Download Mathematical Framework for Dirac's Calculus: Gordon's Binary Relation of Standardness and more Exercises Calculus in PDF only on Docsity!
A mathematical framework for Dirac’s calculus
Yves P´eraire
Abstract The observation and the discussion of the physical reality of phenomena, leads to bring out concepts which have to be described in a non ambiguous mathematical language. Concerning Dirac’s calculus we shall introduce, besides the usual definitions for the concepts of point, number, function etc ... , addi- tional concepts for the physical point, the physical equalities, physical infini- ties and infinitesimals ... etc ... In particular we introduce a new equality =D^ , called Dirac-equality, which differs as well from the classical equality as from the weak equalities introduced in various theories of generalized functions. All these definitions are based on a definition in the language of Relative Set Theory, see [15], of the metaconcepts of improperness used by P.A.M. Dirac in [Di], when he claimed ” Strictly of course, δ(x) is not a proper function of x, ... , ... δ′(x), δ′′(x).... are even more discontinuous and less proper than δ(x) itself ”. We defined this way a concept of observed derivative which extends the usual one to a large class of discontinuous possibly non-standard functions. All the multiplications of improper or very improper functions, including the delta-functions and their observed derivatives, are obviously al- lowed. Now the problem of the multiplication is replaced by another one: under which conditions is the Dirac-equality of two functions preserved by a multiplication term by term?
1 Basic language and definitions
We will indicate first how to define the basic vocabulary in order to develop consis- tently a discourse similar to that of physicists. More precisely, we want to introduce the words improper, very improper etc .... and elements of vocabulary which play the same role in reasoning that the constants dx, dy, etc ... and which generalize them by specifying degrees in infinitesimality. All this can be defined precisely
Received by the editors September 2004. Communicated by J. Mawhin.
Bull. Belg. Math. Soc. 13 (2006), 1007–
1008 Y. P´eraire
in the language of Relative Set Theory, RST. This language is built on two bi- nary predicates : The usual predicate ≪ · ∈ · ≫ and a new predicate denoted ≪ · SR · ≫. Let us summarize some rudiments concerning RST. ( See [15], for further details. )
(a) ≪ x SR y ≫ is to be read : x is standard relatively to y.
(b) Any set defined with uniqueness in the classical mathematics is standard rela- tively to any other set : we summarize saying that it is standard.
(c) The collection of sets is totally preordered by the predicate ≪ · SR · ≫. There are classes of equistandardness. The class of equistandard- ness of a is the collection α = [a] of x such that ( x SR a ∧ a SR x). We say that x is [a]standard, or αstandard if x SR a. Then we use the denotation [a]st(x) or αst(x). Each class of equistandardness has infinitely many members, there are infinitely many classes.
The theory RST is a conservative extension of the usual theory ZFC, the E.Nelson’s IST (see [12]) is a subtheory of RST. E.Gordon also introduced in [7] a binary relation of relative standardness, but his relation is defined inside IST.
1.1 Definition in RST of the levels of improperness.
Let us define finitely many level of improperness 1, 2, ...... w, where w is an integer we can choose as large as we need (for example w = 100). For that we will fix distinct classes of equistandardness
α 0 = [0], α 1 = [a 1 ], α 2 = [a 2 ], .... αw = [aw],
such that the first class α 0 is the standard one and
a 1 SR a 2 , a 2 SR a 3 , · · · aw− 1 SR aw.
If 1 ≦ n ≦ w, we say that a number or a function X is nimproper or that X has the level n of improperness, and we denote nimp(X), if
αn (^) st(X) & ¬ [αn− (^1) st(X)].
In order to extend the former definition we will write
(^0) imp(X) ⇐⇒ st(X), w+1imp(X) ⇐⇒ ¬[aw (^) st(X)].
We denote η(X) the level of improperness of X.
1010 Y. P´eraire
the relativized principle of standardization does not hold, contrary to what happens in RST. The relativized principle of standardization is very useful, the general principles of transfer idealization and choice we have proved in [16] are linked to the existence of such a possibility of standardization. Now a lot of applications can be treated with Gordon’s predicate.
2 Derivations of classically non derivable functions
We need first to develop the concept of point. Of course, none of these definitions completely captures the intuitive notion.
2.1 Various conceptual symbolical patterns for the physical point.
Let a be a point and n a level of improperness such that η(a) ≦ n.
We put ]a[n=]a − hn, a + hn[. If n = η(a) we use the simplified notations ]a[ for ]a[n.
