Lecture Notes in Functional Analysis Lectures by Andrei Khrennikov Note that these are rapidly taken and then even more swiftly typed notes, and as such errors might well occur. Be sure to check any oddities against the course literature [KF20]. Last updated January 22, 2016. Throughout this document, signifies end proof, N signifies end of example, and signifies end of solution. Table of Contents Table of Contents i 1 Lecture I 1 1.1 Metric Spaces ............................. 1 1.2 Continuous Functions in Metric Spaces ............... 2 1.3 Complete Metric Spaces ....................... 3 2 Lecture II & III 5 2.1 More On Contraction Mappings ................... 5 2.2 Applications of the Fixed Point Theorem ............. 6 2.3 Normed Linear Spaces ........................ 9 4 Lecture IV 10 4.1 Norm of Bounded Operator ..................... 10 4.2 Analytic Functions of Bounded Operators ............. 13 5 Lecture V 15 5.1 More On Bounded Operators .................... 15 5.2 Continuous Linear Functionals ................... 16 5.3 Hahn–Banach Theorem ....................... 17 6 Lecture VI 20 6.1 A Special Case of the Hahn–Banach Theorem ........... 20 6.2 Adjoint Operator ........................... 20 6.3 Hilbert Space ............................. 22 6.4 Adjoint Operator in a Hilbert Space ................ 23 6.5 Hermitian Operators ......................... 24 6.6 Hermitian Operators in Quantum Mechanics ........... 24 Notes by Jakob Streipel. i
Transcript
Lecture Notes in Functional Analysis
Lectures by Andrei Khrennikov
Note that these are rapidly taken and then even more swiftly typed notes,and as such errors might well occur. Be sure to check any oddities againstthe course literature [KF20]. Last updated January 22, 2016.
Throughout this document, signifies end proof, N signifies end ofexample, and � signifies end of solution.
Example 1.1.1. Let X = R, the real line. Then ρ(x, y) = |x− y| is a distancemeasure (the ordinary one), with the following properties:
(i) ρ(x, y) ≥ 0,
(ii) ρ(x, y) = 0 if and only if x = y,
(iii) ρ(x, y) = ρ(y, x),
(iv) The triangle inequality; for all x, y, z we have ρ(x, y) ≤ ρ(x, z) + ρ(z, y)(think of the sides of a triangle). N
Note that by setting z = 0 and taking y to be negative in the triangle inequalitywe get the perhaps more common form |x+ y| ≤ |x|+ |y|.
Definition 1.1.2 (Metric, and metric space). Let X be any set and let ρ(x, y)be any function on X×X, then ρ is called a metric and (X, ρ) a metric spaceif ρ satisfies (i)–(iv) above.
Examples 1.1.3. (i) X = Rn, ρ1(x, y) =n∑j=1
|xj − yj |.
(ii) X = Rn, ρ∞(x, y) = max1≤j≤n
|xj − yj |.
(iii) X = Rn, ρ2(x, y) =
√n∑j=1
(xj − yj)2.
(iv) X = C[a, b] = {x : [a, b]→ R | x is a continuous function}, ρ∞(x, y) =maxa≤t≤b
|x(t)− y(t)|, (which exists since [a, b] is closed and bounded). N
Definitions 1.1.4 (Open ball, closed ball, sphere). Given a metric space (X, ρ),with a ∈ X and r a positive real number:
(i) Open ball: Br(a) = {x ∈ X | ρ(x, a) < r},
(ii) Closed ball: Br(a) = {x ∈ X | ρ(x, a) ≤ r},
(iii) Sphere: Sr(a) = {x ∈ X | ρ(x, a) = r}.
Exercises 1.1.5. Let X = R2 (because it’s easy to visualise), and let a be afixed point in X and r be a fixed positive real number.
Draw the open ball, closed ball, and sphere around a with radius r given themetric ρ1, ρ∞, and ρ2, respectively.
Solutions. For ρ2, the open ball of radius r around a is simply the circle centredon a with radius r, since ρ2 is simply the Cartesian distance.
For ρ1 the distance between a and x is the sum of the difference of theircomponents. This sum must be less than r. The open ball then becomes asquare with diagonals 2r, the diagonals being parallel to the axes.
1Date: November 2, 2015.
LECTURE I 2
For ρ∞ we take, for a given x, the biggest of the differences of the componentsof x and a. This will give us a square with sides 2r, with sides parallel to theaxes. �
Exercise 1.1.6. Let X = C[a, b], with ρ∞ as the metric. Draw the ball Br(z).
Solution. The ball of radius r around the function z will be any function definedin the interval [a, b] such that it is contained within an r-corridor around z. �
1.2 Continuous Functions in Metric Spaces
Recall first the definition from single variable calculus: a function f : R→ R iscalled continuous at x0 ∈ R (also written f ∈ C(x0), if for all ε > 0 there existsa δ such that, when |x− x0| < δ, we have that |f(x)− f(x0)| < ε.
We copy this definition into the general settings of metric spaces.
Definition 1.2.1 (Continuity). Given two metric space (X1, ρ1) and (X2, ρ2),a function f : X1 → X2 is called continuous at the point x0 ∈ X1 (f ∈ C(x0))if, for all ε > 0 there exists a δ such that, when ρ1(x, x0) < δ, we also haveρ2(f(x), f(x0)) < ε.
Example 1.2.2. Let us use X = C[a, b] and ρ∞ as the metric. Further letf : X → R, given by a fixed point c ∈ [a, b], such that f(x) = x(c). Then
|f(x)− f(x0)| = |x(c)− x0(c)| ≤ maxa≤t≤b
|x(t)− x0(t)| = ρ∞(x, x0),
whence ε = δ suffices to show continuity. N
Exercise 1.2.3. Let X = `1 ={x = (x1, x2, . . . , xn, . . .)
∣∣ ∞∑j=1
|xj | < ∞}
, the
space of all absolutely summable series, with ρ1(x,y) =∞∑j=1
and T− : `1 → `1 be defined as T−(x1, x2, . . . , xn, . . .) = (x2, x3, . . . , xn, . . .).Is T+ continuous? How about T−?
Solution. We simply plug the known information into the definition of continu-ity, starting with T+:
ρ1(T+(x), T+(x0)) = 0 +
∞∑j=1
∣∣xj − x0j
∣∣ = ρ1(x,x0),
whence ε = δ suffices to show continuity. For T−, on the other hand, we have
ρ1(T−(x), T−(x0)) =
∞∑j=2
∣∣xj − x0j
∣∣ = −|x1 − x01|+ ρ1(x,x0),
whence T− is continuous only if |x1 − x01| = 0, and thereby not continuous in
general. �
LECTURE I 3
Example 1.2.4. The set
X = `∞ ={x = (x1, x2, . . . , xn, . . .)
∣∣ sup1≤j<∞
|xj | <∞}
using ρ∞(x,y) = sup1≤j<∞
|xj − yj | is a metric space. N
Example 1.2.5. The set
X = `2 ={x = (x1, x2, . . . , xn, . . .)
∣∣ ∞∑j=1
x2j <∞
}using
ρ2(x,y) =
√√√√ ∞∑j=1
(xj − yj)2
is a metric space. N
Exercise 1.2.6. Order `1, `2, and `∞ with respect to inclusion.
Solution. We think of it as follows: for `1, the sum of the absolute value of allof the elements in the sequence is finite. In `2, the sum of the square of all ofthe elements in the sequence is finite. Finally, in `∞, the biggest (kind of, atleast the smallest upper bound) element in the sequence is finite.
It is then easy to see that `∞ is bigger than both `1 and `2 (consider, forexample, (1, 1, . . . , 1, . . .) ∈ `∞).
Moreover, if∞∑j=1
|xj | ≤ ∞, then we will also have∞∑j=1
|xj |2 ≤ ∞, (take, for
example, the harmonic series), giving `1 ⊆ `2 , but not the other way around.Thus `1 ⊂ `2 ⊂ `∞. �
1.3 Complete Metric Spaces
Recall from single variable calculus the definitions of a Cauchy sequence andconvergence of a series:
• a1, a2, . . . , an, . . . ∈ R is a Cauchy sequence is |an − am| → 0 when n,m→∞,
• The sequence {an} is called convergent with the limit a (an → a as n →∞) if |an − a| → 0 as n→∞.
Note that, in R, the above are equivalent.Formulated in the terms of general metric spaces (X, ρ):
• The sequence {xn} is a Cauchy sequence if ρ(xn, xm)→ 0 as n,m→∞,
• The sequence Setxn converges to the limit x (xn → x) if, for all ε > 0,there exists an Nε such that for all n ≥ Nvarepsilon we have ρ(xn, x) ≤ ε.
Definition 1.3.1 (Complete metric space). A metric space (X, ρ) is calledcomplete if each Cauchy sequence has its limit in X. Otherwise (X, ρ) isincomplete .
LECTURE I 4
Examples 1.3.2. R is complete using the absolute values. Rn is complete usingall three of ρ1, ρ2, and ρ∞. N
Example 1.3.3. Take X = Rn and ρ∞ as the metric, and let
x(k) = (x(k)1 , x
(k)2 , . . . , x(k)
n )
be a Cauchy sequence in X. We have that
ρ∞(x(k),x(m)) = max1≤j≤n
∣∣∣x(k)j − x
(m)j
∣∣∣→ 0
as k,m→∞. Further we take xj = limk→∞
x(k)j and let ε > 0. Then there exists
an Nε such that if n,m ≥ Nε, we have
limm→∞
(max
1≤j≤n
∣∣∣x(k)j − x
(m)j
∣∣∣) ≤ ε,then max
1≤j≤n
∣∣∣x(k)j − xj
∣∣∣ ≤ ε.In other words, the limit of the n-tuple sequence is the n-tuple of the limits
of the coordinates. N
Exercise 1.3.4. Show that `1, `2, and `∞ are complete.
Solution. We show it for `1. We take a sequence {x(k)} of sequences
(x(k)1 , x
(k)2 , . . . , x(k)
n , . . .).
Since x(k) ∈ `1, then∞∑j=1
∣∣∣x(k)j
∣∣∣ < ∞ for any fixed k, whence certainly
ρ(x(k)n , x
(k)m ) must approach 0 as n and m increase (otherwise the series must
necessarily converge). Therefore {x(k)} is a Cauchy sequence.Since {x(k)} is a Cauchy sequence of real numbers, it must have some limit
{x}. Moreover, this limit is in `1 since
∞∑j=1
xj =
∞∑j=1
limk→∞
x(k)j = lim
k→∞
∞∑j=1
x(k)j <∞,
where switching the order of the limit processes is allowed since x(k) ∈ `1 isabsolutely convergent. �
Definition 1.3.5 (Contration). Let (X, ρ) be a metric space. Then the mapA : X → X is called a contraction if there exists 0 < α < 1 such thatρ(Ax,Ay) ≤ αρ(x, y) for all x, y ∈ X.
Remark 1.3.6. Any contraction is continuous. This is clear by the definition(let ρ(x, y) be sufficiently small, then ρ(Ax,Ay) ≤ αρ(x, y) is even smaller).
Definition 1.3.7 (Fixed point). Let A : X → X be a mapping, then x ∈ X iscalled a fixed point of A if Ax = x.
Theorem 1.3.8 ((Banach’s) Fixed point theorem). Let (X, ρ) be a completemetric space and let A be a contraction. Then there exists a unique fixed pointx ∈ X such that Ax = x.
LECTURE II & III 5
Proof. The following is a constructive proof using the method of iteration.Take any x0 ∈ X and let x1 = Ax0, x2 = Ax1 = A2x0, et cetera. In general,
xn = Axn−1 = Anx0.We claim that {xn} is a Cauchy sequence. Using n ≤ n′, we show this as
where in the first inequality we use that A is a contraction and in the thirdline we use the triangle inequality. The last bracket of the last line is just ageometric sum, however, so
ρ(xn, xn′) ≤ αn1− αn′−n
1− αρ(x0, x1) <
αn
1− αρ(x0, x1).
Since 0 < α < 1, αn will approach 0 as n approaches infinity, whence ρ(xn, xn′)will approach 0 as n and n′ become sufficiently big, meaning that {xn} is aCauchy sequence, and therefore we have some X 3 x = lim
n→∞xn.
Then, since A is continuous (by Remark 1.3.6) we have
Ax = A(
limn→∞
xn
)= limx→∞
Axn = limn→∞
= x,
from which we see that Ax = x, so x is a fixed point of A.It still remains to show that x is the unique fixed point of A. To do this we
simply assume that we have two fixed points x and y, such that Ax = x andAy = y. Then:
ρ(x, y) = ρ(Ax,Ay) ≤ αρ(x, y).
Since ρ(x, y) ≥ 0 and 0 < α < 1, for the above to hold we must have ρ(x, y) = 0,meaning that x = y (since ρ is a metric).
2 Lecture II2 & III3
2.1 More On Contraction Mappings
The Fixed Point Theorem has the following generalisation which will prove veryuseful.
Theorem 2.1.1. Let A be a continuous mapping of a complete metric space(X, ρ) into itself, and suppose that An is a contraction mapping (with n being apositive integer greater than 1). Then A has a unique fixed point.
2Date: November 5, 2015. Kind of; it never took place.3Date: November 9, 2015.
LECTURE II & III 6
Proof. Pick any point x0 ∈ X and let
x = limk→∞
Aknx0,
(which exists and is unique due to the fixed point theorem since An is a con-traction). Then, since A is a continuous mapping, we have
Ax = A limk→∞
Aknx0 = limk→∞
AAknx0.
Since An is a contraction mapping, we have
ρ(AknAx0, Aknx0) ≤ αρ(A(k−1)nAx0, A
(k−1)nx0) ≤ . . . ≤ αkρ(Ax0, x0),
where 0 < α < 1.Since the right-hand side of the inequality is positive, and αk tends to 0 as
k tends to infinity, it follows that
ρ(Ax, x) = limk→∞
ρ(AknAx0, Aknx0) = 0,
whence Ax = x and x is a fixed point of A. To show that this fixed point isunique we show that if any x is a fixed point of A, it must also be a fixed pointof An.
Suppose that we have Ax = x. Then clearly we also have that Anx =An−1(Ax) = An−1x = . . . = Ax = x, whence x is a fixed point of An as well.However since An is a contraction is has exactly one fixed point, so A must aswell.
2.2 Applications of the Fixed Point Theorem
This method of successive approximations can be used to prove both the exis-tence and uniqueness of solutions to integral equations. We will study two suchequations: Fredholm equations, and Volterra equations. However to do this wefirst require the following result.
Theorem 2.2.1. The metric space C[a, b] with the metric
ρ∞(f, g) = maxa≤t≤b
|f(t)− g(t)|
is complete.
Proof. To have fn → f in ρ∞ means that, for all ε > 0, there exists some Nεsuch that for all n ≥ Nε we have
maxa≤x≤b
|f(x)− fn(x)| < ε
and for fn to be a Cauchy sequence means that, for all n,m ≥ Nε, we have
|fn(x)− fm(x)| ≤ ε.
