Past exam of the short options course of the University of Oxford / 2023 / 5 / c

Use the Fourier transform convention $\hat{f}(s)={\int }_{\mathbb{R}}{e}^{-isx}f(x)dx$. First observe that $f(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ satisfies the differential equation, decay, and normalization, using the Gaussian integral. The uniqueness allowed in the question identifies it with $f$. In particular it is a Schwartz function, so the derivative and multiplication rules below are justified.

Taking the Fourier transform of the differential equation givesThe Riemann-Lebesgue lemma gives decay of $\hat{f}$. Every decaying solution $h$ of this equation is proportional to $e^{-s^2/2}$: setting $w=h^{\prime }+sh$ gives $w^{\prime }=sw$, and integration yieldsThe term with $C\ne 0$ grows like $C{e}^{s^2/2}/(2s)$ as $s\to +\infty$, so decay forces $C=0$. Thus $\hat{f}=\lambda f$ for a scalar $\lambda$. This establishes proportionality before using Fourier inversion to calculate the scalar.

At zero, the Fourier transform is $\hat{f}(0)=\int f=1$, so $\lambda f(0)=1$. In particular $\lambda \ne 0$ and $f(0)=1/\lambda$. The Fourier inversion theorem at zero also givesEquating the two values yields $\boxed{{\lambda }^2=2\pi }$. Since the explicit normalized Gaussian function in part (i) has $f(0)>0$, here $\lambda =\sqrt{2\pi }$.

In the space of tempered distributions, this Fourier transform convention gives $\hat{1}=2\pi \delta$ and $\hat{\delta }=1$. ThereforeThe constant distribution and the Dirac delta distribution are linearly independent: testing away from zero first determines the constant coefficient, and then testing at zero determines the delta coefficient. Hence $c=\lambda$ and $\lambda c=2\pi$, giving