If a is a number and n a level of improperness, which have no relation a priori with η(a), we put ≀ a ≀ (^) n = {x ∈ R : x n ∼ a} and we use the simplified notation ≀ a ≀ if n = 0.
Remarks.
(1) The symbol ≀ a≀ (^) n represents a collection which is not a “true set” in the relative set theory.
(2) In RST, it is not possible to define uniformly the open point ]a[ for any a of R by means of a function h : R −→ R which associates to any a an infinitesimal strictly less standard than a itself. The reason is that if such an h existed then an element a ∈ R would exist such that ( h SR a ) because, if a SR h for any a in R, R would be finite (see [15], Theorem 2). Then the axiom of transfer would give h(a) SR a and this is contradictory.
If x ∈ R we shall use the terms :
”Point” to indicate the number x, ”open point” to indicate ]x[, ”analysed point”to name ≀ x ≀.
2.2 The collection R.
The set R below is our basic space of functions. This set could seem very particular. In reality the possibility for a function u ∈ R to be improper or very improper makes it very general: In particular, R contains exact representatives for any standard or improper distribution .... and more ...
Its description requires the next definition.
A mathematical framework for Dirac’s calculus 1011
Definition 1. A subset F of R is said Locally standard-finite if, for any limited x and y, [x, y] ∩ F is finite and its cardinality is standard.
The axioms of RST enable to prove that if F is a locally standard-finite subset of R then a standard subset oF of R exists such that, for any limited x, y ∈ R :
(a) oF ∩ [x, y] is a standard finite set,
(b) for any t ∈ F ∩ [x, y] there exists a standard element ot ∈ oF , called the shadow of t, such that t ∼ ot.
The set oF is called the shadow of F.
Definition 2. We say that a function f R −→ R is regular if
(a) it have a level of improperness among the levels { 1 , 2 , , w} fixed above,
(b) f is C∞^ at any point except the points of a locally standard-finite possibly empty set F (f ),
(c) f has all order left and right derivatives at any point of F (f ).
We denote R the collection of regular functions. In this paper, if f, g ∈ R we put f = g if there exits a finite set F (possibly non standard) such that for any t ∈ R \ F, f (t) = g(t).
Basic examples. In these examples, for any subset A of R, χA denotes the characteristic function of A.
- The Heaviside function Ha at the point a defined by Ha = χ[a,+∞[ is regular and F (Ha) = {a}.
- If we simplify H 0 in H, f = H ◦ sin is regular and F (f ) = π · Z
- Let a ∈ R with the level η(a) of improperness, 0 ≦ η(a) < w. We define the principal evaluation δa of the Dirac-function at the point a by
δa =
2 hη(a)
χ]a[. We will simplify δ 0 in δ.
δa is η(a)+1improper, regular and F (δa) = {a − hη(a), a + h−η(a)}.
- We shall need also ”hyperevaluations” of the Dirac-function.
If 0 ≦ η(a) < n ≦ w, we put
n δa =^ h^1 n χ]a[n. This implies that
1 δa = δa. We shall use the simpler notations δ˜a rather than η(a)+ δa , and ˜δ for
2 δ.
- The functions P and Z
P (x) = 1 − χ]0[, Z(x) = χ]−h,0[ − χ]0,h[.
A mathematical framework for Dirac’s calculus 1013
It will be noticed that the index k in f ] ′[k (x) indicates the level of improperness of the hyper-observed derivative. It would be possible to also define concepts of hyper-observed second derivative, hyper-observed third derivative etc.....
Examples :
(1) For any standard a ∈ R, Ha is standard, 0 improper, so
H] ′[ a =^
2 h
[ Ha+h − Ha−h ] = δa, H] ′[ 2 a =^
2 h 2
[ Ha+h 2 − Ha−h 2 ] = δ˜a,
(2) δa is 1 improper hence
δ]
′[ a =^
2 h 2
[ δa+h 2 − δa−h 2 ], δ]
′[ 3 a =^
2 h 3
[ δa+h 3 − δa−h 3 ]. δ]
′[ a is (^2) improper, δ]′[ 3 a is^ (^3) improper.
Figure 2 below represents δ] ′[
. It is only a synoptic representation : we consider that we are really unable to see precisely inside the point.
