Appel W. Mathematics for Physics and Physicists

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Appendix

A few proofs

This appendix collects a few proofs which were too long to insert in the main

text, but are interesting for various reasons; they may be intrinsically beautifu l , or

illustrate important mathematical techniques which physicists may not have other

opportunities to discover.

THEOREM 4.27 on pag e 96 (Winding number) Let γ be a closed path in C,

U the complement of the image of γ in the complex plane. The winding number

of γ around the point z ,

Ind

(z)

def

2πi

dζ

ζ − z

for all z ∈ U (d.1)

is an integer, and as a function of z it is constant on each connected component of U ,

and is equal to zero on the unbounded connected component of U .

Let z ∈ Ω. Assume that γ is parameterized by [0, 1], so that

Ind

(z) =

2iπ

′

(t)

γ(t) −z

dt. (∗)

If w is a complex number, it is equivalent to show that w is an integer or to show that

2iπw

= 1. Hence, to prove that the winding number Ind

(z) is an integer, it sufﬁces to prove

that ϕ(1) = 1, where

ϕ(t)

def

= exp

Z

′

(s)

γ(s) − z



(0 ¶ t ¶ 1).

598 A few proofs

Now, we can compute easily t h at ϕ

′

(t)/ϕ(t) = γ

′

(t)/



γ(t) − z



, for all t ∈ [0, 1], except

possibly at the ﬁnitely many points S ⊂ [0, 1] where the curve γ is not differentiable. Hence

we ﬁnd



ϕ(t)

γ(t) −z



′

(t)

γ(t) −z

−

ϕ(t)γ

′

(t)



γ(t) − z



ϕ(t)γ

′

(t)



γ(t) −z



−

ϕ(t)γ

′

(t)



γ(t) − z



= 0

on [0 , 1 ] \ S . Because the set S is ﬁnite, it follows that ϕ(t)/



γ(t) − z



is a constant. In

addition, we have obviously ϕ(0) = 1, and hence

ϕ(t) =

γ(t) − z

γ(0) − z

Finally, since γ(1) = γ(0), we obtain ϕ(1) = 1 as desired.

It is clear from the deﬁnition of Ind

as function of z t h at it is a continuous function on

Ω. But a continuous function deﬁned on Ω which takes integral values is well known to be

constant on each connected component of Ω.

Finally, using the deﬁnition (∗) it follows that Ind

(z) < 1 if |z| is large enough, and

therefore the integer Ind

must be zero on t h e unbounded component of Ω .

THEOREM 4.31 on pag e 97 (Cauchy for triangle s) Let ∆ be a closed triangle

in the plane (by this is meant a “ﬁlled” triangle, with boundary ∂ ∆ consisting

of three line segments), entirely contained in an open set Ω. Let p ∈ Ω, and let

f ∈ H



Ω \ {p}



be a function continuous on Ω and holomorphic on Ω exc ept

possibly at the point p. Then

∂∆

f (z) dz = 0.

Let [a, b], [b, c], and [c, a] denote the segments which together form the boundary ∂ ∆ of

the ﬁlled triangle ∆. We can assume that a, b, c are not on a line, since otherwise the result is

obvious. We now consider three cases: ① p /∈ ∆, ② p is a vertex, ③ p is inside th e triangle,

but not a vertex.

① p is not in ∆ . Let c

′

be the middle of [a, b], and a

′

and b

′

the middle, respectively, of

[b , c] and [c, a]. We can subdivide ∆ into four smaller similar triangle s ∆

, i = 1, . . . ,4,

as in the picture, and we have

def

∂∆

f (z) dz =

i=1

∂∆

f (z) dz

′

(the integrals on [a

′

, b

′

], [b

′

, c

′

], and [c

′

, a

′

] which occur on the right-hand side do so

in opposite pairs because of orientation). It follows that for one index i ∈ {1, 2, 3, 4} at

least, we have



∆

f (z) dz



¾ | J /4|. Let ∆

(1)

denote one triangle (any will do) for which

this inequality holds. Repeating the reasoning above with this triangle, and so on, we

