DSpace at VNU: Global Analysis: Functional Analysis Examples c-1

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Global Analysis

Functional Analysis Examples c-1

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Leif Mejlbro

Global Analysis

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Global Analysis

ISBN 978-87-7681-533-2

Disclaimer: The texts of the advertisements are the sole responsibility of Ventus Publishing, no endorsement of them by the author is either stated or implied.

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Contents

Introduction

Ordinary Differential Equations

5 7 13 21 28 34 41 48 53 57 63 67 71

75

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Introduction

This is the first book containing examples from Functional Analysis We shall here deal with the

subject Global Analysis The contents of the following books are

Functional Analysis, Examples c-2

Topological and Metric Spaces, Banach Spaces and Bounded Operators

1 Topological and Metric Spaces

(a) Weierstraß’s approximation theorem

(b) Topological and Metric Spaces

(c) Contractions

(d) Simple Integral Equations

2 Banach Spaces

(a) Simple vector spaces

(b) Normed Spaces

(c) Banach Spaces

(d) The Lebesgue integral

3 Bounded operators

Hilbert Spaces and Operators on Hilbert Spaces

1 Hilbert Spaces

(a) Inner product spaces

(b) Hilbert spaces

(c) Fourier series

(d) Construction of Hilbert spaces

(e) Orthogonal projections and complement

(f) Weak convergency

2 Operators on Hilbert Spaces

(a) Operators on Hilbert spaces, general

(b) Closed operators

Spectral theory

1 Spectrum and resolvent

2 The adjoint of a bounded operator

3 Self-adjoint operators

4 Isometric operators

5 Unitary and normal operators

6 Positive operators and projections

7 Compact operators

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Introduction

Integral operators

1 Hilbert-Schmidt operators

2 Other types of integral operators

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1 Metric Spaces

Example 1.1 Let (X, dX) and (Y, dY) be metric spaces Define

dX ×Y : (X × Y ) × (X × Y ) → R+0

by

dX ×Y ((x1, y1) , (x2, y2)) = max (dX(x1, x2), dY(y1, y2))

1 Show that dX ×Y is a metric on X × Y

2 Show that the projections

pX: X × Y → X, pX(x, y) = x,

pY : X × Y → Y, pY(x, y) = y,

are continuous mappings.

The geometric interpretation is that dX ×Y compares the distances of the coordinates and then chooses

the largest of them

0 0.5 1 1.5 2 2.5 3 3.5

Figure 1: The points (x1, y1) and (x2, y2), and their projections onto the two coordinate axes

1 MET 1 We have assumed that dX and dY are metrics, hence

dX ×Y ((x1, y1), (x2, y2)) = max (dX(x1, x2), dY(y1, y2)) ≥ max(0, 0) = 0

If

dX ×Y ((x1, y1), (x2, y2)) = max (dX(x1, x2), dY(y1, y2)) = 0, then

dX(x1, y1) = 0 and dY(y1, y2) = 0

Using that dX and dY are metrics, this implies by MET 1 for dX and dY that x1 = x2

and y1= y2, thus

(x1, y1) = (x2, y2), and MET 1 is proved for dX ×Y

MET 2 From dX and dY being symmetric it follows that

dX ×Y ((x1, y1), (x2, y2)) = max (dX(x1, x2), dY(y1, y2))

= max (dX(x2, x1), dY(y2, y1))

= dX ×Y((x2, y2), (x1, y1)) , and we have proved MET 2 for dX ×Y

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MET 3 The triangle inequality If we put in (x, y), we get

dX(x1, x2) ≤ dX(x1, x) + dX(x, x2)

≤ dX ×Y((x1, y1), (x, y)) + dX ×Y ((x, y), (x2, y2)) , and analogously,

dY(y1, y2) ≤ dX ×Y((x1, y1), (x, y)) + dX ×Y ((x, y), (x2, y2)) Hence the largest of the numbers

dX(x1, x2) and dY(y1, y2) must be smaller than or equal to the common right hand side, thus

dX ×Y ((x1, y1), (x2, y2)) = max (dX(x1, x2), dY(y1, y2))