]-h[ ]h[
(1/2h). (1/2h 2 )
Figure 2. The function δ]
′[ (x) If we denote, for any function f ∈ R with the level n − 1 of improperness,
f+(x) = f (x + hn) and f−(x) = f (x − hn).
then the observed derivatives have the obvious following properties. For any f, g ∈ R with the same level of improperness
(f + g)]
′[ = f ]
′[
′[ , (1)
(f g)]
′[ = f ]
′[ g+ + f− g]
′[
. (2)
These formulae are false if f and g don’t have the same degree of improperness, for example (H + Hh)]
′[ = δ˜ + δh,
H]
′[
1014 Y. P´eraire
The second formula is not the exact formula of Leibnitz,‘
(f g)] ′[ = f ] ′[ g + f g] ′[ .
However, physicists sometimes freely use formula (1), the formula of Leibnitz as well as the formula of integration by parts even in the presence of jumps, and this does not seems to lead to any concrete physical contradiction.
The introduction, besides the concepts above, for points functions and derivatives, of a concept of equality adapted to the physicist’s discourse, will lighten the mystery.
2.3 Dirac-equalities.
Dirac considers that any improper function whose value is zero outside of the origin and such that the integral is egal to 1, is identical to delta. Starting from this idea, we will define a relation between elements of R which, according to our opinion, is more significant than the classical weak equality of distributions. Its definition uses the below definite notion of reiterated primitives.
If f ∈ R, we denote
∫ (^) x
1 a
f (s) ds =
∫ (^) x
a
f (s) ds,
∫ (^) x
2 a
f (s) ds =
∫ (^) x
a
dt
∫ (^) t
a
f (s) ds,
∫ (^) x
n+1a
f (s) ds =
∫ (^) x
a
dt
∫ (^) t
1 a
f (s) ds...
The next proposition gives useful values and infinitesimal approximations of some reiterated primitives. Let us denote y << a << x if y < a < x, x /∈ ≀ a ≀ and y /∈ ≀ a ≀. Then we can state.
Proposition 1. for any y, a, x such that y << a << x, for any standard k ∈ N⋆, for any n > η(a)
∫ (^) x
k y
Ha =
(x − a)k k!
∫ (^) x
k y
n δa ∼n (x^ −^ a)k−^1 (k − 1)!
∫ (^) x
k y
n δa)^2
n ∼
(x − a)k−^1 2 hn (k − 1)!
∫ (^) x
y
(t − a)
n δa(t)^ dt^ = 0,
∫ (^) x
2 y
(t−a)
n δa(t)^ dt^ =^ −
h^2 n 3
, if k > 1 (6)
∫ (^) x
k y
n δa)]
′[ n ∼
(x − a)k−^2 (k − 2)!
Proof. These results are obtained through elementary calculations. The proof of (1) is obvious. Let us prove (2). We will remark first that
n δa =^
2 hn
[Ha−hn − Ha+hn ].
An application of (1) gives
∫ (^) x
k y
n δa =
(x − (a − hn))k^ − (x − (a + hn))k 2 hnk!
1016 Y. P´eraire
Counterexamples :
1 - Hδ =D^
δ but H(Hδ) = Hδ =D^
δ 6 =D^
δ =D^ H (
δ).
2 - If f is constant, f =
2 h
, g = H−h − Hh then g =D^ 0 but f g = δ 6 =D^ 0.
However, we shall see useful particular cases in section 3, theorem 8 and 9. The next proposition states some useful relations. Proposition 2. Let a ∈ R be such that η(a) ≦ w : (1) If η(a) < n ≦ m ≦ w, then for any λ ∈ R such that η(λ) ≦ n − 1,
λ
n δa =D^ λ
m δ (^) a.^ In particular,^ δa =D^ δ˜a.
(2) Ha δa =D^
δa, (3) δa δ˜a =D^ δ^2 a, (4)
x
δh =D 1 2 δ^2 but
x
δh 6 =D^2 δ^2.
(5) δ] ′[ a =
D 0 0 ,^ (6)
∫ (^) x
−∞
δ] ′[ a =
D (^) δ a,^ (7)^ δ 2 a 6 =
D (^) δ˜ 2 a,^ (8)
=^1
x δ =D^ −
δ] ′[ ,
(9) δa δa+ =D 1
δ^2 a but δaδa+ 6 =D^
δ^2 a, (10)
2 h
Z =D^0 , (11)
2 h
Z]
′[ =D^0 ,
(12)
2 h
) 2 Z =D^
δ
]′[ , (13) Hδ^2 =D 1
δ^2 but Hδ^2 6 =D^
δ^2 ,
(14)
2 h
( δ˜ − δh ) =D^
δ] ′[ , (15) =^1 x δ] ′[ =D^ − 4 δ^2 δ˜ −
δ]”[^ =D 1 − 4 δ^3 −
δ]”[,
(16)
x
)]′[ δ =D^4 δ^2 δ˜ =D 1 4 δ^3.