A few proofs 599

obtain a se quence (∆

(n)

)

n∈N

of triangles such that ∆ ⊃ ∆

(1)

⊃ ∆

(2)

⊃ ··· ⊃ ∆

(n)

⊃ ···

and



∂∆

(n)

f (z) dz



| J |

at each step. Let L be the length of the boundary ∂∆. Then it is clear that the length

of ∂ ∆

(n)

is equal to L/2

for n ¾ 1.

Finally, because the sequence of triangles is a decreasing (for inclusion) sequence of

compact sets with diameter tending to 0, there e xists a unique intersection point:

i¾1

∆

(i)

= {z

By deﬁnition, we have z

∈ ∆ and by assumption, f is therefore holomorphic at z

Now we can show th at J = 0.

Let ǫ > 0. There exists a positive real number r > 0 such that

|f (z) − f (z

) − f

′

)(z − z

)| ¶ ǫ |z − z

for all z ∈ C such that |z − z

| < r. Since the diameter of ∆

(n)

tends to zero, we have

|z − z

| < r for all z ∈ ∆

(n)

if n is large enough. Using Proposition 4.30, we have

∂∆

(n)

f (z

) dz = f (z

)

∂∆

(n)

dz = 0 and

∂∆

(n)

′

) (z − z

) dz = 0.

Hence

∂∆

(n)

f (z) dz =

∂∆

(n)

[ f (z) − f (z

) − f

′

) (z − z

)] dz,

which proves th at



∂∆

(n)

f (z) dz



¶ ǫ

= ǫ

and therefore J ¶ 4



∂∆

(n)

f (z) dz



¶ ǫL

This holds for all ǫ > 0, and consequently we have J = 0 in the case where p /∈ ∆.

② p is a vertex, for instance p = a. Let x ∈ [a, b] and y ∈ [a, c] be two arbitrarily chosen

points. According to the previous case, we have

[x yb]

f (z) dz =

[ yb c]

f (z) dz = 0

and hence

∂∆

f (z) dz =

[a,x]

[x, y]

[ y,a]

f (z) dz,

which tends to 0 as [x → a] and [ y → a], since f is bounded, and the length of the

path of integration tends to 0.

③ p is inside the t riangle, but is not a vertex. It sufﬁces to use the previous result applied to

the three triangles [a, b, p], [b, p, c], and [a, p, c].

600 A few proofs

THEOREM 7.19 on page 18 7 (Cauchy principal value) Let ϕ ∈ D(R) be a

test function. T he quantity

−η

−∞

ϕ(x)

dx +

+∞

+η

ϕ(x)

has a ﬁnite limit as [η → 0

]. M oreover, the ma p

ϕ 7−→ lim

η→0

|x|>η

ϕ(x)

dx (d.2)

is linear and continuous on D(R).

Let ϕ ∈ D, so ϕ is of C

∞

class and has bounded support. I n particular the integral

x|>η

ϕ(x)/x dx exists for all η > 0. Let [−M , M ] be an interval containing the support of ϕ.

For all x ∈ [−M, M ], we have ϕ(x) − ϕ(0) = x



′

(0) + ω(x)



, where ω is a function such

that lim

x→0

ω(x) = 0.

Notice that

η<|x|<M

ϕ(0)

dx = 0 for all η > 0,

so that, by parity, we have

f (η)

def

η<|x|<M

ϕ(x)

dx =

η<|x|<M



′

(0) + ω(x)



η<|x|<M



′

(0) + ω(x)



dx,

and as [η → 0

] this last expression obviously has a limit.

We know must show that the linear map deﬁned for test funct ions by (d.2) is continuous

on D(R).