We fix m and take the limit as n approaches infinity, which yields
limn→∞
|fn(x)− fm(x)| = |f(x)− fm(x)| ≤ ε
LECTURE II & III 7
for all x, whence especially
ε ≥ maxa≤x≤b
|f(x)− fm(x)| = ρ∞(f, fm).
That such a limit f must be continuous if fn are continuous is known fromprevious courses.
Example 2.2.2 (Fredholm equation). A Fredholm equation (of the secondkind) is an integral equation of the form
f(x) = λ
∫ b
a
K(x, y)f(y) dy + ϕ(x), (2.2.1)
which involves two given functions K and ϕ, one unknown function f andan arbitrary parameter λ. The function K is called the kernel of the equa-tion and the equation is said to be homogeneous if ϕ ≡ 0 (otherwise non-homogeneous).
Suppose now that K(x, y) and ϕ(x) are continuous on the square a ≤ x ≤ b,a ≤ y ≤ b, which implies that |K(x, y)| ≤M is bounded by some M > 0 in thesquare, since it is compact.
Consider the mapping g = Af of the complete metric space C[a, b] into itselfgiven by
g(x) = λ
∫ b
a
K(x, y)f(y) dy + ϕ(x).
It is clear that, if g1 = Af1 and g2 = Af2, then
ρ∞(g1, g2) = maxa≤x≤b
|g1(x)− g2(x)| ≤ |λ|M(b− a) maxa≤x≤b
|f1(x)− f2(x)|
= |λ|M(b− a)ρ∞(f1, f2),
so that if
|λ|M(b− a) < 1 ⇐⇒ |λ| < 1
M(b− a), (2.2.2)
then A becomes a contraction.Moreover, by the fixed point theorem, if (2.2.2) holds, the original Fredholm
equation (2.2.1) must have a unique solution for the given λ.More to the point, the successive approximations used to solve the equation
are given by
fn(x) = λ
∫ b
a
K(x, y)fn−1(y) dy + ϕ(x),
for n ≥ 1, and f0 can be chosen as any continuous function on [a, b]. N
Exercise 2.2.3. Solve the Fredholm equation on the interval [0, 1] with thekernel K(x, y) = sin2(x− y), and ϕ(x) = cosx.
Solution. See Assignment I. �
Next we consider the Volterra equation.
LECTURE II & III 8
Example 2.2.4 (Volterra equation). The Volterra equation is given by
f(x) = λ
∫ x
a
K(x, y)f(y) dy + ϕ(x) (2.2.3)
which is identical to the Fredholm equation (2.2.1) apart from the upper limitof the integration interval being the variable x rather than the constant b.
We claim that the Volterra equation can be solved for any λ, not just λsufficiently small, as with the Fredholm equation.
We let A be a mapping on C[a, b] into itself, as before, defined by
Af(x) = λ
∫ x
a
K(x, y)f(y) dy + ϕ(x),
and take any f1, f2 ∈ C[a, b]. Then we have
|Af1(x)−Af2(x)| = λ
∫ x
a
K(x, y)(f1(y)− f2(y)) dy
≤ λM(x− a) maxa≤x≤b
|f1(x)− f2(x)|,
where, as before, M = maxa≤x,y≤b
|K(x, y)|.From this it follows that∣∣A2f1(x)−A2f2(x)
∣∣ ≤ λ2M2 maxa≤x≤b
|f1(x)− f2(x)|∫ x
a
(x− a) dx
= λ2M2 (x− a)2
2maxa≤x≤b
|f1(x)− f2(x)|,
whence in general
|Anf1(x)−Anf2(x)| ≤ λnMn (x− a)n
n!maxa≤x≤b
|f1(x)− f2(x)|
≤ λnMn (b− a)n
n!maxa≤x≤b
|f1(x)− f2(x)|,
since a ≤ x ≤ b. This implies that
ρ(Anf1, Anf2) ≤ λnMn (b− a)n
n!ρ(f1, f2),
in which, regardless of λ, we can always make
λnMn (b− a)n
n!< 1
by choosing n sufficiently large. Therefore it follows from Theorem 2.1.1 that,since An (for n sufficiently large) is a contraction, A must have a unique fixedpoint f . N
This method of iterations is also useful when solving other type of fixed pointproblems, not only integral solutions.
LECTURE II & III 9
Example 2.2.5 (Lipschitz functions and Lipschitz continuity). Lipschitz func-tions are functions f that satisfy |f(x)− f(y)| ≤ L|x− y| for all x and y, whereL is the so-called Lipschitz constant (for f).
It is obvious that these functions are continuous by definition, whence weformulate a sufficient condition for a function in C(1)[a, b] (the space of oncecontinuously differentiable functions) to be Lipschitz.
For such functions we have, by the mean value theorem, that
f(x)− f(y) = f ′(c)(x− y)
for some c ∈ [a, b]. Taking absolute values we therefore get
|f(x)− f(y)| = L|x− y|,
where L = maxa≤c≤b
|f ′(c)|.We see then that if we interpret f as a map between two complete metric
spaces (which it is; from C[a, b] into itself), we have that f almost satisfies thecondition of a contraction; it only remains to require that 0 < L < 1.
It then follows from the fixed point theorem that f(x) = x has a uniquesolution x, and that it can be found numerically by iterating xn = f(xn−1),with x0 as any initial value in [a, b]. N
Exercise 2.2.6. Select a continuous function f on the interval [a, b] such thatits Lipschitz constant L is less than 1. Solve, for this f , the equation f(x) = xby the method of iteration.
Solution. See Assignment II. �
2.3 Normed Linear Spaces
Recall first some basics from linear algebra regarding the definition of a spacebeing linear (meaning that we have two operators, addition of vectors and mul-tiplication by scalars, along with an assortment of properties of these).
We leave it as written that Rn, C[a, b], `1, `2, and `∞ are all examples oflinear spaces.
Definition 2.3.1 (Norm). A map f , usually denoted ‖·‖, from a space E to Ris a norm if it has the following properties:
(i) ‖x‖ ≥ 0 for all x ∈ E,
(ii) ‖x‖ = 0 if and only if x = 0,
(iii) ‖λx‖ = |λ|‖x‖, for all x ∈ E and real (or complex) constants λ,
(iv) Triangle inequality: ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x,y ∈ E.
A space E equipped with a norm is called a normed space .
Any norm also defines a metric by
ρ(x, y) = ‖x− y‖,
which clearly satisfies the conditions for a metric (see Definition 1.1.2), wherebyany normed space is also a metric space. The converse of this is not necessarilytrue.
LECTURE IV 10
Example 2.3.2. All of Rn, C[a, b], `1, `2, and `∞ are normed linear spaces,with norms defined analogously to their metrics. N
A very important class of normed spaces are the following. -
Definition 2.3.3 (Banach space). A normed space that is also complete iscalled a Banach space .
Example 2.3.4. Take E1 = C(1)[a, b] and E2 = C[a, b]. Then ‖f‖(1)= ‖f‖∞+
‖f ′‖∞ is a norm. This can be generalised to E1 = C(n)[a, b], where we get
‖f‖(n)= ‖f‖∞ + ‖f ′‖∞ + . . .+
∥∥∥f (n−1)∥∥∥∞
+∥∥∥f (n)
∥∥∥∞
=
n∑j=0
∥∥∥f (j)∥∥∥∞. N
Definition 2.3.5 (Bounded linear operator). A linear operator A : E1 → E2
is called bounded if there exists some c > 0 such that ‖Ax‖2 ≤ c‖x‖1 for allx ∈ E1, where ‖·‖1 is the norm in E1 and ‖·‖2 is the norm in E2.
Exercise 2.3.6. Check whether the maps for the Fredholm and Volterra equa-tions are bounded linear mappings. In addition, verify that the differentialoperator
p( ddx
),
where
p(ξ) =
n∑j=1
ajξj
is a polynomial of degree n is a bounded linear operator.
Solution. See Assignment II. �
4 Lecture IV4
Recall from last lecture: E is a linear normed space, A is a linear boundedoperator, then A is bounded if there exists a c such that ‖Ax‖ ≤ c‖x‖ for all x.Recall also the definitions of ball and sphere.
4.1 Norm of Bounded Operator
Definition 4.1.1 (Norm of bounded operator). Let A be a linear operator ona linear normed space E, then
‖A‖ = sup‖x‖≤1
‖Ax‖
is called the norm of A.
Remark 4.1.2. Note that ‖A‖ ≤ c, whereby we also have ‖A‖ = inf{c | ‖Ax‖ ≤c‖x‖}.There are different, and often in applications more useful, ways of representingthe norm of an operator.
4Date: 12 November, 2015.
LECTURE IV 11
Theorem 4.1.3. Let A be a linear bounded operator on a linear normed spaceE. Then
(i) ‖A‖ = sup‖x‖=1
‖Ax‖,
(ii) ‖A‖ = supx 6=0
‖Ax‖‖x‖ .
Proof. (i) Clearly we have sup‖x‖=1
‖Ax‖ ≤ sup‖x‖≤1
‖Ax‖, whence it remains to show
the opposite inequality.To do this, take any x such that ‖x‖ < 1. Then, since we have∥∥∥∥ x
‖x‖
∥∥∥∥ = ‖λx‖ = |λ|‖x‖ =1
‖x‖‖x‖ = 1
we have that x‖x‖ belongs to the unit sphere, whence
‖Ax‖ = ‖x‖∥∥∥∥A x
‖x‖
∥∥∥∥ ≤ ∥∥∥∥A x
‖x‖
∥∥∥∥.By taking supremum we are left with
sup‖x‖≤1
‖Ax‖ ≤ sup‖x‖≤1
∥∥∥∥A x
‖x‖
∥∥∥∥ ≤ sup‖y‖=1
‖Ay‖.
(ii) We have
supx 6=0
‖Ax‖‖x‖
= supx 6=0
∥∥∥∥A x
‖x‖
∥∥∥∥ = sup‖y‖=1
‖Ay‖.
Why are we concerned with this? Because given Remark 4.1.2 we must have theinequality ‖Ax‖ ≤ ‖A‖‖x‖. Upper bounds for this are often easy to find, butequalities can prove trickier; this is where the alternate representations abovecome into use.
Example 4.1.4. Let E = `∞ = {x = (x1, x2, . . . , xn, . . .)|‖x‖∞ = sup1≤j<∞
|xj | <
∞}, ans also let T−x = (x2, x3, . . . , xn, . . .) as previously. What is the norm ofT−?
We compute‖T−x‖∞ = sup
2≤j<∞|xj |
and‖x‖∞ = sup
1≤j<∞|xj |,
whence clearly ‖T−‖∞ ≤ 1 = c. Is it equal to c = 1? If we can find an x suchthat ‖x‖∞ = 1 whilst ‖T−x‖∞ = c, we will have shown this to be the case, byTheorem 4.1.3 (i). One example of such an x is x = (1, 1, 1, . . .). N
Exercise 4.1.5. Find the norm of T− in `1 and `2 as well. Also find the normof T+ in `1, `2, and `∞.
Solution. See Assignment III. �
LECTURE IV 12
For a more interesting example, consider the following.
Example 4.1.6. Let E = C[a, b], and let A : E → R be defined as Aϕ = ϕ(t0),for some fixed t0 ∈ [a, b].
We have then that ‖Aϕ‖ = |ϕ(t0)|, whereas
‖ϕ‖∞ = maxa≤t≤b
|ϕ(t)|
whereby obviously ‖A‖ is bounded by 1, since the norm of the image is the valuein a point, and the norm of the function is the maximum of the values in allpoints. Is ‖A‖ = 1? To show this we try to find a ϕ0 such that the maximumof ϕ in [a, b] is 1, and that the value of ϕ in t0 is 1 as well. Clearly the triviallysimple ϕ0(t) = 1 will do just fine. N
Remark 4.1.7. Note that, since sup isn’t max, sometimes it might be necessaryto study a sequence of xn such that ‖xn‖ → 1 and ‖Axn‖ → c as n → ∞. SeeExample 4.2.5.
Theorem 4.1.8 (Properties of operator norm). Let A and B be bounded linearoperators. Then the following hold:
(i) ‖A‖ ≥ 0,
(ii) ‖A‖ = 0 if and only if A = 0,
(iii) ‖λA‖ = |λ|‖A‖,
(iv) ‖A+B‖ ≤ ‖A‖+ ‖B‖.
Proof. These are all basic manipulation either of the definition or one of thederived representations from Theorem 4.1.3.
(i) and (ii) are obvious.(iii) Use Theorem 4.1.3 (ii):
‖λA‖ = supx 6=0
‖λAx‖‖Ax‖
= |λ| supx 6=0
‖Ax‖‖Ax‖
= |λ|‖A‖.
(iv): Use the fact that norms satisfy the triangle inequality:
‖A+B‖ = sup‖x‖≤1
‖Ax+Bx‖ ≤ sup‖x‖≤1
‖Ax‖+ sup‖x‖≤1
‖Bx‖ = ‖A‖+ ‖B‖.
We note that these properties are exactly those of an ordinary norm, whencethe space L (E1, E2) of all bounded operators A : E1 → E2 is a linear normedspace.
Theorem 4.1.9. Let E2 be a Banach space5, then L (E1, E2) is a Banach spaceas well.
Proof. Let (An) be a Cauchy sequence in L (E1, E2), meaning that ‖An −Am‖ →0 as m,n→∞. Therefore, for all x ∈ E1, ‖Anx−Amx‖ ≤ ‖An −Am‖‖x‖ → 0as m,n→∞.
5A complete normed space
LECTURE IV 13
Therefore (Anx) is a Cauchy sequence in E2. However since E2 is complete,this Cauchy sequence must have a limit, so there exists some
limn→∞
Anx = Ax.
Moreover, since the limit is linear, A : E1 → E2 must also be linear.Consider
‖Anx‖ ≤ ‖AnxA1x‖+ ‖A1x‖ ≤ ‖An −A1‖‖x‖+ ‖A1‖‖x‖.
Let n go to infinity and we get
limn→∞
‖Anx‖ = Ax ≤ c‖x‖
for some c, whence A is also bounded.Finally consider the following: for all ε > 0 there is some Nε such that for
all n,m ≥ Nε we have‖Anx−Amx‖ ≤ ε
(since An is a Cauchy sequence). We fix n and let m→∞:
‖Anx−Ax‖ = ‖(An −A)x‖ ≤ varepsilon.
Now take the supremum and we get ‖An −A‖ ≤ ε, whence the limit of theCauchy sequence exists and the space is therefore complete.
In other words, it is enough for the image space to be Banach for the space ofall linear bounded operators into this space to be Banach as well.
Theorem 4.1.10. Let E2 be Banach, let An ∈ L (E1, E2) be bounded, and
let∞∑n=1‖An‖ < ∞. Then
∞∑n=1
An converges in L (E1, E2) and its sum is in
L (E1, E2).