Proof. (1) The functions λ
n δa =D^ λ
n δa and^ λ
n δa =D^ λ
m δ (^) a have the value 0 outside^ ≀^ a^ ≀^. It remains to be shown that the reiterated integrals are infinitesimally close one to the other. In order to prove it let us fix x and y such that x /∈ ≀ a ≀ and y /∈ ≀ a ≀. From the former proposition we get
∫ (^) x
k y
n δa
n ∼
(x − a)k−^1 (k − 1)!
m ∼
∫ (^) x
k y
m δ (^) a hence
∫ (^) x
k y
n δa
n ∼
∫ (^) x
k y
m δ (^) a.
Multiplying term by term by the nstandard number λ we obtain
λ
∫ (^) x
k y
n δa =
∫ (^) x
k y
λ
n δa ∼n
∫ (^) x
k y
λ
m δ (^) a =^ λ
∫ (^) x
k y
m δ (^) a.
(2) and (3) are obvious. Let us prove (4). let us fix x and y such that x /∈ ≀ a ≀ ,
y /∈ ≀ a ≀ and y < a < x. Let us denote ϕ(t) =
∫ (^) t
y
s
δh(s) ds. 1 s
δh(s) = 0 if s /∈ [h] and, if s ∈ [h]
2 h 2 (h + h 2 )
s
δh(s) ≦
2 h 2 (h − h 2 )
Taking the integral we obtain
1 h + h 2
≦ ϕ(x) =
∫
[h]
s
δh(s) ds ≦
h − h 2
A mathematical framework for Dirac’s calculus 1017
As
h + h 2
∼^2
h
∼^2
h − h 2
and
∫ (^) x
y
δ^2 =
2 h
, we obtain
x
δh =D 1 2 δ^2.
This implies
∫ (^) x
2 y
t
δh(t) dt =
∫ (^) x
y
ϕ(t) dt ≦
x − (h − h 2 ) h − h 2
Now, it follows from x /∈ ≀ 0 ≀ and h 2 ∼^2 0 that
x − (h − h 2 ) h − h 2
∼^2 x h
On the other hand,
∫ (^) x
2 y
δ^2 ∼
x 2 h
. Hence
∫ (^) x
2 y
t
δh(t) dt 6 ∼
∫ (^) x
2 y
2 δ^2 (t) dt and this implies
that (^1) x δn 6 =D^2 δ^2.
We let to the reader the proofs of (5), (6)and (7). Let us prove (8).
We deduce from the formula
= x δ =
2 h^2
[−H−h + 2H − Hh] that
∫ (^) x
k y
= x δ =
−(x + h)k^ + 2xk^ − (x − h)k 2 h^2 k!
xk−^2 (k − 2)!
∫ (^) x
k y
δ]
′[ .
Let us prove (9) now. We have δδh =D 0 12 δ^2 because both δδh and 12 δ^2 have the value zero outside of ≀ 0 ≀. Let now x and y be elements of R \ ≀ 0 ≀. If y < 0 < x then an easy calculation gives
∫ (^) x
n y
δδh =
(x − (h − ˜h)n^ − (x − h)n 4 hh 2 n!
∫ (^) x
n y
δ^2 =
(x + h)n^ − (x − h)n 4 h^2 n!
So we obtain :
with n = 1 :
∫ (^) x
1 y
δδh =
∫ (^) x
1 y
δ^2 =
h
with n = 2 :
∫ (^) x
2 y
δδh =
1 2 h x^ −^1 4
∫ (^) x
2 y
δ^2 =
x 8 h
. Hence
δδh =D 1
δ^2 , but δδh 6 =D^
δ^2 because δδh 6 =D 2
δ^2.
the formulas (10) to (16) are left to the reader.
Let us consider now the following functions, which usually play the role of Dirac- functions in physics.
