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The Cauchy-Schwarz Master Class   J. Michael Steele   ★★★★★
This file is personal home work. No one
proofread. Cannot promise correctness.
If you suspect any view point wrong,
please ask a math expert near by.
Freeman 2009-06-19-10-46

Please use MSIE browser to read this file.
Did not test other browser. This file is
written under MSIE 6.0




<a name="docA001"> Index begin Index this file
2009-06-08-19-10 start
This file http://freeman2.com/tute0008.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view 
point wrong, please ask a math expert near by.

<a name="docA002">,<a name="textbook">
This file is a note for reading inequality
book written by Professor J. Michael Steele
The Cauchy-Schwarz Master Class   ★★★★★
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">
Freeman study mechanical engineering. 
Engineering mathematics do not teach 
inequality. Above book is first inequality
book. First time read, it was very hard.
Although high school time learned
  a*a + b*b >= 2*a*b
But this little knowledge do not help.

<a name="docA004">
This file follow textbook chapter section
order, but not continuous, Freeman skip 
those uncertain sections/problems.

This file first function is to learn 
inequality. Second function is to learn
how to use html code to write math 
equations.

<a name="docA005"> Index begin Index this file
On 2009-01-27-10-08 Freeman accessed
the next page
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html
save as sftw.umac.mo_text_math_eqn_good.htm
Above page is the main reference for html
math equation.

In order to let reader to build  html math
equation, previous file tute0007.htm page 
end has math symbol and internal code.

This file third function is to display how
to draw curves in web page. The main engine
is XYGraph v2.3 - Technical Figures

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">
2009-08-23-15-00 start
□ Exercise 1.14 solution

Exercise 1.14 problem (a) is solved by hint
Exercise 1.14 problem (b) is out of LiuHH's
reach. What is "cardinality of a set B⊂Z3" ?
Without clear understand the definition, 
LiuHH is unable to solve problem (b)
LiuHH major in mechanical engineering.
LiuHH is mathematics admirer and outsider.

<a name="docA007">
To solve problems in
The Cauchy-Schwarz Master Class
LiuHH's goal is to solve 50% of the exercise
problems. In reality, solve 40% is good enough.
Please do not be surprise that LiuHH skip some
text section, skip some problems.
2009-08-23-15-12 stop



<a name="ch07c001"> Index begin Index this file
2010-02-13-09-59 start
■ Exercise 7.9 problem statement
  textbook page 118
(Littlewood's Middle Derivative Squeeze)

<a name="ch07c002">
Show that if f:[0,∞)→Real is twice
differentiable and if |f''(x)| is
bounded then
  lim[x→∞]f(x)=0 implies
  lim[x→∞]f'(x)=0 ---eqn.AX001

In his Miscellany, J.E. Littlewood
suggests this "pictorial arguments,
while not so purely conventional
can be quite legitimate" The result
of this exercise is his leading
example, and the picture he offered
is essentially that of Figure 7.2.
2010-02-13-10-09 stop
<a name="exGraph0709">  Index begin Index this file
Figure 7.2 ; 2010-02-13-10-12







<a name="ch07c003">
2010-02-13-15-18 start
■ Exercise 7.9 hint
  textbook page 255

In order to argue by contradiction,
we assume without loss of generality 
that there is a sequence xn→∞ such
that 
  f'(xn)≧ε>0 ---eqn.AX002
//eqn.AX002 is wrong on purpose
Now by Littlewood's figure 7.2 (or by
triangle lower bound of Exercise 7.7)
we note that
<a name="ch07c004">
 
f(xn+δ)-f(xn) =
xn
t=xn
f ' (t)dt
1

2
ε2

B
---page 255
---Line 22
---eqn.AX003
width of above equation
<a name="ch07c005">
2010-02-13-15-36 here
where
  B=sup|f''(x)|<∞ ---eqn.AX004
and
  δ=ε/B ---eqn.AX005
This bound implies that 
  f(x)≠o(1) ---eqn.AX006
so we have our desired 
contradiction.
2010-02-13-15-40 stop




<a name="ch07c006"> Index begin Index this file
2010-02-13-15-42 start
■ Exercise 7.9 solution


First clarify the dimension of
δ, ε, B in  δ=ε/B ---eqn.AX005

Because integration upper bound
is xn+δ, then
  dim(δ)=dim(x) ---eqn.AX007
x is f(x) independent variable.

<a name="ch07c007">
Because they can compare each other
  f'(xn)≧ε>0 ---eqn.AX002
then
  dim(ε)=dim(f'(x))=dim(f/x) ---eqn.AX008
Because the equality in
  B=sup|f''(x)|<∞ ---eqn.AX004
then
  dim(B)=dim(f''(x))=dim(f/x/x) ---eqn.AX009

<a name="ch07c008">
Exam the relation involve
δ, ε, B in  δ=ε/B ---eqn.AX005
  dim(δ) ?=? dim(ε)/dim(B) ---eqn.AX010
that is
  dim(x) ?=? dim(f/x)/dim(f/x/x)
that is
  dim(x) ?=? dim(x)
<a name="ch07c009">
Yes! Confirmed consistency in 
dimension analysis,
δ is same as x
ε is same as f/x
B is same as f/x/x
2010-02-13-15-53 here

