Math212a1403 the Spectral Theorem for Compact Self-Adjoint Operators

Outline Review of last lecture. The Riesz representation theorem. Bessel’s inequality. Self-adjoint transformations. Compact self-adjoint transformations. The spectral theorem for compact self-adjoint operators. Fourier’s Fourier series.

Math212a1403 The spectral theorem for compact self-adjoint operators.

Shlomo Sternberg

Sept. 9, 2014

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. Outline Review of last lecture. The Riesz representation theorem. Bessel’s inequality. Self-adjoint transformations. Compact self-adjoint transformations. The spectral theorem for compact self-adjoint operators. Fourier’s Fourier series.

1 Review of last lecture.

2 The Riesz representation theorem.

3 Bessel’s inequality.

4 Self-adjoint transformations. Non-negative self-adjoint transformations.

5 Compact self-adjoint transformations.

6 The spectral theorem for compact self-adjoint operators.

7 Fourier’s Fourier series. Proof by integration by parts.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. This w is characterized as being the unique element of M which minimizes kv − xk, x ∈ M. The idea of the proof was to use the parallelogram law to conclude that if {xn} is a sequence of elements of M such that kv − xnk approaches the greatest lower bound of kv − xk, x ∈ M then {xn} converges to the desired w, The map v 7→ w is called orthogonal projection of H onto M and is denoted by πM .

Orthogonal projection onto a complete subspace.

If M is a complete subspace of a pre-Hilbert space H then for any v ∈ H there is a unique w ∈ M such that (v − w) ⊥ M.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. The idea of the proof was to use the parallelogram law to conclude that if {xn} is a sequence of elements of M such that kv − xnk approaches the greatest lower bound of kv − xk, x ∈ M then {xn} converges to the desired w, The map v 7→ w is called orthogonal projection of H onto M and is denoted by πM .

Orthogonal projection onto a complete subspace.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. The map v 7→ w is called orthogonal projection of H onto M and is denoted by πM .

Orthogonal projection onto a complete subspace.

If M is a complete subspace of a pre-Hilbert space H then for any v ∈ H there is a unique w ∈ M such that (v − w) ⊥ M. This w is characterized as being the unique element of M which minimizes kv − xk, x ∈ M. The idea of the proof was to use the parallelogram law to conclude that if {xn} is a sequence of elements of M such that kv − xnk approaches the greatest lower bound of kv − xk, x ∈ M then {xn} converges to the desired w,

Orthogonal projection onto a complete subspace.

Linear and antilinear maps.

Let V and W be two complex vector spaces. A map

T : V → W

is called linear if

T (λx + µy) = λT (x) + µT (y) ∀ x, y ∈ V , λ, µ ∈ C

and is called anti-linear if

T (λx + µy) = λT (x) + µT (y) ∀ x, y ∈ V λ, µ ∈ C.

Linear functions.

If ` : V → C is a linear map, (also known as a linear function) then

ker ` := {x ∈ V | `(x) = 0}

has codimension one (unless ` ≡ 0). Indeed, if

`(y) 6= 0

then 1 `(x) = 1 where x = y `(y) and for any z ∈ V , z − `(z)x ∈ ker `.

If V is a normed space and ` is continuous, then ker (`) is a closed subspace. The space of continuous linear functions is denoted by V ∗. It has its own norm deﬁned by

k`k := sup |`(x)|/kxk. x∈V ,kxk6=0 Suppose that H is a pre-Hilbert space. We have an antilinear map

φ : H → H∗, (φ(g))(f ) := (f , g).

The Cauchy-Schwarz inequality implies that

kφ(g)k ≤ kgk

and (g, g) = kgk2 shows that kφ(g)k = kgk. In particular the map φ is injective.

The Riesz representation theorem.

Let H be a pre-hilbert space. Then we have an antilinear map φ : H → H∗, (φ(g))(f ) := (f , g). The Cauchy-Schwarz inequality implies that kφ(g)k ≤ kgk In fact (g, g) = kgk2. This shows that kφ(g)k = kgk. In particular the map φ is injective. The Riesz representation theorem says that if H is a Hilbert space, then this map is surjective: Theorem Every continuous linear function on a Hilbert space H is given by scalar product by some element of H.

