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I'm completely stumped by the problem below because I haven't attended the lectures which used only Riemann integration and am not sure what the author is getting at.

Let $f\in C^1(\mathbb R)\cap L^1(\mathbb R)$. Let $\hat f$ be its Fourier transform. Define $$f_n(x)=\frac 1{2\pi}\int_{-n}^n\hat f(\omega)e^{i\omega x}d\omega$$

  1. Why is $f_n$ well defined?
  2. In what sense does $f_n$ converge to a function?
  3. Is there another sense in which $f_n$ necessarily converges to a function?
  4. Suppose $f\in C_c^\infty (\mathbb R)$. Prove the sequence converges uniformly.
  1. Where does $\hat f$ land assuming $f\in L^1(\mathbb R)\cap C^1 (\mathbb R)$?
  2. Umm... I think it converges pointwise almost everywhere since if $f$ is Lebesgue integrable the limit is just $\frac 1{2\pi}\int _\mathbb{R}\hat f(\omega)e^{i\omega x}d\omega$... What am I missing?
  3. ?
  4. I have no idea..
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  • $\begingroup$ For 4, I think you may be able to use the Riemann-Lebesgue lemma, but I could be totally wrong. For 1, I believe it is because $f$ is continuous, hence $\hat{f}$ exists and is well defined, hence $f_{n}$ is well defined (again, I could be totally wrong). $\endgroup$ – Mattos Jul 13 '16 at 14:51
  • $\begingroup$ @Mattos the Riemann-Lebesgue lemma only needs $f\in L^1$, so I'm not sure what you mean. Meh, I hate this stuff. $\endgroup$ – user153312 Jul 13 '16 at 14:52
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  1. The function $\hat{f}$ is continuous on $\mathbb{R}$ and vanishes at $\infty$ by the Riemann-Lebesgue lemma. So the definition of $f_n$ makes sense as a Riemann integral on the finite interval $[-n,n]$.

  2. The function $f_n$ converges pointwise everywhere to $f$ because $f$ is differentiable at every point of $\mathbb{R}$. You can rewrite $f_n$ in the following form in order to analyze $f_n$: \begin{align} f_n(x) & = \frac{1}{2\pi}\int_{-n}^{n}e^{i\omega x}\left(\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt\right)d\omega \\ & = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin(n(x-t))}{x-t}f(t)dt \\ f_n(x)-f(x) & = \frac{1}{\pi}\int_{-\infty}^{\infty}\sin(n(t-x))\frac{f(t)-f(x)}{t-x}dt \end{align} Using the fact that $f$ is differentiable everywhere on $\mathbb{R}$, the Riemann-Lebesgue lemma gives $\lim_n (f_n(x)-f(x))=0$ because $h_{x}(t)=\frac{f(t)-f(x)}{t-x}$ is absolutely integrable on $\mathbb{R}$ for a fixed $x$.

  3. Because the derivative of $f$ is also continuous, you should be able to use (2) to show that the convergence of $f_n$ to $f$ is uniform on all bounded intervals.

  4. If $f$ is compactly supported in $[-R,R]$ and infinitely differentiable, \begin{align} \hat{f}(\omega) & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt \\ & = \frac{1}{\sqrt{2\pi}(i\omega)^k}\int_{-R}^{R}f^{(k)}(t)e^{-i\omega t}dt \\ |\hat{f}(\omega)| & \le \frac{C}{|\omega|^k}. \end{align} Then, the following uniform estimate gives the uniform convergence of $\{ f_n \}$: $$ |\lim_n f_n(x)-f_n(x)| \le \int_{|x|\ge n}|\hat{f}(\omega)|d\omega \le\int_{|x|\ge n}\frac{C}{|\omega|^k}d\omega $$

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  • $\begingroup$ Could you explain the uniform estimate? $\endgroup$ – user153312 Jul 13 '16 at 21:54
  • $\begingroup$ @Exterior : The estimate on $\hat{f}$ follows using integration by parts. Then the uniform bound on $|\lim_n f_n(x)-f_n(x)|$ follows from the resulting bounded on $|\hat{f}(\omega)|$. $\endgroup$ – DisintegratingByParts Jul 13 '16 at 22:04

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