Proof. This is a simple application of the triangle inequality on partial sums.Let N > M : ∥∥∥∥∥
N∑n=1
An −M∑n=1
An
∥∥∥∥∥ =
∥∥∥∥∥N∑
n=M+1
An
∥∥∥∥∥ ≤N∑
n=M+1
‖An‖,
which approaches 0 if we let M,N →∞.
4.2 Analytic Functions of Bounded Operators
To explore this we must first discuss compositions of bounded operators.
Theorem 4.2.1 (Composition of bounded operators). Let E be Banach andlet A,B ∈ L (E,E). Then ‖AB‖ ≤ ‖A‖‖B‖.
Proof. We use the definition of operator norm:
‖AB‖ = sup‖x‖≤1
‖A(Bx)‖ = supnormx≤1
‖Bx‖∥∥∥∥A Bx
‖Bx‖
∥∥∥∥,where the fraction in the last norm is in the unit sphere, whence
‖AB‖ ≤
(sup‖x‖≤1
‖Bx‖
)(sup‖x‖≤1
‖Ax‖
)= ‖A‖‖B‖.
LECTURE IV 14
Example 4.2.2. Consider the analytic function ex =∞∑n=0
xn
n! , for x ∈ R.
Let A ∈ L (E,E), where E is Banach, and consider
eA =
∞∑n=0
An
n!≤∞∑n=0
‖A‖n
n!= e‖A‖ <∞,
so eA is a well-defined bounded operator. N
Using similar arguments one may define sinA, cosA and ln(1+A). We constructthe latter.
Example 4.2.3. We have from calculus that
ln(1 + x) =
∞∑n=1
(−1)n+1xn
n
which converges for |x| < 1. Thus we also have that ln(1 + A) converges if‖A‖ < 1. N
In general we have that if f is an analytic function, then
f(x) =
∞∑n=0
f (n)(0)
n!xn,
which converges for some |x| ≤ R, then
f(A) =
∞∑n=0
f (n)(0)
n!An
converges for the same ‖A‖ ≤ R.Recall the differential operator from Exercise 2.3.6,
p( ddx
): C(m)[a, b]→ C[a, b].
We would prefer it if we had the same domain and codomain, as we have in all ofthe previous theorems. One obvious candidate for this would be E = C∞[a, b],but there is no norm on this space (since it’d be the sum of infinitely manyderivatives, which needn’t converge) so it is impossible to make this a linearnormed space (although it is a linear topological space; more on this later).
The reason this would be of use is that it would help us solve certain differ-ential equations, which is one of the main goals of functional analysis.
Example 4.2.4. Suppose that E = C∞. Then, if for all t we have ϕ(t) ∈ C∞,it follows that ϕ(t) = ϕ(t, x). Let A = p( ∂
∂x ). Then
∂ϕ
∂t(t, x) = p
( ∂∂x
)ϕ(t, x) =
n∑k=0
ak∂kϕ
∂xk(t, x). N
Example 4.2.5. Let A = ddx : C(1)[0, 1]→ C[0, 1]. Then
‖Aϕ‖∞ = sup0≤t≤1
|ϕ′(t)| ≤ sup0≤t≤1
|ϕ(t)|+ sup0≤t≤1
|ϕ′(t)| = ‖ϕ‖(1),
LECTURE V 15
so clearly ‖A‖ ≤ 1. We would like to show that it is indeed equal to 1, but
clearly we can’t have ‖ϕ‖(1)= 1 and
‖Aϕ‖∞ = ‖ϕ′‖∞ = 1
simultaneously since
‖ϕ‖(1)= ‖ϕ‖∞ + ‖ϕ′‖∞,
meaning that ‖ϕ‖∞ = 0.So instead we look for a sequence ϕδ such that ‖ϕδ‖∞ ≈ 0 whilst ‖ϕ′δ‖∞ ≈ 1.
Consider for example fδ(t) =√δ + t. Clearly the maximum of this is
√1 + δ.
Moreover
f ′δ(t) =1
2√δ + t
the maximum of which is 1/(2δ1/2).We want √
δ11/(2δ1/2)
=√δ + 1 · 2δ1/2
to be small, which it is if δ is much smaller than 1 (i.e. δ approaches 0).Therefore we consider the sequence of functions
ϕδ(t) =fδ(t)
‖fδ‖(1).
This gives us
‖ϕδ‖∞ =1
‖fδ‖(1)‖fδ‖∞ =
‖fδ‖∞‖fδ‖∞ + ‖f ′δ‖∞
=‖fδ‖∞/‖f ′δ‖∞
1 + ‖fδ‖∞/‖f ′δ‖∞=
2√
1 + δ · δ1/2
1 + 2√
1 + δ · δ1/2
which approaches 0 as δ approaches 0, whilst ‖ϕδ‖(1)approaches 1. Therefore
supδ‖Aϕδ‖∞ = 1. N
Exercise 4.2.6. Let A = d2
dx2 : C(2)[0, 1]→ C[0, 1]. Find ‖A‖.
Solution. See Assignment III. �
5 Lecture V6
5.1 More On Bounded Operators
If, in L (E1, E2) from last lecture, E1 = E2 = E, we will denote this simplyL (E). Since this is the space of all linear bounded operators from E to itself,this is of course a linear space. Moreover, it has a multiplication operation;composition of operators. In other words, for A,B ∈ L (E), AB = A ◦B.
6Date: November 16, 2015.
LECTURE V 16
Moreover, this composition is distributive over addition of operators: A(B1+B2) = AB1 +AB2.
An algebraic structure with a linear space and a multiplication operation iscalled an algebra .
Additionally, if E is Banach, then so is L (E), by Theorem 4.1.10. Thismultiplication then also satisfies ‖AB‖ ≤ ‖A‖‖B‖, by Thorem 4.2.1.
We have the following important theorem.
Theorem 5.1.1 (Bounded linear operators are continuous, and vice versa).Let E1 and E2 be linear normed spaces. Then A : E1 → E2 being linear andbounded (A ∈ L (E1, E2)) is equivalent to A being linear and continuous.
Proof. (⇒) Suppose that A ∈ L (E1, E2). Then by its linearity and, for thefinal inequality, boundedness, we have
‖Ax1 −Ax2‖ = ‖A(x1 − x2)‖ ≤ ‖A‖‖x1 − x2‖
which clearly approaches 0 as ‖x1 − x2‖ approaches 0. Ergo boundedness of alinear operator implies continuity.
(⇐) Suppose now that A : E1 → E2 is continuous. Further suppose that Ais not bounded, meaning that
‖A‖ = sup‖x‖≤1
‖Ax‖ =∞.
In other words, if we pick x from the unit ball B1(0) = {x | ‖x‖ ≤ 1}, we wouldlike to show that Ax isn’t bounded in E2.
In particular there must exist some sequence xn ∈ B1(0) such that ‖Axn‖ >n for all positive integers n.
Dividing this last inequality by n (which is positive and so doesn’t changethe inequality) we have ∥∥∥Axn
n
∥∥∥ > 1,
where ∥∥∥xnn
∥∥∥ =‖xn‖n≤ 1
n,
since xn come from the unit ball, making their norm between 0 and 1.We thus have that ∥∥∥Axn
n
∥∥∥ > 1
for all n, but also thatxnn
approaches 0 as n approaches infinity, whence we must approach A0 = 0, whichis a contradiction. Therefore linear operators being continuous implies that theyare bounded.
5.2 Continuous Linear Functionals
We have shown that bounded linear operators and continuous linear operatorsare equivalent, and in the interest of staying consistent with the literature wewill henceforth speak of the latter.
LECTURE V 17
Definition 5.2.1 (Continuous Linear Functionals). Let E be a normed linearspace. Then operators A : E → R (or, alternatively, A : E → C) are calledlinear functionals.
The space of all such linear functionals from E, L (E,R), is denoted E′,which is called the dual space to E.
Typically we will use x ∈ E and y for continuous linear functionals.Note that E′ is a normed linear space, and that since R is Banach, so is E′.With this we are almost ready to attack one of the main results of functional
analysis, however first we require the following definition.
Definition 5.2.2. Let E be a linear space. A function p : E → R is called asemi-norm if, for all x, y ∈ E and λ ∈ R,
(i) p(x) ≥ 0,
(ii) p(λx) = |λ|p(x),
(iii) p(x+ y) ≤ p(x) + p(y).
Note the lack of p(x) = 0⇔ x = 0, distinguishing the semi-norm from norm.
Example 5.2.3. Consider E = C(−∞,∞), the space of all functions continu-ous on the entire real line. Let [a, b], −∞ < a < b <∞ be some interval on thisline, with the semi-norm of a function defined as the maximum of the functionon this interval. Then any function that is zero on this interval has semi-normzero, despite not necessarily being the zero function. N
5.3 Hahn–Banach Theorem
Theorem 5.3.1 (Hahn–Banach theorem). Let L be a linear space and let p bea semi-norm on L. Further let L0 be a subspace of L, L0 ⊂ L, and supposethat we have some linear functional f0 : L0 → R such that f0(x) ≤ p(x) for allx ∈ L0.
Then it is possible to extend f0 onto L such that f : L→ R, with f |L0 = f0,and f(x) ≤ p(x) for all x ∈ L.
Proof. We have L0 ⊂ L, and we assume that L0 6= L (otherwise we are done).We take some z ∈ L \ L0, and define using this z
L1 = {x+ tz | t ∈ R, x ∈ L0}.
Using this we attempt to construct a linear functional f1 : L1 → R:
f1(x+ tz) = f0(x) + tf(z)
such thatf1(x+ tz) ≤ p(x+ tz).
Note that since z is fixed, f1(z) = c for some fixed constant c ∈ R.We thus have
f0(x) + tc ≤ p(x+ tz). (5.3.1)
We consider two scenarios: (i) t being positive, and (ii) t being negative.
LECTURE V 18
(i) If t is positive, we have
f0
(xt
)+ c ≤ p
(xt
+ z),
whencec ≤ −f0
(xt
)+ p(xt
+ z),
where for future reference we’ll refer to x/t, t > 0, as y′.(ii) If t is negative, we have instead that
f0(x)
t≥ p(x+ tz
t,
which if we multiply numerator and denominator in the right-hand side by −1becomes
f0
(xt
)+ c ≥ −p
(− x
t− z),
giving
c ≥ −f0
(xt
)− p(− x
t− z),
in which we label x/t for t < 0 as y′′.Therefore, since y′ and y′′ are simply scaled versions of x, which came from
L0, we have y′, y′′ ∈ L0 as well, which gives us that
−f0(y′′)− p(−y′′ − z) ≤ −f0(y′) + p(y′ + z).
We solve this for f0:
f0(y′)− f0(y′′) ≤ p(y′ + z) + p(−y′′ − z),
in which we rewrite the left-hand side, recalling the triangle inequality for semi-norms:
whence the requisite inequality for f ’s existence holds since there exists somec ∈ R so that (5.3.1) holds.
Using this we can thus construct (L1, f1) such that the extension f1 hasthe properties required. Repeating this on L2 = L \ L1 we can then construct(L2, f2), (L3, f3), et cetera, up to (Ln, fn) for any positive integer n.
This also generalises nicely even if we need to do it countably infinitely manytimes.
How about uncountably infinitely many times? For this we require some settheoretic notions, and so an interlude. . . .
Axiom 5.3.2 (Axiom of choice). Suppose that we have finitely many setsA1, A2, . . . , An. On these we can define a so-called “choice function” f suchthat f(Aj) ∈ Aj for all j. Naturally this works also for countably many sets.
Whether this should work even for uncountably is not obvious, and in factis an axiom (read: choice) made in axiomatic set theory; one decides whetherone accepts it. This is called the Axiom of Choice .
That is, let M be any set. If we accept the axiom of choice, there exists achoice function f such that, for all A ⊂M , f(A) ∈ A.
LECTURE V 19
An equivalent statement is Zorn’s lemma, however to discuss that we first needthe following definitions.
Definition 5.3.3 (Partial order, linear order). A set M is called partiallyordered if between some pairs (a, b) ∈ M × M there can be established arelation ≤ such that it satisfies
(i) Reflexivity; a ≤ a,
(ii) Transitivity; a ≤ b and b ≤ c implies a ≤ c,
(iii) Anti-symmetry; a ≤ b and b ≤ a implies a = b.
Such a relation ≤ is called a partial order relation .If the relation ≤ holds between all pairs (a, b) ∈M×M , ≤ is called a linear
order relation .
Examples 5.3.4. The real numbers R under the ordinary ≤ is a linear order.The space of continuous functions on an interval, C[a, b], with ϕ1 ≤ ϕ2 if,
for all t ∈ [a, b], ϕ1(t) ≤ ϕ2(t), is a partial order. N
Definition 5.3.5 (Chain). Let M be a partially ordered set. Any linearlyordered subset A of M is called a chain .
Definition 5.3.6 (Supremum). Let M be a partially ordered set, and let M ′ ⊂M . Then the smallest m ∈M such that m′ ∈M ′ we have m′ ≤ m is called thesupremum with respect to M ′, denoted supM ′ = m.
Lemma 5.3.7 (Zorn’s lemma). Let M be a partially ordered set. If any chainin M has a supremum, then M has a maximal ordered element.
With this we are ready to continue with the proof of the Hahn–Banach theorem.
Proof continued. We have the following partial order on M , with M being pairsof subspaces of L and a function f as defined earlier: Let (L′, f ′) and (L′′, f ′′)be two such a pairs of subsets of L and a functions as described earlier. Then(L′, f ′) ≤ (L′′, f ′′) if and only if L′ < L′′ and f ′′|L′ = f ′.
Consider under this partial order a chain (Lc, fc), i.e. a linearly orderedsubuset of L, and call the collection of all such M ′ = {(Lc, fc)}.
We then take L to be the union of all such Lc, with f |Lc= fc (which is
possible since (Lc, fc) are all chains. Then (L, f) ≥ (Lc, fc) for all (Lc, fc) ∈M ′,whence it is the supremum.
Thereby, by Zorn’s lemma, there exists a maximal element in M . We claimthat the previously described (L, f) is this maximal element.
That f satisfies the criteria is clear. What about L = L? We see that if thisis not the case, i.e. that L 6= L, we can use the previous construction by takingz ∈ L \ L and constructing L′ using L and this z, whence L wasn’t maximal,giving us a contradiction.
LECTURE VI 20
6 Lecture VI7
Recall from last time the Hahn–Banach theorem which states that, if L0 ⊂ L isa linear space and p is a semi-norm on L, with f0 : L0 → R a linear functional onL0 such that f0(x) ≤ p(x) for all x ∈ L0, then if we assume the axiom of choicewe have that f0 can be extended to f : L → R, where f is a linear functionalon L such that f(x) ≤ p(x) for all x ∈ L and f |L0 = f0.
We now discuss one of the main uses of the theorem.
