The spectra of the unilateral shift and its adjoint
In this note I am writing out some of the material from Paul Halmos, Hilbert Space Problem Book, on shift operators. The reason I’m doing this is because shift operators are standard objects in operator theory and every analyst should know their properties and their spectra. A reference to Problem of Halmos is a reference to Problem in this book. An orthonormal basis for a Hilbert space is an orthonormal set whose span is a dense subset of . The dimension of a Hilbert space is the cardinality of an orthonormal basis for . It is a fact that a Hilbert space is separable if and only if its dimension is countable (Problem 11 of Halmos). Let be the set of nonnegative integers.
2 The Hilbert space ℓ²(𝐼)
If is a set, let be the set of functions such that for all but countably many and such that
and let be the set of functions such that for all but countably many and such that
, and this containment is strict if and only if is infinite. With the inner product
is a Hilbert space, and is an orthonormal basis for it. It follows that is separable if and only if is countable.
It is often useful that for and , we have
and thus (this takes time to prove)
We will be interested in countable orthonormal bases for a Hilbert space, so in this note we take to be a separable complex Hilbert space. Let be an orthonormal basis for , where is countable. is a Hilbert space with orthonormal basis , . We define a map in the following way. For , define , and thus by the definition of norm in and by Parseval’s identity we have
so indeed . One checks that is a linear map and hence, as it is an isometry by the above calculation, for all .11 1 It is a fact that if are Hilbert spaces and is a linear map, then is an isometry if and only if for all . This is proved using the polarization identity and tinkering with real and imaginary parts. A proof is given in John B. Conway, A Course in Functional Analysis, second ed., p. 19, Theorem 5.2. Equivalently, a linear map is an isometry if and only if ; if also , called being a coisometry, then is a unitary isomorphism. Moreover, as is an isometry it is injective.
The image of an isometry is closed: let and say . Then there are , . is a Cauchy sequence, and because is an isometry it follows that is a Cauchy sequence. As is a complete metric space, converges to some . is continuous so . And , so , showing that is closed in . But and so .22 2 Merely defining by and then extending by linearity would have been inadequate. A Hamel basis for a vector space is a maximal linearly independent set in the vector space, and it is a fact that every vector in the vector space is a finite linear combination of elements of the Hamel basis. We can find a Hamel basis for that includes the orthonormal basis . If is infinite then any Hamel basis will have cardinality larger than that of the orthonormal basis; see Problem 5 of Halmos. A linear map can be specified uniquely by defining it on elements of a Hamel basis. Thus the values of a linear map on an orthonormal basis do not uniquely the linear map determine it. For the values of a linear map on an orthonormal basis to uniquely determine the linear map, one must show that the map is continuous on the span of the orthonormal basis, and then since the span is dense this will determine the map on the entire Hilbert space. As is closed and it contains an orthonormal basis for , it follows that . We already found that is linear and injective, so is a linear isomorphism.
We have shown that is a linear isomorphism and for all . We call such a map a unitary isomorphism. Unitary isomorphisms are isomorphisms in the category of Hilbert spaces.33 3 cf. http://math.ucr.edu/home/baez/quantum/node3.html Anything we wish to say about the Hilbert space can be said just as well about the Hilbert space .
Rather than merely saying that has a basis for some countable set , the type of operations that we want to talk about involve ordering and talking about sending a basis element to the next or the previous basis element. Depending on our purpose, we will take either or . In this note we deal with .
3 Definition of the unilateral shift and determination of its adjoint
Define in the following way. For , define by
shifts a sequence one step to the right. We call a unilateral shift.
is linear, and
so is an isometry. A linear map that is an isometry preserves the inner product, so preserves the inner product of , and thus is an isometry. But is not a unitary isomorphism because it is not surjective, as . For , we have, because ,
Thus for all we have .
Whenever we have our hands on a specific operator, we would also like to get a workable expression for its adjoint. satisfies
Now, and and
Hence the adjoint satisfies
If then , and . Thus, for we define for all by and check that this is indeed the adjoint of . The adjoint of the right shift is the left shift .
Since and , is not normal.
4 The spectrum of an operator
Here we review general statements about the spectrum of a bounded linear operator. If is a Hilbert space and , the spectrum of is the set of those such that the map is not bijective, where by we mean . We are often interested in decomposing the spectrum into three disjoint sets. The point spectrum is the set of those such that is not injective. The continuous spectrum is the set of those such that is injective, has dense image, but is not surjective. The residual spectrum is the set of those such that is injective and does not have dense image.
If , it is a fact that is a nonempty compact subset of ; this is not obvious. The spectral radius of , denoted , is defined the be . If then one can define an inverse for using the geometric series, and it follows that . Thus if .44 4 A formula for is , Problem 74 in Halmos.
If is a subset of , let . If then it is straightforward to check, as , that
For any , it is a fact that . If , then , so . Hence , that is, does not have dense image. So either or . Therefore
If then . ; taking orthogonal complements gives . Thus . But
so , showing that . Therefore
5 Spectrum of the unilateral shift and its adjoint
In this section we are going to compute the three parts of the spectrum for each of and . Our technique is to show that are subsets of the closed unit disc; compute and ; use §4 to obtain from this and ; and use the fact that the spectrum is compact. cf. Problems 58 and 67 of Halmos.
is an isometry, so . Hence is a subset of the closed unit disc , and .
Suppose by contradiction that there is some . Then there is some nonzero satisfying . Let . If then because of the minimality of , and by definition of . Thus in any case . and , so it follows that . Then, , contradicting . Therefore
If , then there is some with . For , and , so . Thus . Then
If then we’d get , so . As , it follows that . On the other hand, if , then define by and, for , by . , and , so . As , this means that . Hence
We have found that , and, as , this implies that
is a compact set, is contained in the closed unit disc, and contains which is equal to the open unit disc. Therefore is equal to the closed unit disc. Since and , it follows that .
Now, . On the other hand, . Hence , from which we get
is a compact set that is contained in the closed unit disc and contains the open unit disc, hence is equal to the closed unit disc. As and , it follows that .
We summarize the results of this section in the following.
The spectrum of the right shift :
The spectrum of the left shift :
6 Cyclic vectors
is said to be a cyclic vector for if the span of is a dense subspace of . One proves that for to be a cyclic vector it is equivalent that the set is dense in . Certainly has a cyclic vector: , which is an orthonormal basis for . It turns out that also has a cyclic vector, but this is not obvious. This is shown in Problem 126 of Halmos.