δ1(x) =
2 h
e−^
|x| h (^) , δ2(x) =^1 π
h x^2 + h^2
, δ3(x) =
h
π
e−^
x^2 h^2 , δ4(x) = sin(x h ) πx
Then we have
Proposition 3. δ1 =D^ δ2 =D^ δ3 =D^ δ but δ 4 6 =D^ δ. If a ∼ ±∞ and ξa = (^21) h (Ha− − Ha+ ) then ξa =D^0.
Proof. It is an immediate check
A mathematical framework for Dirac’s calculus 1019
Now, by the definition of R, f ′^ is uniformly continuous over each interval of conti- nuity of f , so
f ′(x + θhn) n ∼ f ′(x), f ′(x + θhn)g(x) n ∼ f ′(x)g(x)
because g is nlimited, and
∫ (^) y′
x′
f ′g ∼
∫ (^) y′
x′
f ]
′[n g whatever x′, y′^ such that [x′, y′] ⊂
([x, y] \ ]F (f )[n.
If a ∈ F (f ), then f ′g and f ]
′[n g are both nlimited on ]a[n so the integrals
∫ (^) y”
x”
f ′g
and
∫ (^) y”
x”
f ] ′[n g are infinitesimals for any x”, y” ∈]a[n, a ∈ F (f ), because |y” − x”| is ninfinitesimal.
We conclude that
∫ (^) y
x
f ′g ∼
∫ (^) y
x
f ]
′[n g. Computing the iterated primitives up to a
standard rank k, we get
∫ (^) y
k x
f ′g ∼
∫ (^) y
k x
f ]
′[n g.
Hence f ′g =D^ f ] ′[n g. (b) Let us expand g ∈ R in the form g = u+
∑
ai∈G
αiHai , with locally standard-finite G
and continuous u ∈ R. An easy proof, making use of the axiom of transfer, see [15], shows that η(u) ≦ η(g) < n and for any i ∈ G, η(ai) ≦ η(g) < n, η(αi) ≦ η(g) < n.
As G is locally standard-finite then for any limited x and y the integrals
∫ (^) x
k y
f g] ′[
and
∫ (^) x
k y
f g]
′[n are respectively equal to
∫ (^) x
k y
f u]
′[η(g)+
∑
ai∈Gx,y
∫ (^) x
k y
f.(αiHai )]
′[η(g)+
and
∫ (^) x
k y
f u]
′[n
∑
ai∈Gx,y
∫ (^) x
k y
f [αiHai ]]
′[n .
This sum being standard-finite, we only have to prove
f u]
′[η(g)+ =D^ f u]
′[n and f.(αiHai )]
′[η(g)+ =D^ f.(αiHai )]
′[n .
The first relation is an application of (a). Item (a) applies because
n ≧ η(g) + 1 > η(g) ≥ η(u), and η(g) ≧ η(f ) ⇒ η(g) + 1 > η(u, f ).
The last inequality is a consequence of the axiom of transfer. The second Dirac-equality follows from
f.(αiHai )]
′[η(g)+ =D^ f (ai)αi
η(g)+ δ (^) ai =D^ f^ (ai)αi
n δai =D^ f^ (αiHai )]
′[n ,
the verification of which is easy.
In order to prove the necessity of the hypothesis, we have to produce counterexam- ples.
1020 Y. P´eraire
Counterexamples.
(1) With f = H and g = H we have, f ′^ = 0 ⇒ f ′g = 0. f ]
′[ = δ, f ]
′[ g =D^
δ. So
f ′g 6 =D^ f ] ′[ g. The missing assumption is the continuity of f
(2) Let be f (x) =
0 if x ≤ h 2 1 ˜h x^ −^
h 2 ˜h if x^ ∈^ [
h 2 ,^
h 2 +^
˜h] 1 if x ≥ h 2 + ˜h
and g = H.
Then f is continuous, f H] ′[ 2 = f ˜δ = 0 and f H]
′[ = f δ =D^
δ. In this example the
problem comes from the relation η(f ) > η(g).
(3) Concerning (b), let f (x) = δ^3 and g(x) = x^3. An easy computation yields g]
′[ (x) = 3x^2 + h^2 , so f (x)g]
′[ (x) − f (x)g′(x) = h^2 δ^3. Now the computation of the first integral prove that h^2 δ^3 6 =D^ 0. We should obtain the Dirac-equality replacing g] ′[ by g] ′[ 2 .
Corollary. For any f ∈ R and any standard integer n ≥ η(f ),
f ] ′[ =D^ f ] ′[n and f ] ′[ =D^ f ′^ if f is continuous.