Proof of the Riesz representation theorem.

The proof follows from theorem about projections applied to

N := ker `.

If ` = 0 there is nothing to prove. If ` 6= 0 then N is a closed subspace of codimension one. Choose v 6∈ N. Then there is an x ∈ N with (v − x) ⊥ N. Let

1 y := (v − x). kv − xk

Then y ⊥ N and kyk = 1. For any f ∈ H, [f − (f , y)y] ⊥ y so f − (f , y)y ∈ N or `(f ) = (f , y)`(y), so if we set g := `(y)y then (f , g) = `(f )

for all f ∈ H.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. Claim: πM exists and X πM (v) = πMi (v). (1)

Projection onto a ﬁnite direct sum.

Suppose that the closed subspace M of a pre-Hilbert space H is the orthogonal direct sum of a ﬁnite number of subspaces M M = Mi i

meaning that the Mi are mutually perpendicular and every element x of M can be written as X x = xi , xi ∈ Mi . (The orthogonality guarantees that such a decomposition is unique.) Suppose further that each Mi is such that the projection

πMi exists.

Projection onto a ﬁnite direct sum.

Suppose that the closed subspace M of a pre-Hilbert space H is the orthogonal direct sum of a ﬁnite number of subspaces M M = Mi i

πMi exists. Claim: πM exists and X πM (v) = πMi (v). (1)

P Proof that πM exists and πM (v) = πMi (v).

Proof. P P Clearly πMi (v) belongs to M. We must show v − i πMi (v) is orthogonal to every element of M. For this it is enough to show that it is orthogonal to each Mj since every element of M is a sum

of elements of the Mj . So suppose xj ∈ Mj . But (πMi v, xj ) = 0 if i 6= j. So X (v − πMi (v), xj ) = (v − πMj (v), xj ) = 0

by the deﬁning property of πMj .

Review: projection onto a one dimensional subspace.

If M is the one dimensional subspace consisting of all (complex) multiples of a non-zero vector y, then M is complete, since C is complete. So πM exists. Let v be any element of H. Since all elements of M are of the form ay, we can write πM (v) = ay for some complex number a. Then (v − ay, y) = 0 or (v, y) = akyk2 so (v, y) a = . kyk2 We called a the Fourier coeﬃcient of v with respect to y. Particularly useful is the case where kyk = 1 and we can write a = (v, y). (2)

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. Suppose that M is a ﬁnite dimensional subspace of a pre-Hilbert space H with an orthonormal basis φi . This implies that M is an orthogonal direct sum of the one dimensional spaces spanned by the φi and hence πM exists and is given by X πM (v) = ai φi where ai = (v, φi ). (3)

Projection onto a ﬁnite dimensional subspace.

We combine the results of the last two slides:

Projection onto a ﬁnite dimensional subspace.

We combine the results of the last two slides: Suppose that M is a ﬁnite dimensional subspace of a pre-Hilbert space H with an orthonormal basis φi . This implies that M is an orthogonal direct sum of the one dimensional spaces spanned by the φi and hence πM exists and is given by X πM (v) = ai φi where ai = (v, φi ). (3)

Bessel’s inequality.

We now look at the inﬁnite dimensional situation and suppose that ∞ we are given an orthonormal sequence {φi }1 . Any v ∈ H has its Fourier coeﬃcients ai = (v, φi ) relative to the members of this sequence. Bessel’s inequality asserts that ∞ X 2 2 |ai | ≤ kvk , (4) 1 In particular Bessel’s inequality asserts that the sum on the left converges.

P∞ 2 2 Proof of Bessel’s inequality: 1 |ai | ≤ kvk .

Proof. Pn Let vn := i=1 ai φi , so that vn is the projection of v onto the subspace Vn spanned by the ﬁrst n of the φi . In any event, (v − vn) ⊥ vn so by the Pythagorean Theorem

n 2 2 2 2 X 2 kvk = kv − vnk + kvnk = kv − vnk + |ai | . i=1 This implies that n X 2 2 |ai | ≤ kvk i=1 and letting n → ∞ shows that the series on the left of Bessel’s inequality converges and that Bessel’s inequality holds.

