<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="test01">
2010-02-06-16-06 start
This file http://freeman2.com/tute0028.htm
include eight external javascript files
jsgraph2.js enable xygraph code
drawcod2.js enable graph interface
mathinc2.js define more math functions
complex2.js define complex functions.
<a name="test02">
If above four files included successfully
click next button program draw
asinh(x)  defined in complex2.js 
acosh(x)  defined in complex2.js 
ratanh(x) defined in mathinc2.js
<a name="test03">
If not draw, you need save
tute0028.htm to your computer 
and download above *.js files and 
save them in same folder as 
tute0028.htm
<a name="test04">
Click next "Draw asinh()"
Wait for two seconds, then move to
two pace below (page down) for figure.
If success, curve show up.
If fail, MSIE status bar show error.
Please try click next "Draw asinh()"
2010-02-06-16-16 stop
<a name="test05">

<a name="ch06g001">　Index begin　Index this file
2010-01-27-16-15 start
■　If G(x) convex, xG(1/x) convex
This is tute0028.htm for 
Chapter 07: Integral Intermezzo. But 
the following is a problem relate to
Chapter 06: Convexity ~~ The Third Pillar
2010-01-22,24,25 three days, LiuHH goto
Internet look for integral inequality
paper. Hope to understand the basic of
integral inequality.

<a name="ch06g002">
2010-01-22-19-14 LiuHH accessed
http://www.cms.math.ca/cjm/v28/cjm1976v28.0793-0804.pdf
save  as  integral_cms.math.ca_cjm1976v28.0793-0804.pdf
This is scanned picture text file. 830,695 bytes
It has G(x) convex then x*G(1/x) convex
Page 2/12 (page 794), lemma 1 is next

<a name="ch06g003">
Lemma 1. If G(x) is convex for x＞0, 
then
  H(x)=xG(1/x)  ---eqn.AV001
is also convex for x＞0.

Proof, Let 0≦λ≦1 and for given positive
x,y define
  p = λ*y/[(1-λ)*x+λ*y] ---eqn.AV002
Evidently 0≦p≦1 with p=0 when λ=0
and p=1 when λ=1. Since
  λ = p*x/[(1-p)*y+p*x] ---eqn.AV003
it follows that λ can be specified for a
given value of p.

Given p, where 0≦p≦1, we have

<a name="ch06g009">
2010-01-27-17-10 here
Therefore by the convexity of G(x) we 
find

<a name="ch06g011">　Index begin　Index this file
2010-01-27-17-39 here
Multiplying by the form (1－λ)x＋λy, 
we obtain

<a name="ch06g012">
and this shows directly that xG(1/x) 
is convex.

A result of this type is stated for
twice differentiable positive convex
function in [9, p.97, Theorem 120].
The transformation or functional
operation 
  H(x)=xG(1/x)  ---eqn.AV001
is involutory,
since
  G(x) = xH(1/x) ---eqn.AV012

Above is a copy from 
http://www.cms.math.ca/cjm/v28/cjm1976v28.0793-0804.pdf
2010-01-27-17-49 stop.

<a name="ch06g013">　Index begin　Index this file
2010-01-27-18-30 start
■　xG(1/x) convex discussion
The proof start from
  p = λ*y/[(1-λ)*x+λ*y] ---eqn.AV002
and
  λ = p*x/[(1-p)*y+p*x] ---eqn.AV003
Limit is
 0≦λ≦1 and 0≦p≦1 
We find p=0 when λ=0
and     p=1 when λ=1

<a name="ch06g014">
eqn.AV002 and AV003 have same structure.
same definition range 0≦λ≦1 and 0≦p≦1 
same value p=0 ↔ λ=0 and p=1 ↔ λ=1
Are they identical function?
2010-01-27-18-35 here

<a name="ch06g015">
2010-01-27-19-08 start
Please goto clickDrawcod2 click [11] 
button.
 Red curve is  p = λ*y/[(1-λ)*x+λ*y] ---eqn.AV002
Blue curve is  λ = p*x/[(1-p)*y+p*x] ---eqn.AV003
Read from drawing, it is easy to
understand that eqn.AV002 and eqn.AV003
are different.
<a name="ch06g016">
eqn.AV004 to eqn.AV008 are simple
calculation. It is easy to follow.
eqn.AV009 use Jensen's inequality.
Review Jensen's inequality 

