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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.

Michela
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  • This is tangentially related, maybe it is useful – Giuseppe Negro Nov 03 '17 at 20:09
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    The "eigenfunctions" of the Laplacian are the plane waves $e^{i a x}$, where $a\in\mathbb R^d$. So it is not a surprise that the Fourier transform pops up: if $f\in L^2(\mathbb R^d)$, the "projection" on the "eigenspace" corresponding to $a\in\mathbb R^d$ is $\hat{f}(a)$. The " " are necessary because plane waves are not true elements of $L^2(\mathbb R^d)$ (they have infinite $L^2$ norm), corresponding to the fact that the Laplacian has continuous spectrum. – Giuseppe Negro Nov 03 '17 at 20:12

2 Answers2

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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.

Disintegrating By Parts
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  • I think I understood. I only have one last comment/remarks. If I consider the Laplacian on $[0,L]$ with, say, periodic boundary conditions, the spectral projector gives a point measure, because the spectrum is discrete. Then, we have $P[0,a]f=P[0,a]\sum_{\lambda_n\geq 0} b_n e^{i\lambda_n x}$=\sum_{\lambda_n\leq a} b_n e^{i\lambda_n}$. Am I right? – Michela Nov 03 '17 at 22:37
  • Actually, I have one more. If I consider the Laplacian with Dirichlet Boundary conditions on the strip $[0,L]\times\mathbb{R}$, I think the spectrum is $[0,\infty)$ and the eigenvalues should be of the form $\lambda^2+\frac{(n\pi)^2}{L^2}$, because I think I can treat the strip as a product manifold. In this case there is no Fourier Transform on the strip , what would then the spectral projector be? Is it some kind of mixture between the Fourier Transform on $\mathbb{R}$ and the discrete Fourier Transform on the interval $[0,L]$? – Michela Nov 03 '17 at 22:49
  • @Michela : In the first case of a discrete sum, yes, $P[0,a]$ is the sum over all discrete projection components where the eigenvalues are in $[0,a]$. in the second case, the sum over all the components within a given range includes discrete sums of truncated integrals where all values of $\lambda^2 + (n\pi)^2/L^2$ are in $[0,a]$. – Disintegrating By Parts Nov 03 '17 at 23:32
  • I am not sure I have understood your answer about the spectral projector on the strip. Does your answer mean that for $g=P[0,a]f$ for $f\in L^2([0,L]\times\mathbb{R})$, the FT (Fourier Transform ) of $g$ on $\mathbb{R}$ has support in $[-a,a]$ and the discrete $FT$ of $g$ is supported in $[-a,a]\cap \frac{1}{L}\mathbb{Z}$? – Michela Nov 04 '17 at 10:54
  • @Michela : $P[0,a]$ is obtained by summing over all $n$ for which $(n\pi)^2/L^2 < a^2$ and the term involving $n$ will also involve a truncated Fourier transform integral which is truncated to $\pm r$ where $r^2+(n\pi)^2/L^2 \le a$. The idea is to sum all eigenvalue components $\lambda$, $n$ for which $\lambda^2+(n\pi)^2/L^2 \le a$. – Disintegrating By Parts Nov 04 '17 at 15:56
  • Ok, thanks for the clarification. It is indeed what I thought. – Michela Nov 04 '17 at 16:31
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You might also have a look at the example on p204 in my book

http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/index.html

where an explicit formula for this projection is derived.

gerald
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