Finding the spectral decomposition of $\Delta= \frac{d^2}{dx^2}$

Question

What is the spectral decomposition of the operator $\Delta= \frac{d^2}{dx^2}$ in $(L^{2}(\mathbb R), dx)$?

Thanks you in advance

Answer 1

The operator $\Delta = -\frac{d^2}{dx^2}$ has a natural domain $\mathcal{D}(L)\subset L^2(\mathbb{R})$ consisting of all twice absolutely continuous functions $f \in L^2$ for which $f'' \in L^2$. On this domain, $\Delta$ is selfadjoint in the strictest sense. The spectral resolution of the identity for this selfadjoint operator is $$ E(S)f= \mathcal{F}^{-1}(\chi_{\{s : s^2 \in S\}}\mathcal{F}(f)), $$ where $\mathcal{F}$ is the Fourier transform and $\mathcal{F}^{-1}$ is the inverse Fourier transform on $L^2$. For example, the spectrum of $\Delta$ is $[0,\infty)$ and \begin{align} E[0,t]f & = \mathcal{F}^{-1}(\chi_{[-\sqrt{t},\sqrt{t}]}\mathcal{F}(f)) \\ &=\frac{1}{\sqrt{2\pi}}\int_{-\sqrt{t}}^{\sqrt{t}}e^{isx}\hat{f}(s)ds \\ &=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(y)\int_{-\sqrt{t}}^{\sqrt{t}}e^{is(x-y)}dsdy \\ &=\frac{1}{\pi}\int_{-\infty}^{\infty}f(y)\frac{\sin(\sqrt{t}(y-x))}{y-x}dy \end{align} This is classically defined for $0 < t < \infty$ and $f \in L^2$.