<a name="ch07c010">
2010-02-13-16-22 start
If reader review Graph0707
That figure may give better
explanation than Graph0709

Exercise 7.9 problem is that
given  f:[0,∞)→Real is twice
differentiable and 
given  |f''(x)| is bounded 
given  lim[x→∞]f(x)=0 
imply  lim[x→∞]f'(x)=0

<a name="ch07c011"> Index begin Index this file
With our direct sense, we feel
given lim[x→∞]f(x)=0 
then  lim[x→∞]f'(x)=0
then  lim[x→∞]f''(x)=0
otherwise it is hard to imagine 
that a f(x)=0_at_x=infinity line
has up/down tendency like wave 
at infinity. Express direct sense
in mathematics, it is
<a name="ch07c012">
 
lim
x→∞
f(xn+δ)-f(xn) = 0 =
xn
t=xn
f ' (t)dt
---page 255
---Line 22 Aux
---eqn.AX011
width of above equation
<a name="ch07c013">
2010-02-13-16-46 here
Because it is hard to argue that
eqn.AX011 integration part = 0
imply f'(x=infinity)=0, textbook 
use method of contradiction.
Assume there is a sequence xn→∞ 
such that
  f'(xn)≧ε>0 ---eqn.AX002
See what ill result we get.

<a name="ch07c014">
We can apply Exercise 7.7 result 
to eqn.AX003 inequality right
half.
Exercise 7.7 say
domain left end function   value
domain left end d[f(t)]/dt value
(two boundary information only)
determine a lower bound for the
whole domain of integration.
It is true even if x approach
to infinity.
<a name="ch07c015">
Here Exercise 7.9, we 
replace f(x)  with f'(x)
replace f'(x) with f''(x)
Here Exercise 7.9, it is
domain RIGHT end d[f(t)]/dt value
Maximum dd[f(t)]/dt/dt in domain
(NOT two boundary information)
determine a lower bound for the
whole domain of f'(x) integration.
(eqn.AX003 use f'(x) integration)
<a name="ch07c016"> Index begin Index this file
Triangle area value here in 
Exercise 7.9 is (1/2)*ε*ε/B
Exercise 7.9 hint made a
contradiction assumption
if x→∞, f'(x→∞)=ε>0
and //above red wrong on purpose
if x→∞, f''(x→∞)=sup|f''(x)|<∞
then ε and B information
created lower bound triangle 
(1/2)*ε*ε/B > 0
Non-zero ε in eqn.AX003 say 
f(xn+δ)-f(xn) should be > 0
<a name="ch07c017">
BUT, when xn approach to infinity
f(xn+δ) and f(xn) should both 
approach to zero and
f(xn+δ)-f(xn) should be = 0 
The assumption
  limit[f'(xn)] ≧ ε>0 ---eqn.AX002
must be wrong.

<a name="ch07c018">
Contradiction result tell us 
the other side story is true, 
that is 
given lim[x→∞]f(x)=0 
then  lim[x→∞]f'(x)=0

Proof is done by method of
contradiction.
2010-02-13-17-21 stop

<a name="ch07c019">
2010-02-13-17-27 start
Exercise 7.7 require that curve 
be convex for positive function
value.
For a curve function value be 
positive and approach to zero
when x approach to infinity, 
This curve must be convex ╰╯

<a name="ch07c020">
For a curve function value be 
negative and approach to zero
when x approach to infinity, 
This curve must be concave ╭╮
We can apply Exercise 7.7 result
by turn positive triangle to 
negative triangle.

<a name="ch07c021">
For a curve function value be 
wave-like and approach to zero
when x approach to infinity, 
This curve is not concave and
is not convex. But it is 
enveloped by a positive convex
and a negative concave and
squeeze f'(x=infinity) to zero
for example
 f(x)=exp(-x)*sin(x) ---eqn.AX012
f'(x)=exp(-x)*(cos(x)-sin(x)) ---eqn.AX013
2010-02-13-17-47 stop


<a name="ch07c022"> Index begin Index this file 2010-02-13-19-16 start ■ Exercise 7.10 problem statement   textbook page 118 (Monotonicity and Integral Estimates) Although the point was not stressed in this chapter, many of the most useful day-to-day estimates of integrals are found with help from monotonicity. Gain some practical experience by proving that
<a name="ch07c023">
for all 0<x<1
 
t=1
t=x
log(1+t)
dt

t
(2*log(2))
1-x

1+x
---page 118
---Line 13
---eqn.AX014
width of above equation
<a name="ch07c024">
2010-02-13-19-28 here
and by showing that 2*log(2) 
cannot be replaced by a smaller 
constant.
Incidentally, this particular 
inequality is one we will see 
again when it helps us with 
Exercise 11.6.
2010-02-13-19-30 stop





<a name="ch07c025">
2010-02-13-19-31 start
■ Exercise 7.10 hint
  textbook page 255/256

Differentiation suffices to confirm
that on (0,1) the map
  t→t-1log(1+t) ---eqn.AX015
is decreasing and 
  t→(1+t-1)log(1+t) ---eqn.AX016
is increasing so we have the bounds
<a name="ch07c026">
 
t=1
t=x
log(1+t)
dt

t
(1-x)*x-1*log(1+x)
---page 256
---Line 2
---eqn.AX017
width of above equation
<a name="ch07c027">  Index begin Index this file
 
