '19 at 16:42
  • @reuns It seems that $$\lim_{R \rightarrow +\infty} \int_{-\infty}^{\infty} f(x_{1})\frac{\sin 2\pi R(x-x_{1})}{R(x-x_{1})}dx_{1} =0 $$ by dominated convergence theorem. – Binjiu Aug 04 '19 at 02:36
  • Since for $R\rightarrow +\infty$, $$f(x_{1})\frac{\sin 2\pi R(x-x_{1})}{R(x-x_{1})}\rightarrow 0$$ for a.e. $x_{1}$ and $$|f(x_{1})\frac{\sin 2\pi R(x-x_{1})}{R(x-x_{1})}|\leq 2\pi |f(x_{1})|,$$ for a.e $x_{1}$. – Binjiu Aug 04 '19 at 02:54
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    Ok but it is because Mostafa Ayaz is wrong it is $f(x_{1})\frac{\sin 2\pi R(x-x_{1})}{ \pi(x-x_{1})}$ – reuns Aug 04 '19 at 02:56
  • @reuns Sorry, I didn't notice it. – Binjiu Aug 04 '19 at 02:58