6.1 A Special Case of the Hahn–Banach Theorem
Let E be a linear normed space. Then, since p(x) = c‖x‖, for some constant c,is of course a semi-norm, we have that f0(x) ≤ c‖x‖ for all x ∈ L0 implies thatf(x) ≤ c‖x‖ for all x ∈ E. Moreover −f0(x) ≤ |f(x)| ≤ c‖x‖ for all x ∈ E.
Thus finally |f0(x)| ≤ ‖f0‖L0‖x‖, which implies that |f(x)| ≤ ‖f‖L‖x‖, and
additionally ‖f0‖L0= ‖f‖L.
Notation 6.1.1. Let E be a linear normed space and let x ∈ E and f ∈ E′.Then x 7→ f(x) = 〈x, f〉 is the so-called duality form , commonly used inphysics and distribution theory. I.e. in general 〈·, ·〉 : E × E′ → R.
This is useful because when E is normed, so is E′, and thus so is (E′)′, et cetera.But in fact often we will have E = E′, or, sometimes, E,E′, E′′ = E′, which wecall a reflexive space .
6.2 Adjoint Operator
Definition 6.2.1 (Adjoint operator). Let E1 and E2 be linear normed spaces,and let A ∈ L (E1, E2). Then whilst A : E1 → E2, its adjoint operator A∗
does E′2 → E′1. In other words, we have a picture like
E1A−→ E2
E′1 ←−A∗
E′2.
In the new duality notation we have, by definition, 〈Ax, f〉 = 〈x,A∗f〉, where,if we pay attention, we note that x ∈ E1, Ax ∈ E2, f ∈ E′2, and A∗f ∈ E′1.
So for each f ∈ E′2 we have some functional on E1 such that x 7→ 〈x,A∗f〉 =〈Ax, f〉, so x : E1 → R.
We ask a few questions:
1. Is x linear? Yes, since A and f are linear, and 〈x,A∗f〉 = f(Ax) is acomposition of two linear maps.
2. Is it continuous? Yes, again because A and f are continuous and x issimply their composition.
Therefore f 7→ A∗f is really from E′2 → E′1.
7Date: November 19, 2015.
LECTURE VI 21
We estimate its norm:
‖A∗‖ = sup‖f‖E′2≤1
‖A∗f‖E′1 = sup‖f‖E′2≤1
sup‖x‖E1
≤1
|〈x,A∗f〉︸ ︷︷ ︸(A∗f)(x)
|
= sup‖f‖E′2≤1
sup‖x‖E1
≤1
|〈Ax, f〉|,
where Ax is a functional on E′2, whence 〈Ax, f〉 ∈ E′′2 , so
‖A∗‖ = sup‖x‖E1
≤1
sup‖f‖E′2≤1
|〈Ax, f〉| = sup‖x‖E1
≤1
‖Ax‖E2= ‖A‖,
so we have that ‖A∗‖ = ‖A‖.There is a striking resemblance between adjoint operators and the transpose
of matrices, as we demonstrate in the following example.
Example 6.2.2. Consider an operator A : Rn → Rm.First of all we claim that (Rn)′ = Rn. To show this, let ej = (0, . . . , 1, . . . , 0)
be the elements of a base in Rn (with the 1 of course occurring in the jthcoordinate). Then naturally we have
x =
n∑j=1
xjej ,
whence for f ∈ (Rn)′ we must, because of the linearity of the operator, have
f(x) = f( n∑j=1
xjej
)=
n∑j=1
xjf(ej),
so for any f ∈ (Rn)′ there exists f = (y1,y2, . . . ,yn) such that
f(x) =
n∑j=1
xjyj ,
whence f is itself an element in Rn, so (Rn)′ = Rn.On Rn we can introduce many norms (‖·‖1, ‖·‖2, ‖·‖∞, et cetera), however
by the following theorem (unfortunately without proof, for it is quite tricky)the choice doesn’t matter.
Theorem. All norms on Rn are equivalent, meaning that for all ‖·‖ and ‖·‖′there exists some constants c1 and c2 such that, for all x, c1‖x‖ ≤ ‖x‖′ ≤ c2‖x‖.
Let us then again consider 〈Ax, y〉 = 〈x,A∗y〉. We have
Ax =
a11 a12 · · · a1n
a21 a22 · · · a2n
......
. . ....
am1 am2 · · · amn
x1
x2
...xn
=
n∑j=1
a1jxja2jxj
...amjxj
,
since 〈Aei, ej〉 = aij .
LECTURE VI 22
Thus〈Ax,y〉 =
∑k
(Ax)kyk =∑k
∑j
akjxjyk,
wherein we can switch the order of the sums since they are all finite, giving us
〈Ax,y〉 =∑j
∑k
akjykxj .
In the first case we have that the inner sum represents A = (ajk) operating, andin the second one, after switching order of the sums, we have that AT = (akj)is operating. So A∗ is a matrix, and in particular it is the matrix AT . N
6.3 Hilbert Space
The Hilbert space is a particularly important kind of normed linear space, how-ever to define it we first need the following definition.
Definition 6.3.1 (Scalar product). Let H be a linear space. Then 〈·, ·〉 :H ×H → C (or R) is a scalar product if it satisfies the following properties:
(i) 〈x,x〉 ≥ 0 for all x ∈ H;
(ii) 〈x,x〉 = 0 if and only if x = 0;
(iii) It is linear with respect to the first argument:
〈λx+ µy, z〉 = λ〈x, z〉+ µ〈y, z〉,
for all λ, µ ∈ C and x,y, z ∈ H;
(iv) It is skew-symmetric: 〈x,y〉 = 〈y,x〉.
Examples 6.3.2. Tale H = Cn, such that x = (x1, x2, . . . , xn). Then 〈x,y〉 =n∑j=1
xjyj is a scalar product.
If instead we take
H = `2 = {x = (x1, x2, . . . , xn) |∞∑j=1
|xj |2 <∞}
with 〈x,y〉 =∞∑j=1
xjyj , we again have a scalar product, but this time on an
infinite dimensional space.Finally consider
H = L2(R) = {f : R→ C |∫ +∞
−∞|f(t)|2 dt <∞},
i.e. the space of complex square differentiable functions, with the scalar product
〈f, g〉 =
∫ +∞
−∞f(t)g(t) dt. N
LECTURE VI 23
Lemma 6.3.3. Let H be a linear space with a scalar product 〈·, ·〉. Then‖x‖ =
√〈x,x〉 defines a norm.
Proof. It is clear that ‖x‖ ≥ 0, since the scalar product is nonnegative andtaking the square root keeps it nonnegative. Likewise it is clear that ‖x‖ = 0 isequivalent with x = 0, since the square root of 0 is 0. We also have that
‖λx‖ =√〈λx, λx〉 =
√λ〈x, λx〉 =
√λ〈λx,x〉
=
√λλ 〈x,x〉 =
√|λ|2〈x,x〉 = |λ|‖x‖.
It remains to verify that the triangle holds:
‖x+ y‖2 = 〈x+ y,x+ y〉 = 〈x,x+ y〉+ 〈y,x+ y〉
= 〈x+ y,x〉+ 〈x+ y,y〉
= 〈x,x〉+ 〈y,x〉+ 〈x,y〉+ 〈y,y〉
= 〈x,y〉2 + ‖x‖2 + ‖y‖2.
The square 〈x,y〉2 is clearly positive, whence if we take it away it becomessmaller, so we have
‖x+ y‖2 ≤ ‖x‖2 + ‖y‖2.Since the square root is a monotone function the inequality remains:
‖x+ y‖ ≤√‖x‖2 + ‖y‖2 ≤ ‖x‖+ ‖y‖.
Therefore on any space with a scalar product there also exists a norm given bythat scalar product.
Definition 6.3.4 (Hilbert space). Let H be a linear Banach space (over R orC, though we will generally use C) on which there is defined a scalar product〈·, ·〉 and a norm ‖x‖ =
√〈x,x〉. Then H is called a Hilbert space .
Theorem 6.3.5. If H is a Hilbert space, then H ′ = H.
6.4 Adjoint Operator in a Hilbert Space
Usually we have the picture
E1A−→ E2
E′1 ←−A∗
E′2
from before, however the dual space of a Hilbert space is again the Hilbert spacein question, whence we now get the much simplified picture
H1A−−⇀↽−−A∗
H2.
Example 6.4.1. Let us once again consider H = Cn, with basis elementsej = (0, . . . , 0, 1, 0, . . . , 0), and A = (aij = (〈Aei, ej〉). Then, due to the skew-
symmetry of the scalar product, aij = 〈Aei, ej〉 = 〈ei, A∗ej〉 = 〈A∗ej , ei〉 = a∗ji,whereby we have that a∗km = amk. N
Thus we also have that
H1A∗∗−→ H2.
LECTURE VI 24
6.5 Hermitian Operators
We have from just above the picture
HA−−⇀↽−−A∗
H.
What if A = A∗? Then A is a so-called Hermitian operator.
Definition 6.5.1 (Hermitian operator). Let H be a Hilbert space and let A :H → H and A∗ its adjoint operator. Then if A = A∗, A is said to be anHermitian operator . Therefore akm = amk.
Example 6.5.2. Let H = C2 and take A to be
A =
(1 i−i 2
). N
In particular we must have aii = aii the diagonal elements aii must be real.
6.6 Hermitian Operators in Quantum Mechanics
Quantum mechanics has the following axioms.(i) Given a complex Hilbert space H, unit vectors ψ ∈ H (i.e. that ‖ψ‖ =√〈ψ,ψ〉 = 1) represent states of quantum systems.(ii) Observables (energy, position, (angular) momentum, polarisation, et
cetera) are represented by Hermitian operators.In classical physics, state space and phase space are the same. Knowing
the position q and momentum p = mv uniquely determines all observables, forexample energy
E =p2
2m+v
2.
In quantum mechanics, on the other hand, ψ determines the value of an observ-able only with some probability. That is to say, quantum mechanics producesaverages of observables with respect to states.
(iii) The average of an observable A is given by 〈A〉ψ = 〈Aψ,ψ〉.For Hermitian operators A, this quantity is real. This follows immediately
from this argument:
〈A〉ψ = 〈Aψ,ψ〉 = 〈ψ,Aψ〉 = 〈Aψ,ψ〉,
wherein the first step is by A being Hermitian, and the second comes from thescalar product by definition being skew-symmetric. Therefore the average of anobservable is equal to its own conjugate, whence it must be real.
We know from probability theory that it isn’t enough knowing the averageµ; we also require knowledge of the standard deviation σ =
√σ2, where σ2 is
the dispersion, defined asσ2 = E[(ξ − µ)2],
where ξ are experimental measurements.
LECTURE VII 25
In quantum mechanics we get
σ2A = 〈(A− 〈A〉ψI)2〉ψ = 〈(A− 〈A〉ψI)2ψ,ψ〉
= 〈(A− 〈A〉ψI)(A− 〈A〉ψI)ψ,ψ〉
= 〈(A− 〈A〉ψI)ψ, (A− 〈A〉ψI)ψ〉 =∥∥∥(A− 〈A〉ψI)ψ
∥∥∥2
≥ 0,
wherein I is the unit operator (Iψ = ψ), whence we can define quantum stan-dard deviation.
7 Lecture VII8
7.1 Schrodinger Inequality
Take two Hermitian operators A and B. Then the Schrodinger inequality statesthat
σ2Aσ
2B ≥
∣∣∣∣12 〈{A,B}〉ψ − 〈A〉ψ〈B〉ψ∣∣∣∣2 +
∣∣∣∣ 1
2i〈[A,B]〉ψ
∣∣∣∣2,where {A,B} = AB + BA is the so-called anti-commutator of A and B (notethat it is 0 if and only if A and B anti-commute, i.e. AB = −BA), and[A,B] = AB − BA is the commutator of A and B, which is 0 if and only ifAB = BA.
If we let fA = (A− 〈A〉ψI)ψ and fB = (B − 〈B〉ψI)ψ, we get
Using this, we find that the Schrodinger inequality reduces to a known in-equality, abbreviated CBS.
Theorem 7.1.1 (Cauchy–Bunyakovsky–Schwartz inequality). Let H be a Hilbertspace and let f, g ∈ H. Then
|〈f, g〉| ≤ ‖f‖‖g‖.
To prove it we will require the following generalisation of the Pythagorean the-orem in general Hilbert spaces.
Lemma 7.1.2. Let H be a Hilbert space and let f, g ∈ H such that 〈f, g〉 = 0
(i.e. that f and g are orthogonal). Then ‖f + g‖2 = ‖f‖2 + ‖g‖2.
Proof. We simply apply the definition of norms in Hilbert spaces and then usethe linearity of its scalar product (along with its skew-symmetry), and also that〈f, g〉 = 0:
‖f + g‖2 = 〈f + g, f + g〉 = 〈f, f + g〉+ 〈g, f + g〉 = 〈f + g, f〉+ 〈f + g, g〉
Proof of the Cauchy–Bunyakovsky–Schwartz inequality. Let
z = f − 〈f, g〉‖g‖2
g
and consider
〈z, g〉 =
⟨f − 〈f, g〉
‖g‖2g, g
⟩= 〈f, g〉 − 〈f, g〉
‖g‖2〈g, g〉 = 0,
whence z and g are orthogonal (meaning that the previous lemma applies).Solving the definition of z for f we have
f = z +〈f, g〉‖g‖2
g,
using which we compute ‖f‖2, in which we may apply the lemma:
‖f‖2 = ‖z‖2 +
∥∥∥∥∥ 〈f, g〉‖g‖2g
∥∥∥∥∥2
= ‖z‖2 |〈f, g〉|2
‖g‖4‖g‖2 = ‖z‖2 +
|〈f, g〉|2
‖g‖2,
in which, if we subtract the nonnegative quantity ‖z‖2, clearly becomes smaller,so we have
‖f‖2 ≥ |〈f, g〉|2
‖g‖2.
Now we multiply both sides by ‖g‖2, which is nonnegative meaning that theinequality remains the same, giving us
‖f‖2‖g‖2 ≥ |〈f, g〉|2.
As an aside it is perhaps of interest to note that Cauchy’s contribution to the in-equality was to consider H = Cn, whereas Bunyakovsky, a student of Cauchy’s,solved it for H = L2(C). It was finally Schwartz who solved it for generalHilbert spaces.
Using fA and fB in the CBS inequality we thus get σ2Aσ
2B = ‖fA‖2‖fB‖2 ≥
|〈fA, fB〉|, which we recognise as zz, if we take z = 〈fA, fB〉, which moreovergives us zz = (Re z)2 + (Im z)2.
Plugging this back into the CBS inequality finally gives us
σ2Aσ
2B ≥ (Re z)2 + (Im z)2 =
∣∣∣∣12 〈{A,B}〉ψ − 〈A〉ψ〈B〉ψ∣∣∣∣2 +
∣∣∣∣ 1
2i〈[A,B]〉ψ
∣∣∣∣2,as sought.