1-x

1+x
(1+x-1)*log(1+x) 2log(2)
1-x

1+x
---page 256
---Line 3
---eqn.AX018
width of above equation
<a name="ch07c028">
2010-02-13-19-48 here
To show 2log(2) cannot be replaced 
by smaller constant, note that
 
lim
x→1
1

1-x
t=1
t=x
log(1+t)
dt

t
= log(2)
---page 256
---Line 5
---eqn.AX019
width of above equation
<a name="ch07c029">
since 
  |log(2)-log(1+t)/t|≦ε  ---eqn.AX020
for all x with
  |1-x|≦δ(ε) ---eqn.AX021
2010-02-13-19-56 stop

<a name="exGraph0710">  Index begin Index this file
Exercise 7.10 ; 2010-02-13-20-22






<a name="ch07c030">
2010-02-13-21-02 start
■ Exercise 7.10 solution


eqn.AX017 inequality is because
change integrand log(1+t)/t to
its maximum value. 

<a name="ch07c031">
Integration domain is t in [x,1] 
and require "for all 0<x<1"
From exGraph0710 we see red
curve is monotone decrease. The
maximum integrand value occur
at left end where t=x. eqn.AX017
less than side change integrand 
[log(1+t)]/t to maximum value and
<a name="ch07c032"> Index begin Index this file
move out of integration, what left
is integral of t from t=x to t=1.
This simple integration has value
1-x. (1-x) reunion with maximized
body [log(1+x)]/x become eqn.AX017 
greater than side
  (1-x)*[log(1+x)]/x ---eqn.AX022

<a name="ch07c033">
From eqn.AX017 greater than side
to eqn.AX018 less than side is a
equality. That is because
(1-x)*[log(1+x)]/x multiplied by 
one. This one is (1+x)/(1+x)
Numerator (1+x) absorbed 1/x and
become (1+1/x). Result is 
[(1-x)/(1+x)] * (1+1/x)*log(1+x) ---eqn.AX023

<a name="ch07c034">
In eqn.AX023, do not bother 
[(1-x)/(1+x)], 
because [(1-x)/(1+x)] is part 
of output. Just look at
  (1+1/x)*log(1+x) ---eqn.AX024
eqn.AX024 is g(x) in exGraph0710 
(1+1/x)*log(1+x) is a monotone
increase function. In eqn.AX017 
and eqn.AX018 chain, g(x) is at
greater than side. To keep same
inequality, we replace g(x) with
its maximum value in domain 
<a name="ch07c035">
which is domain right end t=1
In
  (1+1/x)*log(1+x) ---eqn.AX024
replace x by 1 get constant
 2*log(2).
Because 2*log(2) come from g(x)
evaluated at x=1. We can not 
replace 2*log(2) by smaller value,
<a name="ch07c036">
If we do, then at the high end 
(t=1) greater than condition is
not true any more. Violate exercise
given condition.
2010-02-13-21-28 stop




<a name="ch07c037"> Index begin Index this file 2010-02-14-09-30 start ■ Exercise 7.11 problem statement   textbook page 118 (A Continuous Carleman-Type Inequality) Given an integrable f:[a,b]→[0,∞) and an integrable weight function w:[a,b]→[0,∞) with integral 1 on [a,b], show that one has
<a name="ch07c038">
 
exp
x=b
x=a
{log[f(x)]}w(x)dx ≦ e
x=b
x=a
f(x)w(x)dx
---page 118
---Line 20
---eqn.7.27
width of above equation
2010-02-14-09-41 stop





<a name="ch07c039">
2010-02-14-09-42 start
■ Exercise 7.11 hint
  textbook page 257

If W(x) is the integral of w on 
[a,x], then
  W(a)=0 ---eqn.AX025
  W(b)=1 ---eqn.AX026
and
  W'(x)=w(x) ---eqn.AX027
so we have
<a name="ch07c040">
 
x=b
x=a
{log[W(x)]}w(x)dx
x=1
x=0
{log[v]}dv = -1
---page 257 ---Line 3 ---eqn.AX028
width of above equation
<a name="ch07c041">
We then have the relation
 
exp
x=b
x=a
{log[f(x)]}w(x)dx = e * exp
x=b
x=a
{log[f(x)W(x)]}w(x)dx
---page 257 ---Line 5 ---eqn.AX029
width of above equation
<a name="ch07c042">  Index begin Index this file
 
≦ e * exp
x=b
x=a
{log[f(x)]}w(x)dx ≦ e *
x=b
x=a
f(x)w(x)dx
---page 257 ---Line 6 ---eqn.AX030
width of above equation
<a name="ch07c043">
2010-02-14-10-02 here
where we used first the fact that
0≦W(x)≦1 for all x in [a,b] and
then we applied Jensen's inequality.
2010-02-14-10-04 stop




<a name="ch07c044">
2010-02-14-11-01 start
■ Exercise 7.11 solution


Problem given
[[
an integrable weight function
w:[a,b]→[0,∞) with integral 1 on
[a,b]
]]
that is given 
  ∫[x=a,b]w(x)dx=1 ---eqn.AX031 

