10 Products of functions

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10 Products of functions

10.1 Statement of the problem

One often encounters products of functions such as fg, fA, A B⋅ , or A×B. In electromagnetism, for example, ρv and v×B appear in many equations.

Take the simple product f t g t( ) ( ) . Is there some sort of general inequality that compares the norm of fg to the product norm( ) norm( )f ⋅ g ? Probably not, unfortunately.

Here is why.

10.2 The norm of fg and the product norm( ) norm( ) f ⋅ g

Take our two functions f t( ) and g t( ). We shall see that, in certain cases,

norm(fg)>norm( ) norm( )f ⋅ g (236)

or even

norm(fg)>>norm( ) norm( )f ⋅ g , (237)

and that, in other cases, the reverse is true.

With sinusoidal voltages and currents, for example, it is a well known fact that the rms power dissipated by the circuit, (VI)_rms, can be smaller or larger than product

V_rmsI_rms, according to the phase difference between V and I (see Section 10.4.1).

But consider first a simpler example. Suppose that

f t( )= >A 0 on a certain interval of length l and f t()=0 elsewhere, (239) g t( )= >B 0 on a certain interval of length m and g t( )=0 elsewhere, (240) and that

these two intervals overlap on an interval of length c≤min( , )l m . (241)

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Such functions are pictured in Figure 1.

A B

t y

y=f t( )

y=g( )t

c

l m T

0

Figure 1. The two functions f and g defined in (239)-(241).

If f and g are defined only on the interval from 0 to T, the norm may be defined by Equation (35a). Then

norm( )f A

T A

= ²l = Tl

, norm( )g B m

T B m

= ² = T , (242)

norm( ) ( )

fg AB c

T AB c

= ² = T , (243)

norm( ) norm( )f g AB m

⋅ = Tl

, (244)

norm( ) norm( ) norm( )

fg

f g

cT

⋅ = m

l . (245)

If, for example, c= =l m, the two functions are equal and then

fg

f g

T

⋅ =

l . (246)

This ratio is larger than 1 and can even be much larger than 1, and we have the inequality (236) or (237).

On the other hand, if c is small enough, that is, if the intervals of f and g have only a small part (or none) in common, we can have cT <l , or even cTm <<l . In this casem Equation (245) implies the inverse of the inequalities (236) or (237).

If f and g are defined for all values of t and are nonperiodic, the norm (35b) is used instead of (35a). Equations (242) to (246) remain valid, with T =1. The inequalities (236) or (237) could be true, or their inverses could be true.

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10.3 The mean value of fg

We just saw that the norm of fg cannot, in general, be compared in a simple way to the product of the norms. Consider however the Cauchy-Schwarz inequalities (Section 2.2):

1

T

∫

0^T f t g t dt( ) ( ) ^≤norm( ) norm( )f ^⋅ g (for periodic functions), (247) f t g t dt( ) ( ) norm( ) norm( )f g

−∞

∫

∞ ^≤ ^⋅ (for nonperiodic functions). (248)

Now, taking the absolute value of a function does not change its norm, so that we also have

1

T

∫

0^T f t g t dt( ) ( ) ^≤norm( ) norm( )f ^⋅ g (for periodic functions), (249) f t g t dt( ) ( ) norm( ) norm( )f g

−∞

∫

∞ ^≤ ^⋅ (for nonperiodic functions). (250)

According to the inequality (249), which uses the norm (35a), it is the average value of f t g t( ) ( ) , rather than its norm, that is always inferior or equal to

norm( ) norm( )f ⋅ g .

Replacing F by fg in the inequality (40), we obtain a second inequality similar to (249):

1

T

∫

0^T f t g t dt( ) ( ) ^≤norm(fg) (for periodic functions). (251) So, for periodic functions, the average value of fg is not only smaller than

norm( ) norm( )f ⋅ g (see (249)), but also smaller than norm(fg (see (251)).) With the functions f and g of Figure 1, we have

1

T 0 f t g t dt ABc T

T

( ) ( )

∫

⁼ ^,norm( ) norm( )f g AB m

⋅ = Tl

, norm(fg) AB c

= T , (252)

and the inequalities (249) and (251) are equivalent to the following ones:

ABc

T AB m

≤ Tl

and ABc

T AB c

≤ T , (253)

which are always true, since of course c≤min( , )lm and c T/ ≤1.

Can one find general conditions under which two functions F and G will be such that the norm of FG will be of the same order of magnitude as the product of the norms of

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F and of G? The answer given in Sections 10.4 and 10.5 to this question is partial, but useful.

10.4 The norm of the product of two periodic functions

10.4.1 Some examples and a tentative proposition Note that a function such as

f t( )= Asint (where A > 0) (254)

is most of the time of the same order of magnitude as its amplitude A. Let

g t( )= Bcost (where B > 0). (255)

Then

norm( )f A

= 2 and norm( )g B

= 2, (256)

norm( ) sin

.

fg AB t

dt A B AB

=  AB







⌠

⌡ = = ≈

1 2

2

2 4

1

2 2 2 0 35

2

0

2 2 2

π

, (257)

mean value of fg AB t

( )

⁼ 1 ^⌠_⌡ dt

2

0 2

2

π

π sin

(258)

= ^AB₂ π¹

∫

0 ^{t dt}= ÂB₂ π² = ÂBπ ≈^{0 32}ÂB

πsin . , (259)

norm( ) norm( )f g AB .

⋅ = = AB

2 0 5 . (260)

Here the norm and mean value of fg are both smaller than norm( ) norm( )f ⋅ g , but the three quantities ((257) to (260)) are of the same order!