Theorem 2. For any f ∈ R and y, x /∈ ≀ F (f ) ≀ ,
∫ (^) x
y
f ]
′[ (t) dt ∼ f (x) − f (y).
Proof. Any function of R decomposes in a sum of elementary functions. It is enough to prove the theorem for these elementary functions. The result is obvious if f
is a continuous function. If f = αHa with limited a then f ] ′[ = α
n δa with^ n^ = Max{η(α), η(a)}+1. For any y, x
∫ (^) x
y
f ]
′[
α if y << a << x, 0 if y, x << a, or y, x >> a
= f (x) − f (y).
Theorem 3. For any f, g ∈ R, whatever their level of improperness :
(f + g)] ′[ =D^ f ] ′[
Proof. Let n = Max{η(f ), η(g), η(f +g)}+1. Then by the definition of nhyperobserved derivatives we have (f + g)] ′[n = f ] ′[n
. Now, the corollary above and properties (a), (b) gives,
(f + g)]
′[ =D^ (f + g)]
′[n = f ]
′[n
′[n =D^ f ]
′[
′[ .
1022 Y. P´eraire
we can verify directly that δ δ] ′[ (^6) =D (^) 0, or wait until theorem 6 in order to use the
arguments that δ δ] ′[ =D^
(δ^2 )] ′[ (^6) =D (^) 0 for δ (^2 6) =D (^) 0.
Corollary 2. For any f, g ∈ R, if f is continuous and η(f ) ≦ η(g) then
(f g)]
′[ =D^ f g]
′[
Proof (f g)] ′[ = (f g)] ′[η(g) = f g] ′[η(g)+
- f ] ′[η(g)+ (Theorem 4). Now theorem 1 gives f ]
′[η(g)+ g =D^ f ′g. This end the proof.
Example : Let xq + = H.xq, q being a standard positive integer. Then for any standard integer n, (xq + δ]n[)]
′[ =D^ q xq+− 1 δ]n[^ + xq + δ]n+1[.
Remark.
Let be f ∈ R with f (t) 6 = 0 for all t. Then we can verify the formula
f
)]
′[ = −
f ] ′[
(f−)(f+)
However, (
f
)]
′[ is not generaly Dirac-equal to −
f ] ′[
f 2
. In order to prove it, let us
consider f = 1 + H. We have f ] ′[ = δ,
f
H,
f 2
H,
f
)]
′[ = −
δ, −
f ] ′[
f 2
= −δ (1 −
H) =D^ −
δ.
Now, corollary 1 allows us to write from the equality
f
f = 1,
f
)]
′[ f =D^ −
f
f ]
′[ .
The difficulty lies in the illegality of the multiplication by (^) f^1 of the two members of a Dirac-equality.
Demonstrations of theorems 5 and 6 below, derive directly from the definition of the relations =Dk and =D.
Theorem 5. For any f, g ∈ R and any standard k ∈ N⋆,
( f =Dk g (F ) and a /∈ ≀ F ≀ ) ⇒ (x →
∫ (^) x
a
f ) =Dk− 1 (x →
∫ (^) x
a
g).
Theorem 6 For any f, g ∈ R, and any k ∈ N,
f =Dk g ⇒ f ]
′[ =Dk+1 g]
′[ .
A mathematical framework for Dirac’s calculus 1023
Theorem 7. For any f, ϕ ∈ R, if ϕ is C∞^ on R, if ϕ−^1 (F (f )) and ϕ−^1 (oF (f )) are locally standard-finite then f ◦ϕ ∈ R and for any level n > η((f, ϕ)) of improperness
(f ◦ ϕ)]
′[ =D^ (f ]
′[n ◦ ϕ) × ϕ′.
To avoid complicated denotations, we shall suppose that ϕ is standard. The general result is obtained through a classical shift of the levels of improperness.
Lemma. If ϕ is continuous on R:
(a) For any limited a ∈ R and any locally standard-finite and standard F 0 ⊂ R, [−a, a]
⋂ ϕ−^1 ( ≀ F 0 ≀ ) ⊂ ≀ ϕ−^1 (F 0 ) ≀.
(b) For any locally standard-finite F ⊂ R : ≀ ϕ−^1 (F ) ≀ ⊂ ≀ ϕ−^1 (oF ) ≀.