<a name="ch06g017">
Jensen Inequality for convex function is
  AM_DOMAIN ≦ AM_RANGE ---eqn.AV013
Our target function is
  H(x)=xG(1/x)  ---eqn.AV001
We need to find
  H(x_AM) ≦ AM_RANGE for H(x) ---eqn.AV014
x_AM is 
  x_AM=(1－λ)x＋λy ---eqn.AV015
with 0≦λ≦1
<a name="ch06g018">
If we set λ=1/2, then x_AM is
 (1－1/2)x＋y/2 = (x+y)/2
which is lectured at elementary level.
x_AM is defined in eqn.AV015. then
H(x_AM) = (x_AM)G(1/(x_AM))
that is
  H(x_AM) =  ---eqn.AV016
 ((1－λ)x＋λy)G(1/((1－λ)x＋λy))

<a name="ch06g019">　Index begin　Index this file
eqn.AV016 is AM_DOMAIN part in
  AM_DOMAIN ≦ AM_RANGE ---eqn.AV013
We need AM_RANGE for H(x)=xG(1/x)

generalized AM for domain is
     x_AM = (1－λ)x  ＋λ y ---eqn.AV015
generalized AM for range is
  AM_RANGE=(1－λ)H(x)＋λH(y) ---eqn.AV017
Look at eqn.AV015 and eqn.AV017 one
more time, 
<a name="ch06g020">
do you agree they are
Arithmetic Mean in domain (eqn.AV015)
and
Arithmetic Mean in range (eqn.AV017)
?
x_AM can not compare with AM_RANGE
directly. Because AM_RANGE is function
value, but x_AM is independent variable.
Not on equal foot.
Although x_AM can not but H(x_AM) can.
because H(x_AM) is function H(x)
evaluated at x=x_AM.

<a name="ch06g021">
Jensen Inequality for convex function is
  AM_DOMAIN ≦ AM_RANGE
that is
  H(x_AM) ≦ (1－λ)H(x)＋λH(y) ---eqn.AV018
eqn.AV016 expand H(x_AM)
On the other hand,
  H(x)=xG(1/x)  ---eqn.AV001
tell us that in eqn.AV018
H(x)=xG(1/x) and H(y)=yG(1/y)
Our goal is eqn.AV018, 
<a name="ch06g022">
Its expanded version is
  ((1－λ)x＋λy)*G(1/((1－λ)x＋λy))
  ≦  ---eqn.AV019
  (1－λ)*xG(1/x)＋λ*yG(1/y)
At this moment eqn.AV019 is wait for
prove.

<a name="ch06g023">
G(1/((1－λ)x＋λy)) is target eqn.AV019
major role. Process eqn.AV004 to
eqn.AV008 convert from λ parameter 
to p parameter. Let eqn.AV009 be
Jensen-ready form.
eqn.AV009 has two lines, both are
bold red
eqn.AV009 is true, it come from
the problem assumption that given
G(x) is convex function.

<a name="ch06g024">　Index begin　Index this file
eqn.AV010 change p parameter back 
to λ parameter. Key point is that
we have a "(1－λ)x＋λy" 
Move "(1－λ)x＋λy" cross inequality?
If "(1－λ)x＋λy" is negative, then
inequality change direction !!
What sign "(1－λ)x＋λy" has?

<a name="ch06g025">
Problem require x＞0, then in
"(1－λ)x＋λy" both x and y ＞0
also given 0≦λ≦1 , therefore
"(1－λ)x＋λy" must be positive.

<a name="ch06g026">
Last step move "(1－λ)x＋λy" from
eqn.AV010 greater than side 
denominator to less than side
numerator. We get eqn.AV011
eqn.AV011 is same as eqn.AV019.
Our job is done.

If G(x) is convex, then 
xG(1/x) is convex.
2010-01-27-20-16 stop

<a name="ch06g027">　Index begin　Index this file
2010-01-28-10-07 start
■　Code to draw curve with infinity slope

2010-01-22-19-47 LiuHH access
http://marauder.millersville.edu/~bikenaga/calculus/conc/conc.pdf
Paper title: 
Concavity and the Second Derivative Test
<a name="ch06g028">
Page 3/5 has one example
  f(x)=(9/4)*x4/3-9*x1/3 ---eqn.AV020
We find f' and f'' as following
  f'(x)=(9/4)*(4/3)*x(4/3)-1-9*(1/3)*x(1/3)-1
  f'(x)=3*x1/3-3*x-2/3 ---eqn.AV021
and
  f''(x)=3*(1/3)*x(1/3)-1-3*(-2/3)*x(-2/3)-1
  f''(x)=x-2/3+2*x-5/3
  f''(x)=(x+2)/x5/3 ---eqn.AV022

<a name="ch06g029">
Second Derivative Test tell us that
  f''(x)>0 for x in (-inf., -2)
  f''(x)=0 for x=-2
  f''(x)<0 for x in (-2,0)
  f''(x) undefined for x=0 (it is 1/0)
  f''(x)>0 for x in (0, +inf.)
<a name="ch06g030">
We find
  f(x) is convex ╰╯ for x in (-inf.,-2)
  f(x) is concave╭╮ for x in (-2,0)
  f(x) is convex ╰╯ for x in (0, +inf.)
Please goto clickDrawcod2
and click [12] button.
Blue curve is f''(x) which test concavity.
Black curve is f'(x)
Red curve is  f(x).