Let us see what happens when there is a phase difference between the two sinusoidal functions. Consider, for example, a circuit with applied voltage

V t( )=V0cost (where V₀ > 0) (261)

and current

I t( )=I₀cos(t−ϕ) (where I₀ > 0). (262)

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Then

V t I t( ) ( )=V I^{0 0}cos cos(t t−^ϕ)= ¹²V I^{0 0}

[

cos(²t−^ϕ)+cos^ϕ

]

^, ⁽²⁶³⁾

V t I t( ) ( ) V I cos ( t ) cos( t )cos cos

[ ]

² ⁼ ¹⁴ ⁰² ⁰²

[

² ² ⁻^ϕ ⁺² ² ⁻^ϕ ^ϕ⁺ ²^ϕ

]

^. ⁽²⁶⁴⁾

Taking averages on both sides, we obtain

(VI)_rms² = ¹₄V I₀² ₀²

[

¹₂ + +0 cos²^ϕ

]

^, ⁽²⁶⁵⁾

(VI)_rms= ¹₂V I₀ ₀ ¹₂ +cos²ϕ, (266)

1

2 0 0

1 2

1

2 0 0

3

V I ≤(VI)_rms ≤ V I 2. (267)

Since V_rms is V₀ 2 and I_rms is I₀ 2 , then

V_rmsI_rms ¹₂ ≤(VI)_rms≤V_rmsI_rms ³₂, (268) 0 7. V_rmsI_rms <(VI)_rms<1 3. V_rmsI_rms. (269) The quantity (VI)_rms =norm(VI) is of the same order as the quantity V_rmsI_rms=norm( ) norm( ).V ⋅ I

These observations might possibly be generalized as follows.

A. TENTATIVE PROPOSITION. If f t( ) and g t( ) are periodic functions of period T, or defined only on an interval of length T, then one uses for f and g the norm (35a). Suppose that each of these functions is integrable, has a finite amplitude, and has the property that

it is “most of the time”

“of the same order” as its amplitude. (270) Then, in most cases,

norm( ) ~ amplitude( )f f , (271)

norm( ) ~ amplitude( )g g , (272)

norm(fg) ~ norm( ) norm( )f ⋅ g . (273)

(The sign ~ means “is of the same order as.”) This tentative proposition is very imprecise;

in particular, the expressions “most of the time”, “of the same order”, and “in most cases”

have not been given clear meanings.

Let us test again the validity of Statement A by taking, as a second example, the functions f and g of Figure 1. We have

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fg

f g

cT

⋅ = m

l . (274)

Note that this ratio is equal to 1 if cT =l , that is, ifm

c m

l = T (275)

(see Figure 2).

A B

t y

y=f t( )

y=g( )t

c

l m T

0

Figure 2. The two functions f and g defined in (239)-(241), with c/l=m T/ =1 2./

A B

t y

y=f t( )

y=g( )t

c m

l T

0

Figure 3. The two functions f and g defined in (239)-(241), with l≈m≈c≈T . Suppose now that f and g have property (270); l and m are then both “close” to T, so that c also is necessarily close to T:

since l≈ ≈m T , we have c ≈T and

cT

lm ≈1, (276)

and therefore

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norm(fg)≈norm( ) norm( )f ⋅ g (277) (see Figure 3).

Let us now try to prove the above tentative proposition.

10.4.2 Proof of a first fuzzy proposition

The content of Statement A of Section 10.4.1 can be made more precise through Definitions 4 and 5 below. Let us first define a notion of order of magnitude.

DEFINITION 4. Given a number m>1, let us say that the two real numbers u and v are of the same m-order of magnitude if

1 /m≤u v/ ≤m. (a1)

In such a case, we shall write

u~_mv. (a2)

In physics, m could be 10 , or 10, or any convenient number greater than one.

Let us now define what we mean by a function having a certain property “most of the time.”

DEFINITION 5. Given a number φ ∈[ , ]0 1 , we shall say that a function f of the variable t, defined on the interval from 0 to T, has property P ( )t a fraction φ of the time if the measure of

{

t∈^{[ , ]}⁰ T ^P^{( )}t

}

^(a3)

is φT .

As an example, consider the sine function:

sint ~₁₀ 1 about 94% of the time (a8)

sint ~ ₁₀ 1 about 80% of the time, (a9)

sint ~₂ 1 exactly 2/3 of the time. (a10)

We can now prove the following proposition. It is a fuzzy proposition because all it asserts is that Equation (a5) is true in most cases; this equation can be false, but only in exceptional cases.

FUZZY PROPOSITION 1. Let φ ∈[ , ]0 1 . Let m>1, with m−1 at least

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several times larger than 1−φ. Let f be an integrable function defined on the interval from 0 to T, and let

f t( ) ~ amplitude( ) at least a fraction _m f φ of the time. (a4) Then, in most cases,

norm( ) ~ amplitude( )f _m f . (a5)

PROOF. Let a be the amplitude of such a function f. The lowest possible value of the norm of f will be attained with a function such as

f t

a x

a m x T

x T T

( )

,

/ ( , ],

( , ].

=

∈





 if

if if

0 0 0

φ φ

(a5.5)

The norm of this function is

1 ²

T a

m T a

m







 φ = φ (a6)

and its average is aφ m. Therefore, for any integrable f obeying (a4), a

m φ ≤norm( )f ≤a and a

mφ ≤ f ≤a. (a7)

We have

m− >> −1 1 φ, (a7.1)

a a m

a m

a

− >> −mφ (after multiplying by a m/ ). (a7.2) For integrable functions f obeying (a4), the most likely value of f is at the midpoint of the interval [aφ/m a, ], which, because of (a7.2), is only slightly lower than the midpoint of the interval [ /a m a . The most unlikely values of f are those near a, ] φ m or a.

