I have been reading some notes on spectral theorem for unbounded operator on Hilbert spaces and my toy example is the Laplacian on $\mathbb{R}^d$. The author says that as an "application" one can define the spectral projector of the operator. For the Laplacian on $\mathbb{R}^d$ the spectral projector is $$ \chi_{[0, a]}(-\Delta)$$ where $\chi_A$ denotes the characteristic function of a set $A$, subset of the its spectrum, i.e. of $[0,\infty)$.
Then, I saw the claim $$\chi_{[0, a]}(-\Delta): L^2(\mathbb{R}^d)\rightarrow E_a$$ where $E_a$ is the subspace of all $L^2(\mathbb{R}^d)$-functions with Fourier Transform supported in $[-a,a]^d$.
I am trying to figure out now what $\chi_{[0, a]}(-\Delta) f$ for any $L^2$-function $f$ is but I am lost. Can anybody explain me that?
I understand that in some sense the spectral projector gets rid of all the eigenvalues bigger than $a$, but what this has to do with Fourier Transform? Any help is welcome.
The operator $-\Delta : H^2(\mathbb{R}^d)\subset L^2(\mathbb{R}^d)\rightarrow L^2(\mathbb{R}^d)$ is a closed, densely-defined selfadjoint linear operator with continuous spectrum $[0,\infty)$. The spectral projection associated with $[0,a]$ is given by \begin{align} P[0,a]f & = \frac{1}{(2\pi)^{d}}\int_{|\xi|^2 \le a}\left(\int f(x')e^{-ix'\cdot\xi}dx' \right)e^{ix\cdot\xi}d\xi \\ & = \frac{1}{(2\pi)^{d/2}}\int_{|\xi|^2\le a}\hat{f}(\xi)e^{ix\cdot\xi}d\xi. \end{align} Basically you're summing over all exponentials $e^{i\xi\cdot x}$ where $|\xi|^2 \le a$, which is natural because $$ -\Delta_x e^{i\xi\cdot x} = |\xi|^2e^{i\xi\cdot x}. $$ In other words, $$ P[0,a]f = (\chi_{|\xi|^2\le a}f^{\wedge}(\xi))^{\vee}. $$ Using this last definition as a Fourier multiplier, it is fairly easy to verify that $P$ is a spectral measure.