<a name="ch07c045">
Exercise 7.11 hint suggest to define
  W(x)=∫[t=a,x]w(t)dt ---eqn.AX032
where a≦x≦b.
From eqn.AX032, we see
  W(a)=∫[t=a,a]w(t)dt=0 ---eqn.AX033
(The area from t=a to t=a is zero)
and
  W(b)=∫[t=a,b]w(t)dt=1 ---eqn.AX034
eqn.AX034 come from the given 
relation eqn.AX031 

<a name="ch07c046">
To see the relation between 
uppercase W(x) and lowercase w(x)
we need refer to Leibniz's rule
<a name="ch07c047">  Index begin Index this file
 
d

dx
B(x)
A(x)
f(x,t)dt
B(x)
A(x)
∂f(x,t)

∂x
dt + f(x,B)
dB

dx
- f(x,A)
dA

dx
Leibniz's rule ---eqn.AX035
width of above equation
<a name="ch07c048">
Refer to eqn.AX032, find

W'(x)=
 
d

dx
B(x)=x
t=a
w(t)dt
x
a
∂w(t)

∂x
dt + w(x)
d(x)

dx
- w(a)
d(a)

dx
Apply Leibniz's rule to W(x) ---eqn.AX036
Blue terms are zero, red terms are one.
width of above equation
<a name="ch07c049">
2010-02-14-11-43 here
In eqn.AX036 
  ∂f(x,t)/∂x = ∂w(t)/∂x = 0 ---eqn.AX037
  d(x)/dx = 1 ---eqn.AX038
  d(a)/dx = 0 ---eqn.AX039
eqn.AX037=0 because variable t 
and variable x are independent
eqn.AX038=1 because identical
terms in deno. and numerator
eqn.AX039=0 because 'a' is 
constant.
<a name="ch07c050">
Then eqn.AX036 reduce to
  W'(x)=w(x) ---eqn.AX040 (eqn.AX027)
which is
  d[W(x)]/dx=w(x)
or
  d[W(x)]=w(x)*dx ---eqn.AX041

We need one more preparation
equation, that is eqn.AX028
<a name="ch07c051">
 
x=b
x=a
{log[W(x)]}w(x)dx
W(x)=1
W(x)=0
{log[W(x)]}dW(x)
---page 257 ---Line 3 aux ---eqn.AX042
width of above equation
<a name="ch07c052"> Index begin Index this file
2010-02-14-12-00 here
In eqn.AX042, red term come from 
eqn.AX041.
In eqn.AX042, integration lower
limit change from x=a to W(x)=0
come from eqn.AX033.
In eqn.AX042, integration upper
limit change from x=b to W(x)=1
come from eqn.AX034.

<a name="ch07c053">
In eqn.AX042 right side, change
variable from 'W(x)' to 'v'
then we have Exercise 7.11 hint
eqn.AX028 right end 
  ∫[v=0,1]{log[v]}dv = -1 ---eqn.AX043
How to prove eqn.AX043?
<a name="ch07c054">
Take an integration table, find
  ∫{log(x)}dx = x*log(x) - x ---eqn.AX044
eqn.AX043 has lower and upper 
bound, then eqn.AX044 give us
  ∫[x=0,1]{log[x]}dx =
 = [x*log(x) - x][x=1]
  -[x*log(x) - x][x=0]
 = [1*log(1) - 1]-[0*log(0) - 0]
 = -1 ---eqn.AX045
This is how we get -1 in eqn.AX043

<a name="ch07c055">
Above are all preparation work..
Next begin exercise equation.
Start from eqn.7.27 left side
  exp{∫[x=a,b]log[f(x)]*w(x)*dx} ---eqn.AX046
 =exp{∫[x=a,b]log[f(x)]*w(x)*dx +1-1}
Here insert a zero=+1-1
+1 become exp(1) which is 'e'
-1 become ∫[x=0,1]{log[x]}dx
it is preparation work of eqn.AX043
-1 become ∫[x=a,b]{log[W(x)]}w(x)dx 
it is preparation work of eqn.AX028

<a name="ch07c056">
Continue //blue e from exp(+1)
  exp{∫[x=a,b]log[f(x)]*w(x)*dx} ---eqn.AX047
 =e*exp{∫[x=a,b]log[f(x)]*w(x)*dx
       +∫[x=a,b]{log[W(x)]}w(x)dx
       } //red line is -1

<a name="ch07c057"> Index begin Index this file
Now use log(a)+log(b)=log(a*b) ---eqn.AX048
eqn.AX047 become eqn.AX029 
right side integral
  exp{∫[x=a,b]log[f(x)]*w(x)*dx} ---eqn.AX049
 =e*exp{∫[x=a,b]log[f(x)*W(x)]*w(x)*dx
We know
  W(a)=0 ---eqn.AX025
  W(b)=1 ---eqn.AX026
<a name="ch07c058">
W(x) is in the range 0≦W(x)≦1
In eqn.AX049, if drop W(x), create
an inequality which is greater
than before-drop.
(5*0.6 = 3, drop 0.6 get 5 > 3)
Inequality from eqn.AX029 right
term to eqn.AX030 left term is
the result of drop W(x).
<a name="ch07c059">
Finally come to eqn.AX030 right
inequality.
2010-02-14-12-37 here
2010-02-14-12-48 start
We need apply Jensen Inequality 
for convex function.
Jensen Inequality is next
<a name="ch07c060">
compare with
  e*exp{∫[log[f(x)]*w(x)dx}≦
  e*∫exp{log[f(x)]}*w(x)dx  ---eqn.AX050
g(x) in eqn.6.2 is exp() in
eqn.AX050. 
Function exp(x) is convex. We
can apply Jensen Inequality.
<a name="ch07c061">
The greater than side of eqn.AX050
has exp{log(y)) which is y,
because exp() and log() cancel
each other. Then greater than side 
of eqn.AX050 become
  e*∫f(x)*w(x)dx  ---eqn.AX051
This is eqn.7.27 greater than side
term. Problem solved.
2010-02-14-13-08 stop