<a name="ch06g031">
Interesting thing is that 
  f(x)=(9/4)*x4/3-9*x1/3 ---eqn.AV020
has infinite slope at x=0.
How to draw f(x) in one equation and
include the infinite slope point x=0?
  f'(x)=3*x1/3-3*x-2/3 ---eqn.AV021
  f'(0)=3*01/3-3/0+2/3=-3/0=-infinity
<a name="ch06g032">
Key points are
(1) use gatef() to isolate f(+x) and
    f(-x). Each treat differently.
(2) use abs(x), avoid pow(negative,*)
(3) make up '-' which is removed by
    abs(x)
Please see detail at clickDrawcod2
and click [12] button. Not repeat here.
2010-01-28-10-49 stop




<a name="ch07a001">　Index begin　Index this file
2010-01-31-13-01 start
■■Chapter 07: Integral Intermezzo

Chapter seven is integral inequality.
From 2010-01-28 to 2010-01-30 LiuHH
wrote a program (local)
http://freeman2.com/integral.htm to
calculate function g(x) integration
value. You are welcome to test run.
Link to other author's integration 
page.

<a name="ch07a002">　Index begin　Index this file
■　Problem 7.1 (A Continuum of Compromise)
Show that for an integral f:real to real
one has the bound

<a name="ch07a004">　Index begin　Index this file
2010-01-31-13-24 here
■　Problem 7.1 discussion
If not read textbook, eqn.7.1 is 
definitely a hard problem. Less 
than side is so simple but greater 
than side is so complicated.

<a name="ch07a005">
First thing, and easy thing, we 
can do is to check whether eqn.7.1
both side have physics dimension 
consistency?

<a name="ch07a006">
Assume x is time, assume f(x) is
velocity. Then f(x)dx, velocity
multiply by time get distance.
How about eqn.7.1 right side?
81/2 is a pure number. Not
contribute to dimension.
Right side has |f(x)|2*|f(x)|2
and then take power1/4 return to
|f(x)|ONE
<a name="ch07a007">
Right side has |x|2dx*dx multiply 
to x fourth power, then take 
power1/4 return to xONE
('d' in 'dx' mean differentiation
 'd' do not carry physics dimension)
eqn.7.1 right side is also f(x)*x
Physics dimension consistency check
passed.

<a name="ch07a008">
We start from eqn.7.1 left side
∫|f(x)|dx , x in (-∞,+∞)
Textbook suggest us to divide the
domain from x in (-∞,+∞) to two
sub-domains.
One sub-domain is
  T=(-t, t)  ---eqn.AV023
other sub-domain
  Tc=(-∞,-t]+[t,+∞) ---eqn.AV024
<a name="ch07a009">
In Tc, we change f(x) to f(x)*x/x
Tc not contain 0, 1/0 never happen!

eqn.7.1 left side is same as next

<a name="ch07a011">
2010-01-31-14-07 here
Now we need apply Schwarz inequality.
Schwarz inequality is integral version
of Cauchy inequality.
Schwarz inequality is next

2010-01-31-14-14 stop, 
<a name="ch07a012">
2010-01-31-14-33 start
Apply Schwarz inequality to eqn.7.2a
right side first term ∫|f(x)|dx.
Compare ∫|f(x)|dx with eqn.1.19
left side ∫f(x)g(x)dx, we need two
functions in integrand, but
∫|f(x)|dx has only one |f(x)|,
where is the second g(x) ?
<a name="ch07a013">
If we set
  g(x)≡1 ---eqn.AV025
then 
  f(x)g(x) = f(x)*1 = f(x) ---eqn.AV026
This is called "one trick"
eqn.7.2a right side first term now
become ∫|f(x)|*g(x)dx where g(x)≡1

<a name="ch07a014">
Apply Schwarz inequality to eqn.7.2a
right side first term ∫|f(x)|dx get

<a name="ch07a015">　Index begin　Index this file
2010-01-31-14-59 here
eqn.7.2b contribute to eqn.7.2 
greater than side first term.

