MATH31002/MATH41002: LINEAR ANALYSIS
The Stone-Weierstrass Theorem
Theorem 1.1 is a special case of a much more general theorem valid for compact metric
spaces. If X is a compact metric space, C(X, R) will denote the set of continuous functions
f : X R. W
MATH 31002/41002 Linear Analysis
Solution Sheet 2
1. Consider the set R of the polynomials with rational coecients:
cfw_ n
k
R=
ak x | ak Q, 0 k n, n 0 .
k=0
The set R is clearly countable (as Q is countable) and we claim it to be dense in C[0, 1].
Let u
MATH31002/MATH41002/MATH61002: LINEAR ANALYSIS
Lemma 2.3 (Minkowskis Inequality). For p 1,
( n
)1/p
|ai + bi |p
( n
i=1
)1/p
|ai |p
(
+
i=1
n
)1/p
|bi |p
.
i=1
Proof. Omitted. (Extra reading for MATH41001/MATH61001.)
Remark. The proof uses Hlders Inequali
MATH31002/MATH41002: LINEAR ANALYSIS
Inner Products and Hilbert Spaces
The following is a useful property of Hilbert spaces.
Lemma 2.11 (Parallelogram law). Let H be a Hilbert space. For x, y H, we have
the identity
x + y2 + x y2 = 2x2 + 2y2 .
Proof. We h
MATH31002/MATH41002: LINEAR ANALYSIS
Continuous Linear Functionals
Denition. Let X be a (normed) vector space over R (or C). A linear functional on X is
a map f : X R (or C) such that
f (x + y) = f (x) + f (y),
for all x, y X, , R (or C).
Example. Let X =
MATH31002/MATH41002: LINEAR ANALYSIS
Theorem 2.6. C([0, 1], R) with the norm f = supx[0,1] |f (x)| is a Banach space. (So
is C([0, 1], C) with the same proof.)
Proof. Suppose that cfw_fn is a Cauchy sequence for . This means that, given
n=1
> 0, there e
MATH31002/MATH41002: LINEAR ANALYSIS
Linear Operators and Their Spectra
Denition. Let X and X be normed vector spaces. A linear operator is a map T : X X
such that
T (x + y) = T (x) + T (y),
for all x, y X, , R (or C).
The following is analogous to Propo
MATH31002/MATH41002: LINEAR ANALYSIS
Example 3. Consider l1 = cfw_a = (ai ) :
i=1
and dene
i=1
|ai | < +, ai C. Fix b = (bi ) l
i=1
fb (a) =
ai bi .
i=0
Lemma 3.4. fb (l1 ) and fb = b .
Proof. The sum is nite since
i=1
ai bi
i=1
(
|ai bi |
)(
|ai |
)
su
MATH31002/MATH41002: LINEAR ANALYSIS
Spectrum of Operators
Let X be a Banach space over C. (Here it is important that the eld is C (algebraically
closed) not R.)
We will say that a bounded linear operator T B(X) is invertible if there exists S
B(X) such
4.1. Compact Operators.
Denition 4.1. A subset A of a normed space X is called relatively
compact (or pre-compact) if the closure of A is compact.
Since a compact set is always bounded (i.e., there exists M > 0
such that x M for all x A), so is a relative
MATH 31002/41002: LINEAR ANALYSIS
Example Sheet 2
1. Using the Weierstrass Approximation Theorem, prove that there exists a countable
dense subset of C[0, 1]. (Hint: the polynomials with real coecients are dense but not
countable; how can one modify this
MATH 31002/41002: LINEAR ANALYSIS
Example Sheet 6
1. Let T : C[0, 1] C[0, 1] be dened as follows:
(T f )(x) = f (x2 ).
Show that T is a bounded linear operator on C[0, 1] and compute its norm.
2. Let f C[0, 1] (with the norm ) and dene an operator T : C[0
MATH 31002/41002: LINEAR ANALYSIS
Example Sheet 5
1. Let X = 1 and put f (x) = x1 . Is f a linear functional on X? Justify your answer.
2. Let X = 2 and put f (x) = x0 3x1 for x = (x0 , x1 , . . . ) X. Prove that f X and
compute its norm.
3. Consider X =
MATH 31002/41002: LINEAR ANALYSIS
Example Sheet 4
1. Let X denote the set of n n real matrices. Put for any two matrices A, B X,
A, B = trace(AB T ),
where B T is the transpose of B. Show that , is indeed an inner product on X.
2. Let H = 2 . Compute the
MATH 31001/41001: LINEAR ANALYSIS
Example Sheet 3
1. Let 1 p < q . Show that p
q .
2. Show that the norms 1 and 2 are not equivalent on the space 1 . (Hint: nd a
sequence of elements xn in 1 whose limit x is in 2 but not in 1 .)