<a name="ch07c062">
To solve Exercise 7.11, Liu,Hsinhan
rely upon previous reading notes
at book margin on 2009-02-14 and
2009-06-07. Otherwise, today's
work should spend longer time.
2010-02-14-13-11 stop


<a name="ch07c063"> Index begin Index this file 2010-02-14-15-25 start ■ Exercise 7.12 problem statement   textbook page 119 (Gruss's Inequality--Integral of Products) <a name="ch07c064"> Suppose that -∞<α≦A<∞ and -∞<β≦B<∞ and suppose that function f and g satisfy the bounds α≦f(x)≦A ---eqn.AX052 and β≦g(x)≦B ---eqn.AX053 for all 0≦x≦1 Show that one has the bound
<a name="ch07c065">
 
|
x=1
x=0
f(x)g(x)dx
x=1
x=0
f(x)dx
x=1
x=0
g(x)dx |
≦ (A-α)(B-β)/4 //This is part of eqn.AX054
---page 119 ---Line 6 ---eqn.AX054
width of above equation
and show by example that the 
factor of 1/4 cannot be replaced
by a smaller number.
2010-02-14-15-42 stop





<a name="ch07c066">
2010-02-14-15-47 start
■ Exercise 7.12 hint
  textbook page 257

Setting If to the integral of f 
we have
<a name="ch07c067">
 
x=1
x=0
(f(x)-If)2dx (A-If)(If-α)
x=1
x=0
(A-f(x))(f(x)-α)dx
≦(A-If)(If-α) //This is part of eqn.AX055
---page 257 ---Line 10 ---eqn.AX055
width of above equation
<a name="ch07c068">  Index begin Index this file
and an analogous inequality holds for g.
Schwarz's inequality then gives
|
x=1
x=0
f(x)g(x)dx -If*Ig |
2
 
 
|
x=1
x=0
(f(x)If)(g(x)-Ig)dx |
2
 
 
---page 257 ---Line 13 ---eqn.AX056 continue next line
width of above equation
<a name="ch07c069">
 
x=1
x=0
(f(x)-If)2dx
x=1
x=0
(g(x)-Ig)2dx
---page 257 ---Line 14 ---eqn.AX057 continue next line
width of above equation
≦ (A-If)(If-α)(B-Ig)(Ig-β) ---eqn.AX058
≦ [(A-α)2/4]*[(B-β)2/4] ---eqn.AX059
<a name="ch07c070">
Where in the last step we used
the fact that 
  (U-x)(x-L)≦(U-L)2/4 ---eqn.AX060
for all L≦x≦U
Finally to see that Gruss's inequality
is sharp, 
set f(x)=1 for 0≦x≦1/2 ---eqn.AX061
set f(x)=0 for 1/2<x≦1 ---eqn.AX062
and
set g(x)=1-f(x) for all 0≦x≦1. ---eqn.AX063
2010-02-14-17-31 stop



<a name="ch07c071"> Index begin Index this file
2010-02-14-17-51 start
■ Exercise 7.12 solution


LiuHH do not know how to show
equality in eqn.AX055 is true.
LiuHH skip Exercise 7.12 solution

<a name="ch07c072">
If view from dimension analysis
eqn.AX055 left end integrand is
special. Since 
[[
Setting If to the integral of f 
]]
<a name="ch07c073">
If has dimension of
f*dx which is f*x (dimension)
eqn.AX055 left end integrand is
(f(x) - If) that is
(f - f*x) it can be true only
if independent variable x is
pure number. Even if this is
true, LiuHH still can not find
why equality in eqn.AX055 is true.
2010-02-14-17-58 stop



========= Chapter seven end here =========



<a name="ch08a001"> Index begin Index this file 2010-02-15-14-16 start ■■Chapter 08: The Ladder of   Power Means Please read textbook, textbook has better explanation and more material. This file is just study notes. May contain error ans has less material. <a name="ch08a002"> Index begin Index this file ■ Problem 8.1 (The Geometric   Mean as a Limit) For nonnegative real numbers xk k=1,2,...,n and nonnegative weights pk k=1,2,...,n with total mass p1+p2+...+pn=1 ---eqn.AX064 one has the limit
<a name="ch08a003">
Attention! eqn.8.2 left summation become
right side multiplication. Coefficient pk
become power pk, and t disappear.
lim
t→0
{
k=n
k=1
pkxkt }
1/t
 
 
k=n
k=1
xkpk
define
M0
---page 120 ---Line 18 ---eqn.8.2
width of above equation
Alert! eqn.8.2 is valid for t→0
If t>0, ∑ side > ∏ side, eqn.8.6
If t<0, ∑ side < ∏ side, eqn.8.12
<a name="ch08a004">
2010-02-15-14-38 here

