Section 8.4 Spectral Decomposition
We saw that unbounded self-adjoint operators with compact resolvent have a spectral decomposition identical to that of self-adjoint operators in finite dimension. All these operators have a purely punctual spectrum. What happen with operators whose spectrum is not discrete? Example: spectral decomposition of the multiplication operator. Consider the case of \(M_x:f(x)\mapsto xf(x)\) on \(L^2(\bR)\text{.}\) We saw that \(M_x\text{,}\) with \(D(M_x)=\{f\in L^2(\bR) : M_xf\in L^2(\bR) \}\) is self-adjoint and \(\sigma(M_x) =\bR\text{.}\) As always when trying to pass from discrete sums to integrals, let us approximate \(M_x\) by a locally constant function, namely let us subdivide the real line (i.e. the spectrum of \(M_x\)) in subintervals \(I_n=[nh,(n+1)h], n\in\mathbb Z\text{,}\) and set \(\lambda_n=nh\text{.}\) Then the function
\begin{equation*}
\ell(x)=\sum_{n=-\infty}^\infty\lambda_n 1_{I_n}(x)
\end{equation*}
approximates the identify function and so
\begin{equation*}
M_x(f)(x) \sim \sum_{n=-\infty}^\infty\lambda_n 1_{I_n}(x)f(x).
\end{equation*}
In other words,
\begin{equation*}
M_x \sim \sum_{n=-\infty}^\infty\lambda_n E(I_n) ,
\end{equation*}
where \(E(I_n):L^2(\bR)\to L^2(\bR)\) is the multiplication operator by \(1_{I_n}\text{.}\) These operators \(E(I_n)\) are projections! The reason is simple: \(1_{I_n}^2\text{.}\) Moreover, their sum gives the identity:
\begin{equation*}
(\sum_{n=-\infty}^\infty E(I_n) f) (x) = \sum_{n=-\infty}^\infty 1_{I_n}(x)f(x) = (\sum_{n=-\infty}^\infty 1_{I_n}(x))f(x) = f(x)
\end{equation*}
(the picky reader could notice that the sum will be different from 1 at the endpoints but the value of this sum in a countable set of points is not relevant). Hence, this approximation of \(M_x\) is completely analogue to the compact resolvent case. But this is just an approximation. For \(h\to0\text{,}\) the sum becomes an integral:
\begin{equation*}
\sum_{n=-\infty}^\infty E(I_n) (f) \to \int_\bR \lambda dE(\lambda) (f).
\end{equation*}
Just as in case of real integrals, \(dE(\lambda)\) is just a symbol, does not mean anything in itself (no more than thinking that \(dx\) is an "infinitesimal change in \(x\)"!) Its meaning is given by its integral over a measurable set:
\begin{equation*}
(\int_{[a,b]} dE(\lambda) f) (x)= 1_{[a,b]}(x) f(x)
\end{equation*}
Similarly,
\begin{equation*}
(\int_\bR \lambda dE(\lambda) f) (x) = \lim_{h\to0}\sum_{n=-\infty}^\infty \lambda_n 1_{I_n}(x)f(x) = x f(x).
\end{equation*}
The intuitive meaning is clear: for every \(\lambda\in\sigma(M_x)\text{,}\) the action of \(M_x\) on the portion of \(f(x)\) based at \(x=\lambda\) (which is given by \(dE(\lambda) f\)) is to multiply it by \(\lambda\text{.}\) After that, all these values are "added together" to give the final result, namely \(xf(x)\text{.}\) We can also provide a further heuristic way of looking at \(dE(\lambda)dE(\lambda)\) such that \(M_x=\int_{\sigma(M_x)} \lambda dE(\lambda)\text{.}\) We also know that the Fourier transform exchanges the derivative operator with the multiplication one, namely
\begin{equation*}
\frac{1}{i}\frac{d}{dx} = {\cal F}^{-1} M_k {\cal F}.
\end{equation*}
Recall that
\begin{equation*}
M_k \sim \sum_{n=-\infty}^\infty\lambda_n E(I_n),
\end{equation*}
so
\begin{equation*}
\frac{1}{i}\frac{d}{dx} = {\cal F}^{-1} \sum_{n=-\infty}^\infty\lambda_n E(I_n) {\cal F} = \sum_{n=-\infty}^\infty \lambda_n{\cal F}^{-1} E(I_n) {\cal F} = \sum_{n=-\infty}^\infty \lambda_n F(I_n).