That is to say, from a mathematician’s perspective, the famous Schrodingerinequality from quantum mechanics simply reduces to thee CBS inequality, sincethe operators are Hermitian.
7.2 Robertson Inequality
Another famous inequality from quantum mechanics, which was discovered be-fore Schrodinger’s, is that of Robertson. It states that
σ2Aσ
2B ≥
∣∣∣∣ 1
2i〈[A,B]〉ψ
∣∣∣∣2,which we observe is an immediate consequence of the Schrodinger inequality,since both of the terms in the sum of that inequality are nonnegative.
We will use this instead as a reason to discuss why the physicists insist onthe i in the denominator, even though it disappears when taking the absolutevalues. The reason is that physicists like self-adjoint operators (i.e. operatorsthat are their own adjoint).
Consider C = [A,B] = AB − BA. To consider the adjoint of this we firstneed to know what the adjoint of a composition is.
Lemma 7.2.1. Let A and B be operators on a Hilbert space H. Then (AB)∗ =B∗A∗.
Proof. This is a simple use of the definition of the adjoint and the skew-symmetryof scalar products. Let f, g ∈ H:
〈(AB)∗f, g〉 = 〈f,ABg〉 = 〈ABg, f〉 = 〈Bg,A∗f〉
= 〈g,B∗A∗f〉 = 〈B∗A∗f, g〉,
whence by comparing the first and last steps we see that (AB)∗ = B∗A∗.
LECTURE VII 28
We therefore have C∗ = (AB−BA)∗ = (AB)∗−(BA)∗ = B∗A∗−A∗B∗, which,recalling that A and B are taken to be Hermitian (meaning that A∗ = A andB∗ = B) becomes C∗ = BA−AB = −C = −[A,B].
That is to say, [A,B]∗
= −[A,B], whence [A,B] is not self-adjoint.However if we instead compute(
1
2i[A,B]
)∗=−[A,B]
−2i=
[A,B]
2i,
which is self-adjoint, if we use the following lemma.
Lemma 7.2.2. Let λ ∈ C and let A be a continuous linear operator on a Hilbertspace H. Then (λA)∗ = λA∗.
Proof. Take f, g ∈ H. Then by the definition of adjoint operators, alongwith the linearity of of the first component of scalar products and their skew-symmetry, we have
We now compare the first and last step and observe that (λA)∗ = λA∗.
What does the Robertson inequality mean? If we assume that [A,B] 6= 0, i.e.that A and B do not commute, then σ2
A and σ2B cannot both be arbitrarily
small simultaneously. This then leads us nicely on to a fundamental result ofquantum mechanics.
7.3 Heisenberg’s Uncertainty Principle
The famous Heisenberg’s uncertainty principle is simply the Robertson inequal-ity, taking H = L2 (the space of absolutely square integrable functions) and Aand B to be two particular operators.
The first one is A = q, the position operator, in quantum mechanics, forreasons unknown, taken to be q(ϕ)(x) = xϕ(x), i.e. multiplying by the variable.
The second one is B = p, where p is the momentum operator, defined as
p(ϕ)(x) =~i
∂ϕ
∂x(x),
where ~ is the so-called reduced Planck constant.We plug these into the Robertson inequality and compute:
σ2qσ
2p ≥
1
4
∣∣∣〈[q, p]〉ψ∣∣∣2.Here we get
[q, p]ϕ(x) = (qp− pq)(ϕ)(x),
wherein
(qp)(ϕ)(x) =~iq(ϕ′)(x) =
~ixϕ′(x),
and
(pq)(ϕ(x) = p(xϕ(x)) =~ixϕ′(x) + ϕ(x)
LECTURE VIII 29
by the product rule for derivatives. Thus
[q, p]ϕ(x) = (qp− pq)(ϕ)(x),=~i
(xϕ′(x)− xϕ′(x)−ϕ(x)) = −~iϕ(x) = ~iϕ(x),
whence [q, p] is therefore the operator ~iI. Of course 〈I〉ψ = 〈Iψ, ψ〉 = 1, so
σ2qσ
2p ≥
~2
4.
This is the fundamental limit of uncertainty in quantum mechanics; if one knowsthe momentum well, one cannot know the position well, and vice versa. This isalso known as Bohr’s principle of complementarity.
Exercise 7.3.1. Consider the Hilbert space H = C2 and let
A =
(1 −ii 1
)and B =
(2 11 1
),
(which both obviously Hermitian) and let ψ = 1/√
2(e1 + e2), with e1 = (1, 0)and e2 = (0, 1).
Study for these the Robertson inequality and the Schrodinger inequality.
Solution. See Assignment V. �
8 Lecture VIII9
8.1 Dense Subsets of Metric Spaces
Definition 8.1.1 (Dense subset). Let (X, ρ) be a metric space, and A ⊂ X besome subset of X. We say that A is dense in X if for any x ∈ X and any ε > 0there exists an a ∈ A such that a ∈ Bε(x).
Examples 8.1.2. Let X = R and A = Q. Then A is dense in X, because forany x ∈ R we have the decimal expansion
x = αNαN−1 . . . α1α0.α−1α−2 . . . α−M . . . =
N∑j=−∞
αj10j .
Take xM to be the truncation of this to M decimal places, i.e.
xM =
N∑j=−M
αj10j = αNαN−1 . . . α1α0.α−1α−2 . . . α−M .
Then clearly xM is a rational number, and |x− xM | ≤ 10−M → 0 as M →∞.Consider now X = Rn instead, with ρ(x,y) being any of the distances we
are used to (‖x − y‖∞, ‖x − y‖1, or ‖x − y‖2). Then A = Qn is dense in X,by the same argument as above, componentwise. N
Example 8.1.3. Consider instead X = C[a, b], with the distance ρ∞(ϕ,ψ) =‖ϕ− ψ‖∞. We then need the following theorem, stated without proof.
9Date: November 26, 2015.
LECTURE VIII 30
Theorem (Weierstrass approximation theorem). Any continuous function de-fined on a closed interval [a, b] can be uniformly approximated by a polynomial.
This means that for any ϕ ∈ C[a, b] and for any ε > 0 there exists somepolynomial p ∈ P [a, b],
p(t) = cktk,
ck being real numbers, such that maxa≤t≤b
|ϕ(t) − p(t)| < ε, in other words P [a, b]
is dense in C[a, b]. N
8.2 Separable Metric Spaces
Definition 8.2.1 (Separable metric space). The metric space (X, ρ) is calledseparable if there exists a countable dense subset. Otherwise the space is callednon-separable .
Examples 8.2.2. All of the previous examples are separable spaces. In thefirst case because Q is countable, in the second case because Qn is a finite unionof countable sets, which is again countable.
The space of polynomials P [a, b], on the other hand, is not countable, becausethe coefficients are real. However we can approximate any real polynomial witha rational polynomial from PQ[a, b], which is countable and also dense in C[a, b].
That is to say, for any ϕ ∈ C[a, b] and for any ε > 0, by Weierstrass theorem
there exists some pε =N∑k=0
cktk ∈ P [a, b], ck ∈ R, such that
maxa≤t≤b
|ϕ(t)− pε(t)| < ε,
but there also for every coefficient ck exists some qk ∈ Q such that for everyδ > 0 we have |qk − ck| < δ, whence
maxa≤t≤b
|pε(t)− pδε(t)| =N∑k=0
|ck − qk|bk < δ
N∑k=0
bk,
where pδε is the polynomial with qk as its coefficients, so we take
δ <ε∑N
k=0 bk,
whence ‖pε − pδε‖∞ < ε, finally giving us ‖ϕ − pδε‖∞ < 2ε, whence C[a, b] isseparable since PQ[a, b] is countable and dense. N
Example 8.2.3. Take the space X = `1 = {x = (xj), xj ∈ R | ‖x‖1 < ∞} ofall absolutely convergent. Since
‖x‖1 =
∞∑j=1
|xj | <∞,
there must for every ε > 0 exist some N such that
∞∑j=N+1
|xj | < ε.
LECTURE VIII 31
(This is simply because of the absolute convergence of the series.)Take therefore xε = (x1, x2, . . . , xN , 0, . . .). Then clearly ‖x− xε‖1 < ε.Moreover let qj ∈ Q be rational numbers chosen in such a way that |qj −
xj | < δ for some small δ, and construct the sequence qδε = (q1, q2, . . . , qN , 0, . . .).The collection of all such qδε is countable since there are finitely many nonzeropositions, and each of them have countably many options from Q. In addition wehave that ‖xε−qδε‖1 < Nδ, whence if we take δ < ε/N , we have ‖x−qδε‖1 < 2ε,so it is dense in `1, and therefore since it’s countable `1 is separable. N
Example 8.2.4. With an almost identical proof (just replace the norm, butuse the same technique) we see that `2 is separable as well. N
On the other hand, not all spaces we are used to are separable.
Example 8.2.5. The space X = `∞ = {x = (xj)|‖x‖∞ <∞} is non-separable,meaning that there exists no dense and countable subset. This relies on R notbeing countable, indeed not even [0, 1] is countable.
Consider the binary expansions of real numbers on this interval between 0and 1. They have the form x = 0.β1β2 . . . βj . . ., where βj ∈ 0, 1 for all j, and theset of all such binary strings is uncountable. We translate these into sequencesbelonging to `∞; let Y = {x = (β1, β2, . . . , βj , . . .)} ⊂ `∞. That this is a subsetof `∞ is clear since the supremum of terms in such a sequence is 1 (or 0, if it’sconstantly 0).
Moreover, for any x,y ∈ Y , we clearly have ‖x−y‖∞ = max1≤j<∞
|xj−yj | = 1,
if x 6= y. Therefore if we take ε = 1/2 we clearly have B1/2(x) ∩B1/2(y) = ∅.Suppose now that `∞ is separable. This means that there must exist some
A = {a1,a2, . . . ,an, . . .}, where aj ∈ `∞ for all j, such that for all z ∈ `∞ andall ε > 0 there exists some aj ∈ A such that aj ∈ Bε(z).
In particular, then, this must be true if we pick ε = 1/2 and take z ∈ Y , ourset of binary strings. But we have shown that Y is uncountable, and since A iscountable there aren’t enough members of A for this to be true, so we have acontradiction. Therefore `∞ cannot be separable. N
Exercise 8.2.6. Let X = c0 ={x = (x1, x2, . . . , xn, . . .)
∣∣ limj→∞
xj = 0}
with
the norm‖x‖∞ = max
1≤j<∞|xj |.
Investigate whether c0 is separable or not.
Solution. Since the sequence x = (x1, x2, . . . , xn, . . .) decays to 0 as n increases,then by the very definition of the limit there must for every ε > 0 exist some Nsuch that |xn| < ε for all n > N .
Therefore the maximum of the sequence, that is ‖x‖∞, must be xk for some1 ≤ k ≤ N . This serves to tell us that there are only countably many positionsk where this maximum can be.
We then for any δ > 0 construct the sequence xδε = (0, . . . , 0, qk, 0, . . .),where qk is in the kth position of xδε, just like the maximum of x, and moreoverqk ∈ Q is chosen in such a way that |xk − qk| < δ.
Since there are countably many positions 1 ≤ k ≤ N to choose from, andcountably many rational numbers in Q, the space of all such sequences Q =
LECTURE VIII 32
{xδε = (0, . . . , 0, qk, 0, . . .)} is countable (since the countable union of countablesets is again countable).
It then remains to show that this set Q is dense in c0. This is clear byconstruction: we have, if we let qj denote the jth term in xδε, that
‖x− xδε‖∞ = max1≤j<∞
|xj − qj | = xk − qk < δ,
where δ can be made arbitrarily small. Thus the for any radius δ, there existssome xδε ∈ Q such that xδε ∈ Bδ(x), meaning that Q is dense in c0.
Thus since Q ⊂ c0 is dense and countable, c0 is separable. �
8.3 Unbounded Linear Operators
By Theorem 5.1.1, this is of course the same as discontinuous linear operators.The aim here is to explore differential operators, like d/dx : C(1) → C,
but instead d/dx : H → H for some Hilbert space H. What people generally
use in applications is H = L2[a, b] = {f : [a, b] → C |∫ ba|f(x)|2 dx < ∞}. Of
course not all of these need be differentiable, so we consider C∞ ⊂ L2, andd/dx : C∞ → C∞.
Definition 8.3.1 (Domain of definition). Let H be a Hilbert space and let Dbe an uncountable linear subspace (meaning that for all x, y ∈ D we have thatλx+µy ∈ D) which is dense in H. Moreover let A : D → H be a linear operator.Then we call A a linear operator with the domain of definition D.
Remark 8.3.2. Note that this definition is interesting only for operators whichare not continuous, because otherwise, since D is dense in H, and A is contin-uous, we have for any x ∈ H some sequence xn → x, xn ∈ D. We then takeyn = Axn, which since xn → x converges to some y ∈ H, since A is continuous.Thus for any x ∈ H we can extend A to be defined also for this x, so D cansimply be taken to be the whole H.
For this reason, when we in the future consider A : D → H, D ⊂ H being dense,we always assume that A is not continuous.
For such operators we naturally cannot have an inequality of the form‖Ax‖ ≤ c‖x‖, where c = ‖A‖, which we are so used to, since this wouldn’tbe finite.
Example 8.3.3. Consider the Hilbert spaceH = L2[−∞,∞], with the operatorx : ϕ(x)→ xϕ(x). This is unbounded, because having∫ ∞
−∞|ϕ(x)|2 dx <∞
does not necessarily imply that∫ ∞−∞
x2|ϕ(x)|2 dx <∞,
since taking for example ϕ(x) = 1/(1 + |x|) we have that
|ϕ(x)|2 ∼ 1
|x|2
LECTURE VIII 33
as x approaches either infinity, whereas
x2|ϕ(x)|2 ∼ 1
as x approaches plus or minus infinity, whence the latter integral doesn’t con-verge. N
A question which remains is what D is. There is a natural choice, as shownbelow, but it is not always the best choice.
Definition 8.3.4 (Natural domain of definition). Let H be a Hilbert space.For a linear operator A : D → H, the natural domain of definition isDA = {x ∈ H |Ax ∈ H}.
Examples 8.3.5. If, for example, H = L2, and x is as above, we have Dx ={ϕ ∈ L2 | xϕ(x) ∈ L2}. Similarly, if A = d/dx, we have the natural domain ofdefinition DA = {ϕ ∈ L2 | ∃ ϕ′ ∈ L2}. N
Example 8.3.6 (Hermitian unbounded operators). Recall that for an operatorA to be Hermitian we require that 〈Ax, y〉 = 〈x,Ay〉, i.e. that A = A∗, andconsider again the operator x on L2. Is this Hermitian? Yes, since
〈xϕ1, ϕ2〉 =
∫ ∞−∞
xϕ1(x)ϕ2(x) dx =
∫ ∞−∞
xϕ1(x)xϕ2(x) dx = 〈ϕ1, xϕ2〉.