If two functions f(t) and g(t) 
ratio f(t)/g(t) has a limit value
P as variable t approach to zero.
If limit value P is zero
<a name="ch08a005">
lim
t→0
f(t)

g(t)
0
---page 121
---Line 1
---eqn.AX065
width of above equation
<a name="ch08a006">
We use lowercase symbol o()
  f(t)=o(g(t)) as t→0 ---eqn.AX066
to represent this zero limit 
case. 
Similarly, if limit value P is
finite (or bounded)
<a name="ch08a007">  Index begin Index this file
lim
t→0
f(t)

g(t)
P (finite)
---page 121
---Line 3
---eqn.AX067
width of above equation
<a name="ch08a008">
We use uppercase symbol O()
  f(t)=O(g(t)) as t→0 ---eqn.AX068
to represent this finite limit
case. 
This little o() and big O() method
is called Landau's Notation.

<a name="ch08a009">
Big O help us at the following
solution. Big O relate to two 
functions.

First function, exam 
  h(x)=log(1+x) ---eqn.AX069
From integration table, find
  log(1+x)=x -x^2/2 +x^3/3
          -x^4/4 +-+- ...  ---eqn.AX070
require x^2<1
(Peirce table page 91, item 768)
<a name="ch08a010">
Our focus is x approach to zero,
requirement x^2<1 match our goal.

In eqn.AX070, when x approach to
zero, x^1 become dominant, higher
power term (x^2 or higher) are
smaller than x^1. We write
  h(x)=log(1+x)=x + f(x)  ---eqn.AX071
  f(x)=-x^2/2 +x^3/3-x^4/4 +-+- ...   ---eqn.AX072
  g(x)= x^2  ---eqn.AX073
<a name="ch08a011">
Because
  lim[x→0]{f(x)/g(x)}=-1/2 ---eqn.AX074
it is finite (qualified for o,O)
and nonzero (qualified for O)
we write
  h(x)=x + O(x^2)  as x→0 ---eqn.AX075
or write
  log(1+x)=x + O(x^2) as x→0 ---eqn.8.3

<a name="ch08a012"> Index begin Index this file
Second function, exam 
  h(x)=exp(x) ---eqn.AX076
Peirce table page 89, item 759
give us
  exp(x)=1 +x +x^2/2! +x^3/3!
         +x^4/4! +... ---eqn.AX077
<a name="ch08a013">
Again,
Our focus is x approach to zero,
In eqn.AX077, when x approach to
zero, x^1 become dominant, higher
power term (x^2 or higher) are
smaller than x^1. 
We write
  h(x)=exp(x)=1 +x + f(x)  ---eqn.AX078
  f(x)=+x^2/2! +x^3/3!+x^4/4! ...   ---eqn.AX079
  g(x)= x^2  ---eqn.AX080
<a name="ch08a014">
Because
  lim[x→0]{f(x)/g(x)}=+1/2! ---eqn.AX081
it is finite (qualified for o,O)
and nonzero (qualified for O)
we write
  h(x)=1+x + O(x^2) as x→0 ---eqn.AX082
or write
  exp(x)=1+x+O(x^2) as x→0 ---eqn.8.4

<a name="ch08a015">
We will use eqn.8.3 (log())
and eqn.8.4 (exp()) in the
following discussion.
2010-02-15-15-36 here

<a name="ch08a016">
Now back to our target eqn.8.2
To take down the power (1/t)
in eqn.8.2, we can take log
for the whole eqn.8.2.  To 
simplify the following math
equation, we omit "lim t→0"
"lim t→0" is not present in 
the following equation, but
"lim t→0" is still working!

<a name="ch08a017"> Index begin Index this file
Take log and hide "lim t→0"
eqn.8.2 become the following
log {(
k=n
k=1
pkxkt )
1/t
 
 
}
1

t
log {
k=n
k=1
pket*log(xk) }
---page 121 ---Line 20 ---eqn.AX083
width of above equation
<a name="ch08a018">
We used twice
  log(a^b)=b*log(a) ---eqn.AX084
We used once
  exp(log(c))=c ---eqn.AX085

eqn.AX083 left side whole term
apply eqn.AX084 once and move
power (1/t) to right side leading
term.

<a name="ch08a019">
eqn.AX083 right side treat xkt
as 'c' in eqn.AX085. Push xkt
to power above 'e'. Then second
time apply eqn.AX084 to xkt and
move 't' to become 't*log(xk)'
All operation result equality.

<a name="ch08a020">
eqn.AX083 right side exp() is
exp(t*log(xk)) and big O() is
  exp(x)=1+x+O(x^2) as x→0 ---eqn.8.4
we have "t*log(xk)" as x
then 1+x become 1+t*log(xk)
eqn.8.4 has x approach to zero,
"t*log(xk)" has t approach to zero.

