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I have read that for $f\in\mathcal{L}_2(\mathbb R)$, its Fourier transform need not coincide with the familiar form for $\mathcal{L}_1(\mathbb R)$-functions, on account that this integral might not exist. For $g\in\mathcal{L}_1(\mathbb R)$ the 'familiar form' of its Fourier transform I take to be: $$\hat g(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp(-ist)g(s)\text{d}s,$$ for $t\in\mathbb R$.

  1. What are examples of $f\in \mathcal{L}_2(\mathbb R)$ for which this idea is exemplified? That is to say, for whom the above integral representation need not exist; the more accessible the example, the better.

To circumvent this, for $f\in \mathcal{L}_2(\mathbb R)$ we can take as the expression for its Fourier transform $$\hat f(t)=\lim_{N\to\infty}\int_{-N}^N\exp(-ist)f(s)\text{d}s.$$ The limit here is with respect to the $\mathcal{L}_2(\mathbb R)$-norm. With regards to this, I have read that this representation is on account of the Dominated Convergence Theorem.

  1. How is it, exactly, that the Dominated Convergence Theorem has been applied to give us this workable form of the Fourier transform for $f\in\mathcal{L}_2(\mathbb R)$? Also, how does it allow us to avoid the convergence issues of the 'familiar form' first considered?
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  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa Jun 4 '19 at 19:31
  • $\begingroup$ Hi there - I will survey these and bear them in mind, thanks! Is there anything in particular you think I should change/add to in my question? $\endgroup$ – FourierFan69 Jun 4 '19 at 19:33
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    $\begingroup$ You can forget the dominated convergence theorem. Let $\hat{f}_N(t) =\int_{-N}^N\exp(-ist)f(s)\text{d}s$, if $f \in L^2$ then $\hat{f}_N$ is continuous and bounded. The theorem is that $\int_{-\infty}^\infty |\hat{f}_N(t)-\hat{f}_M(t)|^2dt = 2\pi \int_{-N}^{-M}+\int_{M}^{N} |f(s)|^2ds$. Thus $(\hat{f}_N)_N$ is a Cauchy sequence in $L^2$ and its limit is called $\hat{f}$. $\endgroup$ – reuns Jun 4 '19 at 19:40
  • $\begingroup$ @FourierFan69: looks like a great question. Too often, users down vote or vote to close without leaving feedback. While the guides reflect collective standards, individual responses vary greatly. Hopefully, you can use the links to better make sense of the responses and to enjoy your interactions. $\endgroup$ – dantopa Jun 4 '19 at 19:41
  • $\begingroup$ Not only you can forget about the dominated convergence theorem (cfr. reuns), but you should forget about it. The point is exactly that you cannot compute that integral via dominated convergence, because $f\notin L^1$. $\endgroup$ – Giuseppe Negro Dec 9 '19 at 14:54
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Take as an example $f(t)=1/\sqrt{1+t^2}$, which is a function in $L^2(\mathbb{R})$ that is not in $L^1(\mathbb{R})$. So the Fourier transform of $f$ is not absolutely convergent. However, the following limit does exist as a function in $L^2$: $$ \hat{f}(s)=\lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}\frac{1}{\sqrt{1+t^2}}e^{-ist}dt \\ = \lim_{R\rightarrow\infty}\sqrt{\frac{2}{\pi}}\int_{0}^{R}\frac{1}{\sqrt{1+t^2}}\cos(st)dt. $$ The above limit not only converges in the $L^2$ norm as $R\rightarrow\infty$, but it also converges pointwise everywhere except at $s=0$, which can be seen by looking at this integral in light of the alternating series test for convergence, keeping in mind that $1/\sqrt{1+t^2}$ is strictly monotone and converges to $0$ as $t\rightarrow\infty$. However, the integral does not converge absolutely for any $s\in\mathbb{R}$.

The above integral converges pointwise using the same reasoning if $1/\sqrt{1+t^2}$ is replaced by $1/(1+t^2)^{p}$ for any $p > 0$. However, the result is a Fourier transform that is not in $L^2(\mathbb{R})$ if $0 < p < 1/4$ because $1/(1+t^2)^p \notin L^2(\mathbb{R})$ for such $p$. So it cannot converge in $L^2$ for $0 < p < 1/4$.

I don't believe that the dominated convergence theorem says much about these simple cases.

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