Parseval’s equality.

n 2 2 2 2 X 2 kvk = kv − vnk + kvnk = kv − vnk + |ai | . i=1 Continuing the above argument, observe that 2 X 2 2 kv − vnk → 0 ⇔ |ai | = kvk . 2 But kv − vnk → 0 if and only if kv − vnk → 0 which is the same as saying that vn → v. Now vn is the n-th partial sum of the series P ai φi , and in the language of series, we say that a series P converges to a limit v and write ai φi = v if and only if the partial sums approach v. So X X 2 2 ai φi = v ⇔ |ai | = kvk . (5) i

P In general, we will call the series i ai φi the Fourier series of v (relative to the given orthonormal sequence) whether or not it converges to v. Thus Parseval’s equality says that the Fourier P 2 2 series of v converges to v if and only if |ai | = kvk .

Orthonormal bases.

We still suppose that H is merely a pre-Hilbert space. We say that an orthonormal sequence {φi } is a basis of H if every element of H is the sum of its Fourier series. For example, one of our tasks will inx ∞ be to show that the exponentials {e }n=−∞ form a basis of C(T). If the orthonormal sequence φi is a basis, then any v can be approximated as closely as we like by ﬁnite linear combinations of the φi , in fact by the partial sums of its Fourier series. We say that the ﬁnite linear combinations of the φi are dense in H.

Conversely, suppose that the ﬁnite linear combinations of the φi are dense in H. So for any v and any > 0 we can ﬁnd an n and a set of n complex numbers bi such that X kv − bi φi k ≤ .

But we know that vn is the closest vector to v among all the linear combinations of the ﬁrst n of the φi . So we must have

kv − vnk ≤ .

But this says that the Fourier series of v converges to v, i.e. that the φi form a basis.

Orthonormal bases in a Hilbert space.

In the case that H is actually a Hilbert space, and not merely a pre-Hilbert space, there is an alternative and very useful criterion for an orthonormal sequence to be a basis: Let M be the set of all limits of ﬁnite linear combinations of the φi . Any Cauchy sequence in M converges (in H) since H is a Hilbert space, and this limit belongs to M since it is itself a limit of ﬁnite linear combinations of the φi (by the diagonal argument for example). Thus ⊥ H = M ⊕ M , and the φi form a basis of M. So the φi form a basis of H if and only if M⊥ = {0}. But this is the same as saying that no non-zero vector is orthogonal to all the φi . So we have proved Theorem

In a Hilbert space, the orthonormal set {φi } is a basis if and only if no non-zero vector is orthogonal to all the φi .

Eigenvectors and their eigenvalues.

Let V denote a pre-Hilbert space. Let T be a linear transformation of V into itself. This means that for every v ∈ V the vector Tv ∈ V is deﬁned and that Tv depends linearly on v i.e. that T (av + bw) = aTv + bTw for any two vectors v and w and any two complex numbers a and b. We recall from linear algebra that a non-zero vector v is called an eigenvector of T if Tv is a scalar times v, in other words if Tv = λv where the number λ is called the corresponding eigenvalue.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. Notice that if v is an eigenvector of a symmetric transformation T with eigenvalue λ, then

λ(v, v) = (λv, v) = (Tv, v) = (v, Tv) = (v, λv) = λ(v, v),

so λ = λ. In other words, all eigenvalues of a symmetric transformation are real.

Symmetric transformations on a pre-Hilbert space.

A linear transformation T on V is called symmetric if for any pair of elements v and w of V we have

(Tv, w) = (v, Tw).

Symmetric transformations on a pre-Hilbert space.

A linear transformation T on V is called symmetric if for any pair of elements v and w of V we have

(Tv, w) = (v, Tw).

Notice that if v is an eigenvector of a symmetric transformation T with eigenvalue λ, then

λ(v, v) = (λv, v) = (Tv, v) = (v, Tv) = (v, λv) = λ(v, v),

so λ = λ. In other words, all eigenvalues of a symmetric transformation are real.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. d 1 For example, the operator dx is not bounded on C (T).

Bounded linear transformations on a pre-Hilbert space.

