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Lecture 1: Historical remarks about Hilbert, his 23 problems, his spaces, relevance to integral equations, quantum theory, Fourier series. The difficulty of giving an example of a discontinuous linear operator between Banach spaces, and the role of completeness.
Theorem 1.1. (Baire’s Category Theorem) Let M be a complete metric space, and (Gn) be a sequence of dense open subsets of M. Let G =
n Gn. Then^ G^ is dense in M.
Proof. Sutherland’s topology book (first edition, or the supplementary material on OUP Website), for example
Lecture 2:
Corollary 1.2. A complete metric space M is not a countable union of nowhere dense subsets.
Proof. This was easily deduced from Theorem 1.1. Direct proof in Kreyszig (Baire’s Category Theorem)—basically the proof of Theorem 1.1 adapted to this corollary.
Examples 1.3. 1. Any countable complete metric space has an isolated point.
- An infinite-dimensional Banach space does not have a countable algebraic basis.
Theorem 1.4. (Uniform Boundedness Theorem; Banach-Steinhaus Theo- rem) Let X be a Banach space, Y be a normed space, and E ⊆ B(X, Y ) be such that, for each x ∈ X, supT ∈E ‖T x‖ < ∞. Then supT ∈E ‖T ‖ < ∞.
Proof. Kreyszig and many other functional analysis books
Example 1.5. There exists a continuous function f on [−π, π], with f (π) = f (−π), whose Fourier series does not converge at 0.
Lecture 3:
Corollary 1.6. Let X be a Banach space, Y be a normed vector space, E ⊆ B(X, Y ). Suppose that sup T ∈E
|φ(T x)| < ∞
for each x ∈ X, φ ∈ Y ′. Then supT ∈E ‖T ‖ < ∞.
Example 1.7. Completeness of X is essential in Theorem 1.4 and Corollary 1.6.
A map T : X → Y is open if T maps open sets onto open sets. 1
Theorem 1.8. (Open Mapping Theorem) Let X and Y be Banach spaces and T be a bounded linear mapping of X onto Y. Then T is open.
Proof. (tricky): Rynne & Youngson, Kreyszig, etc
Corollary 1.9. (Inverse Mapping Theorem; Banach’s Isomorphism Theo- rem) Let X and Y be Banach spaces and T : X → Y be a bounded linear bijection. Then T −^1 is continuous (so X and Y are isomorphic).
Corollary 1.10. Let X be a Banach space with respect to two norms ‖ · ‖ 1 and ‖ · ‖ 2 , and suppose that there exists c such that ‖x‖ 1 ≤ c‖x‖ 2 for all x ∈ X. Then there exists c′^ such that ‖x‖ 2 ≤ c′‖x‖ 1 for all x ∈ X.
Definition of the graph G(T ) of T : X → Y as a subspace of X × Y. If X and Y are Banach spaces then X × Y is a Banach space in any of several equivalent norms inducing the product topology.
Lecture 4:
Any continuous map T : X → Y has closed graph.
Theorem 1.11. (Closed Graph Theorem) Let X, Y be Banach spaces, and T : X → Y be a linear operator whose graph is closed. Then T is continuous.
Proof. (short) Rynne & Youngson, Kreyszig, etc
Remarks that applications of CGT are usually abstract or theoretical, the theorem indicates why it is so hard to find discontinuous linear operators between Banach spaces, but there are many important unbounded linear operators whose domains are dense subspaces of Banach spaces and whose graphs are closed in X × Y. The OMT, IMT and CGT are more or less equivalent.
Example 1.12. Let h : R → R be a measurable function such that h.f ∈ L^1 (R) for all f ∈ L^1 (R). The operator f 7 → h.f on L^1 (R) has closed graph (by facts from integration theory), so it is bounded. Hence h is bounded a.e.
Example 1.13. Let L be the left-shift operator on ^1. Then limn→∞ ‖Lnx‖ 1 = 0 for each x ∈ ^1 , but ‖Ln‖B( (^1) ) = 1 for each n.
Proposition 1.14. Let X and Y be Banach spaces and Tn ∈ B(X, Y ) (n ≥ 1). The following are equivalent:
- There exists T ∈ B(X, Y ) such that limn→∞ Tnx = T x for all x ∈ X.
- For each x ∈ X, limn→∞ Tnx exists in Y.
- There exist a constant C and a subset Z of X such that (a) ‖Tn‖ ≤ C for all n, (b) The linear span of Z is dense in X, (c) limn→∞ Tnx exists for each x ∈ Z.
When these conditions are satisfied, T is the unique bounded linear operator such that T x = limn→∞ Tnx for each x ∈ Z. Moreover ‖T ‖ ≤ C.