Therefore, because of Equation (40), the most likely values of norm( )f are around or above the midpoint of [ /a m a , and the most unlikely values are near a, ] φ m or a.

Now

a m

a

φ≤ φ ≤ m (a7.3)

and Equation (a7.2) makes it obvious that the vast majority of the functions f we are considering will have a norm larger than a m/ . ■

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Should we try to give a precise, quantitative meaning to the expression “in most cases”? In my view, this is not necessary; I find the above statement precise enough for our purposes, and the demonstration given sufficient. It proves a part of statement A, namely Equations (271)-(272). Can we also prove the last part, Equation (273)?

10.4.3 Towards a proof of a second fuzzy proposition

B. TENTATIVE FUZZY PROPOSITION. Let φ ∈[ , ]0 1 . Let m>1, with m−1 at least several times larger than 1−φ. Let f and g be

integrable functions defined on the interval from 0 to T, with finite amplitudes. Assume that

f t( ) ~ amplitude( ) at least a fraction φ of the time,_m f (a11) g t( ) ~ amplitude( ) at least a fraction φ of the time._m g (a12) Then, in most cases,

norm(fg) ~ norm( ) norm(g)_m f . (a13)

If this could be proven in the same fashion as Fuzzy Proposition 1, this would complete the proof — to my satisfaction at least — of Statement A of Section 10.4.1.

Unfortunately, I have been able to prove Statement B only under the condition that φ is

“close enough” to 1, not under the more explicit condition that m−1 be at least several times larger than 1−φ as in Fuzzy Proposition 1. Here is how I approach the problem.

Let a and b be the respective amplitudes of f and g. Equation (a7) can be applied to both functions:

a

m φ ≤norm( )f ≤a and b

m φ ≤norm( )g ≤b. (a14)

Therefore ab

m₂φ ≤norm( ) norm( )f g ≤ab. (a15)

Let ε= −1 φ. If ε ≤ ¹₂, the lowest possible value of the norm of fg is

1 ₂ 1 2 1 2

2

T 2

ab

m T ab

m







 ( − ε) = − ε. (a16)

Therefore ab

m₂ 1 2− ε ≤norm(fg)≤ab. (a17)

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Now

1 2− ε ε+ ² = − =(1 ε) φ, (a18)

so that

1 2− ε φ≤ . (a19)

We then conclude from Equations (a15) and (a17) that, in all cases,

norm(fg), norm( ) norm(g)f ∈

[

ab ^{1 2}− ^ε m ab²,

]

^. ^(a19.5)

Now, if ε is closer to 0 than to 1 2/ , the quantity 1 2− ε will be close to φ. Then, because of our hypotheses concerning φ and m, we shall have, in most cases,

norm(fg) ~_m2 norm( ) norm(g)f . (a20)

But what we want to prove is that, in most cases,

norm(fg) ~ norm( ) norm(g)_m f , (a21)

which is a much stronger statement.

Let us start by replacing this last equation by a simpler, equivalent one. Recall that the norm of a function F defined on [ , ]0 T is the square root of the average value of F² over [ , ]0 T :

norm( )F = F² . (a22)

Equation (a21) is therefore equivalent to the following ones:

norm (² fg) ~_m2 norm ( ) norm (g)² f ² , (a23)

f g² ² ~_m2 f² g .² (a24)

Now

f²( ) ~t _m2 amplitude(f²) at least a fraction φ of the time, (a25) g t²( ) ~_m2 amplitude(g²) at least a fraction φ of the time. (a26) What we want to prove, finally, is the following fuzzy proposition.

C. TENTATIVE FUZZY PROPOSITION. Let φ ≤1 and φ ≈1. Let M >1, with M −1 at least several times larger than 1−φ. Let F and G be nonnegative integrable functions defined on the interval from 0 to T, with finite amplitudes. Assume that

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F t( ) ~ amplitude( ) at least a fraction _M F φ of the time, (a27) G t( ) ~ amplitude( ) at least a fraction _M G φ of the time. (a28) Then, in most cases,

FG~_M F G. (a29)

The following subsection presents a rigorous proof of this statement, but only with φ =1 and some further conditions on M.

10.4.4 A less general, but strict proposition

Let me first state more clearly what I am trying to prove.

D. PROPOSITION TO BE PROVEN. Let F and G be positive integrable functions defined on the interval from 0 to T, with finite

amplitudes. Assume that

F t( ) ~ amplitude( ) for all t_M F ∈[ , ]0 T , (a30) G t( ) ~ amplitude( ) for all t_M G ∈[ , ]0 T . (a31) Then, in all cases (at least for M in a certain interval),

FG~_M F G. (a32)

The proof will proceed by steps and, at the end, this statement will be made more detailed.

Before proving Statement D, I shall prove a stronger version of it in the particular case in which F=G:

E. PROPOSITION TO BE PROVEN. Let M>1. Let F be a positive integrable function defined on the interval from 0 to T, with a finite amplitude. Assume that

F t( ) ~ amplitude( ) for all t_M F ∈[ , ]0 T . (a33) Then, in all cases,

F² ~_µ F², where µ = (M + ) M

1 4

2

. (a34)

This µ is in ( ,1 M .)

Here any value of M larger than 1 will work. Let us verify that the µ given above is indeed smaller than M:

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(M ) ( ) M

M M

+1 < + =

4 4

2 2

. (a35)

When M is large, µ is close to M / 4.

The proof of Statement E will also proceed by steps.

The first thing to realize is that, for any integrable function F defined on [ , ]0 T , we have F² ≥ F² (Equation (38)).