We will let S = S(V ) denote the “unit sphere” of V , i.e. S denotes the set of all φ ∈ V such that kφk = 1. A linear transformation T is called bounded if kT φk is bounded as φ ranges over all of S. If T is bounded, we let

kT k := l.u.b.φ∈SkT φk.

Then kTvk ≤ kT kkvk for all v ∈ V . A linear transformation on a ﬁnite dimensional space is automatically bounded, but not so for an inﬁnite dimensional space.

Bounded linear transformations on a pre-Hilbert space.

kT k := l.u.b.φ∈SkT φk.

Then kTvk ≤ kT kkvk for all v ∈ V . A linear transformation on a ﬁnite dimensional space is automatically bounded, but not so for an inﬁnite dimensional space.

d 1 For example, the operator dx is not bounded on C (T).

The null space (kernel) and the range of a linear transformation.

For any linear transformation T , we will let N(T ) denote the kernel of T , so N(T ) = {v ∈ V |Tv = 0} and R(T ) denote the range of T , so

R(T ) := {v|v = Tw for some w ∈ V }.

Both N(T ) and R(T ) are linear subspaces of V .

A subtle diﬀerence we will encounter later.

For bounded transformations, the phrase “self-adjoint” is synonymous with “symmetric”. Later on we will need to study non-bounded (not everywhere deﬁned) symmetric transformations, and then a rather subtle and important distinction will be made between self-adjoint transformations and those which are merely symmetric. But for the rest of this section we will only be considering bounded linear transformations, and so we will freely use the phrase “self-adjoint”, and (usually) drop the adjective “bounded” since all our transformations will be assumed to be bounded. We denote the set of all (bounded) self-adjoint transformations by A, or by A(V ) if we need to make V explicit.

Non-negative self-adjoint transformations. Non-negative self-adjoint transformations.

If T is a self-adjoint transformation, then

(Tv, v) = (v, Tv) = (Tv, v)

so (Tv, v) is always a real number. More generally, for any pair of elements v and w, (Tv, w) = (Tw, v). Since (Tv, w) depends linearly on v for fixed w, we see that the rule which assigns to every pair of elements v and w the number (Tv, w) satisfies the first two conditions in our definition of a semi-scalar product.

Non-negative self-adjoint transformations.

Since (Tv, v) might be negative, condition 3 in our definition of a semi-scalar product need not be satisfied. This leads to the following definition:

A self-adjoint transformation T is called non-negative if

(Tv, v) ≥ 0 ∀ v ∈ V .

Non-negative self-adjoint transformations. The semi-scalar product associated to a non-negative self-adjoint transformation.

If T is a non-negative self-adjoint transformation, then the rule which assigns to every pair of elements v and w the number (Tv, w) is a semi-scalar product to which we may apply the Cauchy-Schwarz inequality and conclude that

1 1 |(Tv, w)| ≤ (Tv, v) 2 (Tw, w) 2 .

Now let us assume in addition that T is bounded with norm kT k. Let us take w = Tv in the preceding inequality. We get

2 1 1 kTvk = |(Tv, Tv)| ≤ (Tv, v) 2 (TTv, Tv) 2 .

Non-negative self-adjoint transformations. A very useful inequality.

2 1 1 kTvk = |(Tv, Tv)| ≤ (Tv, v) 2 (TTv, Tv) 2 . Now apply the Cauchy-Schwarz inequality for the original scalar product to the last factor on the right:

1 1 1 1 1 1 1 (TTv, Tv) 2 ≤ kTTvk 2 kTvk 2 ≤ kT k 2 kTvk 2 kTvk 2 = kT k 2 kTvk,

where we have used the deﬁning property of kT k in the form kTTvk ≤ kT kkTvk. Substituting this into the previous inequality we get

2 1 1 kTvk ≤ (Tv, v) 2 kT k 2 kTvk.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. We will make much use of this inequality. For example, it follows from (6) that (Tv, v) = 0 ⇒ Tv = 0. (7)

It also follows from (6) that if we have a sequence {vn} of vectors with (Tvn, vn) → 0 then kTvnk → 0 and so

(Tvn, vn) → 0 ⇒ Tvn → 0. (8)

Non-negative self-adjoint transformations.

