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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.