How about the operator A = d/dx? We compute the scalar product
〈Aϕ)1, ϕ2〉 =
∫ ∞−∞
ϕ′1(x)ϕ2(x) dx,
which using integration by parts we handily rewrite as
ϕ1(x)ϕ2(x)∣∣∣∞−∞−∫ ∞−∞
ϕ1(x)ϕ′2(x) dx = ϕ1(x)ϕ2(x)∣∣∣∞−∞− 〈ϕ1, Aϕ2〉.
So on the natural domain of definition DA we have two problems, first the eval-uation of ϕ1(x)ϕ2(x) which we would like to be zero, and secondly the minussign in front of the scalar product. So we take instead D = {f ∈ L2 | f ′ ∈L2, f(x) → 0, x → ±∞}. On this domain of definition we therefore have〈Aϕ)1, ϕ2〉 = −〈ϕ1, Aϕ2〉.
But since i = −i, we consider instead the operator C = iA = id/dx, whence〈Cϕ)1, ϕ2〉 = 〈ϕ1, Cϕ2〉, so C is Hermitian on the domain of definition D. N
Problem 8.3.7. If H is a Hilbert space and D ∈ H is dense, and moreoverA : D → H is continuous, then A ∈ L(H), as we have explored in detail before.
If, on the other hand, A is unbounded on D, then it cannot be extended toH.
Having said that, is it possible to construct a linear operator A : E → E,where E is a Banach space, such that A is unbounded?
Solution. Let B = {ej} be a Hamel basis of E (see [KF20, p. 128]), meaningthat any x ∈ E can be written as a linear combination of finitely many basiselements
x =
N∑j=1
cjenj. (8.3.1)
LECTURE VIII 34
We assume that ‖ej‖ = 1, because if this were not the case we could insteadhave taken B′ = {ej/‖ej‖} to be our basis.
We now define the map ϕ : B → C by
ϕ(ej) = j (8.3.2)
for all j, and extend it to the entirety of E and make it linear in the naturalway:
ϕ(x) =
N∑j=1
cjϕ(enj ).
(Here the linear combination (8.3.1) using the Hamel basis being made up offinitely many elements is crucial; if the amount of elements were infinite, wewould generally not have convergence of the above sum. Since the sum is nowfinite, this is not a problem.)
That this is linear is fairly easy to show. Let us take some z = λx + µy,where x =
∑Nj=1 cjenj
and y =∑Nj=1 djenj
are any elements in the Banachspace and λ and µ are any constants. Then
z = λx+ µy = λ
N∑j=1
cjenj + µ
N∑j=1
djenj =
N∑j=1
(λcj + µdj)enj ,
whence if we plug this into ϕ we get
ϕ(z) =
N∑j=1
(λcj+µdj)ϕ(enj) = λ
N∑j=1
cjϕ(enj)+µ
N∑j=1
djϕ(enj) = λϕ(x)+µϕ(y).
Moreover that this is unbounded is quite clear by (8.3.2); since ‖ej‖ = 1, butat the same time |ϕ(ej)| = j, for all j, we cannot have |ϕ(x)| ≤ C‖x‖, for aconstant C, for all x ∈ E. Taking x = ej and letting j tend to infinity makesthe left-hand side of the inequality arbitrarily large.
Finally we want to make this a map not from E to C, but an operator fromE to E. To do this we simply multiply ϕ(x) by any nonzero x0 ∈ E (takingx0 = 0 would of course make it the new operator bounded).
So finally we have the linear operator A : E → E defined by
Ax = ϕ(x)x0,
which is unbounded.That this new operator A preserves the linearity of ϕ is clear by the same
calculations as above; just multiply by the fixed x0. �
Remark 8.3.8. To prove that a Hamel basis always exists we require Zorn’slemma. With it, we can use the same line of reasoning we used to prove theHahn–Banach theorem, since by Zorn’s lemma there must amongst the set of alllinearly independent subsets of the linear space E exist a maximal one. Thatthis maximal set generates the whole space follows from precisely the samecontradiction proof used in the proof of Hahn–Banach.
LECTURE VIII 35
8.4 Separable Hilbert Spaces
To discuss these, we require the notion of an (orthonormal) basis in generalHilbert spaces.
Definition 8.4.1 (Orthonormal basis). Let H be a Hilbert space. The set {ej}is an orthonormal basis in H if 〈ei, ej〉 = δij , where δii = 1 and δij = 0, i 6= j,and
x =
∞∑j=1
cjej ,
where cj ∈ C, in the sense that we have convergence in H:
∥∥∥x− N∑j=1
cjej
∥∥∥2
→ 0
as N →∞.
Note that we then have, since ‖·‖ is continuous and the series converges, that
〈x,x〉 = ‖x‖2 =⟨ ∞∑j=1
cjej ,
∞∑j=1
cjej
⟩=
∞∑j=1
∞∑k=1
cjckδjk =
∞∑j=1
|cj |2, (8.4.1)
which is known as Parseval’s equality .Since the norm is well-defined for all x ∈ H, we must have that the above
sum converges for all x. Is the converse true?
That is to say, if we have any sequence (cj) such that∞∑j=1
|cj |2 < ∞, does
there exist an x ∈ H such that
x =
∞∑j=1
cjej?
The answer is yes, and to show it we finally make use of Hilbert spaces being
complete. Set xN =N∑j=1
cjej ∈ H be partial sums, and then let N ≤ M . Then
we have by Parseval’s equality (8.4.1)
‖xN − xM‖2 =∥∥∥ M∑j=N+1
cjej
∥∥∥2
=
M∑j=N+1
|cj |2 →∞
as N,M →∞ since the series converges. Therefore {xN} is a Cauchy sequencein H, and because of the completeness of H there must exist a within H thelimit of the sequence, i.e. some x ∈ H such that
limN→∞
xN = x,
meaning that
x =
∞∑j=1
cjej ,
as claimed.
LECTURE IV 36
9 Lecture IV10
9.1 Inverse Operators
In what follows we will require the following sufficient condition for the existenceof inverse operators.
Theorem 9.1.1. Let A ∈ L (E), where E is a Banach space, and let ‖A‖ < 1.
Then there exists an operator (I −A)−1 =∞∑n=0
An.
Proof. Note first that
(I −A)−1 =
∞∑n=0
An, (9.1.1)
if it exists, follows from the ordinary proof for the sum of a geometric series.We must now show that the sum in (9.1.1) is in L (E). To do this, we show
that the sequence of partial sums is a Cauchy sequence. Let N ≤M , then
∥∥∥ N∑n=0
An −M∑n=0
An∥∥∥ =
∥∥∥ M∑n=N+1
An∥∥∥ ≤ M∑
n=N+1
‖A‖n,
wherein the last step is due to the triangle inequality. Clearly this approaches0 as N and M approach infinity, since ‖A‖ < 1. Thus
C = (I −A)−1 =
∞∑n=0
An ∈ L (E).
We must now show that C is indeed the inverse of I −A, meaning that we wishto show that C(I −A) = (I −A)C = I:
(I −A)C = (I −A)
∞∑n=0
An =
∞∑n=0
An −∞∑n=0
An+1 = I,
and
C(I −A) =( ∞∑n=0
An)
(I −A) =
∞∑n=0
An −∞∑n=0
An+1 = I,
whence C is indeed the inverse of I −A.
Theorem 9.1.2. Let E be a Banach space and let A,B ∈ L (E) such thatA−1 ∈ L (E), and further let ‖A−1B‖ < 1. Then there exists an (A+B)−1 inL (E).
Proof. We factor the inside of (A + B)−1, and recall that for any operators Cand D we have as a rule that (CD)−1 = D−1C−1:
(A+B)−1 = (A(I −A−1B))−1 = (I −A−1B)−1A−1.
We must thus ask ourselves whether (I − A−1B)−1 is in L (E), but since wehave by assumption that ‖A−1B‖ < 1, we have by the previous theorem thatthis is the case.
10Date: November 30, 2015.
LECTURE IV 37
9.2 Spectrum and Resolvent
Definition 9.2.1 (Resolvent). Let A : D → E be an operator, possibly un-bounded, where E is a normed space and D is a dense subspace of E. If thereexists an operator Rλ = (A−λI)−1 ∈ L (E), then we say that λ belongs to theresolvent set of A, written R(A) = {λ | ∃Rλ ∈ L (E)}.
The motivation for the particular form A − λI naturally comes from linearalgebra and the study of eigenvalues and eigenvectors.
Definition 9.2.2 (Spectrum). The spectrum of A is the complement to theresolvent set, denoted S(A). Thus in the real case we have S(A) = R \ R(A),and in the complex case we have S(A) = C \R(A).
That is to say, if λ ∈ S(A), then Rλ = (A− λI)−1 does not exist.
Example 9.2.3. Let us for a moment consider eigenvalues and eigenvectors,letting λ0 being an eigenvalue. Then λ0 belongs to the spectrum, since A−λ0Ihave an inverse, since there exists some nonzero eigenvector ϕ such that (A −λ0I)ϕ = 0, whence λ is mapped to 0 by A−λ0I, and so is of course 0, wherebywe do not have injectivity, and so no inverse map can exist. N
Definitions 9.2.4 (Kernel and image). Let C : D → E be some mapping.Then we denote by kerC = {ϕ |Cϕ = 0} the kernel of C and by imC = {ψ =Cϕ | ϕ ∈ D} the image of C.
Example 9.2.5. Consider the finite dimensional case where C : Rn → Rm, andlet kerC = {0}. This implies that n = m and that imC = Rn, which in turnimplies that there exists some C−1.
In this finite dimensional case S(A) is identical to the set of eigenvalues,since if λ is not an eigenvalue, we have that there does not exist any ϕ suchthat (A − λI)ϕ = 0, whence ker(A − λI) = {0}, and so Rλ = (A − λI)−1
exists. N
Example 9.2.6. We now consider the infinite dimensional case. Here it canhappen that ker(A−λI) = {0}, but despite this we have im(A−λI) 6= E, whencethere does not exist an Rλ = (A− λI)−1. Ergo in the infinite dimensional caseS(A) is not just the set of eigenvalues. N
Let us consider some concrete example.
Example 9.2.7. Let our space be E = `2 and take T+(x1, x2, . . . , xn, . . .) =(0, x1, x2, . . . , xn, . . .) be our operator.
Clearly we have that kerT+ = {0}, since every component must be 0 forthe shifted ∞-tuple to be 0. On the other hand we have that imT+ is thesubspace of all sequences that start with a 0, but otherwise can behave howeverthey like. Thus T+ is an injection (i.e. one-to-one, which is easy to see), butnot a surjection. (Note here that the sequences in ell2 being infinitely long isimportant; if they’re not, then it won’t even be injective.)
If we now take A− I = T+, meaning that A = T+ + I, we see that λ = 1 ∈S(A), but it is not an eigenvalue, because there exists no nonzero x such thatT+x = 0. N
LECTURE IV 38
9.3 Some Elements of Topology
In the discussion that follows we will require some basic topology. Recall that if(X, ρ) is a metric space, then we have the ball Br(x0) = {x ∈ X | ρ(x0, x) < r}.
Definition 9.3.1 (Open set). Let (X, ρ) be a metric space. Then a set O iscalled open is for every x0 ∈ O there exists some r > 0 such that Br(x0) ⊂ O.
Examples 9.3.2. Let X = R, and O = (a, b). Then O is open, but for example[a, b] is not, since any ball around the endpoints will lie partly outside theinterval.
Consider not X = R2. Then any set without boundary is open, whereas ifwe include the boundary it is not. N
Definition 9.3.3 (Closed set). Let (X, ρ) be a metric space. Then a set M iscalled closed if it is the complement to some open set.
Example 9.3.4. Let X = R and O = (a, b) again. Is O closed? No, since itscomplement Oc = (−∞, a] ∪ [b,∞) isn’t open, due to the endpoints at a andb. N
Exercise 9.3.5. Find a metric space in which there exist subsets that are bothopen and closed at the same time.
Solution. Take X to be any nonempty set and let ρ be a metric such thatρ(x, x) = 0, and ρ(x, y) = 1 if x 6= y, for all x, y ∈ X. That this is in fact ametric is easy to verify; the only part that isn’t blindingly obvious is the triangleinequality, which holds true since ρ(x, y) = 1 ≤ 2 = ρ(x, y) + ρ(y, z).
Now take any x ∈ X, and pick some radius r < 1. Then clearly by thedefinition of our metric we have that the ball Br(x) contains only x, whenceevery single subset of X is open, because around any point x in a set O we canconstruct balls of radius r < 1 such that everything in the ball, which is just x,is in O.
Therefore since every subset is open, there complements must be as well,whence all subsets are closed as well.
(Of course this is also true, but not as interesting, if we take X = {x} tobe a singleton with any metric ρ. Or indeed (if we allow just one such subset)(X, ρ) to be any metric space, because O = ∅ is both open and closed.) �
With this in hand we are equipped to tackle the next theorem regarding resol-vents.
Theorem 9.3.6. The resolvent set for an operator in a Banach space is alwaysopen.
Proof. We have that A : D → E, where E is some Banach space. By definitionthen we have that λ0 ∈ R(A), meaning that Rλ0 = (A − λ0I)−1 ∈ L (E).Consider now some ∆λ such that |∆λ| < ε. We would like to show that λ0 +∆λ ∈ R(A), for some sufficiently small ε.
We therefore consider the operator
Rλ0+∆λ = (A− (λ0 + ∆λ)I)−1 = ((A− λ0I)−∆λI)−1,
whence by Theorem 9.1.2 we have that this inverse exists if A − λ0I has aninverse (which it does, since we assumed that λ0 ∈ R(A)), and
‖(A− λ0I)−1∆λI‖ < 1.
LECTURE X 39
Let us therefore investigate that norm. Taking the constant ∆λ outside of thenorm and solving for it we have that
|∆λ| = 1
‖(A− λ0I)−1I‖=
1
‖Rλ0‖,
so we take this to be our ε, which shows that the resolvent set is open, sinceRλ0+∆λ ∈ L (E) for all |λ0 −∆λ| < ε.
Corollary 9.3.7. The spectrum of an operator in a Banach set is always closed.
Proof. This follows from the definition of a set being closed, the definition ofthe spectrum, and the resolvent set being closed.
10 Lecture X11
10.1 More On Open and Closed Sets
Recall from last lecture that the set O is open if and only if for each x0 ∈ Othere exists some ε > 0 such that Bε(x0) ⊂ O, and that a set F is closed ifF = X \O, where O is open.
Theorem 10.1.1. Let (X, ρ) be a metric space. Then the set F is closed if andonly if for each convergent sequence (xn) where xn ∈ F , its limit also belongsto F .
