Advanced Analysis, Notes 4: Hilbert spaces (bounded operators, Riesz Theorem, adjoint)
by Orr Shalit
Up to this point we studied Hilbert spaces as they sat there and did nothing. But the central subject in the study of Hilbert spaces is the theory of the operators that act on them. Paul Halmos, in his classic paper “Ten Problem in Hilbert Space“, wrote:
Nobody, except topologists, is interested in problems about Hilbert space; the people who work in Hilbert space are interested in problems about operators.
Of course, Halmos was exaggerating; topologists don’t really care much for Hilbert spaces for their own sake, and functional analysts have much more to say about the structure theory of Hilbert space then what we have learned. Nevertheless, this quote is very close to the truth. We proceed to study operators.
1. Bounded operators
A linear transformation mapping between two Hilbert spaces is said to be bounded if the operator norm of , defined as
satisfies . An immediate observation is that for all we have .
A bounded linear transformation is usually referred to as a bounded operator, or simply as an operator. The set of bounded operators between two Hilbert spaces and is denoted and is usually abbreviated as .
In this course we will almost exclusively deal with bounded operators, so we will not bother using the adjective “bounded” over and over again. Although unbounded operators defined on linear subspaces of Hilbert spaces are of great importance and are studied a lot, unbounded operators defined on the entire Hilbert space are hardly ever of any interest.
Example: Every linear transformation on a finite dimensional space is bounded. Let be a finite dimensional Hilbert space (with the standard inner product). Every linear transformation on is of the form given by , where is an matrix. By elementary analysis, is continuous, and by compactness of the unit sphere , is bounded.
Alternatively, one can find an explicit bound for by using the definition. Let . Then
By Cauchy-Schwarz, the right hand side is less than
Exercise A: Prove that the above bound for may be attained in some cases, but that in general it is not sharp.
A simple and useful way of constructing bounded operators on Hilbert spaces is by applying the following exercise.
Exercise B: Let be a dense linear subspace of a Hilbert space . Let be a linear transformation defined on , and suppose that , where runs over all with , exists and is equal to . Then there is a unique bounded operator defined on that agrees with on and satisfies . Moreover, prove that if for all , then for all .
Example: Let , and let . We define the multiplication operator by
The operator is clearly defined and bounded on , so one may use Exercise B to extend to a bounded operator on all of (if one knows measure theory, then this extension procedure is not needed). It is not difficult to compute .
Exercise C: Prove that there exists an unbounded operator on .
Proposition 1: For a linear transformation mapping between two inner product spaces, the following are equivalent:
- is bounded.
- is continuous.
- is continuous at some .
Exercise D: Prove Proposition 1. (Of course, Proposition 1 holds in any normed space, and has nothing in particular to do with inner products).
Definition 2: For every , we define the kernel of to be the space . We define the range of to be the space .
Proposition 3: Let . Then if and only if for all and . In case , then if and only if for all .
Exercise E: Prove the proposition, and show that the second part fails for real Hilbert spaces.
2. Linear functionals and the Riesz representation theorem
Definition 4: A linear operator is said to be a linear functional.
Example: Let be a Hilbert space, and let . We define a linear functional on as follows:
By the linearity of the inner product in the first component, is a linear functional. For every , Cauchy-Schwarz implies that , thus . When , we apply to and obtain , thus . In fact, this example covers all the possibilities.
Theorem 5 (Riesz representation theorem): If is a bounded linear functional on a Hilbert space , then there exists a unique with , such that for all .
Proof: If then we choose and the proof is complete, so assume that . Then is a closed subspace which is not . We have . Since the restriction of to is a linear functional with trivial kernel, the dimension of is equal to . Choose that is a unit vector (i.e., ). We claim that does the trick. Indeed, let . Let be the decomposition of with respect to . Then by Theorem 15 in Notes 2, . Then we have
The equality has been worked out in the above example. Finally, is unique, because is another candidate for this role, then for all , whence (because, in particular, ).
Exercise F: Where did we use the fact that is a Hilbert space? Where did we use the fact that is bounded? Construct examples showing that neither of these assumptions can be dropped.
Exercise G: Prove that a linear functional on a Hilbert space is bounded if and only if its kernel is closed.
3. The adjoint of a bounded operator
Proposition 6 (existence of the adjoint): For every there exists a unique such that for all .
Proof: Fix , and consider the functional . It is clear that is linear, and the estimate
shows that is bounded and . By Theorem 5, there exists a unique element in , call it , such that for all . What remains to be proved is that the mapping is a bounded linear map. Linearity is easy. For example, if , then
so . As for boundedness, Theorem 5 also tells us that , thus is bounded and . The uniqueness of also follows, from the uniqueness of for each .
Definition 7: For every , we denote by the unique operator for which for all . The operator is said to be the adjoint of .
Example: Let , let and let be multiplication by . If , then denote by the matrix with th entry . Then . This should be well known from a course in linear algebra, but is also immediate:
Example: Let , and let . Then .
Proposition 8: For and the following hold:
- If is bijective and is bounded then is invertible and .
Remark: We will later learn that when is invertible, then is automatically bounded.
Exercise H: Prove Proposition 8.
Proposition 9: Let . We have the following:
Proof: 1 and 2 are equivalent, and so are 3 and 4. To prove 1, let . Then if and only if for all , , and this happens if and only if .
To prove 4, apply to both sides of 1 to obtain . Thus the result follows from the following lemma.
Lemma 10: For every linear subspace in a Hilbert space, .
Proof: It is easy to see that , thus (as we know already from Lecture 2 that for a closed subspace , it holds that ) .
4. Special classes of operators
Definition 11: An operator is said to be
- normal if ,
- self adjoint if ,
- isometric (or an isometry) if for all ,
- unitary if it is a bijection satisfying for all ,
- positive if for all .
- Let denote that identity operator on , that is for all . Then belongs to all of the above categories.
- Let be a closed subspace. Then is positive and self adjoint. Indeed, if , write where . Then . By Proposition 12 below, is self adjoint.
- Let be the right shift on : that is, . Then is an isometry. A computations shows that is given by . Further computations show that , and , where . So is not normal.
Proposition 12: is self adjoint if and only if .
Proposition 13: For , the following are equivalent:
- is an isometry.
- for all .
Proposition 14: For , the following are equivalent:
- is unitary.
Exercise I: Prove the above propositions.