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 $t\in\mathbb R$.
- 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.
- 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?
Best Answer
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.