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I'm trying to gain some intuition for the usefullness of the spectral theory for bounded self adjoint operators. I work in PDE and any interesting applications/examples I've ever encountered are concerning compact operators and unbounded operators. Here I have the examples of $-\Delta$, the laplacian and $(-\Delta)^{-1}$, the latter being compact.

The most common example I see of a bounded non-compact operator is the shift map on $l_2$ given by $T(u_1,u_2,\cdots) = (u_2,u_3,\cdots)$. While this nicely illustrates the different kind of spectra, I don't see why this is useful or where this may come up in practice.Why does knowing things about the spectrum of the shift operator help you in any practical way?

Secondly, concerning the spectral theorem for bounded, self adjoint operators. All useful applications I have encountered concern compact or unbounded operators. Is there an example arising in PDE (preferably) or some other applied field where knowing the spectral representation for a bounded, non-compact operator is useful? I have yet to encounter one that didn't just reduce to the compact case. Any insight/suggestions are appreciated.

Best, dorian

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6 Answers 6

I think Helge's answer cuts to the historical heart of the matter: solution operators for various differential equations tend to be bounded, non-compact operators (obtained in many cases from an unbounded differential operator via the functional calculus), and it is often quite useful from that point of view to know something about their spectra. This is one reason why the theory of elliptic operators over non-compact spaces is more complicated than the corresponding theory for compact spaces: one has to deal with the fact that the eigenvalues of the solution operators can accumulate at 0.

What I find more persuasive, however, are the ways in which spectral theory mediates the relationship between functional analysis and geometry. In many cases you will miss this relationship unless you ignore compact operators entirely. The celebrated Atiyah-Singer Index Theorem provides a particularly dramatic example of this phenomenon, but I'll focus on more digestible examples (A-S is more about spectral theory for unbounded operators anyway).

First, consider the classical Toeplitz operators. Given a complex valued function $g \in C(S^1)$, the Toeplitz operator $T_g$ with symbol $g$ is defined as follows. Form the Hardy space $H^2(S^1)$ by considering the $L^2$-closure of the space of polynomial functions on $S^1$, regarding $S^1$ as a subspace of $\mathbb{C}$, and let $P$ denote the orthogonal projection from $L^2(S^1)$ to $H^2(S^1)$. Then define $T_g: H^2(S^1) \to H^2(S^1)$ to be $T_g(f) = T(fg)$. This is a bounded operator, and the classical Toeplitz index theorem asserts that its Fredholm index (the dimension of its kernel minus the dimension of its cokernel) is precisely the winding number of $g$. Thus an analytic invariant of the Toeplitz operator with symbol $g$ calculates a topological invariant of $g$.

That result by itself isn't heavy on the spectral theory. The connection with spectral theory is revealed by a more refined statement. Recall that the essential spectrum of a bounded operator $T$ on a Hilbert space $H$ is the spectrum of the image of $T$ in the Calkin algebra $Q(H)$ (which is the space of bounded operators on $H$ modulo the space of compact operators). A consequence of the Toeplitz index theorem (and its proof) is the fact that the essential spectrum of the Toeplitz operator $T_g$ is precisely the range of $g$. While this statement alone is scant evidence, this suggests a deep relationship between the essential spectrum of a bounded operator and geometry. This line of thinking culminates in the Brown-Douglas-Fillmore theorem, which makes the following startling assertion. Let $X$ be a nonempty subset of the complex plane, and define $Ext(X)$ to be the space of essential unitary equivalence classes of essentially normal operators with essential spectrum $X$ (here "essential" always means "modulo compact operators"). Direct sum of operators gives $Ext(X)$ the structure of a commutative semigroup, and the BDF theorem asserts that $Ext(X)$ is naturally isomorphic to the space $Hom(\pi^1(X), \mathbb{Z})$ of group homomorphisms between the first cohomotopy group of $X$ and $\mathbb{Z}$. (Note: it is not even obvious that $Ext(X)$ has a zero element!) Thus spectral theory helps to classify certain kinds of bounded operators mod compacts in a particularly beautiful way (via algebraic topology).

As mentioned above, there are also fruitful interactions between functional analysis and geometry - mediated by spectral theory - which flow from analysis to geometry. Aside from Atiyah-Singer, there are fruitful generalizations of the Toeplitz index theorem along these lines. But let me give a different sort of example in the theory of hyperbolic diffeomorphisms.

Informally, a diffeomorphism $f: M \to M$ on a smooth manifold $M$ is said to be Anosov (or uniformly hyperbolic) if $M$ admits transverse stable and unstable foliations for the action of $f$. Prototypical examples of Anosov diffeomorphisms on the 2-torus can be obtained by considering $2 \times 2$ matrices with integer entries and irrational eigenvalues. It turns out that spectral theory has a great deal to say about smooth dynamical systems in general and Anosov diffeomorphisms in particular.

Given any diffeomorphism $f: M \to M$, consider the bounded operator $f_*$ on the Banach space $\Gamma^0(TM)$ of continuous vector fields on $M$ defined by

$(f_*v)(x) = df(v)(f^{-1}(x))$

If the non-periodic orbits of $f$ are dense in $M$, then a theorem of Mather asserts that the spectrum of $f_*$ is a disjoint union of finitely many annuli centered at the origin. If $H_i$ is the invariant subspace for $f_*$ corresponding to the $i$th annulus then the subspaces $E_i(x)$ of $T_x M$ consisting of the vectors $v(x)$ for $v \in H_i$ form a $df$ invariant continuous distribution on $M$, and the direct sum of the $E_i(x)$'s gives the whole tangent space $T_x M$. So the Mather spectral theory of $f$ is very closely related to its dynamics. Indeed, one can characterize the Anosov diffeomorphisms as precisely those $f$ for which $1$ is not in the spectrum of $f_*$. Pesin used this idea to prove that Anosov diffeomorphisms are structurally stable, meaning they form an open subset of the full diffeomorphism group of $M$ (so that a small perturbation of an Anosov diffeomorphism is still Anosov). The same strategy also works for partially hyperbolic dynamical systems, which have a slightly different spectral characterization.