2 1 1 kTvk ≤ (Tv, v) 2 kT k 2 kTvk. If kTvk= 6 0 we may divide this inequality by kTvk to obtain

1 1 kTvk ≤ kT k 2 (Tv, v) 2 . (6)

This inequality is clearly true if kTvk = 0 and so holds in all cases.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. It also follows from (6) that if we have a sequence {vn} of vectors with (Tvn, vn) → 0 then kTvnk → 0 and so

(Tvn, vn) → 0 ⇒ Tvn → 0. (8)

Non-negative self-adjoint transformations.

2 1 1 kTvk ≤ (Tv, v) 2 kT k 2 kTvk. If kTvk= 6 0 we may divide this inequality by kTvk to obtain

1 1 kTvk ≤ kT k 2 (Tv, v) 2 . (6)

This inequality is clearly true if kTvk = 0 and so holds in all cases. We will make much use of this inequality. For example, it follows from (6) that (Tv, v) = 0 ⇒ Tv = 0. (7)

Non-negative self-adjoint transformations.

2 1 1 kTvk ≤ (Tv, v) 2 kT k 2 kTvk. If kTvk= 6 0 we may divide this inequality by kTvk to obtain

1 1 kTvk ≤ kT k 2 (Tv, v) 2 . (6)

This inequality is clearly true if kTvk = 0 and so holds in all cases. We will make much use of this inequality. For example, it follows from (6) that (Tv, v) = 0 ⇒ Tv = 0. (7)

It also follows from (6) that if we have a sequence {vn} of vectors with (Tvn, vn) → 0 then kTvnk → 0 and so

(Tvn, vn) → 0 ⇒ Tvn → 0. (8)

Non-negative self-adjoint transformations.

Notice that if T is a bounded self adjoint transformation, not necessarily non-negative, then rI − T is a non-negative self-adjoint transformation if r ≥ kT k: Indeed,

((rI − T )v, v) = r(v, v) − (Tv, v) ≥ (r − kT k)(v, v) ≥ 0

since, by Cauchy-Schwarz,

(Tv, v) ≤ |(Tv, v)| ≤ kTvkkvk ≤ kT kkvk2 = kT k(v, v).

So we may apply the preceding results to rI − T .

Compact self-adjoint transformations.

We say that the self-adjoint transformation T is compact if it has the following property: Given any sequence of elements un ∈ S, we

can choose a subsequence uni such that the sequence Tuni converges to a limit in V . Some remarks about this complicated looking definition: In case V is finite dimensional, every linear transformation is bounded, hence the sequence Tun lies in a bounded region of our finite dimensional space, and hence by the completeness property of the real (and hence complex) numbers, we can always find such a convergent subsequence. So in finite dimensions every T is compact. More generally, the same argument shows that if R(T ) is finite dimensional and T is bounded then T is compact. So the definition is of interest essentially in the case when R(T ) is infinite dimensional.

Compact implies bounded.

Also notice that if T is compact, then T is bounded. Otherwise we could ﬁnd a sequence un of elements of S such that kTunk ≥ n

and so no subsequence Tuni can converge.

Eigenvalues of compact self-adjoint operators.

We now come to a key result which we will use over and over again: Theorem Let T be a compact self-adjoint operator. Then R(T ) has an orthonormal basis {φi } consisting of eigenvectors of T and if R(T ) is inﬁnite dimensional then the corresponding sequence {rn} of eigenvalues converges to 0.

Important remark. I want to emphasize that in the statement and proof of this theorem, all we assume is that V is a pre-Hilbert space. We make no completeness assumptions about V .

Proof step 1.

We know that T is bounded. If T = 0 there is nothing to prove. So assume that T 6= 0 and let

m1 := kT k > 0.

By the definition of kT k we can find a sequence of vectors un ∈ S such that kTunk → kT k. By the definition of compactness we can

ﬁnd a subsequence of this sequence so that Tuni → w for some w ∈ V . On the other hand, the transformation T 2 is self-adjoint and bounded by kT k2. Hence kT k2I − T 2 is non-negative, and 2 2 2 2 ((kT k I − T )un, un) = kT k − kTunk → 0. So we know from (8) that 2 2 kT k un − T un → 0.

