# Functional Analysis/Hilbert spaces

←Chapter 2: Banach spaces | Functional Analysis Chapter 3: Hilbert spaces |
Chapter 4: Geometry of Banach spaces→ |

- The chapter is almost done, but there are still some errors in the proofs that have to be rectified. (Also, we could add a discussion of the polar decomposition of unbounded operators.) |

A normed space is called a *pre-Hilbert space* if for each pair of elements in the space there is a unique complex (or real) number called an *inner product* of and , denoted by , subject to the following conditions:

- (i) The functional is linear.
- (ii)
- (iii) for every nonzero

The inner product in its second variable is not linear but antilinear: i.e., if , then for scalars . We define and this becomes a norm. Indeed, it is clear that and (iii) is the reason that implies that . Finally, the triangular inequality follows from the next lemma.

**3.1 Lemma (Schwarz's inequality)** * where the equality holds if and only if we can write for some scalar .*

If we assume the lemma for a moment, it follows:

since for any complex number

Proof of Lemma: First suppose . If , it then follows:

where the equation becomes if and only if . Since we may suppose that , the general case follows easily.

**3.2 Theorem** *A normed linear space is a pre-Hilbert space if and only if .*

Proof: The direct part is clear. To show the converse, we define

- .

It is then immediate that , and . Moreover, since the calculation:

,

we have: . If is a real scalar and is a sequence of rational numbers converging to , then by continuity and the above, we get:

**3.3 Lemma** *Let be a pre-Hilbert. Then in norm if and only if for any and as .*
Proof: The direct part holds since:

- as .

Conversely, we have:

- as

**3.4 Lemma** *Let be a non-empty convex closed subset of a Hilbert space. Then admits a unique element such that*

- .

Proof: By denote the right-hand side. Since is nonempty, . For each , there is some such that . That is, . Since is convex,

- and so .

It follows:

as |

This is to say, is Cauchy. Since is a closed subset of a complete metric space, whence it is complete, there is a limit with . The uniqueness follows since if we have

where the right side is for the same reason as before.

The lemma may hold for a certain Banach space that is not a Hilbert space; this question will be investigated in the next chapter.

For a nonempty subset , define to be the intersection of the kernel of the linear functional taken all over . (In other words, is the set of all that is orthogonal to every .) Since the kernel of a continuous function is closed and the intersection of linear spaces is again a linear space, is a closed (linear) subspace of . Finally, if , then and .

**3.5 Lemma** *Let be a linear subspace of a pre-Hilbert space. Then if and only if .*
Proof: The Schwarz inequality says the inequality

is actually equality if and only if and are linear dependent. (TODO: the proof isn't quite well written.)

**3.6 Theorem (orthogonal decomposition)** *Let be a Hilbert space and be a closed subspace. For every we can write*

where and , and and are uniquely determined by .
Proof: Clearly is convex, and it is also closed since a translation of closed set is again closed. Lemma 3.4 now gives a

*unique*element such that . Let . By Lemma 3.5, . For the uniqueness, suppose we have written:

where and . By Lemma 3.5, . But, as noted early, such must be unique; i.e., .

**3.7 Corollary** *Let be a subspace of a Hilbert space . Then*

*(i) if and only if is dense in .**(ii) .*

Proof: By continuity, . (Here, denotes the image of the set under the map .) This gives:

- and so

by the orthogonal decomposition. (i) follows. Similarly, we have:

- .

Hence, (ii).

**3.8 Theorem (representation theorem)** *Every continuous linear functional on a Hilbert space has the form:*

- with a unique and

Proof: Let . Since is continuous, is closed. If , then take . If not, by Corollary 3.6, there is a nonzero orthogonal to . By replacing with we may suppose that . For any , since is in the kernel of and thus is orthogonal to , we have:

and so:

The uniqueness follows since for all means that . Finally, we have the identity:

where the last inequality is Schwarz's inequality.

**3.9 Exercise** *Using Lemma 1.6 give an alternative proof of the preceding theorem.*

In view of Theorem 3.5, for each , we can write: where , a closed subspace of , and . Denote each , which is uniquely determined by , by . The function then turns out to be a linear operator. Indeed, for given , we write:

- and

where and for . By the uniqueness of decomposition

- .

The similar reasoning shows that commutes with scalars. Now, for (where and ), we have:

That is, is continuous with . In particular, when is a nonzero space, there is with and and consequently . Such is called an *orthogonal projection* (onto ).

The next theorem gives an alternative proof of the Hahn-Banach theorem.

**3 Theorem** *Let be a linear (not necessarily closed) subspace of a Hilbert space. Every continuous linear functional on can be extended to a unique continuous linear functional on that has the same norm and vanishes on .*

Proof: Since is a dense subset of a Banach space , by Theorem 2.something, we can *uniquely* extend so that it is continuous on . Define . By the same argument used in the proof of Theorem 2.something (Hahn-Banach) and the fact that , we obtain . Since on , it remains to show the uniqueness. For this, let be another extension with the desired properties. Since the kernel of is closed and thus contain , on . Hence, for any ,

- .

The extension is thus unique.

**3 Theorem** *Let be an increasing sequence of closed subspaces, and be the closure of . If is an orthogonal projection onto , then for every .*

Proof: Let . Then is closed. Indeed, if and , then

and so . Since , the proof is complete.

Let be Hilbert spaces. The direct sum of is defined as follows: let and define

- .

It is then easy to verify that is a Hilbert space. It is also clear that this definition generalizes to a finite direct sum of Hilbert spaces. (For an infinite direct sum of Hilbert spaces, see Chapter 5.)

Recall from the previous chapter that an isometric surjection between Banach spaces is called "unitary".