Let (ϕ

)

n∈N

be a sequence of test functions converging to 0. By deﬁnition, there exists a

ﬁxed interval [−M , M ] containing the support of all the ϕ

 Let ǫ > 0. There exists an integer N such that, if n ¾ N, we have kϕ

∞

¶ ǫ/M,

kϕ

′

∞

¶ ǫ, and kϕ

′′

∞

¶ ǫ. Moreover, the Taylor-Lagrange inequality shows that for n ∈ N

and x ∈ R, we have

|ϕ

(x) −ϕ

(0) − x ϕ

′

(0)| ¶

kϕ

′′

∞

Let n ¾ N be an i nteger. Choose, arbitrarily for the moment, a real number η ∈ ]0, 1], and

consider

|x|>η

(x)

dx =

η<|x|<1

(x)

dx +

|x|>1

(x)

dx.

The second integral is bounded by 2M ×kϕ

∞

¶ 2ǫ. Using the symmetry of the interval in

the ﬁrst integral, we can write

η<|x|<1

(x)

dx =

η<|x|<1

(x) −ϕ

(0)

dx.

But we have



η<|x|<1

(x) −ϕ

(0)



η<|x|<1

(x) −ϕ

(0) − xϕ

′

(0)



η<|x|<1

xϕ

′

(0)



¶ kϕ

′′

∞

η<|x|<1



dx + kϕ

′

∞

¶ 2ǫ.

A few proofs 601

This is valid for any η ∈ ]0, 1]. Letting then [η → 0

] for ﬁxed n, we ﬁnd th at



lim

η→0

|x|>η

(x)



¶ 4ǫ ∀n ¾ N . 

This now proves that the map (d.2) is continuous.

THEOREM 8.18 on pag e 232 Any Dirac sequence converges weakly to the Dirac

distribution δ in D

′

(R).

Let ( f

)

n∈N

be a Dirac sequence and let A > 0 be such that f

is non-negative on [−A, A]

for all n.

Let ϕ ∈ D be given. We need to show t h at 〈f

, ϕ〉 −→ ϕ(0). First, let ψ

def

= ϕ − ϕ(0), so

that ψ is a function of C

∞

class, with bounded support, with ψ(0) = 0. Let K be the support

of ϕ.

 Let ǫ > 0. Since ψ is continuous, there exists a real number η > 0 such that

|ψ(x)| ¶ ǫ

for all real numbers x with |x| < η. We can of course assume that η < A. Because of

Condition ➁ in the deﬁnition 8.15 on page 231 of a Dirac sequence, there exists an integer N

such that



|x|>η

(x) ψ(x) dx



¶ ǫ

for all n ¾ N . Because f

is non-negative on [−η, η], we can also write



|x|¶η

(x) ψ(x) dx



¶ ǫ

|x|¶η



(x)



dx = ǫ

|x|¶η

(x) dx.

Again from the deﬁnition, the last i ntegral tends to 1, so that if n is l arge enough we have



|x|¶η

(x) ψ(x) dx



¶ 2ǫ, hence



(x) ψ(x) dx



¶ 3ǫ.

Now, there only remains to notice that

(x) dx −−→

n→∞

1 (since K is a bounded set), and

then for all n large enough, we obtain



(x) ϕ(x) dx −ϕ(0)



¶ 4ǫ. 

Since ϕ i s an arbitrary test function, this ﬁnally shows that f

′

−−→ δ.

602 A few proofs

THEOREM 9.28 on page 26 1 The space ℓ

, with the hermitian product

(a|b) =

∞

n=0

is a Hilbert space.

The inequality |αβ| ¶

(α

+ β

) shows that ℓ

is a vector space. It is also i mmediate to

check that the formula above deﬁnes a hermit ian product on ℓ

. The hard part is to prove that

ℓ

is a complete space for the norm k·k

associated to this hermitian product.

