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I know there is a version of Fejer's theorem stating:"If $f$ is a function in $L^1(\mathbb{(- \pi, \pi)})$ then its Fejer's sums converge to $f$ in $L^1$ norm".

The question is: is this still true for Fourier transforms? I mean, is it true that $$\lim_{N\to \infty}\frac{1}{2 \pi} \int_{-N}^{N} \left(1- \frac{|\phi|}{N} \right) \widehat{f(\phi)} e^{ix \phi} d\phi = f(x)$$ in $L^1(\mathbb{R})$ for $f \in L^1(\mathbb{R})$?

I know this is true for continuous functions, and that the proof is very similar for both Fejer's sums and for this integral (you still use a convolution).

The problem here is that for the proof above (both for Fourier series and Fourier transform) you evaluate the $f$ at some point and say it is bounded. So how do you do for $L^1$ functions, whose value is not defined in single points?

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The answer is yes if $f,\widehat{f} \in L^1(\Bbb R)$

The following results are well known:

Theorem: 1- If $f,\widehat{f} \in L^1(\Bbb R)$ Then, $$ \frac{1}{2 \pi} \int_{\Bbb R} \widehat{f(\phi)} e^{ix \phi} d\phi = f(x)$$ 2-$f\in L^1(\Bbb R)$ then $\widehat{f} $ is continuous.

3-similarly, $f, \widehat{f} \in L^1(\Bbb R)$ then $f$ is continuous.

Let $f_N(\phi)= \left(1- \frac{|\phi|}{N} \right) \widehat{f(\phi)} e^{ix \phi}\mathbf1_{(-N,N)}(\phi) \to \widehat{f(\phi)} e^{ix \phi} $

We have, $$\lim_{N\to\infty}f_N(\phi)= \lim_{N\to\infty}\left(1- \frac{|\phi|}{N} \right) \widehat{f(\phi)} e^{ix \phi}\mathbf1_{(-N,N)}(\phi) =\widehat{f(\phi)} e^{ix \phi} $$ pointwise and $$|f_N(\phi)| = |\left(1- \frac{|\phi|}{N} \right) \widehat{f(\phi)} e^{ix \phi}\mathbf1_{(-N,N)}(\phi)|\le |\widehat{f(\phi)}| \in L^1(\Bbb R). $$ Then by convergence dominated theorem and the above theorem we have,

$$\lim_{N\to\infty} \frac{1}{2 \pi} \int_{-N}^{N} \left(1- \frac{|\phi|}{N} \right) \widehat{f(\phi)} e^{ix \phi} d\phi \to \frac{1}{2 \pi} \int_{\Bbb R} \widehat{f(\phi)} e^{ix \phi} d\phi = f(x).$$

Counterexample: Now if $\widehat{f} \not\in L^1(\Bbb R)$ then it is not always true take

$$f(x) =\mathbf1_{(-1,1)}(x) $$

then, $$\widehat{f}(\phi) =c\frac{\sin\phi}{\phi}\not\in L^1(\Bbb R)$$

You can check that The property fails here.

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  • $\begingroup$ Tyhanks, but in the highlighted rectangle shouldn't you also suppose $f$ continuous in x? Otherwise how can you evaluate $f$ in that point? $\endgroup$ – tommy1996q Nov 21 '17 at 11:33
  • $\begingroup$ @tommy1996q of course . if the Fourier transform is integralble then the function is continuous. as we know that Fourier transforms of àn integralble function is continuous $\endgroup$ – Guy Fsone Nov 21 '17 at 11:45
  • $\begingroup$ Yeah, but this is a bit different. You take $f$ in $L^1$ such that its Fourier transform is in $L^1$ too. Then you define $f(x)$ through the integral, but in general the $f$ you define is a continuous function (let's call that $f_1$ ) such that (for how you have constructed it) the $L^1$ norm of $f-f_1$ is 0 and thus $f$ and $f_1$ are the same function in the $L^1$ space. Am I right? $\endgroup$ – tommy1996q Nov 21 '17 at 11:52
  • $\begingroup$ @tommy1996q I see your problem that is a theorem : If a function and it's Fourier transform are integralble then the Fourier transform invertible. this is precisely the result I used there. $\endgroup$ – Guy Fsone Nov 21 '17 at 13:22
  • $\begingroup$ @tommy1996q have a look to the edit. I put all the details $\endgroup$ – Guy Fsone Nov 21 '17 at 16:31

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