Proof. Let x ∈ [−a, a]
⋂ ϕ−^1 ( ≀ F 0 ≀ ) then there exits u 0 ∈ F 0 such that ϕ(x) ∼ u 0. From x limited and ϕ standard we deduce that x has a shadow ox and ϕ(x) is limited. Also u 0 is limited. Now from F 0 standard and locally standard finite we deduce that u 0 is standard. From the continuity of ϕ we get ϕ(x) ∼ ϕ(ox). Hence
ϕ(ox) = u 0 , ox ∈ ϕ−^1 (F 0 ), x ∈ ≀ ϕ−^1 (F 0 ) ≀.
This prove the inclusion of (a). Let us prove (b). If x ∈ ≀ ϕ−^1 (F ) ≀ then there exists a such that x ∼ a and ϕ(a) ∈ F. We have ϕ(oa) = oϕ(a) ∈o^ F. From x ∼ oa, and oa ∈ ϕ−^1 (oF ) we derive x ∈ ≀ ϕ−^1 (oF ) ≀. This ends the proof of (b).
Proof of Theorem 7.
Case where f is continuous.
Let us prove that (f ◦ ϕ)] ′[ =D^ (f ] ′[n ◦ ϕ)ϕ′^ ( ϕ−^1 (oF (f )) ). It makes sense because ϕ−^1 (oF (f )) is locally standard-finite. The function f being in R its derivatives exists on R \ F. Hence, (f ◦ ϕ)(p)^ exists on ϕ−^1 (F (f )) at any order p. The function f ◦ ϕ is continuous so we have, from the corollary of theorem 1
(f ′^ ◦ ϕ) × ϕ′^ = (f ◦ ϕ)′^ =D^ (f ◦ ϕ)] ′[ ( ϕ−^1 (F (f )) ).
As ≀ ϕ−^1 (F ) ≀ ⊂ ≀ ϕ−^1 (oF ) ≀ (lemma, (b))
(f ′^ ◦ ϕ) × ϕ′^ = (f ◦ ϕ)′^ =D^ (f ◦ ϕ)] ′[ ( ϕ−^1 (oF (f )) ).
It remains to prove that
(f ′^ ◦ ϕ) × ϕ′^ =D^ (f ]
′[n ◦ ϕ) × ϕ′^ ( ϕ−^1 (oF (f )) ).
(a) Proof of : (f ′^ ◦ ϕ) × ϕ′^ =D 0 (f ]
′[n ◦ ϕ) × ϕ′^ ( ϕ−^1 (oF (f )) ).
A mathematical framework for Dirac’s calculus 1025
If ϕ is locally decreasing
f ◦ ϕ = μ H ◦ ϕ = μ (1 − Ha) μ H−hn ◦ ϕ = μ (1 − Hα) μ Hhn ◦ ϕ = μ (1 − Hβ )
If ϕ is locally increasing
(f ◦ ϕ)]
′[ = (μ Ha)]
′[ = μ δa
(f ] ′[n ◦ ϕ) × ϕ′^ =
μ (H−hn ◦ ϕ − Hhn ◦ ϕ) × ϕ′ 2 hn
μ(Hα − Hβ ) 2 hn
× ϕ′
Now,
2 hn
μ (Hα − Hβ ) × ϕ′^ is Dirac-equal to μ, δa for it is zero outside of ≀ a ≀ , it
have a constant sign, and if y << a << x then
∫ (^) x
y
2 hn
μ (Hα − Hβ ) × ϕ′) =
2 hn
μ,
∫ (^) β
α
ϕ′^ =
2 hn
μ(ϕ(β) − ϕ(α))
= μ
2 hn
· 2 hn = μ. Hence: (f ◦ ϕ)]
′[ =D^ (f ]
′[n ◦ ϕ) × ϕ′.
The proof is quite similar if ϕ is locally decreasing.
If a is a local extremum then f ◦ ϕ is constant, its value is μ or 0 so (f ◦ ϕ)] ′[ is 0. So is (f ] ′[n ◦ ϕ) × ϕ′, because H−hn ◦ ϕ = Hhn ◦ ϕ = 1 or 0. The reader is now able to proceed alone to the more general case.
If η(f ) ≥ η(ϕ), the formula of the theorem becomes
(f ◦ ϕ)]
′[ =D^ (f ]
′[ ◦ ϕ)ϕ′.
If η(ϕ) > η(f ) the formula (f ◦ ϕ)] ′[ =D^ (f ] ′[ ◦ ϕ)ϕ′^ could be false.