≤dX ×Y ((x1, y1), (x, y)) + dX ×Y ((x, y), (x2, y2)) , and MET 3 is proved

Summing up, we have proved that dX ×Y is a metric on X × Y

2 Since pX: X × Y → X fulfils

dX(pX((x, y)), pX((x0, y0))) = dX(x, x0) ≤ dX ×Y ((x, y), (x0, y0)) ,

we can to every ε > 0 choose δ = ε Then it follows from dX ×Y ((x, y), (x0, y0)) < ε that

dX(pX((x, y)), pX((x0, y0))) ≤ dX ×Y ((x, y), (x0, y0)) < ε,

and we have proved that pX is continuous

The proof of pY : X × Y → Y also being continuous, is analogous

Example 1.2 Let (S, d) be a metric space For every pair of points x, y ∈ S, we set

d(x, y) = d(x, y)

1 + d(x, y).

Show that d is a metric on S with the property

0 ≤ d(x, y) < 1 for all x, y ∈ S.

Hint: You may in suitable way use that the function ϕ : R+

0 →R+0 defined by

ϕ(t) = t

1 + t, t ∈ R+

0,

is increasing.

MET 1 Obviously,

d(x, y) = d(x, y)

1 + d(x, y) ≥0, and if d(x, y) = 0, then d(x, y) = 0, hence x = y

MET 2 From d(x, y) = d(y, x) follows that

d(x, y) = d(x, y)

1 + d(x, y) =

d(y, x)

1 + d(y, x) = d(y, x).

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0 0.2 0.6 1 1.2

1 2 3 4 5 6

Figure 2: The graph of ϕ(t) and its horizontal asymptote

MET 3 We shall now turn to the triangle inequality,

d(x, y) ≤ d(x, z) + d(z, x)

Now,

d(x, y) ≤ d(x, z) + d(z, y),

and

ϕ(t) = t

1 + t = 1 −

1

1 + t ∈[0, 1[ for t ≥ 0,

is increasing Since a positive fraction is increased, if its positive denominator is decreased

(though still positive), it follows that

d(x, y) = d(x, y)

1 + d(x, y) = ϕ(d(x, y))

≤ ϕ(d(x, z) + d(z, x)) = d(x, z) + d(z, y)

1 + d(x, z) + d(z, y)

= d(x, z)

1 + d(x, z) + d(z, y)+

d(z, y)

1 + d(x, z) + d(z, y)

= d(x, z)

1 + d(x, z)+

d(z, y)

1 + d(z, y)

= d(x, z) + d(z, y), and we have proved that d is a metric

Now, ϕ(t) ∈ [0, 1] for t ∈ R+

0, thus d(x, y) = ϕ(d(x, y)) ∈ [0, 1[ for all x, y ∈ S,

hence

0 ≤ d(x, y) < 1 for all x, y ∈ S

Remark 1.1 Let ϕ : R+

0 →R+0 satisfy the following three conditions:

1 ϕ(0) = 0, and ϕ(t) > 0 for t > 0,

2 ϕ is increasing

3 0 ≤ ϕ(t + u) ≤ ϕ(t) + ϕ(u) for all t, u ∈ R+

0

If d is a metric on S, then ϕ ◦ d is also a metric on S

The proof which follows the above, is left to the reader ♦

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Example 1.3 Let K be an arbitrary set, and let (S, d) be a metric space, in which 0 ≤ d(x, y) ≤ 1

for all x, y ∈ S.

Let F (K, S) denote the set of mappings f : K → S.

Define D : F (K, S) × F (K, S) → R+

0 by

D(f, g) = sup

t∈K

d(f (t), g(t))

1 Show that D is a metric on F (K, S).

2 Let t0∈ K be a fixed point in K and define

Evt0 : F (K, S) → S by Evt0(f ) = f (t0)

Show that Evt 0 is continuous.

(Evt0 is called an evolution map.)

