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While trying to understand a proof in a paper, I came upon the following a calculation needing the following identity: $$\lim_{t\to 0} \int_{-\infty}^\infty \left(e^{-\log(4\pi i t)/2} e^{ik^2/4t} -\delta(k)\right)f(k)\,dk=0.$$ for $f\in\mathcal{S}(\mathbb{R})$ and $t>0$. Of course, this means that the exponential kernel converges to the Dirac measure in the limit $t \to 0$ in the sense of distributions. I'm able to show this is the case if $t$ approaches $0$ along the negative imaginary axis in the complex plane, since then, we can apply the dominated convergence theorem after the substitution $k\mapsto k\sqrt{t}$ and exchange limit and integral. But in this case, this doesn't help us because $e^{-ik^2/4}$ lies on the unit circle for all $k\in\mathbb{R}$, and thus its modulus isn't integrable. We can also try shifting the contour of integration by $\pi/4$ radians and attempt to apply Cauchy's formula to take the contour back to the real axis, but I run into issues where $f$, extended to the complex plane, may not be bounded on the rays/curves we are interested in.

Perhaps this equality does not hold the way I've asked it. What is a way to derive such an equality and the context in which it is true?

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Allow me to replace $k$ by $x$. The kernel $$G(x,t)=(4\pi it)^{-1/2}e^{ix^2/4t}$$ is the Green function of the Schrödinger equation, which can be written in the integral form $$G(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}e^{-ik^2 t}dk.$$ In the limit $t\rightarrow 0$ we then have an integral representation of the delta function, $$2\pi\delta(x)=\int_{-\infty}^\infty e^{-ikx}dk.$$ Both these integral representations need to be understood in the distributional sense (they are formally divergent for real $x$ and $t$; they need to be multiplied by a test function and integrated for a finite answer).

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  • $\begingroup$ Thank you for the answer. How do you justify switching the limit and the integral though? $\endgroup$
    – Dispersion
    Jun 22, 2021 at 19:57
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    $\begingroup$ If $\omega_t \to \omega_0$ as tempered distributions, then testing on any Schwartz function you get $\langle \omega_t - \omega_0, f\rangle \to 0$ (by definition). What Carlo's answer suggests doing is to take the Fourier transform on both sides. Using that Fourier transform is a continuous mapping from Schwartz space to itself, and hence dually from the space of tempered distribution to itself, you get that $\hat{\omega}_t \to \hat{\omega}_0 \iff \omega_t \to \omega_0$. The reason for taking Fourier transforms is that the convergence of $\hat{\omega}_t \to \hat{\omega}_0$ is a lot more obvious $\endgroup$
    – Willie Wong
    Jun 22, 2021 at 20:03
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    $\begingroup$ By definition $\hat{\omega}_t\to \hat{\omega}_0$ iff for every $f$ Schwartz $\langle \hat{\omega}_t - \hat{\omega}_0, f\rangle \to 0$. This you can justify by dominated convergence. (Here $\omega_t$ is the tempered distribution given by $G(t,x)$ and $\omega_0 = \delta$..) $\endgroup$
    – Willie Wong
    Jun 22, 2021 at 20:06
  • $\begingroup$ Thank you for the explanation Willie Wong. I think I now understand the idea. $\endgroup$
    – Dispersion
    Jun 23, 2021 at 15:00

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