### Spectral Theory and Wake Operators for the Schrödinger Equation

I am a bit stuck trying to show that directly using Cauchy-Schwarz CS. It is a consequence of CS that for any measurable subset ,. Hmm, I see now that to get the sharp inequality one needs to already possess the functional calculus, since with that calculus one can write as for some piecewise constant multiplier bounded in magnitude by 1.

## radial schrodinger equation: Topics by beqyxywobyce.cf

So to avoid circularity one may have to revert back to the depolarisation argument that loses an absolute constant something like 4 , since one does not need the sharp inequality at this stage of the construction. Can you please give a reference for the spectral theorem multiplication form for unbounded normal operators. A wave equation approach to automorphic forms in analytic number theory What's new. Would you say something about what would people do if the unbounded linear operator is not densely defined on a Hilbert space?

Can it be reduced to the densely defined case? If the operator is linear and densely defined, but the domain of definition is not a linear space, the ex.

### Asymptotic analysis of harmonic series using Calculus

Let be densely defined on a set , a complex Hilbert space, and let and. We call linear iff implies. Dear Prof. Tao, in the proof of Prop.

## Notes Spectral Theory

In ex. At this point, after several hours of thought, I raise the white flag. To which harmonic function do I have to apply Liouville? You are commenting using your WordPress. You are commenting using your Google account. You are commenting using your Twitter account. You are commenting using your Facebook account. Notify me of new comments via email. Notify me of new posts via email. Create a free website or blog at WordPress. Ben Eastaugh and Chris Sternal-Johnson. Subscribe to feed.

What's new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. In particular, given any function on the spectrum of , one can then define the linear operator by the formula which then gives a functional calculus , in the sense that the map is a -algebra isometric homomorphism from the algebra of bounded continuous functions from to , to the algebra of bounded linear operators on.

Indeed, for any vectors , there is a complex measure on with the property that indeed, one can set to be the discrete measure on defined by the formula One can also view this complex measure as a coefficient of a projection-valued measure on , defined by setting Finally, one can view as unitarily equivalent to a multiplication operator on , where is the real-valued function , and the intertwining map is given by so that.

Then is symmetric and positive, but the operator does not have dense image for any complex , since for all test functions , as can be seen from a routine integration by parts. Then the resolvent can be uniquely defined for in the upper half-plane, but not in the lower half-plane, due to the obstruction for all test functions note that the function lies in when is in the lower half-plane.

Unitary equivalence with a multiplication operator. Self-adjointness and resolvents — To begin, we study what we can abstractly say about a densely defined symmetric linear operator on a Hilbert space. However, all such operators are closable , in that they have a closure: Lemma 1 Closure Let be a densely defined symmetric linear operator. But then by symmetry On the other hand, as ,. Here is a simple example of this: Exercise 2 Let be the Laplacian , defined on the dense subspace of.

Show that the closure is defined on the Sobolev space , defined as the completion of under the Sobolev norm and that the action of is given by the weak distributional derivative,. However, it is still closed: Exercise 3 Let be a densely defined linear operator, and let be its adjoint.

We caution that the adjoint of a symmetric densely defined operator need not be itself symmetric, despite extending the symmetric operator : Exercise 5 We continue Example 2. Now we can define essential self-adjointness. Note that this implies in particular that is symmetric and closed. We first need some basic properties of the resolvent: Exercise 8 Let be densely defined, symmetric, and closed.

Hint: compute two different ways. In particular, and hence, by definition of , lies in. Now from 7 we have for all ; since is dense, we have as required. From Exercise 8 ii and 7 one has for all. Hint: use Neumann series. Define the Jacobi operator from the space of compactly supported sequences to the space of square-summable sequences by the formula with the convention that vanishes for. Self-adjointness and spectral measure — We have seen that self-adjoint operators have everywhere-defined resolvents for all strictly complex.

We will need a useful tool from complex analysis, which places a one-to-one correspondence between finite non-negative measures on and certain analytic functions on the upper half-plane: Theorem 4 Herglotz representation theorem Let be an analytic function from the upper half-plane to the closed upper half-plane , obeying a bound of the form for all and some.

Hint: apply the maximum principle to the difference between and , viewed as a harmonic function of and. We consider the function on the upper half-plane. We can use this function and the Herglotz representation theorem to construct spectral measures : Exercise 15 Let , , and be as above. Hint: You will find Exercise 8 to be useful. Hint: first show this for , writing for some.

We can depolarise these measures by defining to obtain complex measures for any such that for all in either the upper or lower half-plane. In particular, if is real-valued, then is self-adjoint.

Hint: use the resolvent identity. In particular, and commute for all. Improve the to. Hint: to get this improvement, use the identity and the tensor power trick. Show that is a countably additive projection-valued measure , thus for any sequence of disjoint Borel , where the convergence is in the strong operator topology. Show that is a dense subspace of and hence of , and that maps to.