Let

amplitude( )F = A. (a36)

The values of F are in the interval [A M A ./ , ]

Now we shall find functions F that have the largest possible ratio F² F². In order to do this, let us first find functions F that, for a given F ∈[A M A/ , ], have the largest ratio F² F²; in other words, functions F that, for a given F ∈[A M A/ , ], have the largest variance

Variance( )F =

(

F −F

)

² ⁼ F² ⁻ F²^. ^(a37)

I must first underline an important fact about the functions F we are studying. We require that amplitude( )F = A and that the values of F lie in the interval [A M A ./ , ] However, the inverse image of A , the set F⁻¹ A ⊆[ , ], can be of measure zero.0 T Therefore,

since we are looking at averages, it is sufficient, given a certain A>0, to require that the values of F lie in the interval [A M A ;/ , ] the additional requirement that amplitude( )F = A is unnecessary.

In addition,

to simplify calculations, assume from now on that T =1.

PROPOSITION 2. Let A> >B 0. Among the step functions F defined on [ , ]0 1 that have a given fixed average F and that have only two values, both in the interval [ , ]B A , the function

F t B t c

A c t

( ) ,

= ≤ ≤ ,

< ≤



 if if 0

1 (a38)

with

c A F A B

= −

− , (a39)

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is one whose variance attains the highest possible value, which is Variance( )F =(F −B A)( −F). (a40) PROOF. Consider the function depicted in Figure 4. Since its average is F , then

kc=l(1−c), (a41)

(k+l c) =l, (a42)

c l

k l

= + , (a43)

Variance( )F k l

k l l k

k l klk l k l kl

= + +

+ = +

+ =

2 2

. (a44)

This is a maximum when k and l have the largest permitted values. ■

A

B t

y

c 1

0 F _

k

l

Figure 4. A step function F with two values and a given F .

The following proposition states that no function with values in [ , ]B A and given mean F has a variance that is larger than the one given in Equation (a40).

PROPOSITION 3. Let A> >B 0. Any function G defined on [ , ]0 1 , with a given fixed average G =F and with values in the interval

[ , ]B A has a variance

Variance( )G ≤(F −B A)( −F). (a45)

PROOF. Let F be the step function described in Proposition 2. Let G be any function defined on [ , ]0 1 , with average G =F and with values in the interval [ , ]B A (Figure 5).

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A

B t

y

c 1

0 F _

F G

F

Figure 5. A step function F and a function G, with values in [ , ]B A and with F =G. Let D= −G F, so that

G= +F D. (a46)

Let I₁=[ , ) and I0 c ₂ =( , ]. Thenc1

D t A B t I

B A t I

( ) [ , ] ,

[ , ] .

∈ − ∈

− ∈



 0

0

1

2

if

if (a47)

Of course D =0 and

D t dt D t dt

I ( ) I ( )

2 1

∫

^{= −}

∫

^≤0^. ^(a48)

Finally, since G =F and Variance( )G = G² −G², we need only show that G² ≤ F². Using (a47) and (a48), we have

G² G dt² F DF D dt

0

1 2 2

0 1

=

∫

=

∫

⁽ +2 + ⁾ ^(a49)

= ^F² +²^B

∫

^I1^{D dt}+²^A

∫

^I2^{D dt}+

∫

⁰¹^{D dt}²

(a50)

= ^F² −² ^A−^B

∫

^I1^{D dt}+

∫

^I1^{D dt}² +

∫

^I2^{D dt}²

( ) (a51)

≤ ^F² −² ^A−^B

∫

^I1^{D dt}+ ^A−^B

∫

^I1^{D dt}+ ^A−^B

∫

^I2 ^{D dt}

( ) ( ) ( ) (a52)

= ^F² −² ^A−^B

∫

^I1^{D dt}+² ^A−^B

∫

^I1^{D dt}

( ) ( ) (a53)

= F² , (a54)

which proves the proposition. ■

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We can now prove the following proposition, which contains Statement E above.

PROPOSITION 4. Let A>0 and M>1. For functions F defined on [ , ]0 1 , with a given fixed average F and with values in the interval [ /A M A , the ratio F, ] ² F² has the following bounds:

1

2 2

2

≤ F ≤ + − 2

F

A B F

B

F . (a55)

The upper bound in (a55) is largest when

F A

= M + 2

1, (a56)

in which case

1 1

4

2 2

2

≤ F ≤ = +

F

M µ ( M)

(this µ is in ( ,1 M ).) (a57)

PROOF. For a given F , the ratio F² F² is maximum when the quantity

F² =Variance( )F + F² (a57,5)

is maximum. The maximum value of Variance( )F , for a given F , is given in Proposition 3, Equation (a45). Therefore the maximum value of F² F² for a given F is

R F F A M A F F

F

A A M F A M

( ) ( / )( ) ( / F) /

= − − + ² = + −

2

2 , (a58)

=AM+ −

M F

A M F

1 1 ² 1

2 . (a59)

Now

dR x

dx AM

M x

A M x ( )= − +1 1 +2 1

2 2

3. (a60)

Since x>0 here, the sign of this derivative is the same as the sign of

−AM+ +

M x A

M

1 2 ²

. (a61)

The above expression is a decreasing function of x. It is equal to zero when

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AM

M x A

M +1 = 2 ²

, (a62)

x A

= M + 2

1. (a63)

Therefore R x′( ) is positive when x is smaller than 2A/ (M+1) and it is negative when x is larger. For x>0, the function R x( ) has an absolute maximum at this point. This maximum is

R A

M AM

M M

A A M

M A 2

1

1 1

2

1 4

2 2

+ 2







 = + + − ( + )

(a64)

=(M+ ) −( + ) =( + ) M

M M

M M 1

2

1 4

2 2 2

. (a65)

Recall from Equation (a35) that this bound is smaller than M. ■ We can now prove a more precise version of Statement D.

