7.1 Various Kinds of Spaces

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Remark 7.1.1. C[−_, _] is defined to be the set of continuous maps from

the interval {x 2 R : −_ _ x _ _} to R. It is easy to verify that C[−_, _] is

a vector (linear) space, where we add maps and scale them pointwise:

8f, g 2 C[−_, _], 8x 2 [−_, _](f + g)(x) , f(x) + g(x)

and

8f 2 C[−_, _], 8t 2 R, 8x 2 [−_, _] (tf )(x) , tf (x)

Exercise 7.1.1. Prove that with these operations, C[−_, _] is a vector space.

(A soothing exercise in axiom bashing.)

Remark 7.1.2. It is rather well known that Rn is a vector space for each

positive integer n and has dimension n. It is obvious that C[−_, _] does not

have dimension n for any positive integer n.

Exercise 7.1.2. Prove this claim

Remark 7.1.3. One of the reasons for studying abstract vector spaces is so

that we can transfer intuitions about R2 and R3 to function spaces. We can

carry this further:

125

126 CHAPTER 7. FOURIER THEORY

Definition 7.1.1. Normed Vector Space A norm on a vector space V is

a map

k k : V −! R

x   kxk

such that:

8x 2 V, kxk _ 0 and kxk = 0 ) x = 0 (7.1)

8x 2 V, 8t 2 R, ktxk = |t|kxk (7.2)

8x, y 2 V, kx + yk _ kxk + kyk (7.3)

A pair (V, kk) where kk is a norm is called a normed vector space.

Remark 7.1.4. In Rn

kxk =

q

x21

+ x22

+ · · · + x2

n

is a norm.

Exercise 7.1.3. prove the above claim.

Definition 7.1.2. Inner Product Space A vector space V has an inner

product <,> iff

<,>: V × V −! R

x, y   < x, y >

is such that:

8 x, y, z 2 V < x, y + z > = < x, y > + < x, z > (7.4)

8 x, y 2 V 8t 2 R < x, ty > = t < x, y > (7.5)

8 x, y, z 2 V < x + y, z > = < x, z > + < y, z > (7.6)

8 x, y 2 V 8t 2 R < tx, y > = t < x, y > (7.7)

8 x, y 2 V < x, y > = < y, x > (7.8)

8 x 2 V < x, x >_ 0 and < x, x >= 0 ) x = 0. (7.9)

The pair (V,<,>) is called an inner product space

Remark 7.1.5. lines 7.4 to 7.7 are summarised by saying that the inner

product is bilinear line 7.8 says it is symmetric, and line 7.9 says it is positive

definite.

7.1. VARIOUS KINDS OF SPACES 127

Example 7.1.1. The good old dot product on Rn is an inner product.

Exercise 7.1.4. Prove the last claim. (This sort of thing is called axiom

bashing and is supposed to be good for the soul. If it is, that would explain

why I am so saintly.)

Proposition 7.1.1. Schwartz Inequality If (V,<,>) is an inner product

space then:

8 x, y 2 V, (< x, y >)2 _ (< x, x >)(< y, y >)

Proof

8 s, t 2 R, 8 x, y 2 V < sx − ty, sx − ty > _ 0

) < sx, sx > − < sx, ty > − < ty, sx > + < ty, ty > _ 0

) s2 < x, x > −2st < x, y > +t2 < y, y > _ 0

) 2st < x, y > _ s2 < x, x > +t2 < y, y >

In particular, putting s =

p

< y, y > and t =

p

< x, x > we get

2

p

< x, x >

p

< y, y > < x, y > _ 2(< x, x >< y, y >)

from which we deduce that

< x, y > _ kxkkyk

If we put s = −

p

< y, y > and t =

p

< x, x > we deduce

− < x, y > _ kxkkyk

Putting the two deductions together we obtain

(< x, y >)2 _ < x, x >< y, y >

_

Remark 7.1.6. If we define cos(_) for the angle between two vectors by the

rule:

< x, y >= kxkkyk cos(_)

then the above inequality tells us that −1 _ cos(_) _ 1 which is nice to know.

Alternatively we could use that fact to remember the Schwartz inequality.