Let (u

(p)

)

p∈N

be a Cauchy sequence in ℓ

. We need to prove that it converges in ℓ

. There are

three steps: ﬁrst a candidate u is found that should be the limit of (u

(p)

)

p∈N

; then it is proved

that u is an element of ℓ

; and ﬁnally, the convergence of (u

(p)

)

p∈N

to u in ℓ

is established.

• Looking for a limit Since the sequence (u

(p)

)

p∈N

is a Cauchy sequence, it is obvious

that for any p ∈ N, the sequences of complex numbers (u

(p)

)

n∈N

is a Cauchy sequence in C.

Since C is complete, this sequence converges, and we may deﬁne

def

= lim

p→∞

(p)

Obviously, the sequence (u

)

n∈N

of complex numbers should be the limit of (u

(p)

)

• The sequence u is in ℓ

The Cauchy sequence (u

(p)

)

p∈N

, like any other Cauchy se-

quence, is bounded in ℓ

: there exists A > 0 such that



(p)



∞

n=0



(p)



¶ A

for all p ∈ N. Let N be an integer N ∈ N. We obtain in particular

∀p ∈ N,

n=0



(p)





(p)



¶ A,

and as [p → ∞], we ﬁnd

n=0



lim

p→∞

(p)



n=0

¶ A

(since the sum over n is ﬁnite, exchanging limit and sum is obviously permitted). Now this

holds for all N ∈ N, and it follows th at u ∈ ℓ

and i n fact kuk

¶ A.

• The sequence u

( p)

tends to u in ℓ

that is, we have



(p)

− u



−−→

p→∞

 Let ǫ > 0. There exists an integer N ∈ N such that



(p)

− u

(q)



¶ ǫ i f p, q ¾ N . Let

p ¾ N . Then for all q ¾ p and all k ∈ N, we have

n=0



(p)

− u

(q)





(p)

− u

(q)



¶ ǫ.

We ﬁx k, and let [q → ∞], deriving that

n=0



(p)

− u



¶ ǫ

(as before, t he sum is ﬁnite, so exchanging limit and sum is possible). Again, this holds for al l

k ∈ N, so i n the limit [k → ∞], we obtain



(p)

− u



¶ ǫ. 

All this shows that



(p)

− u



−−→

p→∞

0, and concludes the proof.

A few proofs 603

THEOREM 12.36 on page 349

−1



sin kc t



4πc r

δ(r − c t) −δ(r + c t)

−1

[cos kc t] =







4πr

′

(r − c t) if t > 0,

δ( r) if t = 0.

We prove the second formula. It is obviously correct for t = 0, and if t > 0, we compute

−1

[cos kc t] ( r ) =

(2π)

ZZZ

cos(kc t) e

i k·r

(2π)

2π

dϕ

dθ

+∞

dk cos(kt) e

ikr cos θ

sin θ

(2π)

+∞

cos(kt)

ikr

−e

−ikr

)

ikr

8π

+∞



ik(t−r)

−e

−ik(t−r)

+ e

ik(t+r)

−e

−ik(t+r)



ik dk

8π

+∞

−∞



ik(t−r)

+ e

ik(t+r)



ik dk by parity

4πr

−1

δ(r − t) +

δ(r + t)

4πr



′

(r − t) + δ

′

(r + t)



by th e properties of Fourier transforms of derivatives. Since t > 0, the variable r is positi ve

and the distribution δ

′

(r + t) is identically zero, leading to the result stated.

The ﬁrst formula is obtained in the same manner.

604 A few proofs

THEOREM 16.18 o n page 439 (E xistence and uniqueness of the tensor

product) Let E and F be two ﬁnite-dimensional vector spaces over K = R or C.

There exists a K-vector space E ⊗ F such that, for any vector space G, the space of

linear maps from E ⊗ F to G is isomorphic to the space of bilinear maps from

E × F to G, that is,

L (E ⊗ F , G) ≃ Bil(E × F , G).