PROPOSITION 5. Let M>1. Let F and G be positive integrable functions defined on the interval from 0 to T, with finite amplitudes. Assume that

F t( ) ~ amplitude( ) for all t_M F ∈[ , ]0 T , (a66) G t( ) ~ amplitude( ) for all t_M G ∈[ , ]0 T . (a67) Then, with µ =(M+ )

M 1 4

2

(this µ is in ( ,1 M ), we have)

FG~_µ F G whenever FG ≥ F G (a68) and, if 1<M≤5,

FG~_M F G whenever FG <F G. (a69)

PROOF. We have Variance( )F =0 if, and only if, F is almost everywhere constant.

With such an F we obviously have FG = F G. Assume then that F and G both have nonzero variances. With our hypotheses, we can define the covariance and the correlation coefficient of F and G just as we can define the covariance and the correlation coefficient of a finite family of ordered pairs ( , )x y_i _i

i

( )

n=1. Now, with nonzero variances, the correlation coefficient is

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r F G FG F G

F G F G

= Covariance( , ) = −

σ σ σ σ (a70)

(where σF = Variance( ) , F σG = Variance( ) ), so thatG

FG−F G =rσ σ_F _G =r F² −F² G² −G² , (a71) FG

F G r F

F

G

− =1 −1 G −1

2 2

2

2 , (a72)

FG

F G r F

F

G

= +1 −1 G −1

2 2

2

2 . (a73)

Since r∈ −[ 1 1 , we have, ]

1 1 1 1 1 1

2 2

2

− F − − ≤ ≤ + − 2 −

F

G G

FG F G

F F

G

G . (a74)

Using Proposition 4, Equation (a57), we obtain

1 1

4 1 1 1

4 1

2 2

− + −

 

 ≤ ≤ + + −

 



(M ) ( )

M

FG F G

M

M , (a75)

1 1

4 1 1

4

1 4

2 2 2

−(M− ) ≤ ≤ +( − ) = ( + ) M

FG F G

M M

M

M (a76)

(remember that M >1). The above right-hand bound is larger than 1 and smaller than M for all M>1 (Equation (a35)). The left-hand bound is smaller than 1. Under what conditions is this left-hand bound larger than 1 / M ? We want

1 1

4

2 1

−(M− ) ≥

M M, (a77)

4M−(M−1)² ≥4, (a78)

M² −6M+ ≤5 0, (a79)

(M−1)(M− ≤5) 0, (a80)

1<M≤5. (a81)

■

Here is an immediate corollary.

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PROPOSITION 6. Let m>1. Let f and g be integrable functions defined on the interval from 0 to T, with finite amplitudes. Assume that

f t( ) ~ amplitude( ) for all t_m f ∈[ , ]0 T , (a82) g t( ) ~ amplitude( ) for all t_m g ∈[ , ]0 T . (a83) Then, with µ =m +

m

2 1

2 (this µ is in ( , )1 m ), we have

norm(fg) ~ norm( ) norm( )_µ f g (a84) whenever the former is LARGER than the latter, and, if 1< ≤m 5,

norm(fg) ~ norm( ) norm( )_m f g (a85) whenever the former is smaller than the latter.

The upper bound of 5 for m is not large. In principle, in physics, any value of m larger than 1 could be useful. In the following sections, we shall be able to lift this restriction, but only with a lot of work.

10.4.5 Roadmap

Only the case in which FG < F G (see Proposition 5) is of concern now. In order to prove that Equation (a69) of Proposition 5 holds for all values of M larger than 1, we shall use a method similar to the one applied to obtain Proposition 4. We shall first find simple step functions F and G that have the lowest possible covariance FG− F G, for given F and G (Sections 10.4.6 to 10.4.8). Then we shall verify that no other pairs of permitted functions have a lower covariance, for the same given F and G (Section 10.4.9). Then we shall find the lowest possible value of the ratio FG

( )

F G ^{among all} permitted F and G (Section 10.4.10). Many pages will be required to accomplish this.

We must now work with two different functions at a time, F and G; as a consequence, our task will now become appreciably more complex. I was not able to find a shorter path to the desired result than the one described in Sections 10.4.6 to 10.4.10.

10.4.6 Two step functions with lowest possible covariance

Consider the step functions F and G illustrated in Figure 6 or 7 below. As before, because of the given mean values, we have

c l

k l

= + and d n m n

= + . (b1)

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If c≤d (Figure 6), then

Covariance( , )F G =(F −F G)( −G)= −kmc+lm d( − −c) ln(1−d) (b2)

= − + +

+ − +







− km l +

k l lm n m n

l

k l ln m

m n (b3)

= − + −

+ = −

+ + = −

kml k l

l m k l

lm

k l k l lm

2

( ) . (b4)

A

t y

c 1

0 F _

k

l

A′

Function F

t y

0 1 _

′

Function G B

B

d m

n G

Figure 6. Two step functions F and G with c≤d.

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A

t y

c 1

0 F _

k

l

A′

Function F

t y

0 1 _

′

Function G B

B

d m

n G

Figure 7. Two step functions F and G with c≥ d. If c≥d (Figure 7), then

Covariance( , )F G =(F −F G)( −G)= −kmd +kn c( −d)−ln(1−c) (b5)

= − + +

+ − +







− km n +

m n kn l k l

n

m n ln k

k l (b6)

= − + −

+ = −

+ + = −

kmn m n

kn m n

kn

m n m n kn

2

( ) . (b7)

Note that, if c=d, then k l/ =m n/ and kn=lm. Finally,

PROPOSITION 7. The two step functions F and G of Figures 6 or 7 have the following covariance:

Covariance( , ) ,

F G lm c d. kn c d

= − ≤

− ≥



 if

if (b8)

In other words, the covariance of the two functions of Figure 6 or 7 is equal to minus the product of the two deviations from the mean that have the same sign and that overlap.