Proof. Very standard—in every book
Example 3.2. Let (γr)r≥ 1 be a sequence of strictly positive numbers, and let `^2 γ be
the space of all sequences x = (αr) of scalars such that
r γr|αr|
(^2) converges. Then ` (^2) γ
is a Hilbert space in the inner product 〈x, y〉` (^2) γ =
r=1 γrαrβr.
Theorem 3.3. (Riesz-Fischer Theorem). L^2 (R) is a Hilbert space.
Proof. Most functional analysis books avoid proving this, because they do not assume that the reader knows about Lebesgue Integration. For proofs, see books on integration, such as those by Capinski & Kopp, Priestley, or Stein & Shakarchi (Vol. III).
Lecture 7:
Similarly, L^2 (a, b), L^2 (R^2 ) are Hilbert spaces. Indeed L^2 (Ω, F, μ) is a Hilbert space for any measure space.
Examples 3.4. 1. `^2 = L^2 (μ 1 ) where μ 1 is counting measure on N.
- `^2 γ = L^2 (μγ ), where μγ (E) =
r∈E γr^ for^ E^ ⊆^ N.
- The weighted∫ L^2 -space L^2 (R, gdt) is the L^2 -space of the measure mg, where mg(E) =
E g(t)^ dt^ for measurable subsets^ E^ of^ R.
Example 3.5. The Sobolev space W 1 ,^2 (R) is a Hilbert space for the inner product
〈f, g〉W 1 , 2 =
R
f g +
R
f ′g′.
Any closed subspace of a Hilbert space is a Hilbert space (for the same inner product).
Examples 3.6. 1. The space L^2 even(R) of all even functions in L^2 (R) (those with f (−t) = f (t) a.e.) is a closed subspace of L^2 (R) so it is a Hilbert space.
- Let D be the open unit disc in C equipped with two-dimensional Lebesgue measure (area). The subspace A^2 (D) of L^2 (D) consisting of those holomorphic functions which are in L^2 (D) is closed, so it is a Hilbert space for the L^2 -inner product. It is the Bergman space.
- Let∑ Y be the space of all those f ∈ L^2 (−π, π) with Fourier series of the form ∞ n=0 ane
int (^) (so an = 0 for all n < 0). Then Y is a closed subspace of L (^2) (−π, π). Sometimes Y is denoted by H^2 (T), and known as the Hardy space.
Lecture 8:
Reminder of definition of orthonormal set and orthogonal complement.
Proposition 4.1. Let Y be a subset of an inner product space X. Then
- Y ⊥^ is a closed linear subspace of X,
- Y ⊆ Y ⊥⊥,
- If Y ⊆ Z ⊆ X, then Z⊥^ ⊆ Y ⊥,
- (span Y )⊥^ = Y ⊥.
Proposition 4.2. Let Y be a subspace of an inner product space X, and let x ∈ X. Then x ∈ Y ⊥^ if and only if ‖x − y‖ ≥ ‖x‖ for all y ∈ Y , i.e. 0 is the nearest point of Y to x.
Theorem 4.3. Let Y be a non-empty closed convex subset of a Hilbert space X, and let x ∈ X. Then there exists unique y 0 ∈ Y such that ‖x − y‖ ≥ ‖x − y 0 ‖ for all y ∈ Y.
Proofs. The above results may be found in any book on Hilbert spaces. Look for orthogonal complement or closest point.
Theorem 4.4. (Projection Theorem). Let Y be a closed subspace of a Hilbert space X. Then X = Y ⊕ Y ⊥.
Proof. This is also in every book on Hilbert spaces.
Lecture 9:
Hence there is a projection PY on X with ker PY = Y ⊥^ and Ran PY = Y. Then PY x is the nearest point of Y to x. To calculate PY , either (a) find by inspection, a decomposition x = y + z where y ∈ Y and z ∈ Y ⊥; or (b) use orthonormal bases (see Section 6).
Examples 4.5. 1. Let X = L^2 (−π, π), and Y be the space of all f ∈ X such that f (t) = 0 for almost all t ∈ (0, π). Then Y ⊥^ is the space of all f ∈ X such that f (t) = 0 for almost all t ∈ (−π, 0), and PY f = f.χ(−π,0).
- Let X = L^2 (Ω, F, P) be a probability space, G be a sub-σ-algebra of F, and Y = L^2 (Ω, G, P). For an F-measurable random variable ξ, PY ξ is the conditional expectation E(ξ|G) of ξ with respect to G.
Corollary 4.6. Let Y be a subspace of a Hilbert space X.
- The following are equivalent: (i) Y is closed in X, (ii) X = Y + Y ⊥, (iii) Y = Y ⊥⊥.
- Y ⊥⊥^ = Y.
- The following are equivalent: (i) Y is dense in X, (ii) Y ⊥^ = { 0 }.