More precisely, there exists a bilinear map

ϕ : E × F −→ E ⊗ F

such that, for any vector space G and any bilinear map f from E × F to G, there

exists a unique linear map f

∗

from E ⊗ F into G such that f = f

∗

◦ϕ, which is

summarized by the following diagram:

E × F

E ⊗ F

∗

The space E ⊗ F is called the tensor product of E and F . It is only unique up to

(unique) isomorphism.

The bilinear map ϕ from E × F to E ⊗ F is denoted “⊗” and also ca lled the

tensor product. It is such that if (e

)

and (f

)

are bases of E and F , respectively,

the family (e

⊗f

)

i, j

is a basis of E ⊗ F .

All this means that the tensor product “linearizes what was bilinear,” or also that

“to any bilinear ma p on E × F , one can associate a unique linear map on the

space E ⊗ F .”

• Construction of the tensor product First, consider the set E × F of pairs (x, y) with

x ∈ E, y ∈ F . Then consider the free vector space M generated by E × F , that is, the set of

“abstract” linear combinations of th e type

(x, y)∈E×F

(x, y)

[(x, y)] ,

where [(x, y)] is just some symbol associated with the pair (x, y) and λ

(x, y)

is in K, with the

additional condition that λ

(x, y)

= 0 except for ﬁnitely many pairs (x, y).

The set M has an obvious structure of a vector space (inﬁnite-dimensional), with basis the

symbols [(x, y)] for (x, y) ∈ E × F : indeed we add up elements of M “term by term”, and

multiply by scalars in the same way:

(x, y)∈E×F

(x, y)

[(x, y)] +

(x, y)∈E×F

(x, y)

[(x, y)] =

(x, y)∈E×F

(λ

(x, y)

+ µ

(x, y)

)[(x, y)],

µ ·



(x, y)∈E×F

(x, y)

[(x, y)]



(x, y)∈E×F

µλ

(x, y)

[(x, y)] .

A few proofs 605

Now, in order to transform bilinear maps on E × F into linear maps, we will use M and

identify in M certain el ements (for instance, we want the basis elements [(x, 2 y)] and [(2x, y)]

to be equal, and to be equal to 2 [ (x, y)]).

For this purpose, let N be the subspace of M generated by al l elements of the following

types, whi ch are obviously in M:

• [(x + x

′

, y)] −[(x, y)] −[(x

′

, y)];

• [(x, y + y

′

)] −[(x, y)] −[(x, y

′

)];

• [(ax, y)] −a[(x, y)] ;

• [(x, a y)] −a[(x, y)];

where x ∈ E, y ∈ F and a ∈ K are arbitrary.

Consider now T = M/N , the quotient space, and let π be the c anonical projection map

M → M/N , that is, the map which associates to z ∈ M the equivalence class of z, namely, the

set o f elements of the form z + n with n ∈ N . This map is surjective and linear.

Now, E × F can be identiﬁed w ith the “canonical” basis of M , and we can therefore

construct a map

ϕ : E × F −→ T = M/N

by sending (x, y) to the class of the basis vector [(x, y)].

We now claim that this map ϕ and T as E ⊗ F satisfy the properties stated for the tensor

product.

Indeed, let u s ﬁrst show that ϕ is bilinear. Let x, x

′

∈ E and y ∈ F. Then ϕ(x + x

′

, y) is the

class of [(x + x

′

, y]). But notice that the element [(x + x

′

, y)] −[(x, y)] −[(x

′

, y

′

)] is in N by

deﬁnition, w h ich means that t he class of [(x + x

′

, y)] is the class of [(x, y)] + [(x

′

, y)]. This

means that ϕ(x + x

′

, y) = ϕ(x, y) + ϕ(x

′

, y). Exactly i n the same manner, the other elements

we have put in N ensure that all properties required for the bilinearity of ϕ are valid.

We now write E ⊗ F = T , and we simply write ϕ(x, y) = x ⊗ y.