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10.4.7 Maximum values of k, l, m, and n, for given values of F , G , c, and d Consider the same two step functions F and G as in Figures 6 or 7. We want two such functions that, for given F and G , have the smallest covariance possible. As a first step toward this goal, let us find the maximum values that k, l, m, and n can have, for given c and d.

We assume that the values of F are restricted to the interval [A A′, ] and those of G to the interval [B B , where ′, ] A , A, ′ B , and B are′ fixed positive constants, with that A′ <A and B′ <B .

Let us define a few useful symbols, illustrated in Figure 8:

′ = −

c 1 c, d′ = −1 d, (b10)

K= − ′F A , L= −A F, (b11)

M = −B G, N= − ′G B . (b12)

A

t y

c 1

0 F _

k

l

A′

Function F

K L

c′

t y

0 1 _

′

Function G B

B

d m

n G

M

N

d′

Figure 8. Our two step functions, F and G.

Now, if F is equal to A or A, this means that F is constant, except possibly on a set of′ measure zero. Similarly, if G equals B or B, then G is constant, except possibly on a set′ of measure zero. If F or G is almost everywhere constant, then obviously FG = F G. We

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need not consider such cases. Therefore

we also assume that K, L, M, and N are all greater than zero.

For given values of K and L, we observe that:

• if c c/ ′ ≤L K/ , then, for this c, the maximum k is K and the maximum l is Kc c/ ′ ≤L;

• if c c/ ′ =L K/ , then, for this c, the maximum k is K and the maximum l is L;

• if c c/ ′ ≥L K/ , then, for this c, the maximum k is Lc′/c≤K and the maximum l is L.

Therefore, for given values of K and L, we have the following:

for a given c, the maximum value of k is min{ ,K Lc′/ }c , (b13) for a given c, the maximum value of l is min{Kc c L/ ′, }. (b14) Similarly, for given values of M and N, we observe that:

• if d d/ ′ ≤N M/ , then, for this d, the maximum m is M and the maximum n is Md d/ ′ ≤N;

• if d d/ ′ =N M/ , then, for this d, the maximum m is M and the maximum n is N;

• if d d/ ′ ≥N M/ , then, for this d, the maximum m is Nd′/d≤M and the maximum n is N.

Therefore, for given values of M and N, we have the following:

for a given d, the maximum value of m is min{ ,M Nd′/ }d , (b15) for a given d, the maximum value of n is min{Md d N/ ′, }. (b16)

10.4.8 Minimum covariance of our two step functions, for given values of K, L, M, and N

Equations (b8), (b14), and (b15) imply that, for given K, L, M, N, c, and d, with c≤d, the smallest possible value of Covariance( , )F G is

Covariance( , )F G = −min{Kc c L/ ′, }min{ ,M Nd′/ }d . (b17) In this equation the value Kc c/ ′ is used if c c/ ′ ≤L K/ (and then Kc c/ ′ ≤L),

the value L is used if c c/ ′ ≥L K/ , the value M is used if d d/ ′ ≤N M/ ,

and the value Nd′/ is used if d dd / ′ ≥N M/ (and then Nd′/d ≤M).

What is the smallest possible covariance for given K, L, M, and N, and for given c and d such that c≤d?

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1. Suppose first that L K/ ≤N M/ (that is, LM ≤KN). The smallest covariance possible, with any c and d such that c≤d, is then −LM , obtained when

L K/ ≤c c/ ′ ≤d d/ ′ ≤N M/ . (b18)

2 (a). Suppose now that L K/ ≥N M/ (that is, LM ≥KN). Suppose first that

c c/ ′ ≤d d/ ′ ≤N M/ ≤L K/ . (b19)

The smallest covariance possible, with any c and d obeying (b19), is

− ⋅K (largest possible c c/ ′ under (b19) )⋅M = −K N M M( / ) = −KN . (b20) 2 (b). Still with L K/ ≥N M/ (that is, LM ≥KN), suppose now that

c≤d and N M/ ≤d d/ ′ ≤L K/ . (b21)

− ⋅K (largest possible c c/ ′ under (b21) )⋅Nd′/d = −K d d Nd( / ′) ′/d = −KN . (b22) 2 (c). Still with L K/ ≥N M/ (that is, LM ≥KN), suppose finally that

c≤d and N M/ ≤L K/ ≤d d/ ′. (b23)

−LN (largest possible ⋅ d′/ under (b23) )d = −LNK L/ = −KN . (b24) 3. In conclusion, if LM≤KN, the smallest covariance possible for any c and d such that c≤d is −LM . And, if LM ≥KN, the smallest covariance possible for any c and d such that c≤d is −KN . In short,

the smallest covariance possible

for any c and d such that c≤dis −min{LM KN ., } (b25)

❆ ❆ ❆

It turns out that we have the very same minimum covariance, under the condition that c≥d. Intuitively, it should be possible to prove this “by symmetry.” Unfortunately, such a proof seems about as intricate as the more pedestrian proof that follows.

Equations (b8), (b13), and (b16) imply that, for given K, L, M, N, c and d, with c≥d, the smallest possible covariance is

Covariance( , )F G = −min{ ,K Lc′/ }min{c Md d N/ ′, }. (b26) In this equation the value K is used if c c/ ′ ≤L K/ ,

the value Lc′/ is used if c cc / ′ ≥L K/ (and then Lc′/c≤K), the value Md d/ ′ is used if d d/ ′ ≤N M/ (and then Md d/ ′ ≤N),

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and the value M is used if d d/ ′ ≥N M/ .