Examples 4.7. 1. Let X = L^2 (−π, π), and Y and Z be the subspaces of all even and odd functions, respectively. Then Z ⊆ Y ⊥^ and X = Y + Z. Hence, Y ⊥^ = Z and Z⊥^ = Y. In particular, Y and Z are closed.
- Let X = `^2 , γ ∈ F and Y be the linear span of {en − γen+1 : n ≥ 1 }, where en are the standard (orthonormal) basis vectors. Then Y is dense in X if and only if |γ| ≤ 1.
Theorem 4.8. (Riesz Representation Theorem; aka Riesz-Fr´echet Theorem). Let X be a Hilbert space. There is a conjugate-linear isometry J of X onto X′, given by: J(y)(x) = 〈x, y〉 (x, y ∈ X).
Recall from B4a the notion of the spectrum σ(T ) of a bounded operator T on a complex Banach space. In Hilbert space, the eigenvalues of T ′^ are the conjugates of the eigenvalues of T.
Proposition 5.5. Let T ∈ B(X) where X is a complex Hilbert space X. Then σ(T ∗) = {λ : λ ∈ σ(T )}.
Recall that an operator between Hilbert spaces is unitary if it is isometric and surjective.
Proposition 5.6. Let T, U : X → Y be bounded linear operators between Hilbert spaces.
- The following are equivalent: (i) T is isometric; (ii) 〈T x, T y〉 = 〈x, y〉 for all x, y ∈ X; (iii) T ∗T = IX.
- The following are equivalent: (i) U is isometric and surjective (i.e., U is unitary); (ii) U ∗U = IX and U U ∗^ = IY (i.e. U is invertible and U −^1 = U ∗); (iii) U and U ∗^ are both isometric.
Examples 5.7. 1. The left shift is unitary on `^2 (Z).
- A multiplication operator Mh is unitary on L^2 if and only if |h(t)| = 1 a.e.
- The map (αn) 7 → (γ n^1 / 2 αn) is unitary from
^2 γ to^2 if γn > 0 for all n. - For measurable, non-negative g, the map f 7 → g^1 /^2 f is isometric from L^2 (R, gdt) to L^2 (R). It is unitary if and only if g(t) > 0 a.e.
Proposition 5.8. If U : X → X is unitary then σ(U ) ⊆ T.
Proofs. All are quite simple and they can be found in books, e.g. Rynne & Youngson.
Lecture 12:
Definition of a (bounded) self-adjoint operator T.
Proposition 5.9. 1. If T ∈ B(X, Y ), then T ∗T and T T ∗^ are self-adjoint (on X and Y , respectively).
- If S and T are self-adjoint on X(= Y ) and ST = T S then ST is self-adjoint.
- The set of all self-adjoint operators on X is a norm-closed real subspace of B(X).
- Let T ∈ B(X). Then there are unique operators S 1 and S 2 such that T = S 1 + S 2 , S 1 is self-adjoint and S 2 is skew-adjoint.
Example 5.10. A multiplication operator Mh on L^2 is self-adjoint if and only if h(t) is real a.e.
For self-adjoint T , 〈T x, x〉 is real. Define mT = inf{〈T x, x〉 : ‖x‖ = 1}, MT = sup{〈T x, x〉 : ‖x‖ = 1}, wT = sup{|〈T x, x〉| : ‖x‖ = 1} = max(−mT , MT ).
Proposition 5.11. For any self-adjoint operator T ,
- ‖T ‖ = max(−mT , MT ) = wT ,
- ‖T n‖ = ‖T ‖n^ for all n ∈ N.
Remarks and examples that Proposition 5.11 fails for non-self-adjoint operators, but wT ≤ ‖T ‖ ≤ 2 wT in the complex case.
Theorem 5.12. Let T be a self-adjoint operator on a complex Hilbert space X. Then σ(T ) ⊆ {λ ∈ R : mT ≤ λ ≤ MT }. Moreover mT ∈ σ(T ) and MT ∈ σ(T ).
Proofs. Proposition 5.9 is standard (as in Part A). For other results, see Rynne & Youngson or Kreyszig.
Lecture 13:
Theorem 5.13. (Spectral theorem). If T is a self-adjoint operator on a Hilbert space X then there exist a measure space (Ω, F, μ), a measurable function h : Ω → R and a unitary U : X → L^2 (μ) such that T = U ∗MhU.
The Spectral Theorem is quite deep—well beyond this course—but the result makes it easy to see that certain things will be true, such as Theorem 5.12.
Proposition 5.14. Let X = Y ⊕ Z where Y and Z are both closed subspaces of X. Let P be the operator P (y + z) = y. Then the following are equivalent:
(i) Z = Y ⊥; (ii) P ∗^ = P ; (iii) ‖P ‖ ≤ 1.