<a name="ch08a021">
Now use eqn.8.4 to simplify
exp(). This simplification is
valid only if t approach to zero.
eqn.AX083
right side
1

t
log {
k=n
k=1
pk ( 1 + t*log(xk) +O(t2) )}
red term applied eqn.8.4
---page 121 ---Line 21 ---eqn.AX086
width of above equation
<a name="ch08a022"> Index begin Index this file
2010-02-15-16-26 stop
2010-02-15-17-07 start
In eqn.AX086 change {term} as 
following
  ∑[k=1,n]pk[1 + t*log(xk) +O(t2)]
 =∑[k=1,n]pk*1 //blue term = 1
 +∑[k=1,n]{pk*t*log(xk)} ---eqn.AX087
 +∑[k=1,n]pk*O(t2) //blue term = 1
<a name="ch08a023">
We have
  p1+p2+...+pn=1 ---eqn.AX064
so two blue terms in eqn.AX087
are one. eqn.AX087 become
  1+∑[k=1,n]{pk*t*log(xk)}+O(t2) ---eqn.AX088
t is independent from k summation
move t out of k summation, get
  1+t*∑[k=1,n]{pk*log(xk)}+O(t2) ---eqn.AX089
<a name="ch08a024">
Put eqn.AX089 back to {term} in
eqn.AX086 
eqn.AX086+
eqn.AX089
1

t
log { 1 + t*
k=n
k=1
pk*log(xk) +O(t2) }
---page 121 ---Line 22 ---eqn.AX090
width of above equation
<a name="ch08a025">
In eqn.AX090, we have
log(1 + something), eqn.8.3 is
called here. Log{} in eqn.AX090 
become
eqn.AX090
+ eqn.8.3
1

t
{ t*
k=n
k=1
pk*log(xk) +O(t2) }
red term applied eqn.8.3
---page 121 ---Line 22 ---eqn.AX091
width of above equation
<a name="ch08a026">
In eqn.AX091, denominator 't'
cancel numerator 't' in summation
term and cancel one 't' in big O
We find
  eqn.AX091=∑[k=1,n]pk*log(xk) +O(tone)
Complete equation start from
left side of eqn.AX083 and end
at eqn.AX091. Put two end into 
one equation as following
<a name="ch08a027">  Index begin Index this file
log {(
k=n
k=1
pkxkt )
1/t
 
 
}
k=n
k=1
pk*log(xk) + O(t)
---page 121 ---Line 20/23 ---eqn.AX092
width of above equation
<a name="ch08a028">
2010-02-15-17-48 here
We always have t approach to
zero, eqn.AX092 right side
O(t) drop out when t→0.
eqn.AX092 right side, let pk
re-gain power, change
from pk*log(xk)
 to  log(xkpk)
eqn.AX092 become
<a name="ch08a029">
log {(
k=n
k=1
pkxkt )
1/t
 
 
}
k=n
k=1
log(xkpk)
---page 121 ---Line 20/23 ---eqn.AX093
width of above equation
<a name="ch08a030">
2010-02-15-17-56 here
Take exp() for eqn.AX093 whole
equation.
Left side exp(log(y)) become y
right side exp(summation) become
product of exp(single), and 
exp(single) has one more time
exp(log(z)) become z.
The result is our target eqn.8.2

Is it an easy problem ?
Yes, only if thorough understand !!
2010-02-15-18-00 stop

<a name="ch08a031"> Index begin Index this file
2010-02-15-19-37 start
■ Problem 8.2 (Power Mean Bound
  for the Geometric Mean)

Follow in Siegel's footsteps and 
prove that for any nonnegative
weights pk k=1,2,...,n with total
mass
  p1+p2+...+pn=1 ---eqn.AX094
and for any nonnegative real
numbers xk k=1,2,...,n, one has
the bound
<a name="ch08a032">
k=n
k=1
xkpk {
k=n
k=1
pkxkt }
1/t
 
 
  for all t>0
---page 122 ---Line 24 ---eqn.8.6
width of above equation
Alert: eqn.8.2, t→0, ∏ term = ∑ term
Alert: eqn.8.6, t>0, ∏ term ≦ ∑ term
Alert: eqn.8.12, t<0, ∏ term ≧ ∑ term
<a name="ch08a033">
2010-02-15-19-58 here
Textbook defined Mt as following
Mt = Mt[x;p] ≡ {
k=n
k=1
pkxkt }
1/t
 
 
---page 120 ---Line 3 ---eqn.8.1
width of above equation
<a name="ch08a034">
With this definition, we have the following variation.
Mtt =
k=n
k=1
pkxkt
M2t = {
k=n
k=1
pkxk2t }
1/(2t)
 
 
M2tt = {
k=n
k=1
pkxk2t }
t/(2t)
 
 
---eqn.AX095 ---eqn.AX096 ---eqn.AX097
width of above equation
<a name="ch08a035">
2010-02-15-20-26 here
Read eqn.8.6 and think this
way. Assume t=2>0, what happen
if t is halved?
Mt=M2, t=2 is far away from t=0,
eqn.8.6 two side gap is bigger.
Mt=M1, t=1 is closer to t=0,
eqn.8.6 two side gap is smaller.
Mt=M0, t=0 is same as t→0,
eqn.8.6 two side gap is zero.
<a name="ch08a036">
With above observation, we can
see 
  M0≦M1≦M2 ---eqn.AX098
We hope to prove
  Mt≦M2t ---eqn.8.7
How can we create a 1t and 2t
relation? Cauchy ! Cauchy !
<a name="ch08a037"> Index begin Index this file
Cauchy's splitting trick will
help us. Write
  pkxkt=pk1/2*pk1/2xkt ---eqn.AX099
Apply Cauchy's inequality for 
two sequences pk1/2 and pk1/2xkt
Use eqn.AX095 for Mtt, we find
  Mtt=∑[k=1,n]pkxkt ---eqn.AX100 begin
  Mtt=∑[k=1,n]pk1/2*pk1/2xkt
<a name="ch08a038"> Mt definition
(
k=n
k=1
pk2/2 )
1/2
 