What is the smallest possible covariance for given K, L, M, and N, and for given c and d such that c≥d?

1. Suppose first that L K/ ≥N M/ (that is, LM ≥KN). The smallest covariance possible, with any c and d such that c≥d, is then −KN , obtained when

N M/ ≤d d/ ′ ≤c c/ ′ ≤L K/ . (b27)

2 (a). Suppose now that L K/ ≤N M/ (that is, LM ≤KN). Suppose first that

L K/ ≤N M/ ≤d d/ ′ ≤c c/ ′. (b28)

− ⋅L (largest possible c′/ under (b28) )c ⋅ = −N L M N N( / ) = −LM . (b29) 2 (b). Still with L K/ ≤N M/ (that is, LM ≤KN), suppose now that

c≥d and L K/ ≤d d/ ′ ≤N M/ . (b30)

− ⋅L (largest possible c′/ under (b30) )c ⋅Md d/ ′ = −L d( ′/ )d Md d/ ′ = −LM . (b31) 2 (c). Still with L K/ ≤N M/ (that is, LM ≤KN), suppose finally that

c≥d and d d/ ′ ≤L K/ ≤N M/ . (b32)

−KM (largest possible d d⋅ / ′ under (b32) )= −KML K/ = −LM . (b33) 3. In conclusion, if LM≤KN, the smallest covariance possible for any c and d such that c≥d is −LM . And, if LM ≥ KN, the smallest covariance possible for any c and d such that c≥d is −KN . In short,

the smallest covariance possible

for any c and d such that c≤dis −min{LM KN ,, } (b34) which is the same as (b25).

❆ ❆ ❆

Finally,

PROPOSITION 8. For given positive K, L, M, and N, the smallest possible covariance, for any c and d, of the step functions F and G of Figure 8 (the values of F being restricted to the interval [A A′, ] and the values of G to the interval [B B ) is ′, ] −min{LM KN ., }

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Examples of such step functions are given in Figures 9 and 10.

A

t y

c 1

0 F _

A′

Function F

K L

c′

FunctionΦ

t y

0 1 _

′

Function G B

B

d G

M

N

d′

FunctionΓ

Figure 9. Two step functions F and G with L K/ ≤N M/ (i. e. LM ≤KN) and with minimum covariance −LM (Proposition 8). Here we have

L K/ =c c/ ′ =d d/ ′ ≤N M/ . The function Φ has values in [A A and the′, ] same mean as F. The function Γ has values in [B B and the same mean′, ] as G.

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t y

0 1 _

′

Function F A

A

c F

L

K

′ c FunctionΦ

B

t y

d 1

0 G _

B′

Function G

N M

d′

FunctionΓ

Figure 10. Two step functions F and G with L K/ ≥N M/ (i. e. LM ≥KN) and with minimum covariance −KN (Proposition 8). Here we have

N M/ ≤d d/ ′ =c c/ ′ =L K/ . The function Φ has values in [A A and the′, ] same mean as F. The function Γ has values in [B B and the same mean′, ] as G.

10.4.9 Minimum covariance of any two functions with values in [A A and′, ] [B B respectively, for given values of K, L, M, and N′, ]

Consider first the two functions F and G of Figure 9, for which L K/ ≤N M/ (i. e.

LM ≤KN). The covariance of F and G is −LM . Consider now two functions Φ and Γ, defined on [ , ]0 1 , with values in [A A and [′, ] B B respectively and with averages′, ]

Φ =F and Γ =G (b35)

(see Figure 9). We wish to prove that

Covariance( , )Φ Γ ≥Covariance( , )F G = −LM , (b36) or, equivalently, that

ΦΓ ≥ FG . (b37)

Let

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I₁=[ , ) and I0 c ₂ =( , ] (here c cc 1 / ′ =L K/ , i. e., c=L/ (L+K)), (b38)

Φ = +F D and Γ = +G E . (b39)

Then, on I₂, the value of G−G is −Md d/ ′ = −ML K/ . The values of D and E are as follows:

on I₁, D t( )∈[ ,0 K +L] and E t( )∈ − −[ M N, ]0 , (b40) on I₂, D t( )∈ − −[ K L, ]0 and E t LM

K N LM

K M

( )∈ − , +

 

. (b41)

We have

ΦΓ =

∫

⁰¹⁽^F +^{D G}⁾⁽ +^{E dt}⁾ =

∫

⁰¹⁽^FG+^FE+^DG+^{DE dt}⁾ ^(b42)

= ^FG +

∫

^I1^{FE dt}+

∫

^I2^{FE dt}+

∫

^I1^{DG dt}+

∫

^I2^{DG dt}+

∫

0^{DE dt} 1

(b43)

= + ′ + + + − −





 +

∫ ∫ ∫ ∫ ∫

FG A E dt A E dt B D dt B M LM

K D dt DE dt

I₁ I₂ I₁ I₂ 0

1

. (b44) Now, because of (b40), the first integral in the last line above is equal to a nonpositive number −p. The fact that E =0 then implies that the second integral in line (b44) is equal to the nonnegative number p. Similarly, because of (b40), the third integral in line (b44) is equal to a nonnegative number q; and the fact that D =0 implies that the fourth integral in line (b44) is equal to the nonpositive number −q . Then

ΦΓ = + − ′ + +





 +

∫

+

∫

FG A A p M LM

K q DE dt DE dt

I I

( )

1 2

(b45)

= + − ′ + +





 −

∫

− −

∫

−

FG A A p M LM

K q D E dt DE dt

I I

( ) ( )

1 2

. (b46)

Now, over I₁, the function −E is positive and D t( )∈[ ,0 A− ′A]. Therefore

D E dt A A E dt A A p

I (− ) ≤( − ′) I − =( − ′)

∫

1

∫

1

. (b47)

Similarly, over I₂, the function −D is positive and E t( )≤(LM K/ )+M. Therefore

− ≤ +





 − = +





∫

I₂ ^{DE dt} ^LM_K ^M

∫

I₂ ^{D dt} ^LM_K ^{M q} . (b48) Finally, Equations (b45)-(b48) imply that

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ΦΓ ≥ FG , (b49) which is what we wanted to prove.