Conclude in particular that maps to for all. Show that is a closed set which is the union of the supports of the as range over. In particular, , and when is positive. Show that is the identity map. Show that for every there is a unique linear operator such that whenever. Exercise 20 i Show that if is an invariant subspace of , then so is , and furthermore the orthogonal projections to and commute with for every.

## SIAM Review

Furthermore, one has for all. Furthermore, in this case we have. Self-adjointness and flows — Now we relate self-adjointness to a variety of flows, beginning with the heat flow. The latter two properties are asserting that is a one-parameter semigroup. We now establish a converse to the above statement: Theorem 5 Let be a separable Hilbert space, and suppose one has a family of bounded self-adjoint operators of norm at most for each , which is continuous in the strong operator topology, and such that and for all. The basic idea is to somehow use the identity which suggests that which should allow one to recover from the.

Exercise 23 Let the notation be as in the above theorem. Hint: use Exercise The above exercise gives an important way to establish essential self-adjointness, namely by solving the heat equation 1 : Exercise 24 Let be a densely defined symmetric positive definite operator. Exercise 26 Let the notation be as in the above theorem. We can now see a clear link between essential self-adjointness and completeness, at least in the case of scalar first-order differential operators: Exercise 28 Let be a smooth manifold with a smooth measure , and let be a smooth vector field on which is divergence-free with respect to the measure.

More precisely, we will show the following version of Exercise 27 : Theorem 7 Let be a densely defined positive symmetric operator. We introduce a new inner product on by the formula By hypothesis, this is a Hermitian inner product on. We then define an inner product on by the formula then this is a Hermitian inner product on. Then if we define the energy then one easily computes using 3 that , and so In particular, if , then , and so twice continuously differentiable solutions to 3 are unique.

Then does not have dense image, and so there exists a non-zero such that for all. For any , define the Littlewood-Paley projection where is the Schwartz function. Also, from functional calculus one sees that and in particular maps to as well. In particular, we can write for some operators and. For any , we have combining this with 12 we see that and for all. We can then expand as and thus In particular for all.

By duality noting that is a bounded real function of we conclude that But by sending and using the spectral theorem, this implies from monotone convergence that the spectral measure is zero, and thus vanishes in , a contradiction. Essential self-adjointness of the Laplace-Beltrami operator — We now discuss how one can use the above criteria to establish essential self-adjointness of a Laplace-Beltrami operator on a smooth complete Riemannian manifold , viewed as a densely defined symmetric positive operator on the dense subspace of. For instance, suppose one is trying to solve the Helmholtz equation for some large real , and some.

If was a Euclidean space , then we have an explicit formula for the solution; note that the exponential decay of will keep in as well. Inspired by this, we can try to solve the Helmholtz equation in curved space by proposing as an approximate solution This is a little problematic because develops singularities after a certain point, but if one has a uniform lower bound on the injectivity radius here we are implicitly using the hypothesis of completeness , and also uniform bounds on the curvature and its derivatives, one can truncate this approximate solution to the region where is small, and obtain an approximate solution whose error can be made to be smaller in norm than that of if the spectral parameter is chosen large enough; we omit the details.

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There is another small missprint in Exercise 17 item x : should be replaced with [Corrected, thanks — T.

A 28 4 , , Applied Mathematics Research eXpress 1 , , Communications in Partial Differential Equations 39, , Communications in Partial Differential Equations 31 5 , , Communications in Partial Differential Equations 40 8 , , Journal of Geometric Analysis 22 2 , , Journal of Functional Analysis 3 , , Communications in Partial Differential Equations 34 5 , , Articles 1—20 Show more. Maybe we also have the Laplace operator plus some sort of controlled potential say a smooth, bounded potential with bounded derivatives. There is one case of this that is important but too straightforward.

So you can derive the existence of a fancy spectrum that is not really explicit, but the non-compactness is handled using an explicit method. Too big to fit well as comment: There is a seeming-technicality which is important to not overlook, the question of whether a symmetric operator is "essentially self-adjoint" or not. As I discovered only embarrasingly belatedly, this "essential self-adjointness" has a very precise meaning, namely, that the given symmetric operator has a unique self-adjoint extension, which then is necessarily given by its graph- closure.

In many natural situations, Laplacians and such are essentially self-adjoint.

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• But with any boundary conditions, this tends not to be the case, exactly as in the simplest Sturm-Liouville problems on finite intervals, not even getting to the Weyl-Kodaira-Titchmarsh complications. The broader notion of Friedrichs' canonical self-adjoint extension of a symmetric edit! This most-trivial example already shows that the spectrum, even in the happy-simple discrete case, is different depending on boundary conditions. Since this has not been mentioned, let me point to the Weyl-Stone-Titchmarsh-Kodaira theorem which gives the generalized Fourier transform and Plancherel formula of a selfadjoint Sturm-Liouville operator.

See also the nice original paper Kodaira In several variables, scattering theory provides Plancherel theorems.