We know that

2 2 2 2 m1un − T un = kT k un − T un → 0.

Applied to the subsequence uni this says that

2 m1uni → Tw

since Tuni → w and T is continuous. So we have proved that 1 uni → 2 Tw. m1

Proof step 2.

1 uni → 2 Tw. m1 Applying T to this we get

1 2 Tuni → 2 T w m1 or 2 2 T w = m1w.

Also kwk = kT k = m1 6= 0. So w 6= 0. So w is an eigenvector of 2 2 T with eigenvalue m1. We have

2 2 0 = (T − m1I )w = (T + m1I )(T − m1I )w.

Proof step 3.

2 2 We know that 0 = (T − m1I )w = (T + m1I )(T − m1I )w. If (T − m1I )w = 0, then w is an eigenvector of T with eigenvalue m1 and we normalize it by setting 1 φ := w. 1 kwk

Then kφ1k = 1 and T φ1 = m1φ1.

If (T − m1I )w 6= 0 then y := (T − m1I )w is an eigenvector of T with eigenvalue −m1 and again we normalize by setting 1 φ := y. 1 kyk

So we have found a unit vector φ1 ∈ R(T ) which is an eigenvector of T with eigenvalue r1 = ±m1.

Proof step 4.

Now let ⊥ V2 := φ1 .

If x ∈ V2 so that (x, φ1) = 0, then

(Tx, φ1) = (x, T φ1) = r1(x, φ1) = 0.

In other words, T (V2) ⊂ V2

and we can consider the linear transformation T restricted to V2 which is again compact. If we let m2 denote the norm of the linear transformation T when restricted to V2 then m2 ≤ m1 and we can apply the preceding procedure to ﬁnd a unit eigenvector φ2 with eigenvalue ±m2.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. So there are two alternatives:

Proof step 5.

We proceed inductively, letting

⊥ Vn := {φ1, . . . , φn−1}

and ﬁnd an eigenvector φn of T restricted to Vn with eigenvalue ±mn 6= 0 if the restriction of T to Vn is not zero.

Proof step 5.

We proceed inductively, letting

⊥ Vn := {φ1, . . . , φn−1}

and ﬁnd an eigenvector φn of T restricted to Vn with eigenvalue ±mn 6= 0 if the restriction of T to Vn is not zero. So there are two alternatives:

after some finite stage the restriction of T to Vn is zero. In this case R(T ) is finite dimensional with orthonormal basis φ1, . . . , φn−1. Or The process continues indefinitely so that at each stage the restriction of T to Vn is not zero and we get an infinite sequence of eigenvectors and eigenvalues ri with |ri | ≥ |ri+1|. The first case is one of the alternatives in the theorem, so we need to look at the second alternative.

Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. To complete the proof of the theorem we must show that the φi form a basis of R(T ).

Proof step 6.

We ﬁrst prove that |rn| → 0. If not, there is some c > 0 such that |rn| ≥ c for all n (since the |rn| are decreasing). If i 6= j,then by the Pythagorean theorem we have

2 2 2 2 2 2 kT φi − T φj k = kri φi − rj φj k = ri kφi k + rj kφj k .

Since kφi k = kφj | = 1 this gives

2 2 2 2 kT φi − T φj k = ri + rj ≥ 2c .

Hence no subsequence of the T√φi can converge, since all these vectors are at least a distance c 2 apart. This contradicts the compactness of T .

Proof step 6.

We ﬁrst prove that |rn| → 0. If not, there is some c > 0 such that |rn| ≥ c for all n (since the |rn| are decreasing). If i 6= j,then by the Pythagorean theorem we have

2 2 2 2 2 2 kT φi − T φj k = kri φi − rj φj k = ri kφi k + rj kφj k .

Since kφi k = kφj | = 1 this gives

2 2 2 2 kT φi − T φj k = ri + rj ≥ 2c .

Hence no subsequence of the T√φi can converge, since all these vectors are at least a distance c 2 apart. This contradicts the compactness of T . To complete the proof of the theorem we must show that the φi form a basis of R(T ).

Proof step 7.