128 CHAPTER 7. FOURIER THEORY

Definition 7.1.3. Metric Space A metric on a set X is a map

d : X × X −! R

such that

8x, y 2 X d(x, y) _ 0 and d(x, y) = 0 ) x = y (7.10)

and 8x, y 2 X d(x, y) = d(y, x) (7.11)

and 8x, y, z 2 X d(x, z) _ d(x, y) + d(y, z) (7.12)

A pair (X, d) where d is a metric on X is called a metric space.

Remark 7.1.7. This abstracts our idea of what a distance function has to

be like. The first says that the distance between places is always positive

except when the places are the same, when it is zero. The second says that

the distance between here and the pub is the same as the distance between

the pub and here. And the third, the triangle inequality, says that going to

the pub via some other place can never make for a shorter distance all up.

This clearly works for pubs.

There is nothing in here that says that X has to be a vector space and no

reason why it should be. It is fairly easy to think of a sensible metric on S1,

the unit circle.

Exercise 7.1.5. Go on then.

Remark 7.1.8. We can define continuity for maps between metric spaces:

Definition 7.1.4. If f : (X, d) −! (Y, e) is a map between metric spaces,

then f is continuous at a 2 X iff

8 " 2 R+ 9 _ 2 R+ : 8 x 2 X, d(x, a) < _ ) e(f(x), f(a)) < "

Definition 7.1.5. If f : (X, d) −! (Y, e) is a map between metric spaces,

then f is continuous iff 8 a 2 X, f is continuous at a.

Remark 7.1.9. Note that this is equivalent to the usual definition of continuity

on the old familiar spaces Rn but that it now makes sense to say when

a function f : S1 −! S1 is continuous provided we specify a way to say what

a distance is for points in S1. Easiest is to just give the distance in R2.

Remark 7.1.10. Recall that inner product spaces and normed spaces have

to be vector spaces.

7.1. VARIOUS KINDS OF SPACES 129

Proposition 7.1.2. If (X,<,>) is an inner product space then it is also a

normed vector space with:

8x 2 X kxk =

p

< x, x >

Proof: Calling something a norm doesn’t make it one, so we need to show

that with kxk defined as above, the axioms for a norm hold.

8x 2 X, kxk _ 0 since < x, x > _ 0 and

kxk = 0 )

p

< x, x > = 0 ) < x, x > = 0 ) x = 0

and

8t 2 R, 8 x 2 X, ktxk =

p

< tx, tx > =

p

t2

p

< x, x > = |t|kxk

Finally,

8 x, y 2 X kx + yk =

p

< x + y, x + y >

) k(x + yk2 = kxk2 + 2 < x, y > +kyk2

) k(x + yk2 _ kxk2 + 2kxkkyk + kyk2

) k(x + yk2 _ (kxk + kyk)2

) k(x + yk _ kxk + kyk

_

Remark 7.1.11. Note the use of the Schwartz inequality in the last part.

Proposition 7.1.3. If (V, kk) is a normed vector space then there is a metric

d derived from the norm by

8 x, y 2 V, d(x, y) = kx − yk

Proof It is immediate that 8 x, y 2 V d(x, y) _ 0 and

8 x, y 2 V d(x, y) = 0 ) kx − yk = 0 ) x − y = 0 ) x = y

Further:

8x, y, z 2 V, d(x, z) = kx − zk

and d(x, y) = kx − yk

and d(y, z) = ky − zk

130 CHAPTER 7. FOURIER THEORY

and from the last property of a norm

k(x − y) + (y − z)k _ kx − yk + ky − zk

) kx − zk _ kx − yk + ky − zk

_

Remark 7.1.12. This means that in the case of Rn we have been using a

metric without knowing it. Like Moliere’s hero who was very impressed to

discover he had been speaking prose for years and years.

Exercise 7.1.6. Prove that Pythagoras’ Theorem holds in any inner product

space

Remark 7.1.13. Why am I doing this abstraction? Because we want to

work in function spaces. If we can say that a space has an inner product

we can go around doing what we were doing in Rn with a good conscience,

all the stuff about lines and hyperplanes and projections and orthogonality

makes sense in a function space. This is a way of pushing intuitions based on

the plane and the space we live in up to infinite dimensional function spaces,

quite a smart trick.

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