**3 Lemma (Hilbert adjoint)** *Define by . (Clearly, is a unitary operator.) Then is a graph (of some linear operator) if and only if is densely defined.*

Proof:
Set . Let . Then

- for every .

That is to say, , which is a graph of a *linear* operator by assumption. Thus, . For the converse, suppose . Then

and so for every in the domain of , dense. Thus, , and is a graph of a function, say, . The linear of can be checked in the similar manner.

Remark: In the proof of the lemma, the linear of was never used.

For a densely defined , we thus obtained a linear operator which we call . It is characterized uniquely by:

- for every ,

or, more commonly,

- for every .

Furthermore, is defined if and only if

is continuous for every . The operator is called the *Hilbert adjoint* (or just adjoint) of . If is closed in addition to having dense domain, then

Here, . By the above lemma, is densely defined. More generally, if a densely defined operator has a closed extension (i.e., ), then and are both densely defined. It follows: . That is, is densely defined and exists. That follows from the next theorem.

**3 Theorem** *Let be a densely defined operator. If is also densely defined, then*

for any closed extension of .

Proof: As above,

Here, the left-hand side is a graph of . For the second identity, since is a Hilbert space, it suffices to show . But this follows from Lemma 3.something.

The next corollary is obvious but is important in application.

**3 Corollary** *Let be Hilbert spaces, and a closed densely defined linear operator. Then if and only if there is some such that:*

- for every

**3 Lemma** *Let be a densely defined linear operator. Then *

Proof: is in either the left-hand side or the right-hand side if and only if:

- for every .

(Note that for every implies .)

In particular, a closed densely defined operator has closed kernel. As an application we shall prove the next theorem.

**3 Theorem** *Let be a closed densely defined linear operator. Then is surjective if and only if there is a such that*

- for every .

Proof: Suppose is surjective. Since has closed range, it suffices to show the estimate for . Let with . Denoting by the inverse of restricted to , we have:

The last inequality holds since is continuous by the closed graph theorem. To show the converse, let be given. Since is injective, we can define a linear functional by for .,

- for every .

Thus, is continuous on the range of . It follows from the Hahn-Banach theorem that we may assume that is defined and continuous on . Thus, by Theorem 3.something, we can write in with some . Since is continuous for ,

- for every .

Hence, .

**3 Corollary** *Let be as given in the preceding theorem. Then is closed if and only if is closed.*

Proof: Define by . It thus suffices to show is surjective when has closed range (or equivalently is surjective.) Suppose is convergent. The preceding theorem gives:

- as .

Thus, is Cauchy in the graph of , which is closed. Hence, converges *within* the range of . The converse holds since .

We shall now consider some concrete examples of densely defined linear operators.

**3 Theorem** * is continuous if and only if is continuous. Moreover, when is continuous,*

- .

Proof: It is clear that is defined everywhere, and its continuity is a consequence of the closed graph theorem. Conversely, if is continuous, then is continuous and . For the second part,

- for every .

Thus, is continuous with . In particular, is continuous, and so:

- for every .

That is to say, . Applying this result to in place of completes the proof.

The identity in the theorem shows that is a -algebra, which is a topic in Chapter 6.

**3 Lemma** *Let . If for , then .*
Proof: Let . We have and . Summing the two we get: for . Taking gives for all or .

Remark: the above lemma is false if the underlying field is .

Recall that an isometric surjection is called unitary.

**3 Corollary** *A linear operator is unitary if and only if and are identities.*

Proof:
Since , we see that is the identity. Since , is the identity on the range of *U*, which is by surjectivity. Conversely, since
,
is an isometry.

Curiously, the hypothesis on *linearity* can be omitted:

**3 Theorem** *If is a *function* such that *

for every *x* and *y* and , then is a linear operator (and so unitary).
Proof: Note that

*U*is continuous. Since , we have:

- .

Thus,

It now follows:

for any and scalar .

There is an analog of this result for Banach space. See, for example, http://www.helsinki.fi/~jvaisala/mazurulam.pdf)

**3 Exercise** *Construct an example so as to show that an isometric operator (i.e., a linear operator that preserves norm) need not be unitary. (Hint: a shift operator.)*

A densely defined linear operator is called "symmetric" if . If the equality in the above holds, then is called "self-adjoint". In light of Theorem 3.something, every self-adjoint is closed and densely defined. If is symmetric, then since is an extension of ,

- .

**3 Theorem** *Let be densely defined linear operators for . Then where the equality holds if and is closed and densely defined.*

Proof: Let . Then

- for every .

But, by definition, denotes . Hence, is an extension of . For the second part, the fact we have just proved gives:

- .

**3 Theorem** *Let be a Hilbert spaces. If is a closed densely defined operator, then is a self-adjoint operator (in particular, densely defined and closed.)*

Proof: In light of the preceding theorem, it suffices to show that is closed. Let be a sequence such that converges to limit .
Since

- ,

there is some such that: . It follows from the closedness of that . Since and is closed, .

**3 Theorem** *Let be a symmetric densely defined operator. If is surjective, then is self-adjoint and injective and is self-adjoint and bounded.*

Proof: If ,

- and

if has a dense range (for example, it is surjective). Thus, is injective. Since is closed (by Lemma 2.something) and , is a continuous linear operator. Finally, we have:

- .

Here, , and the equality holds since the domains of and coincide. Hence, is self-adjoint. Since we have just proved that the inverse of a self-adjoint is self-adjoint, we have: is self-adjoint.

**3 Theorem** *Let be a closed linear subspace of a Hilbert space . Then is an orthogonal projection onto if and only if and the range of is .*

Proof: The direct part is clear except for . But we have:

since and are orthogonal. Thus, is real and so self-adjoint then. For the converse, we only have to verify for every . But we have: and .

We shall now turn our attention to the spectral decomposition of a compact self-adjoint operator. Let be a compact operator.