I am now studying Quantum Mechanics, and I am facing some issue about the convergence of Fourier Transform. I want to know when I can have the following equality:
$$
\lim_{n \to \infty}(\mathcal{F}f_n)(\omega)= \mathcal{F}(\lim_{n \to \infty}f_n)(\omega)
$$
Where $\mathcal{F}f(\omega)=\int_\mathbb{R}f(x)e^{-2\pi ix\omega}$. Both the limit on the LHS and RHS means convergent pointwise.
Of course, I can apply Lebesgue Dominated convergence theorem here when $f(x)e^{-2\pi ix\omega}$ is integrable. However, even if it is not integrable, the fourier transformation could exist. I try to find a theorem about whether we can swap lim and $\mathcal{F}$, but I have not get anything.
Let's take an example. Take $f_n(x)=\cos{x}$ when $x\in [-\frac{(2n-1)\pi}{2},\frac{(2n-1)\pi}{2}]$, and $f_n(x)=0$ otherwise. Then $f_n \to \cos$ pointwise. The fourier transform of $f_n$ is something like $(-1)^n\frac{2\cos(\frac{(2n-1)\pi}{2}x)}{x^2-1}$. But the Fourier transform of $\cos x$ is $\sqrt{\frac{\pi}{2}}\delta(\omega-1)+\sqrt{\frac{\pi}{2}}\delta(\omega+1)$. Clearly $\lim_{n \to \infty}(\mathcal{F}f_n)(\omega)= \mathcal{F}(\lim_{n \to \infty}f_n)(\omega)$ is not true in this case. Why it is not true?
Could anyone list some general theorems about this topic?
Best Answer
Let $g_R(x) = \chi_{[-R,R]}(x) \cos x.$ Then the Fourier transform can easily be calculated: $$ \widehat{g_R}(\xi) = \int_{-\infty}^{\infty} g_R(x) \, e^{-i \xi x} \, dx = \int_{-R}^{R} \cos x \, e^{-i \xi x} \, dx = \int_{-R}^{R} \frac12 \left( e^{ix} + e^{-ix} \right) \, e^{-i \xi x} \, dx \\ = \frac12 \int_{-R}^{R} \left( e^{i(1-\xi)x} + e^{-i(1+\xi)x} \right) \, dx = \frac12 \left[ \frac{1}{i(1-\xi)} e^{i(1-\xi)x} + \frac{1}{-i(1+\xi)} e^{-i(1+\xi)x} \right]_{-R}^{R} \\ = \frac12 \left[ \frac{1}{i(1-\xi)} e^{i(1-\xi)R} + \frac{1}{-i(1+\xi)} e^{-i(1+\xi)R} - \frac{1}{i(1-\xi)} e^{i(1-\xi)(-R)} - \frac{1}{-i(1+\xi)} e^{-i(1+\xi)(-R)} \right] \\ = \frac{1}{1-\xi} \frac{e^{i(1-\xi)R} - e^{-i(1-\xi)R}}{2i} + \frac{1}{1+\xi} \frac{e^{i(1+\xi)R} - e^{-i(1+\xi)R}}{2i} \\ = \frac{\sin (1-\xi)R}{1-\xi} + \frac{\sin (1+\xi)R}{1+\xi} $$ After a small rewrite we end up with $$\widehat{g_R}(\xi) = R \frac{\sin (\xi-1)R}{(\xi-1)R} + R \frac{\sin (\xi+1)R}{(\xi+1)R} = K_R(\xi-1) + K_R(\xi+1),$$ where $$K_R(\xi) = R \frac{\sin \xi R}{\xi R}.$$
As a function, $K_R$ does not converge as $R \to \infty,$ neither pointwise nor in $L^p.$ But as a distribution it does. For any $\phi \in C_c^\infty(\mathbb R)$ or $\mathscr S(\mathbb R)$ we have $$ \langle K_R, \phi \rangle = \int_{-\infty}^{\infty} R \frac{\sin \xi R}{\xi R} \phi(\xi) \, d\xi = \{ \text{set } \eta = \xi R \} = \int_{-\infty}^{\infty} \frac{\sin\eta}{\eta} \phi(\eta/R) \, d\eta \\ \to \int_{-\infty}^{\infty} \frac{\sin\eta}{\eta} \phi(0) \, d\eta = \int_{-\infty}^{\infty} \frac{\sin\eta}{\eta} \, d\eta \, \phi(0) = \pi \, \phi(0) = \langle \pi \, \delta, \phi \rangle, $$ i.e. $K_R \to \pi \, \delta$ as a distribution. Thus, $\widehat{g_R}(\xi) = K_R(\xi-1) + K_R(\xi+1) \to \pi \, \delta(\xi-1) + \pi \, \delta(\xi+1)$ as a distribution.
And since $$ \cos x = \frac12 (e^{ix} + e^{-ix}) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left( \pi \, \delta(\xi-1) + \pi \, \delta(\xi+1) \right) e^{i \xi x} \, d\xi $$ we have $$\widehat{\cos}(\xi) = \pi \, \delta(\xi-1) + \pi \, \delta(\xi+1).$$
Thus, as a distribution, $g_R \to \cos$ and $\widehat{g_R} \to \widehat{\cos},$ i.e. limit commutes with Fourier transform.