1
3. Let X be the space of
MATH 31002/41002: LINEAR ANALYSIS
Example Sheet 1
1. Let (X, ) be a normed space. Verify that the function d(x, y) = x y is a metric.
2. Compute the nth Bernstein polynomial Bn (f ; x) for the following functions:
(a) f (x) = x;
(b) f (x) = x2 .
(c)* f (x
MATH 41002 Linear Analysis
The University of Manchester, 2011
Solutions
Section A
A1. Let V be a normed vector space (with norm ). Then we dene a
norm of a linear functional f by
f = sup |f (x)|.
x=1
Alternatively,
|f (x)|
.
x=0 x
f = sup
(Either denition
MATH 31002/41002 Linear Analysis
Solution Sheet 3
1. Let us prove that p q ; we need to show that for any sequences (x0 , x1 , . . . )
rst
such that |xi |p < , one has |xi |q < . Indeed, since |xi |p < , we
i=0
i=0
i=0
have |xi | 0 as i , whence there exi
MATH 31001/41001 Linear Analysis
Solution Sheet 1
1. It is obvious that d(x, y) = x y 0 and = 0 if and only if x = y. Also, clearly,
d(x, y) = d(y, x). Let us check the triangle inequality:
d(x, y) + d(y, z) = x y + y z x y + y z = x z = d(x, z).
2. (a) H
MATH31002/MATH41002: LINEAR ANALYSIS
Example 3. Consider l1 = cfw_a = (ai ) :
i=1
and dene
i=1
|ai | < +, ai C. Fix b = (bi ) l
i=1
fb (a) =
ai bi .
i=0
Lemma 3.4. fb (l1 ) and fb = b .
Proof. The sum is nite since
i=1
ai bi
i=1
|ai bi |
(
)(
|ai |
)
su
MATH31002/MATH41002: LINEAR ANALYSIS
Theorem 2.15. Every innite dimensional separable Hilbert space contains a complete
orthonormal set cfw_en . Furthermore, each x H can be written uniquely in the form
n=1
x=
an en , where an = x, en .
n=1
Proof. Since
MATH31002/MATH41002: LINEAR ANALYSIS
Chapter 1: Approximating Continuous Functions
Notation and Basic Facts
Notation. We shall use R to denote the set of real numbers and C to denote the set of
complex numbers.
The modulus or absolute value on R is denote
Given T : X X, we can consider powers T 2 , T 3 , . . . , T n , . . . We
can also form polynomial combinations: if P (x) = an xn + + a1 x + a0
then we write P (T ) = an T n + + a1 T + a0 I.
Proposition 4.1. If P (x) is a polynomial then
spec(P (T ) = cfw_
MATH31002/MATH41002/MATH61002: LINEAR ANALYSIS
Lemma 2.3 (Minkowskis Inequality). For p 1,
( n
)1/p
|ai + bi |p
( n
i=1
)1/p
|ai |p
+
( n
i=1
)1/p
|bi |p
.
i=1
Proof. Omitted. (Extra reading for MATH41001/MATH61001.)
Remark. The proof uses Hlders Inequali
MATH31002/MATH41002: LINEAR ANALYSIS
Denition. If T : X X is a bounded linear operator then we dene its norm T by
T = sup T (x).
x=1
By Theorem 4.1, this is nite and an equivalent denition is
T (x)
.
x=0 x
T = sup
(N.B. it still has to be proved that this
MATH31002/MATH41002: LINEAR ANALYSIS
Exercise. Show that if is an eigenvalue of T then spec(T ).
Solution. Suppose that C is an eigenvalue of T . Then, by denition, there exists
x X, x = 0, such that
T x = x
or, equivalently,
(I T )x = 0.
Suppose that spe
MATH31002/MATH41002: LINEAR ANALYSIS
Inner Products and Hilbert Spaces
Corollary. The map , : H H C is continuous (with respect to the associated
norm).
Proof. Suppose that (xn , yn ) (x, y) in H H, as n +. Then limn+ xn x = 0
and limn+ yn y = 0. Using th
MATH31002/MATH41002: LINEAR ANALYSIS
Lemma 2.5. On Rn (and Cn ), all norms are equivalent.
Proof. Let | |1 be the 1-norm on Rn and let | | be an arbitrary norm. We shall show
that | |1 and | | are equivalent.
n
Write x = i=1 ai ei and let M = max1in |ei |
MATH 31002/41002 Linear Analysis
Solution Sheet 4
1. Let us verify all the axioms of an inner product:
(1) Let A = (ai,j )n . We have A, A = trace(AAT ) = n a2 0 and = 0 if and
i,j=1
i,j=1 i,j
only if A = 0.
(2) Since the trace of the transpose of a matri
MATH 31002/41002 Linear Analysis
Solution Sheet 6
1. The fact that T is linear is a straightforward check. Since the map x x2 is a bijection
on [0, 1], we have
f = max |f (x)| = max |f (x2 )| = T f
x[0,1]
x[0,1]
for any f C[0, 1]. Hence T = 1. (Moreover,