Consider a K-vector spac e G. If we have a linear map T = E ⊗F → G, then the composite

E × F −→ E ⊗ F −→ G

is a bilinear map E ×F → G. Conversely, let B : E ×F → G be a bilinear map. We construct

a map E ⊗ F → G as follows. First, let

B : M → G be the map deﬁned by

(x, y)∈E×F

(x, y)

[(x, y)] 7−→

(x, y)∈E×F

(x, y)

B(x, y).

It is obvious that

B is linear. Moreover, it is immediate that we have

B(n) = 0 for all n ∈ N

because B is bilinear; for instance,

B([(x + x

′

, y)] −[(x, y)] −[(x

′

, y)]) = B(x + x

′

, y) − B(x, y) − B(x

′

, y) = 0.

This means that we can unambiguously use

B to “induce” a map T → G by x ⊗ y 7→ B(x, y),

because i f we c h ange th e representative of x ⊗ y in M , the value of B(x, y) does not change.

Thus we have constructed a line ar map E ⊗ F → G.

It is now easy (and an excellent exercise) to c heck that the applications just described,

L (E ⊗ F , G) −→ Bil(E × F ,G) and Bil(E × F ,G) −→ L (E ⊗ F ,G),

are reciprocal to eac h other.

• Uniqueness Let ϕ : E × F → H and ϕ

′

: E × F → H

′

be two maps satisfying the

property stated in the theorem, where H and H

′

are K-vector spaces.

Since ϕ

′

is itself bilinear, there exists (by the property of ϕ) a unique linear map ψ : H →

′

such t h at ϕ

′

= ψ ◦ ϕ. Similarly, exchanging the roles of ϕ and ϕ

′

, there exists a unique

′

: H

′

→ H such that ϕ = ψ

′

◦ϕ

′

. Now notice that we have both ϕ = Id

◦ϕ (a triviality)

and ϕ = ψ

′

◦ψ ◦ϕ. The universal property of ϕ again implies that ψ

′

◦ψ = Id

. Similarly,

we ﬁnd ψ ◦ψ

′

= Id

′

, and hence ψ is an isomorphism from H to H

′

This shows that the tensor product is unique up to isomorphism, and in fact “up to unique

isomorphism.”

606 A fe w proofs

THEOREM 19.18 on page 516 (Poincaré formula) Let n ∈ N

∗

and let

, . . . , A

be arbitrary events, not necessarily disjoint. Then we have

[

i=1

k=1

(−1)

k+1

1¶i

<···<i

¶n

P(A

∩···∩ A

Following Pascal,

consider the random variables 1

which are characteristic functions of

the events A

, that is,

: Ω −→ R,

ω 7−→

0 if ω ∈ A,

1 if ω /∈ A.

Then we have E(1

) = P(A

). Moreover, these random variables satisfy

∩A

= 1

·1

and 1

= 1

Ω

−1

Using both properties, we ﬁnd

P(A

∪··· ∪ A

) = E



Ω

−

i=1

Ω

−1

)



Now expand the product on the right-hand side, noting that 1

Ω

·1

= 1

; we get

i=1

Ω

−1

) = 1

Ω

k=1

(−1)

1¶i

<···<i

¶n

·1

. . . 1

It only remains to compute the expectation of both sides to derive the stated formula.

Blaise Pascal (1623—1662) studied mathematics when he was very young, and investigated

the laws of hydrostatics and hydrodynamics (having his brothe r in law climb to the top of

the Puy de Dôme with a barometer in order to prove that atmospheric pressure decreases with

altitude). He found many results in geometry, arithmetics, and inﬁnitesimal calculus, and was

one of the very ﬁrst to study probability theory. In addition, he invented and built the ﬁrst

mechanical calculator. In 1654, Pascal abandoned mathematics and scie nce for religion. In his

philosophy there remain, however, some traces of his scientiﬁc mind.

This may be proved by induction, for instance, in the same manner th at one shows the

that

i=1

(1 − x

) = 1 +

k=1

(−1)

1¶i

<···<i

¶n

. . . x