–1 –0.5 0 0.5 1

Figure 3: The metric D measures the largest point-wise distance d between the graphs of two functions

over each point in the domain t ∈ K

First notice that since 0 ≤ d(x, y) ≤ 1, we have

D(f, g) = sup

t∈K

d(f (t), g(t)) ≤ 1 for all f, g ∈ F (K, S)

Without a condition of boundedness the supremum could give us +∞, and D would not be defined

on all of F (K, S) × F (K, S)

1 MET 1 Clearly, D(f, g) ≥ 0 Assume now that

D(f, g) = sup

t∈K

d(f (t), g(t)) = 0

Then

d(f (t), g(t)) = 0 for all t ∈ K, thus f (t) = g(t) for all t ∈ K This means that f = g, and MET 1 is proved

MET 2 is obvious, because

D(f, g) = sup

t∈K

d(f (t), g(t)) = sup

t∈K

d(g(t), f (t)) = D(g, f )

MET 3 It follows from

d(f (t), g(t)) ≤ d(f (t), h(t)) + d(h(t), g(t)) for all t ∈ K, that

D(f, g) = sup

t∈K

d(f (t), g(t)) ≤ sup

t∈K

{d(f (t), h(t)) + d(h(t), g(t))}

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The maximum/supremum of a sum is of course at most equal to the sum of each of the

maxima/suprema, so we continue the estimate by

D(f, g) ≤ sup

t∈K

d(f (t), h(t)) + sup

t∈K

d(h(t), g(t)) = D(f, h) + D(g, h), and MET 3 is proved

Summing up, we have proved that D is a metric on F (K, S)

2 Since

d(Evt0(f ), Evt0(g)) = d (f (t0), g(t0)) ≤ sup

t∈K

d(f (t), g(t)) = D(f, g),

we can to every ε > 0 choose δ = ε, such that if

D(f, g) < δ = ε,

then

d(Evt0(f ), Evt0(g)) ≤ D(f, g) < ε,

and the map Evt0 : F (K, S) → D is continuous

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Example 1.4 Example 1.1 (2) and Example 1.3 (2) are both special cases of a general result Try to

formulate such a general result.

Let (X, dX) and (T, dY) be two metric spaces, and let ϕ : R+

0 → R+

0 be a continuous and strictly increasing map (at least in a non-empty interval of the form [0, a]) with ϕ(0) = 0 Then the inverse

map ϕ−1: [0, ϕ(a)] → [0, a] exists, and is continuous and strictly increasing with ϕ−1(0) = 0

Theorem 1.1 Let f : X → Y be a map If

dY(f (x), f (y)) ≤ ϕ (dX(x, y)) for all x, y ∈ X,

then f is continuous.

Proof We may without loss of generality assume that 0 < ε < a Choose δ = ϕ−1(ε) If x, y ∈ X

satisfy

dX(x, y) < δ = ϕ−1(ε),

then we have for the image points that

dY(f (x), f (y)) ≤ ϕ (dX(x, y)) < ϕϕ−1(ε) = ε,

and it follows that f is continuous

Examples

1 In the previous two examples, ϕ(t) = t, t ∈ R+

0 Clearly, ϕ is continuous and strictly increasing, and ϕ(0) = 0

2 Another example is given by ϕ(t) = c · t, t ∈ R+

0, where c > 0 is a constant

3 Of more sophisticated examples we choose

ϕ(1) =√

t, ϕ(t) = exp(t) − 1, ϕ(t) = ln(t + 1), ϕ(t) = sinh(t), ϕ(t) = tanh t, ϕ(t) = Arctan t,

etc etc

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2 Topology 1

Example 2.1 Let (S, d) be a metric space For x ∈ S and r ∈ R+ let Br(x) denote the open ball in

S with centre x and radius r Show that the system of open balls in S has the following properties:

1 If y ∈ Br(x) then x ∈ Br(y).

2 If y ∈ Br(x) and 0 < s ≤ r − d(x, y), then Bs(y) Br(x).

3 If d(x, y) ≥ r + s, where x, y ∈ S, and r, s ∈ R+, then Br(x) and Bs(y) are mutually disjoint.

We define as usual

Br(x) = {y ∈ S | d(x, y) < r}

Figure 4: The two balls Br(x) and Br(y) and the line between the centres x and y Notice that this

line lies in both balls

1 If y ∈ Br(x), then it follows from the above that d(x, y) < r Then also d(y, x) < r, which we

interpret as x ∈ Br(y)

Figure 5: The larger ball Br(x) contains the smaller ball Bs(y), if only 0 < s ≤ r − d(x, y)

2 If z ∈ Bs(y), then it follows from the triangle inequality that

d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + s ≤ d(x, y) + {r − d(x, y)} = r,

which shows that z ∈ Br(x) This is true for every z ∈ Bs(y), hence

Bs(y) Br(x)

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