\end{equation*}
The operators
\begin{equation*}
F(I_n) = {\cal F}^{-1} E(I_n) {\cal F}
\end{equation*}
are projectors just like the \(E(I_n)\) since
\begin{equation*}
F(I_n)F(I_n) = {\cal F}^{-1} E(I_n) {\cal F} {\cal F}^{-1} E(I_n) {\cal F} = {\cal F}^{-1} E(I_n)E(I_n) {\cal F} = {\cal F}^{-1} E(I_n) {\cal F} = F(I_n).
\end{equation*}
Passing to the limit for \(h\to0\text{,}\) we get that
\begin{equation*}
\frac{1}{i}\frac{d}{dx} = \int_{\mathbb R}\lambda dF(\lambda) = \int_{\sigma(\frac{1}{i}\frac{d}{dx})}\lambda dF(\lambda).
\end{equation*}
Even in this case we can also provide a further heuristic way of looking at \(dE(\lambda)\text{.}\) The operator \(-i d/dx\) does have eigenfunctions in \(C^\infty(\bR)\text{:}\)
\begin{equation*}
\frac{1}{i}\frac{d}{dx} e^{ikx} = k e^{ikx}.
\end{equation*}
The problem is that these functions are not in \(L^2(\bR)\text{.}\) Nevertheless, set
\begin{equation*}
e_k = e^{ikx}
\end{equation*}
and
\begin{equation*}
E(k) = e_k\otimes \varepsilon^k,
\end{equation*}
where
\begin{equation*}
\varepsilon^k(f) = \langle e_k,f\rangle_{L^2(\bR)}.
\end{equation*}
In physicists' notation, this would look like
\begin{equation*}
E(k) = |e_k\rangle \langle e_k|
\end{equation*}
Then
\begin{equation*}
dE(\lambda) = E(\lambda)d\lambda = e_\lambda\otimes \varepsilon^\lambda d\lambda.
\end{equation*}
and
\begin{equation*}
(\int_\bR \lambda dE(\lambda) f) (x) = (\int_\bR e_\lambda \varepsilon^\lambda(f) d\lambda)(x) =
\end{equation*}
\begin{equation*}
= \int_\bR \lambda e^{i \lambda x} (\int_\bR e^{-i \lambda y} f(y) dy) d\lambda
= \int_\bR e^{i \lambda x} \lambda \hat f(\lambda) d\lambda = \frac{1}{i}\frac{d}{dx}f(x)
\end{equation*}
so we really get back the derivative operator. This argument, though, sufferrs of the same problems of the one for the multiplication operator.
Example: spectral decomposition of the Laplacian operator. In case of the Laplacian operator \(-\Delta\) on \(\mathbb R^m\) we saw that
\begin{equation*}
-\Delta = {\cal F}^{-1} M_{|k|^2} {\cal F}.
\end{equation*}
In momentum space, which is now \(m\)-dimensional, we subdivide the space into hypercubes \(I_n\text{,}\) where now \(n\in\mathbb Z^m\text{,}\) and we get that
\begin{equation*}
M_{|k|^2} \sim \sum_{n=-\infty}^\infty\lambda_n E(I_n),
\end{equation*}
where now
\begin{equation*}
E(I_n) = M_{1_{I_n}}\text{ and }\lambda_n = |n|^2.
\end{equation*}
Hence, just as we did for the derivative operator,
\begin{equation*}
-\Delta = {\cal F}^{-1} \sum_{n=-\infty}^\infty\lambda_n E(I_n) {\cal F} = \sum_{n=-\infty}^\infty \lambda_n{\cal F}^{-1} E(I_n) {\cal F} = \sum_{n=-\infty}^\infty \lambda_n F(I_n).
\end{equation*}
Passing to the limit for \(h\to0\text{,}\) we get that
\begin{equation*}
-\Delta = \int_{[0,\infty)}\lambda dF(\lambda) = \int_{\sigma(-\Delta)}\lambda dF(\lambda).
\end{equation*}
Subsection 8.4.1 Applications of the spectral decomposition
As we pointed out already, the main reason why we care about the spectral decomposition is that it allows us to evaluate functions of operators: given
\begin{equation*}
A = \int_{\sigma(A)}\lambda dE(\lambda)
\end{equation*}
and given a continuous real or complex 1-variable function \(\phi\text{,}\)
\begin{equation*}
\phi(A) = \int_{\sigma(A)}\phi(\lambda) dE(\lambda).
\end{equation*}
The most frequent case is the exponential:
\begin{equation*}
e^{c A} = \int_{\sigma(A)}e^{c \lambda} dE(\lambda).