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Simple answer: bounded operators are simpler than unbounded ones, so it's better to study them.

Discretized models lead to bounded operators. For example consider the discrete Laplacian $\Delta$ on $\mathbb{Z}^d$ given by $$ \Delta u(n) = \sum_{| m - n|_1 = 1} u(m), $$ where $u:\mathbb{Z}^d\to \mathbb{C}$. Boundedness just follows from the triangle inequality.

Furthermore, if one considers the time evolution operator $U(t)$ (for example in quantum mechanics $U(t) = e^{- i tH}$). Then $U(t)$ is a bounded even unitary operator. ($H$ might not be).

Edit: Another important example is if $A$ is a (possibly unbounded) operator with non-trivial essential spectrum, then the resolvents $$ (A - z)^{-1} $$ are bounded but not compact.

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Ok that's a nice answer Helge. Can you tell me why $U(t)$ is bounded? Let's say $H = -\Delta$ for simplicity. –  Dorian Sep 5 '10 at 17:14
    
Oh it's just because $e^{-it\Delta}u = \sum_n e^{-it \lambda_n} \phi_n$ and so it maps $L^2$ to $L^2$. –  Dorian Sep 5 '10 at 17:17
    
Unitary => Bounded. In the QM framework, U(t) being unitary corresponds to the conservation of probability. In the mathy framework, it's just $H$ is self-adjoint. –  Helge Sep 5 '10 at 17:27
    
Is this still linear though? –  Dorian Sep 5 '10 at 17:45
    
Yes ... (of course that depends on your base dynamics ...) –  Helge Sep 5 '10 at 18:02

In Ergodic theory unitary operators naturally arise as $U_T:f\mapsto f\circ T$ where $T$ is a measure preserving transformation on a probability space. The spectral theory of the operator $U_T$ carries some information on the dynamics of $T$.

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I have some familiarty with measure preserving transformations/ergodic theory. Can you please elaborate on what kind of information it tells you about the dynamics? Does spectra somehow represent fixed points or limit cyles? –  Dorian Sep 5 '10 at 22:44
    
For instance, $T$ is ergodic if and only if 1 is a simple eigenvalue of $U_T$ (note that it is always an eigenvalue, since constant functions are obviously invariant under $U_T$). Similarly, $T$ is mixing if and only if the only eigenvectors of $T$ are the trivial ones, i.e. the constant functions (that is, the discrete spectrum of $U_T$ consists only of the point 1). Many other properties of $T$ are reflected in spectral properties of the operator $U_T$ and since $U_T$ is a unitary operator (if $T$ is invertible), the spectral theorem comes in handy. –  Mark Sep 6 '10 at 9:49
    
There is a minor error in my former comment which I've just noticed. "mixing" should be "weakly mixing". –  Mark Feb 7 '11 at 21:41

Pseudodifferential operators are bounded (Theorem of Calderon-Vaillancourt) in contrast to Differential operators which are unbounded (they are defined on a dense subspace of $L^2$). We can try to use the spectral to obtain explicitely the spectral function $E(\lambda)$.

Another example is given by the paper of Safarov http://www2.imperial.ac.uk/~alaptev/Papers/Berez.pdf which gives another application of the spectral theorem to obtain the Berezin inequality which is important in the study of spectral properties of differential and pseudodifferential operators.

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Maybe the corresponding functional calculus (FC) addresses your question about why it is interesting to know the spectrum of an operator? Many applications use it, see for example Pedersen's book on Functional Analysis for the various types (Holomorphic, Continuous, Measurable FC).

For an intuitive approach -and keeping in mind that you are familiar with compact operators- you might like to learn some more about generalised eigenvectors.

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The original answer was deleted this is a refined answer :

The Fourier transform $F$ is a bounded operator and non compact see http://en.wikipedia.org/wiki/Fourier_transform where some properties are given. $F$ is defined on $S$ the Schwartz space which is dense in the hilbert $L^2$, hence $F$ can be extended as a bounded operator on $L^2$. $F$ is not selfadjoint but unitary, its spectrum lies on tne unit circle $abs(z)=1$ of the complex plane. However the real (and imaginary too) part of $F$ is selfadjoint (the Cosine operator) wich we note $C$. $F$ is not a compact operator because $F^4=I$ , so if $F$ was compact the product must be compact which is not the case for the identity operator, so the Cosine operator $C$ is a bounded selfadjoint operator. $F$ has 4 eigenvalues +1,-1;+i,-i and the eigenfunctions are gaussian functions. If I am not mistaken, the Cosine operator $C$ has eigenvalues +1 and -1 which belong to the discrete spectrum and the continuous spectrum is the real interval ]-1,+1[. As a bounded operator it is a good example for the spectral theorem and has many applications, for example Discrete Cosine Transform is a discete version of $C$.

A good reference is the book "Leçons d'analyse fonctionnelle" of F. Riesz and B. Nagy.

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