Proof. (⇒) We let F be closed, meaning that we must show that for everysequence (xn) such that xn → x, where xn ∈ F , we must have x ∈ F . To showthis we assume that x 6∈ F , meaning that since F = X \O, where O is open, wemust have x ∈ O. However since O is open and x ∈ O, there exists some ε > 0such that Bε(x) ⊂ O. Therefore, since x is the limit of xn, there must also existsome sufficiently large N such that, for all n > N , we have xn ∈ Bε(x) ⊂ O,whence xn 6∈ F , which is a contradiction.
(⇐) We now instead assume that for all convergent sequences (xn), xn ∈ F ,we have xn → x ∈ F . Using this we must show that F is closed, that is thatO = X \ F is open. This means that if we have some y 6∈ F , we cannot havethat xn → y for any sequence of xn ∈ F . Thus there must exist some ε > 0such that Bε(y) ⊂ O since sequences (xn) can never approach y.
Definition 10.1.2 (Closure). Let (X, ρ) be a metric space, and let A be asubset of X. Its closure is the minimal closed set that contains A, denoted[A] = min{F ⊃ A | F closed}.
A fairly obvious, yet useful, consequence of the definition is that if A ⊂ B, then[A] ⊂ [B]. Moreover, pairing the definition with the previous theorem we seethat the closure of a set must also be the union of the set and the set of alllimits of sequences in the set.
Exercise 10.1.3. Consider the metric space (R, ρ(x, y) = |x − y|). What isthe closure of the rational numbers, of the irrational numbers, and the naturalnumbers?
11Date: December 3, 2015.
LECTURE X 40
Solution. We know already that sequences of rational numbers have real numberlimits (see Dedekind sections). Therefore by the previous theorem [Q] = R.Similarly the closure of the irrational numbers is the set of real numbers, forthe same reason.
The closure of the natural numbers is the set of natural numbers, that is tosay [N] = N. This is easy to see since, for any 0 < ε < 1 we have that the ballBε(n) contains only n, for all n ∈ N, whereby the natural numbers are closed,and clearly the closure of a closed set is itself. Of course it is also easy to seethat any sequence of natural numbers can have only a natural number as itslimit. �
Exercise 10.1.4. Consider now X = C[a, b], with A = C(1)[a, b]. What is [A]?
Solution. Recall that by Weierstrass theorem (page 30) we have that any elementin C[a, b] can be uniformly approximated using polynomials from P [a, b]. More-over we clearly have P [a, b] ⊂ C(1)[a, b] ⊂ C[a, b], and since polynomials canapproximate continuous functions uniformly, it means that [P [a, b]] = C[a, b].Therefore we have that C[a, b] = [P [a, b]] ⊂ [C(1)[a, b]] ⊂ [C[a, b]] = C[a, b],whereby we must have that [C(1)[a, b]] = C[a, b] as well. �
Problem 10.1.5 (p-adic metric space). The p-adic natural numbers are definedas follows. Let p be a fixed prime. Then for all n in N, we define the p-adicabsolute value as |n|p = p−α, where n = pαk, p - k. For example, if we take
p = 2 we have |2|2 = 1/2 since 2 = 21, and |4|2 = 1/4 since 4 = 22. On theother hand, |5|2 = 1 since 5 = 20 · 5. We define p-adic distance analogously tothe Euclidean distance: ρp(n1, n2) = |n1 − n2|p.
Consider now the ball Bε(x) = {y |ρp(x, y) ≤ ε}. Show that this ball is bothopen and closed. Moreover, show that the same thing is true for the sphereSε(x) = {y | ρp(x, y) = ε}.
Solution. We will spend some time not only to solve the problem, but to exploreinteresting features of the p-adic metric.
Let us show that the p-adic absolute value not only satisfies the triangleinequality (which of course it does, since it defines a metric) but also the strongtriangle inequality , |x+ y|p ≤ max{|x|p, |y|p}.
Take x = pαm and y = pβn, where α, β,m, n ∈ N, and neither m nor n aredivisible by p. Let us in addition assume without loss of generality that α ≤ β.Then we can write x+ y = pαm+ pβn = pα(m+ pβ−αn).
Given this we must therefore have
|x+ y|p = |pα(m+ pβ−αn)|p ≤ p−α = |x|p.
Had we on the other hand had β ≤ α we would get |x + y|p ≤ |y|p, whence ingeneral we get |x + y|p ≤ max{|x|p, |y|p}. Of course this maximum is smallerthan or equal to the sum of the two since they are nonnegative, whence it alsosatisfies the triangle inequality.
Let us show using this the complement of the original ball Bε(x) is open. Todo so we pick some y outside of the ball, necessarily meaning that ρp(x, y) > ε.We now pick some δ < ε and form the ball Bδ(y) and show that the intersectionbetween this ball and Bε(x) is empty, meaning that the complement of theoriginal ball is open.
LECTURE X 41
Take any z in the new ball Bδ(y). Then
ε < |x− y|p = |x− z + z − y|p ≤ max{|x− z|p, |z − y|p},
the first step being true since y is outside Bε(x), and the last step being thestrong triangle inequality. Note now that the second term in the maximumconcerns z and y, whence this is less than δ, which was taken to be less than ε,whereby it all reduces to ε < |x− z|p, whereby the entire ball Bδ(y) is outside
of Bε(x), and so Bε(x) is closed.That the ball Bε(x) is open follows from the remarkable claim: due to the
strong triangle inequality we have the curious property that any (open or closed)ball has every point in it as its centre, which we show thus: take x, y ∈ Bε(z).Then by definition ρp(x, z) < ε. Moreover, by the strong triangle inequality, wehave ρp(x, y) ≤ max{ρp(x, z), ρp(y, z)} < ε since z is the defined centre of theball. Therefore x is also in the ball Bε(y), which is true for any x and y in theoriginal ball.
Thus every ball of any radius r ≤ ε centred on any point in Bε(x) is containedin Bε(x) and so it is open, and all together both closed and open.
Let us now consider the sphere Sε(x) = {y | ρp(x, y) = ε} instead. We takeany y on this sphere and, for δ < ε, form the ball Bδ(y) of radius δ around y.If we can show that all points of this ball are on the sphere, the sphere is open.
We do this as follows. Let z be an arbitrary point in Bδ(y). Then considerthe distance between z and x, which we would like to be ε:
|z − x|p ≤ max{|z − y|p, |y − x|p} = ε
since |z − y|p < δ < ε and |y − x|p = ε. On the other hand we have
ε = |y − x|p ≤ max{|z − y|p, |x− z|p},
wherein we can guarantee that |x−z|p ≤ ε is the bigger of the two, |z−y|p < δ,since the latter can be made sufficiently small by choosing δ < ε appropriately.We thereby have ε ≤ |x − z|p ≤ ε, whence |x − z|p = ε, and so z ∈ Bδ(y),making Sε(x) open. To see that it is also closed we note that its complement isthe union of the ball Bε(x) and the same complement we considered for Bε(x),whereby we are done. �
10.2 More On Separability
With the advent of the notion of closure, we can discuss separability in a newlight.
Theorem 10.2.1. The metric space (X, ρ) is separable if and only if there existsa countable set A = {a1, a2, . . . , an, . . .} such that [A] = X.
With this we get the following result.
Theorem 10.2.2. Let H be a Hilbert space. Suppose that there exists a count-able orthonormal basis (em)∞n=1 in H, meaning that 〈en, em〉 = δnm. Then His separable.
LECTURE X 42
Proof. Let us first note that if H = HN is a finite N -dimensional space, thenwe are done, because then HN = CN . There then exists a dense and countablesubset (Q×Q)N such that the distance between
ϕ =
N∑n=1
cnen ∈ HN and ϕε =
N∑n=1
(q1n + q2ni)en
is less than ε, for every ε.For infinite dimensional spaces we use almost the same strategy, with one
small modification. We know from previously that Parseval’s equality must besatisfied:
‖ϕ‖2 =
∞∑n=1
|cn|2.
Due to convergence this then means that for every ϕ ∈ H we have that thepartial sums
ϕN =
N∑n=1
cnen
become arbitrarily close to ϕ, meaning that
‖ϕ−ϕN‖ =
∞∑n=N+1
|cn|2 → 0
as N → ∞. This in addition means that for every ε > 0 there exists some Nsuch that ‖ϕ−ϕN‖ < ε. We then have by the above argument in HN that forevery such ϕN there exists some vector that is arbitrarily close to it, and sincethese are countable, we are done.
The converse is in fact true as well.
Theorem 10.2.3. Let H be a separable Hilbert space. Then there exists acountable orthonormal basis (en)∞n=1 in H.
Proof. Since H is separable we know that there exists some countable set A ={a1,a2, . . . ,an, . . .} which is dense in H.
Recall from linear algebra that any finite set of vectors can be made orthonor-mal by, for example the Gram–Schmidt process. We apply this repeatedly tothe set A, producing B = {e1, e2, . . . , en, . . .} where 〈en, em〉 = δnm.
Since A is dense in H, and each ej by Gram–Schmidt is a linear combinationof members from A, the set B is dense in H as well.
Suppose now that there exists some ϕ ∈ H that cannot be written as
ϕ =
∞∑k=1
ckek.
This means that ϕ is orthogonal to every element of B, and in turn to everyelement of A. However this implies that ϕ = 0, since if 〈aj ,ϕ〉 = 0 for every j,we must have that for every ε > 0 some akε such that ‖akε −ϕ‖ < ε, which byParsevals equality means that the sum of the squares of the coordinates ck of ϕmust be arbitrarily small, whence ϕ = 0.
LECTURE XI 43
Combining the two previous results we get the following.
Theorem 10.2.4. A Hilbert space has a countable orthonormal basis if andonly if it is separable.
Remark 10.2.5. If we have an uncountable basis (eα), α ∈ A not countable, thenthe sum
∑α∈A
cαeα has meaning only if only countably many cα are nonzero.
10.3 Something More About Spectrums
We recall from the end of last lecture that the spectrum S(A) is always a closedsubset. So if λn ∈ S(A), and there exists some λ = lim
n→∞λn, then λ ∈ S(A) as
well.
Theorem 10.3.1. Let H be a Hilbert space (or indeed a Banach space), andlet A ∈ L (H). Then S(A) ⊂ Br(0), where r = ‖A‖.
Proof. Recall from last lecture that the resolvent set is
R(A) = {λ |Rλ = (A− λI)−1 ∈ L (H)},
and that the spectrum is its complement. Therefore S(A) is the set of all λ suchthat (A − λI)−1 6∈ L (H). Further recall from Theorem 9.1.1 that a sufficientcondition for (I − C)−1 existing is that ‖C‖ < 1. We have
A− λI = λ(Aλ− I)
= −λ(I − A
λ
),
whence C = A/λ and therefore we must have ‖A/λ‖ = 1/|λ|‖A‖ ≥ 1. Multi-plying either side by the nonnegative |λ| we thus get ‖A‖ ≥ |λ|.
This then means that if we pick any λ such that ‖A‖ ≥ |λ|, then (A−λI)−1
does not exist, whence λ ∈ S(A).
11 Lecture XI12
11.1 Something More On Metric Spaces
Recall from previous lectures the notion of separability of metric spaces. Wehave the following result.
Theorem 11.1.1. Suppose that (X, ρ) is a separable metric space. Let Y ⊂ X.Then the metric space (Y, ρ) is separable as well.
Moreover we have the following.
Theorem 11.1.2. Let (X, ρ) be a complete metric space and let Y be a closedsubset of X. Then (Y, ρ) is complete as well.
Proof. Since (X, ρ) is complete, any Cauchy sequence (yn) from Y must havea limit, and since Y is closed, this limit must be in Y , and therefore (Y, ρ) iscomplete.
12Date: December 7, 2015.
LECTURE XI 44
Finally let us have a think about Hilbert spaces once more. A consequence ofthe two above theorems is the following, recalling the definition of Hilbert space.
Corollary 11.1.3. Let L be a linear subspace of a Hilbert space H, and let Lbe closed. Then L is a Hilbert space. Moreover, if H is separable, then so is L.
Moreover, recalling what we have proved previously regarding orthonormal basesof Hilbert spaces, we have this, again following from the previous theorem.
Corollary 11.1.4. Any closed subspace of a separable Hilbert space has a count-able orthonormal basis.
11.2 Orthogonal Complement of a Closed Linear Subspace
Following the last corollary we will henceforth always work with separableHilbert spaces.
Definition 11.2.1 (Orthogonal complement). Let H be a separable Hilbertspace and let L be a subspace of H. Then we denote by L⊥ the orthogonalcomplement of L, defined as L⊥ = {x ∈ H | 〈y, x〉 = 0 ∀y ∈ L}.
Theorem 11.2.2. For any L ⊂ H, H being a separable Hilbert space, L⊥ is aclosed linear subspace of H.
Proof. That L⊥ is linear is clear, since the scalar product is linear with respectto its first component.
Next we show that it is closed. Let xn ∈ L⊥, meaning that 〈xn, y〉 = 0 forall y ∈ L. Moreover suppose that xn → x as n → ∞. Then 0 = 〈xn, y〉 →〈x, y〉 = 0, so the limit x also belongs to L⊥.
Theorem 11.2.3. Let L be a closed linear subspace of a Hilbert space H. Thenany z ∈ H can be uniquely represented as z = a+ b where a ∈ L and b ∈ L⊥.
Proof. From Corollary 11.1.4 we have that, say, (en)∞n=1 is an orthonormal basisof L and that (fn)∞n=1 is an orthonormal basis of L⊥. We would like to showthat (en, fn) is a basis of the whole space H.
To do this, suppose that it is not. Then there must exist some y ∈ H suchthat 〈y, en〉 = 0 and 〈y, fn〉 = 0 for all n. The first scalar product implies thaty ∈ L⊥, whereas the second one implies that y ∈ (L⊥)⊥ = L. Therefore y mustbe orthogonal to itself, whence y = 0 is the only option, and so all x ∈ H canbe written uniquely as
x =
∞∑n=1
anen +
∞∑n=1
bnfn.
With the notion of orthogonal complements comes the notion of orthogonalprojections (cf. linear algebra).
Definition 11.2.4 (Orthogonal projection). Let L be a closed linear subspaceof a Hilbert space H. By the previous theorem we have that any z ∈ H can bewritten as z = a+b, where a ∈ L and b ∈ L⊥. We denote by PL the orthogonalprojection on L; PLz = a.
LECTURE XI 45
It is clear that we may write
PLz =
∞∑n=1
〈z, en〉en,
where (en) is some basis in L.
Theorem 11.2.5. Let P be an orthogonal projection on some closed linearsubspace L of a Hilbert space H. Then P is idempotent, meaning that P 2 = P ,and Hermitian, meaning that P ∗ = P .
Proof. That it is idempotent is clear from the definition. That it is Hermitianis more interesting. We show it in two ways, using z = a + b and y = c + d,where these are decompositions in the way described earlier:
〈Pz, y〉 =⟨ ∞∑n=1
〈z, en〉en, y⟩
=
∞∑n=1
〈z, en〉〈en, y〉 =
∞∑n=1
〈en, y〉〈z, en〉
=
∞∑n=1
〈y, en〉〈z, en〉 =⟨z,
∞∑n=1
〈y, en〉en⟩
= 〈z, Py〉.