\end{equation*}
We list below several cool things we can accomplish because of that.
1. Solving PDEs Consider the Schrodinger equation
\begin{equation*}
\frac{1}{i}\frac{d}{dt}\psi = \Delta\psi, \psi(0)=\psi_0.
\end{equation*}
Rewrite it as
\begin{equation*}
\dot\psi = -i A\psi,\quad A=-\Delta.
\end{equation*}
Then
\begin{equation*}
\psi(t) = e^{-i At}\psi_0
\end{equation*}
Functional calculus gives
\begin{equation*}
e^{-i At} = \int_{\sigma(A)}e^{-i \lambda t} dE(\lambda).
\end{equation*}
By passing again to discrete sums and then taking the limit, one can see that
\begin{equation*}
\int_{\sigma(A)}e^{-i \lambda t} dE(\lambda)\psi_0 =
\cF^{-1}(e^{-i t|k|^2}\hat \psi_0(k)),
\end{equation*}
so the solution is ultimately given, in momentum space, by
\begin{equation*}
\hat\psi(t,k) = e^{-i t|k|^2}\hat\psi_0(k).
\end{equation*}
In case of the heat equation
\begin{equation*}
\dot u = \Delta u, u(0)=u_0
\end{equation*}
we get
\begin{equation*}
u(t) = e^{At}u_0
\end{equation*}
Functional calculus gives
\begin{equation*}
e^{At} = \int_{\sigma(A)}e^{\lambda t} dE(\lambda).
\end{equation*}
By passing again to discrete sums and then taking the limit, one can see that
\begin{equation*}
\int_{\sigma(A)}e^{\lambda t} dE(\lambda)u_0 =
\cF^{-1}(e^{t|k|^2}\hat u_0(k)),
\end{equation*}
so the solution is ultimately given, in momentum space, by
\begin{equation*}
\hat u(t,k) = e^{t|k|^2}\hat u_0(k).
\end{equation*}
Consider finally the wave equation
\begin{equation*}
\ddot u = \Delta u, u(0)=u_0, \dot u(0) = u_1.
\end{equation*}
Here we get
\begin{equation*}
u(t) = \cos(t\sqrt{-\Delta})u_0 + \frac{\sin(t\sqrt{-\Delta})}{\sqrt{-\Delta}}u_1.
\end{equation*}
and the solution is ultimately given, in momentum space, by
\begin{equation*}
\hat u(t,k) = \cos(t|k|)\hat u_0(k) + \frac{\sin(t|k|)}{|k|}\hat u_1(k).
\end{equation*}
2. Defining nonlocal operators Funtional calculus makes it easy defining operators such as
\begin{equation*}
(-\Delta)^{s/2} := \int_{[0,\infty)}\lambda^{s/2}dE(\lambda).
\end{equation*}
In momentum space, this writes as
\begin{equation*}
\cF((-\Delta)^{s/2}u)(k) = |k|^s\hat u(k).
\end{equation*}
Similarly, one can define
\begin{equation*}
(1-\Delta)^{s/2}:=\int_{[0,\infty)}(1+\lambda)^{s/2}dE(\lambda),
\end{equation*}
so that
\begin{equation*}
\cF((1-\Delta)^{s/2}u)=(1+|k|^2)^{s/2}\hat u(k).
\end{equation*}
This is used to define non-integer Sobolev spaces such as
\begin{equation*}
H^s(\bR^n) = \{u\in L^2(\bR^n) : (1+|k|^2)^{s/2}\hat u(k)\in L^2(\bR^n)\}
\end{equation*}
3. Norms estimates If \(\phi\) is continuous and bounded, so is \(\phi(A)\text{:}\)
\begin{equation*}
\|\phi(A)\|_{op}=\sup_{\lambda\in\sigma(A)}|\phi(\lambda)|
\end{equation*}
For instance, in case of the Schrodinger equation,
\begin{equation*}
\|e^{it\Delta}\|=\sup_{\lambda\in[0,\infty)|e^{-it\lambda}|=1.
\end{equation*}
In case of the heat equation,
\begin{equation*}
\|e^{t\Delta}\|=\sup_{\lambda\in[0,\infty)|e^{-t\lambda}|\leq1
\end{equation*}
4. Writing resolvents in terms of integrals. Functional calculus shows that the resolvent can be written as
\begin{equation*}
(z-A)^{-1} = \int_{\sigma(A)}\frac{dE(\lambda)}{z-\lambda}
\end{equation*}
This formula can be re-written as a complex integral and is the starting point of an important application of complex integrals to functional analysis.