❆ ❆ ❆

Consider now the two functions F and G of Figure 10, for which L K/ ≥N M/ (i. e. LM≥KN). Once again, we avoid intricate considerations of symmetry and take the pedestrian way. The covariance of F and G is −KN . Consider again two functions Φ and Γ, defined on [ , ]0 1 , with values in [A A and [′, ] B B respectively and with averages′, ]

Φ =F and Γ =G (b50)

(see Figure 10). We wish to prove that

Covariance( , )Φ Γ ≥Covariance( , )F G = −KN , (b51) or, equivalently, that

ΦΓ ≥ FG . (b52)

Let

I₁=[ , ) and I0 c ₂ =( , ] (again c cc 1 / ′ =L K/ , i. e., c=L/ (L+K)), (b53)

Φ = +F D and Γ = +G E . (b54)

Then, on I₁, the value of G−G is Nd′/d= NK L/ . The values of D and E are as follows:

on I₁, D t( )∈[ ,0 K +L] and E t N NK

L M NK

( )∈ − − , − L

 

, (b55)

on I₂, D t( )∈ − −[ K L, ]0 and E t( )∈[ ,0 M +N]. (b56) As before,

ΦΓ =

∫

⁰¹⁽^F +^{D G}⁾⁽ +^{E dt}⁾ =

∫

⁰¹⁽^FG+^FE+^DG+^{DE dt}⁾ ^(b57)

= ^FG +

∫

^I1^{FE dt}+

∫

^I2^{FE dt}+

∫

^I1^{DG dt}+

∫

^I2^{DG dt}+

∫

0^{DE dt} 1

(b58)

= + ′ + + ′ + +





 + ′ +

∫ ∫ ∫ ∫ ∫

FG A E dt A E dt B N NK

L D dt B D dt DE dt

I₁ I₂ I₁ I₂ 0

1

. (b59) Now, because of (b56), the second integral in the above equation is equal to a nonnegative number p. The fact that E =0 then implies that the first integral in line

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(b59) is equal to the nonpositive number −p. Similarly, because of (b55), the third integral in line (b59) is equal to a nonnegative number q; and the fact that D =0 implies that the fourth integral in line (b59) is equal to the nonpositive number −q . Then

ΦΓ = + − ′ + +





 +

∫

+

∫

FG A A p N NK

L q DE dt DE dt

I I

( )

1 2

(b60)

= + − ′ + +





 −

∫

− −

∫

−

FG A A p N NK

L q D E dt DE dt

I I

( ) ( )

1 2

. (b61)

Now, over I₁, the function D is positive and −E t( )≤ +N (NK L/ ). Therefore

D E dt N NK

L D dt N NK

L q

I (− ) ≤ + I





 = +





∫

1

∫

1 

. (b62)

Similarly, over I₂, we have −D t( )∈[ ,0 A− ′A ] and E is positive. Therefore

− ≤ − ′ = − ′

∫

I₂ ^{DE dt} (^A ^A)

∫

I₂^{E dt} (^A ^{A p}) . (b63) Finally, Equations (b60)-(b63) imply that

ΦΓ ≥ FG , (b64)

which is what we wanted to prove.

❆ ❆ ❆

We have proved the following proposition.

PROPOSITION 9. Let F and G be any two integrable functions defined on [ , ]0 1 , with values restricted to [A A and [′, ] B B′, ] respectively. Then, for given fixed values of the means F and G , the smallest possible covariance of any two such functions is

−min{(A−F B)( −G),(F − ′A )(G − ′B)}. (b65) Figures 9 and 10 show examples of such functions F and G that have this minimum covariance.

10.4.10 Minimum ratio FG F G

Let F and G be as in Figure 9 or 10. For given F and G , and under the usual conditions, these two functions have the smallest possible covariance, which is given in

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Equation (b65).

We now choose any number

γ >1. (b66)

This γ is the M of Sections 10.4.3 and 10.4.4. Since Section 10.4.7, we have been using M with a different meaning: M= −B G (Figures 9 and 10).

Now let

′ =

A A /γ and B′ =B /γ (where of course A B, >0 ). (b67) We shall examine the ratio

FG F G

F G F G

= Covariance( , )F G +

(b68)

= F G− A−F B−G F − ′A G − ′B F G

min{( )( ),( )( )}

(b69)

= −  −





 −





  − ′





 − ′













1 min A 1 1 , 1 1 

F

B G

A F

B

G . (b70)

Since the covariance is negative, this ratio is smaller than 1. We want to show that it is also greater than 1 /γ.

Now let

x= A F/ and y=B G/ . (b71)

Of course

x y, ∈[ , ]1γ . (b72)

We have FG

F G x y x y

= − − −  −





 −











 1 min ( 1)( 1), 1 1 

γ γ (b73)

= −¹ ^{min (}

{

^x−¹⁾⁽^y−¹^),(^γ −^x⁾⁽^γ −^y⁾ ^γ²

}

^. ^(b74)

We must find the maximum value of the function

h x y( , )=min (

{

x−¹)(y−¹),(^γ −x)(^γ −y) ^γ²

}

^(b75)

under the constraint (b72).

We shall find this maximum by examining the contour lines of h inside the square [ , ] [ , ]1γ × 1γ . First consider the curve whose equation is