Alternatively, we have that Pz = a and Py = c. Therefore by the linearity ofthe scalar product
〈Pz, y〉 = 〈a, c+ d〉 = 〈a, c〉+ 〈a, d〉,
however 〈a, d〉 = 0 since a ∈ L and d ∈ L⊥. Similarly,
〈P ∗z, y〉 = 〈z, Py〉 = 〈a+ b, c〉 = 〈a, c〉+ 〈b, c〉,
where again 〈b, c〉 vanishes since they are in L and L⊥ respectively. Thus〈Pz, y〉 = 〈P ∗z, y〉 whence P = P ∗.
The converse is true as well.
Theorem 11.2.6. Let P : H → H be a linear operator in a Hilbert space H,such that P is idempotent and Hermitian. Then P is an orthogonal projection.
Proof. Let us first denote by L the image of P on the whole space H. For anya ∈ L we of course have Pa = a. Hence if we take any H 3 z = a+ b, this beingthe same sort of decomposition as above, we want to have Pz = Pa+ Pb = a,whence we must show that Pb = 0.
To do this we first realise that Pb ∈ L, and since b ∈ L⊥, we have 〈Pb, b〉 = 0.Since P is idempotent we also have 〈P 2b, b〉 = 0, but since P is Hermitian thisis the same as 〈Pb, Pb〉 = 0, whence the norm of Pb is 0, which is true if andonly if Pb = 0.
Combining the two previous theorems we get the following.
Theorem 11.2.7. Let H be a Hilbert space and P : H → H. Then P is anorthogonal projector if and only if it is both idempotent and Hermitian.
LECTURE XII 46
11.3 Operator Representation of Quantum States
In quantum mechanics a Schrodinger equation is an equation of the form
~i
∂ψ
∂t= Hψ,
where ψ|t=0 = ψ0. Here ψ0 is a quantum state, meaning that 〈ψ0, ψ0〉 = 1.Taking ψ 7→ ρψ as the (one dimensional) projection on ψ, meaning that
ρψϕ = 〈ϕ,ψ〉ψ, we aim to rewrite the Schrodinger equation as an equation ofoperators.
We first solve the differential equation above as per usual, by first rewritingit as
∂ψ
∂t=( iH
~
)ψ,
giving us
ψ(t) = eit~Hψ0,
where the exponential is usually denoted Ut, called the evolution operatorand H is called the generator of evolution , which lets us write it shortly asψ(t) = Utψ0.
Recalling that H is a quantum observable and that they are always Hermi-tian, and that exponentials can be written as sums quite neatly, we computeU∗t :
〈Utψ0, ψ0〉 =⟨ ∞∑n=0
(it~)nn!Hnψ0, ψ0
⟩=
∞∑n=0
(it~)nn!〈Hnψ0, ψ0〉
=⟨ψ0,
∞∑n=0
(−it~)n
n!Hnψ0
⟩= 〈ψ0, e
−it~ Hψ0〉,
where in the last step conjugates have been taken of the terms in the sum. Thus
U∗t = e−it~ H,
whereby we have U∗t Ut = UtU∗t = I, meaning that U∗t = U−1
t . From this we alsohave that 〈Utϕ1, Utϕ2〉 = 〈ϕ1, U
∗t Utϕ2〉 = 〈ϕ1, ϕ2〉, showing that Ut preserves
scalar products.
Definition 11.3.1 (Unitary operator). Let H be a Hilber space and let U :H → H be an operator that has an inverse and preserves scalar products. ThenU is called a unitary operator .
This leads to quantum physicists saying things like “quantum evolution is uni-tary.”
12 Lecture XII13
12.1 Spectral Decomposition of Self-adjoint Operators
Consider the finite dimensional case dimH = n, and let L (H) 3 A = A∗ bean Hermitian operator. Then as discussed Example 9.2.5, page 37 together
13Date: December 10, 2015.
LECTURE XII 47
with some basic linear algebra, the eigenvalues are real and the eigenvectors fordistinct eigenvalues λj are all orthogonal. Moreover if we have λ1 < λ2 < . . . <λk with dimLλj
= mj , Lλjbeing the eigen subspace for the eigenvalue λj , then
k∑j=1
mk = n.
Let Pλ denote the projector on the subspace Lλ, then
A =
k∑j=1
λjPλj, (12.1.1)
in other words
Aϕ =
k∑j=1
λjPλjϕ,
for all ϕ ∈ H, which we motivate as follows. From the right-hand side we have
ϕ =
k∑j=1
Pλjϕ,
since, because distinct eigenvalues are orthogonal, the eigen subspaces for dis-tinct eigenvalues are orthogonal as well.
Let(e
(j)s
)be an orthonormal basis of Lλj
, then
Pλjϕ =
k∑s=1
⟨ϕ, e(j)
s
⟩e(j)s ,
whereby
ϕ =
k∑j=1
Pλjϕ =
k∑j=1
k∑s=1
⟨ϕ, e(j)
s
⟩e(j)s ,
or equivalently I =k∑j=1
Pλj. Thus by the linearity of the operator A,
Aϕ = A
k∑j=1
Pλjϕ =
k∑j=1
APλjϕ =
k∑j=1
λjPλjϕ.
We would like to rewrite the expression in (12.1.1) in the form of an integral,since when we attempt to generalise it to the infinite dimensional case, the sumwill naturally become an integral.
To do so, let
EA(λ) =∑λj<λ
Pλj, (12.1.2)
which we claim is again a projector (i.e. by Theorem 11.2.7 last lecture, it isboth Hermitian and idempotent). It being Hermitian is clear by the linearity of
LECTURE XII 48
scalar products with respect to the first component and Pλjthemselves being
projectors and so Hermitian:⟨EA(λ)ϕ,ψ
⟩=⟨∑λj<λ
Pλjϕ,ψ
⟩=∑λj<λ
〈Pλjϕ,ψ〉 =
∑λj<λ
〈ϕ, Pλjψ〉
=⟨ϕ,∑λj<λ
Pλjψ⟩
=⟨ϕ,EA(λ)ψ
⟩,
whereas the idempotency follows from the next theorem.
Theorem 12.1.1. Let H be a Hilbert space and L1 and L2 two subspaces ofH. Then L1 and L2 are orthogonal if and only if the projectors PL1 and PL2
on the two subspaces commute.
Proof. (⇒) We assume that L1 and L2 are orthogonal subspaces, and we lookto prove that PL1PL2 = PL2PL1 .
Let (ei) be an orthonormal basis of L1 and let (fj) be an orthonormal basisof L2. Since L1 and L2 are orthonormal, 〈ei, fj〉 = 0 for all i and j. Moreoverwe have
PL1ϕ =
∑i
〈ϕ, ei〉ei and PL2ϕ =
∑j
〈ϕ, fj〉fj ,
from which by the linearity of scalar products we get
PL2PL1
ϕ =∑j
⟨∑i
〈ϕ, ei〉ei, fj⟩fj =
∑j
∑i
〈ϕ, ei〉〈ei, fj〉fj = 0
since 〈ei, fj〉 = 0, and similarly
PL1PL2ϕ =∑i
⟨∑j
〈ϕ, fj〉fj , ei⟩ei =
∑i
∑j
〈ϕ, fj〉〈fj , ei〉ei = 0,
for the same reason, whence PL1PL2
= PL2PL1
= 0.(⇐) If we now instead assume that PL1PL2 = PL2PL1 , meaning that we
have equality between the two double sums above, for all ϕ, we must have〈ei, fj〉 = 〈fj , ei〉 = 0 since the remaining terms in the sums can take on arbitraryvalues.
Recalling (12.1.2), we now use it to express Pλj in a different way: Pλj =EA(λj+1)− EA(λj), using which we reformulate (12.1.1) as
A =
k∑j=1
λj(EA(λj+1)− EA(λj)),
which after an interlude on integration theory we will recognise as somethingmeaningful.
12.1.1 Riemann–Stieltjes Integral
Recalling that we define the Riemann integral as∫ b
a
f(λ) dλ = lim∆→0
N−1∑j=1
f(λj)(λj+1 − λj),
LECTURE XII 49
where ∆ = max1≤j<N
{λj + 1− λj}, and a = λ1 < λ2 < . . . < λN = b is a partition
of the interval [a, b], assuming the limit exists.The Riemann–Stieltjes integral is defined analogously, with a slight modifi-
cation. Take E(λ) to be a real-valued function (for now) that is monotonouslyincreasing, i.e. that E(λ′) ≥ E(λ) whenever λ′ ≥ λ. In other words, theso-called increment E(λ′)− E(λ) is nonnegative.
The Riemann-Stieltjes integral is then defined as∫ b
a
f(λ) dE(λ) = lim∆→0
N−1∑j=1
f(λj)(E(λj+1)− E(λj)), (12.1.3)
where now ∆ = max1≤j<N
{E(λj+1)− E(λj)}, again provided the limit exists.
Note that if we take E(λ) = λ, this becomes the Riemann integral.We now wish to generalise the concept of a Riemann–Stieltjes integral to
have E(λ) take its values in the space of linear, bounded operators on a Hilbertspace: E(λ) ∈ L (H) for all λ.
The conditions on E are then the following:
(i) For all λ, E(λ) is Hermitian (this to mimic E being real-valued above);
(ii) For all λ′ ≥ λ, we have E(λ′) ≥ E(λ).
Of course for this to make sense we must define the notion of nonnegativeoperators.
Definition 12.1.2 (Nonnegative operator). Let H be a Hilbert space and letC ∈ L (H). Then C is called nonnegative , denoted C ≥ 0, if 〈Cϕ,ϕ〉 ≥ 0 forall ϕ ∈ H.
We discuss the connection to linear algebra in the following example.
Example 12.1.3. Consider a finite dimensional Hilbert space H, say dimH =n. Then we can represent any linear operator in this space as a matrix
C =
c11 c12 · · · c1nc21 c22 · · · c2n...
.... . .
...cn1 cn2 · · · cnn
,
which by Sylvester’s criterion from linear algebra we know is positive semidef-inite if and only if the determinants of the upper left m × m matrices are allnonnegative, for all 1 ≤ m ≤ n. N
Example 12.1.4. Taking H = L2(R), the operator Cϕ(x) = x2ϕ(x) is non-negative since
〈Cϕ,ϕ〉 =
∫x2|ϕ(x)|2 dx ≥ 0.
Indeed, taking Cϕ(x) = γ(x)ϕ(x) for any γ(x) ≥ 0 is nonnegative. N
Thereby for E(λ) satisfying conditions (i) and (ii) above and f : R → R wedefine the Riemann-Stieltjes integral as per (12.1.3), if this limit exists.
LECTURE XII 50
Remark 12.1.5. Note that if E(λ) is discrete (i.e. it’s a jump function), thenthe integral is just the same as the sum we produced earlier.
Thereby our now long studied expression (12.1.1) becomes
A =
∫ λk
λ1
λ dEA(λ),
where λ1 and λk as before are the smallest and largest eigenvalues, respectively,since we wrote A as precisely the sort of sum seen in our definition of theRiemann–Stieltjes integral from before.
This is the simplest case of spectral decomposition of A.
12.2 The Infinite Dimensional Case
Let us now consider infinite dimensional Hilbert spacesH, again with A ∈ L (H)and A Hermitian. Select from H an orthonormal basis, call it (ej). Moreoverconsider the subspace generated by the first K basis elements,
HK ={x =
K∑j=1
xjej
},
with πK = PHKbeing the projector on this space. Let us now study
AK = πKAπK ,
an operator from H to HK (although we can always restrict it to AK : HK →HK), called a finite dimensional dressing of A.
Now recalling that (ABC)∗ = C∗B∗A∗ (from Lemma 7.2.1, page 27), we seeimmediately that AK is Hermitian, whence we may call
AK =
∫S(AK)
λ dEAK (λ),
since AK is a finite dimensional operator.This leads us nicely to one of the main results of functional analysis, on the
spectral decomposition of arbitrary bounded Hermitian operators in Hilbertspaces, although presented without proof.
Theorem 12.2.1. Let H be an infinite dimensional Hilbert space and let A ∈L (H) be an Hermitian operator. Then the limit
limK→∞
EAK (λ) = EA(λ)
exists and moreover is
(i) an orthogonal projection,
(ii) is monotonously increasing,
and finally (iii) is
A =
∫S(A)
λ dEA(λ).
Note that the integral in the infinite dimensional case doesn’t have the sameelegant representation as a sum as the finite dimensional case.
REFERENCES 51
12.3 Functions of Hermitian Operator
Recall how previously we discussed analytic functions of bounded operators. Us-ing the previous result we are able to study more general functions of Hermitianoperators.
Take f : S(A)→ R, using which we define
f(A) =
∫S(A)
f(λ) dEA(λ)
as the limit of the corresponding Riemann–Stieltjes integral.
Example 12.3.1. Take gµ(λ) to be the step function
gµ(λ) =
{1, if λ ≤ µ,0, otherwise.
Then
gµ(A) =
∫S(A)
gµ(A) dEA(λ) =
∫ µ
−∞dEA(λ) = EA(µ)− EA(−∞) = EA(µ)
since EA(−∞) = 0 due to A ∈ L (H) implying that S(A) ⊂ B‖A‖(0) (seeTheorem 10.3.1, page 43). N
This is one of the most useful formulae in function analysis, in particular oncewe during next lecture generalise it to the unbounded case.
References
[KF20] A. N. Kolmogorov, S. V. Fomin. Introductory Real Analysis. Dover Pub-lications, 416 pages, New edition, 2000.
NOTATIONS 52
Notations
|·| Absolute value
{·, ·} Anti-commutator
Br(a) Closed ball of radius r centred on a
x Vector
C Complex numbers
[·] Closure of a set
[·, ·] Commutator
z Conjugate of z
D Domain of definition
DA Natural domain of definition
〈·, ·〉 Duality notation
〈A〉ψ Average of A with respect to state ψ
`p Space of sequences with finite p-norm
~ Planck constant
H Generator of evolution
im Image
ker Kernel
≤ Partial order relation
L (E1, E2) Space of linear operators between E1 and E2
N Natural numbers
‖·‖ Norm
Q Rational numbers
R Real numbers
ρ(x, y) Metric; distance between x and y
〈·, ·〉 Scalar product
σ2 Dispersion
A∗ Adjoint operator of A
Br(a) Open ball of radius r centred on a
C[a, b] Space of smooth functions on [a, b]
NOTATIONS 53
C(n)[a, b] Space of n times differentiable functions on [a, b]
c0 Space of null sequences
E′ Dual space to E
L⊥ Orthogonal complement of L
L2[a, b] Space of absolutely square integrable functions on [a, b]