So if v is any element of our pre-Hilbert space V , and we set w := Tv, we must show that the Fourier series of w with respect to the φi converges to w. We begin with the Fourier coeﬃcients of v relative to the φi which are given by

an = (v, φn).

Then the Fourier coeﬃcients of w are given by

bi = (w, φi ) = (Tv, φi ) = (v, T φi ) = (v, ri φi ) = ri ai .

So n n n X X X w − bi φi = Tv − ai ri φi = T (v − ai φi ). i=1 i=1 i=1

Proof step 7.

n n n X X X w − bi φi = Tv − ai ri φi = T (v − ai φi ). i=1 i=1 i=1 Pn Now v − i=1 ai φi is orthogonal to φ1, . . . , φn so belongs to Vn+1. So n n X X kT (v − ai φi )k ≤ |rn+1|k(v − ai φi )k. i=1 i=1 By the Pythagorean theorem,

n X k(v − ai φi )k ≤ kvk. i=1

Proof step 8, ﬁnal step.

We know that n n X X kT (v − ai φi )k ≤ |rn+1|k(v − ai φi )k. i=1 i=1 and n X k(v − ai φi )k ≤ kvk. i=1 Putting the two previous inequalities together we get n n X X kw − bi φi k = kT (v − ai φi )k ≤ |rn+1|kvk → 0. i=1 i=1 This proves that the Fourier series of w converges to w concluding the proof of the theorem. Shlomo Sternberg Math212a1403 The spectral theorem for compact self-adjoint operators. Outline Review of last lecture. The Riesz representation theorem. Bessel’s inequality. Self-adjoint transformations. Compact self-adjoint transformations. The spectral theorem for compact self-adjoint operators. Fourier’s Fourier series.

A converse for Hilbert spaces.

The “converse” of the above result is easy. Here is a version: Suppose that H is a Hilbert space (not merely a pre-Hilbert space) with an orthonormal basis {φi } consisting of eigenvectors of an operator T , so T φi = λi φi , and suppose that λi → 0 as i → ∞. Then T is compact. Indeed, for each j we can ﬁnd an N = N(j) such that 1 |λ | < ∀ r > N(j). r j

We can then let Hj denote the closed subspace spanned by all the eigenvectors φr , r > N(j), so that ⊥ H = Hj ⊕ Hj ⊥ is an orthogonal decomposition and Hj is ﬁnite dimensional, in fact is spanned the ﬁrst N(j) eigenvectors of T .

Now let {ui } be a sequence of vectors with kui k ≤ 1 say. We decompose each element as

0 00 0 ⊥ 00 ui = ui ⊕ ui , ui ∈ Hj , ui ∈ Hj .

We can choose a subsequence so that u0 converges, because they ik

all belong to a finite dimensional space, and hence so does Tuik since T is bounded. We can decompose every element of this ⊥ subsequence into its Hj and Hj components, and choose a subsequence so that the first component converges. Proceeding in this way, and then using the Cantor diagonal trick of choosing the k-th term of the k-th selected subsequence, we have found a subsequence such that for any fixed j, the (now relabeled) ⊥ subsequence, the Hj component of Tuj converges. But the Hj component of Tuj has norm less than 1/j, and so the sequence converges by the triangle inequality.

Plan.

We want to apply the theorem about compact self-adjoint operators that we just proved to conclude that the functions einx form an orthonormal basis of the space C(T). In fact, a direct proof of this fact is elementary, using integration by parts. So we will pause to given this direct proof. Then we will go back and give a (more complicated) proof of the same fact using our theorem on compact operators. The reason for giving the more complicated proof is that it extends to far more general situations.

Proof by integration by parts. Integration by parts.

We have let C(T) denote the space of continuous functions on the real line which are periodic with period 2π. We will let C1(T) denote the space of periodic functions which have a continuous ﬁrst derivative (necessarily periodic) and by C2(T) the space of periodic functions with two continuous derivatives. If f and g both belong to C1(T) then integration by parts gives

1 Z π 1 Z π f 0gdx = − f g 0dx 2π −π 2π −π since the boundary terms, which normally arise in the integration by parts formula, cancel, due to the periodicity of f and g.