Ytukyg

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 $mathcal{F}f(omega)=int_mathbb{R}f(x)e^{-2pi ixomega}$. Both the limit on the LHS and RHS means convergent pointwise.

Of course, I can apply Lebesgue Dominated convergence theorem here when $f(x)e^{-2pi ixomega}$ 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 $xin [-frac{(2n-1)pi}{2},frac{(2n-1)pi}{2}]$, and $f_n(x)=0$ otherwise. Then $f_n to cos$ pointwise. The fourier transform of $f_n$ is something like $(-1)^nfrac{2cos(frac{(2n-1)pi}{2}x)}{x^2-1}$. 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?

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
$$
langle K_R, phi rangle
= int_{-infty}^{infty} R frac{sin xi R}{xi R} phi(xi) , dxi
= { text{set } eta = xi R }
= int_{-infty}^{infty} frac{sineta}{eta} phi(eta/R) , deta \
to int_{-infty}^{infty} frac{sineta}{eta} phi(0) , deta
= int_{-infty}^{infty} frac{sineta}{eta} , deta , phi(0)
= pi , phi(0) = langle pi , delta, phi rangle,
$$
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
$$
cos x
= frac12 (e^{ix} + e^{-ix})
= frac{1}{2pi} int_{-infty}^{infty} left( pi , delta(xi-1) + pi , delta(xi+1) right) e^{i xi x} , dxi
$$
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.

One simple criteria would be dominated convergence. That is, if ${ f_n }$ are functions such that $|f_n| le g$ with $gin L^1$, and if ${ f_n }$ converges pointwise to $f$, then what you wrote would hold. The Lebesgue dominated convergence theorem would give it.

Just having $fin L^1$ is not enough to have dominated convergence if ${ f_n }$ only converges poinwise, is it?