 
(
k=n
k=1
pk2/2xk2t )
1/2
 
 
=M2tt
Blue 2,t before Cauchy, Red 2, 1/2 from Cauchy
---page 123 ---Line 4 ---eqn.AX100 end
width of above equation
<a name="ch08a039">
2010-02-15-21-07 here
Refer to eqn.AX094
"eqn.AX100 end" (left term)=1
Refer to eqn.AX097
"eqn.AX100 end" (right term)=M2tt
From "eqn.AX100 begin"
 to  "eqn.AX100 end" the whole
equation is
  Mtt≦M2tt ---eqn.AX101
eqn.AX101 take t-th root get
  Mt≦M2t ---eqn.8.7 (Mt definition)

<a name="ch08a040">
Up to now, proved Mt≦M2t, we can
iterate prove, for all real t
we have the following halves
inequality. 
First present a numerical example,
then present a symbolic equation.
Assume t=16, then
  M0.5≦M1≦M2≦M4≦M8≦M16 ---eqn.AX102
in symbol, eqn.AX102 become
  Mt/2j≦Mt/2j-1≦...≦Mt/2≦Mt ---eqn.AX103
<a name="ch08a041">
In eqn.AX103, increase j to infinity
Push Mt/2j to M0
In the limit, we have
  lim[j→∞]Mt/2j=M0
From eqn.8.2, we know
  M0 = ∏[k=1,n]xkpk
eqn.8.6 is proved. (Mt definition)
2010-02-15-21-33 stop



<a name="ch08a042"> Index begin Index this file
2010-02-16-10-33 start
■ Problem 8.3 (Power Mean
  Inequality)
Consider positive weights pk 
k=1,2,...,n which have total mass
  p1+p2+...+pn=1 ---eqn.AX104
and show that for nonnegative
real numbers xk k=1,2,...,n the 
mapping t→Mt is a nondecreasing 
function on all of Real. 
<a name="ch08a043">
That is show that for all
  -∞<s<t<∞ 
one has
{
k=n
k=1
pkxks }
1/s
 
 
{
k=n
k=1
pkxkt }
1/t
 
 
---page 123 ---Line 26 ---eqn.8.10
width of above equation
<a name="ch08a044">
Finally, show that then one has 
equality in the bound (8.10) if 
and only if x1=x2=...=xn
2010-02-16-10-51 stop
<a name="figure0801">  Index begin Index this file
Figure 8.1 ; 2010-02-15-22-42


<a name="ch08a045">
2010-02-16-10-58 start
Please compare next two equations
{
k=n
k=1
pkxk }
p
 
 
{
k=n
k=1
pkxkp }
---page 124 ---Line 5 ---eqn.AX105
width of above equation
and //Alert: power index p is NOT pk
<a name="ch08a046">
Jensen Inequality need a base 
function and this function must
be convex on domain. How to find
the base function from given?
From both equation greater than
side, we see directly that
  f(xk)=xkp ---eqn.AX106
From both equation less than
side, we see
  f(∑px)={∑px}p ---eqn.AX107

<a name="ch08a047"> Index begin Index this file
eqn.AX106 and eqn.AX107 suggest
that 
  f(x)=xp ---eqn.AX108

Jensen Inequality for convex 
function require that
  p>1 ---eqn.AX109
Please pay attention to p and pk
they are different, because
P>1 but 0≦pk≦1.
2010-02-16-11-28 stop
//debug HelloJensen() correction at 9902161158
<a name="ch08a048">
2010-02-16-12-14 start
  f1(x)=xp and p>1 ---eqn.AX110
  f2(x)=xp and p=1 ---eqn.AX111
  f3(x)=xp and p<1 ---eqn.AX112
f1(x) is convex; shape like "╰╯" 
f2(x) convex and concave; shape "/"
f3(x) is concave; shape like "╭╮" 

<a name="ch08a049">
We will solve Problem 8.3 use
Jensen Inequality and convex
function 
  f(x)=xp ---eqn.AX108

Let us assume that
  0<s<t ---eqn.AX113
Define
  p=t/s>1 ---eqn.AX114
This definition make sure that p>1
and f1(x)=xp is convex.

<a name="ch08a050">
In eqn.AX105, replace
  p → t/s>1 ---eqn.AX115
replace nonnegative real numbers 
  xkyks ---eqn.AX116
We find 
<a name="ch08a051"> Jensen Inequality
red match red, blue match blue.
{
k=n
k=1
pkyks }
t/s
 
 
{
k=n
k=1
pkyks*t/s }
---page 124 ---Line 8 ---eqn.8.11
width of above equation
<a name="ch08a052"> Index begin Index this file
In eqn.8.11, take t-th root, 
write yk as xk (nothing to do
with eqn.AX116), the result 
is our goal eqn.8.10.

Equality condition follow
When Jensen's Inequality become equality?
It says for equality, we must have
  x1=x2=...=xn ---eqn.AQ022
Problem solved.
2010-02-16-12-50 stop

2010-02-16-19-15 done proofread
2010-02-16-20-40 done spelling check


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