Counterexample.
f = H and ϕ(x) = hx^2 give : (f ◦ ϕ)] ′[ = 0. For any locally standard-finite F ⊂ R, there exists x such that −x, 2 x ∈/ ≀ F ≀ , −x << 0 << 2 x ≦ 1. (f ] ′[ ◦ ϕ)ϕ′(x) = (H] ′[ (hx^2 )) × 2 h x =
δ(hx^2 ) 2hx 6 =D^0 , because
∫ (^2) x
−x
δ(ht^2 ) × 2 h t dt =
2 h
h
[ t^2
] 2 x −x
3 x^2 2
Corollary. for any f ∈ R and any derivable standard ϕ such that ϕ′^ have isolated zeros then f ◦ ϕ ∈ R and
(f ◦ ϕ)] ′[ =D^ (f ] ′[ ◦ ϕ)ϕ′.
Proof. The reason is that under the assumption on ϕ, ϕ−^1 (F (f )) and ϕ−^1 (oF (f )) are locally standard-finite. Let us prove it. Let [x, y] be a standard compact interval then ϕ([x, y]) is a standard compact interval too so, it contains a finite-standard
1026 Y. P´eraire
number of element of F (f ). If ϕ−^1 (F (f )) ∩ [x, y] or ϕ−^1 (oF (f )) contains non finite-standard many elements then c ∈ F (f ) exists such that card(ϕ−^1 ({c})) is not standard-finite. This implies that the standard set Z = (ϕ′)−^1 ({ 0 }))∩ [x, y] is infinite and contradicts the hypoth- esis that the zero of ϕ′^ are isolated. Hence card(ϕ−^1 (F (f )) ∩ [x, y]) and ϕ−^1 (oF (f )) are standard-finite.
Examples.
- H(x^2 ) = 1 gives H(x^2 )] ′[ = 0. With de denotations of theorem 7 we have F (f ) =o^ F (f ) = ϕ−^1 (oF (f )) = { 0 } so H(x^2 )] ′[ =D^ δ(x^2 ) × x^2. Hence δ(x^2 ) × x^2 =D^ 0, which can be obtained also through a direct calculation.
- δ(x^2 )] ′[ =D^ δ] ′[ (x^2 ) × x^2.
Open problem. What happens if it is only supposed that ϕ ∈ R. Under which conditions is the formula (f ◦ ϕ)] ′[ =D^ (f ] ′[ ◦ ϕ)ϕ] ′[ true?
The next theorems, 8 and 9, approach the problem of the conservation of the Dirac- equality after a multiplication term by term.
Theorem 8. For any f, g ∈ R and any standard integer n
f =D^ g ⇒ xn^ f =D^ xn^ g.
Proof. It is enough to prove that f =D^ g ⇒ x f =D^ x g, and one obtains the theorem through an induction. Let us prove inductively that the property P (n) : ∀ f, g ∈ R ( f =Dn g ⇒ x f =Dn x g ), is satisfied for any standard n. P (0) is obvious. The reason is that for any x /∈ ≀ F (f ) ≀ ∪ ≀ F (g) ≀ , and any limited order k ≧ 0 of derivation, (xf (x))(k)^ = xf (k)(x)+ kf (k−1)(x). The numbers x and k being limited, the relations f (k)(x) ∼ g(k)(x) and f (k−1)(x) ∼ g(k−1)(x) (if k > 0) involve xf (k)(x) ∼ xg(k)(x) and, if k > 0, kf (k−1)(x) ∼ kg(k−1)(x). Hence (xf (x))(k)^ ∼ (xg(x))(k). Let us suppose P (n) for a fixed standard n. Let f, g be elements of R Let us
fix y /∈ ≀ F (f ) ≀ ∪ ≀ F (g) ≀. Let us put F (x) =
∫ (^) x
y
f (t) dt and G(x) =
∫ (^) x
y
g(t) dt.
Theorem 5 gives F =D^ G, so F =Dn G. P (n) being true, we obtain x F (n) =Dn x G(x), and by theorem 6, (x F (n))]
′[ =Dn+1 (x G(x))]
′[
. From the corollary 2 of theorem 4 we get
x f (x) + F (x) =Dn+1 x g(x) + G(x). As F =Dn+1 G, we obtain x f (x) =Dn+1 x g(x). Hence P (n) is true for any standard n.
Of course, in theorem 8 we can replace xn^ by (x − a)n. We shall prove now a similar theorem where xn^ is replaced by a non polynomial function.