Then ‖P ‖ = 1 or P = 0.
Proof. Fairly elementary; see Rynne & Youngson, for example.
Reminder of Gram-Schmidt process
Examples 6.1. Consider the space of all polynomials, with the usual algebraic basis 1 , t, t^2 ,... , tn,.... Take any inner product, and use Gram-Schmidt to obtain orthonor- mal polynomials pn(t) of degree n (unique up to scalar multiples of absolute value 1).
- For the inner product 〈f, g〉 =
− 1 f g, the orthonormal polynomials are the^ Legendre polynomials which are normalised Legendre polynomials.
- For the inner product 〈f, g〉 =
0 f^ (t)g(t)e
−t (^) dt, the orthonormal polynomials are the Laguerre polynomials Ln(t). The Laguerre functions e−t/^2 Ln(t) are orthonormal in L^2 (0, ∞).
- For the inner product 〈f, g〉 =
R f^ (t)g(t)e
−t^2 dt, the orthonormal polynomials are normalised versions of the (physicists’) Hermite polynomials Hn(t). The Hermite functions, obtained by multiplying the polynomials by e−t
(^2) / 2 , are orthonormal in L^2 (R).
Lecture 15:
Example 6.
- The normalised Haar functions form a complete orthonormal sequence of step functions in L^2 (0, 1). They are χ(0,1), χ(0, 1 2 )^
− χ( 1 2 ,1)
2(χ(0, 1 4 )^
− χ( 1 4 ,^
1 2 )
Theorem 6.6. Let X be a separable, infinite-dimensional, Hilbert space, and (en) be a complete orthonormal sequence in X. There is a unitary U : `^2 → X given by
U (αn) =
∑^ ∞
n=
αnen, U −^1 (x) = (〈x, en〉).
Proof. Apply Proposition 6.2 and Theorem 6.4.
There is a version of Theorem 6.6 for an inseparable Hilbert space X. Then X has an orthonormal family {eλ : λ ∈ Λ} whose span is dense in X, and there is a unitary from∑ ^2 (Λ) onto X where^2 (Λ) is the Hilbert space of all families (αλ)λ∈Λ such that
λ∈Λ |αλ|
Proposition 6.7. Let X be a separable Hilbert space with complete orthonormal se- quence (en), and let φ ∈ X′. Then
φ(x) = 〈x, y〉 (x ∈ X),
where y =
n=1 φ(en)en.
Proof. Apply Theorem 4.8 and Theorem 6.4.
To prove that the trigonometric orthonormal set is complete for L^2 (−π, π), consider the Poisson kernel
Pr(t) =
n∈Z
r|n|eint^ =
1 − r^2 1 − 2 r cos t + r^2
(0 < r < 1 , −π < t < π).
For f ∈ L^1 (−π.π), let
an =
2 π
∫ (^) π
−π
f (s)e−ins^ ds, fr(t) =
n∈Z
anr|n|eint.
Proposition 6.8. For f ∈ L^1 (−π, π),
lim r↗ 1
‖f − fr‖L 1 = 0.
Proof. Stein & Shakarchi prove a version of this with L^1 -convergence replaced by a.e. convergence. Their version requires more sophisticated techniques.
Lecture 16:
Theorem 6.9. The trigonometric orthonormal set is complete in L^2 (−π, π).
Corollary 6.10. For any f ∈ L^2 (−π, π), the Fourier series of f converges to f in L^2 -norm.
Proofs. Almost immediate from Theorem 6.4 and Proposition 6.8.
Remark that another way to Corollary 6.10 is via the classical (pointwise) theory of Fourier series.
Theorem 6.11. (Pointwise convergence) If f : [−π, π] → R is either monotonic or piecewise differentiable, then the Fourier series of f converges at each point to { (^) f (t+)+f (t−) 2 (|t|^ < π), f ((−π)+)+f (π−) 2 (|t|^ =^ π).
Theorem 6.12. (Fej´er’s Theorem) Let f : [−π, π] → C be continuous with f (−π) = f (π). Let
sm(t) =
∑^ m
n=−m
areirt, σk(t) =
k
∑^ k
m=
sm(t).
Then σk(t) → f (t) uniformly for t ∈ [−π, π].
Proofs. Not given in lectures, but they can be found in Young, Stein & Shakarchi (Vol 1), and many other books
Fej´er’s Theorem can also be used to prove the Weierstrass Approximation Theorem, but there are more direct proofs.
Theorem 6.13. (Weierstrass Approximation Theorem) The polynomials are dense in C[a, b] for the sup norm.
Proof. The direct proof given in lectures can be found in W. Rudin, Principles of Mathematical Analysis (Theorem 7.26 in the third edition).