Next, apply Schwarz inequality 
to eqn.7.2a right side second 
term ∫{|xf(x)|/|x|}dx get

2010-01-31-15-18 stop
<a name="ch07a017">
2010-01-31-16-25 start
eqn.7.2c greater than side first
term (red) is
  ∫[x∈Tc]{dx/|x|2}
or
  ∫[x∈Tc]{dx/x2)} ---eqn.AV027
or
  ∫[x∈Tc]{d(-1/x)} ---eqn.AV028
where
  Tc=(-∞,-t]+[t,+∞) ---eqn.AV024
<a name="ch07a018">
for x in [t,+∞)
  ∫[x∈[t,+∞)]{d(-1/x)}=-1/x
    {x at [t,+∞) two end points}
  = -1/(+∞) - [-1/(+t)]
  = 1/t ---eqn.AV029
<a name="ch07a019">
for x in (-∞,-t]
  ∫[x∈(-∞,-t]{d(-1/x)}=-1/x
    {x at (-∞,-t] two end points}
  = -1/(-t) - [-1/(-∞)]
  = 1/t ---eqn.AV030
<a name="ch07a020">
Sum eqn.AV029 and eqn.AV030 get
  ∫[x∈Tc]{d(-1/x)}=1/t+1/t
  =2/t ---eqn.AV031
eqn.AV031 is
 "eqn.7.2c Red term is √(2/t)"
2010-01-31-16-32 here

<a name="ch07a021">　Index begin　Index this file
eqn.7.2b and eqn.7.2c we applied 
Schwarz Inequality twice.
Add eqn.7.2b and eqn.7.2c get 
textbook eqn.7.2

<a name="ch07a024">
2010-01-31-16-57 here
eqn.7.2 is true. However integration
domain are different
"eqn.7.2a 2nd" left side x∈(-∞,+∞)
eqn.7.2 left x∈T =(-t, t)
eqn.7.2 right x∈Tc =(-∞,-t]+[t,+∞)
<a name="ch07a025">
Not on equal foot. Hard to compare.
Work with inequality, we are free
to increase greater than side value
and free to decrease less than side 
value. Since these two operation do
not change inequality.

<a name="ch07a026">　Index begin　Index this file
Because
"eqn.7.2a 2nd" left side x∈(-∞,+∞)
then 
change from right side shorter domain
[[
eqn.7.2 left x∈T =(-t, t)
eqn.7.2 right x∈Tc =(-∞,-t]+[t,+∞)
]]
<a name="ch07a027">
to full size domain
[[
eqn.7.2 left x∈T =(-∞,+∞)
eqn.7.2 right x∈Tc =(-∞,+∞)
]]
<a name="ch07a028">
that is to increase greater than side 
value. Let all integration stand on
equal domain foot. With this change 
in mind, we define the following

<a name="ch07a031">
2010-01-31-17-23 here
From eqn.AV032 and eqn.AV033 we
know eqn.7.2 greater than side is
bounded by φ(t), define φ(t) below
 φ(t)=√(2)*√(t)*√(A)
    + √(B)*√(2)/√(t) ---eqn.AV034
<a name="ch07a032">
Red √(2)*√(t) come from eqn.7.2 red.
Blue √(2)/√(t) come from eqn.7.2 blue.
√(A) ≧ eqn.7.2 left black term.
√(B) ≧ eqn.7.2 right black term.
We find φ(t) bound eqn.7.2 less than
side term ∫[x=-∞,+∞]{|f(x)|dx}

<a name="ch07a033">　Index begin　Index this file
φ(t) has t as variable, both A and B
are not function of t. φ(t) is
 φ(t)=√(2)*√(t)*√(A) + √(B)*√(2)/√(t) 
re-write as
 φ(t)=Constant_M*√(t) + Constant_N/√(t) 
See the curve φ(t) what shape it has.
write 't' as 'x', take Constant_M=2
take Constant_N=3 
Change φ(t) line to next h(x) line
 h(x)=2*sqrt(x)+3/sqrt(x)
<a name="ch07a034">
Please goto drawing board
Click [dA] delete all other equation.
paste next line
2*sqrt(x)+3/sqrt(x)
to f0(x) box, set the following numbers
x min: 0 , x max: 10 ; y min: 0 , y max: 10
x/t begin 0 , x/t end 10
Click [Draw f0(x) to f3(x)] button.
Graph show up below click button.
Minimum point is about (1.5, 4.8989)

<a name="ch07a035">
In φ(t) we can vary t and find a 
minimum value point. Let
  m=√(2)*√(A) ---eqn.AV035
  n=√(2)*√(B) ---eqn.AV036
write
 φ(t)=√(2)*√(t)*√(A) + √(B)*√(2)/√(t) 
as
 φ(t)=m*√(t) + n/√(t)  ---eqn.AV037
 dφ(t)/dt = 0.5*m*t^(-0.5)
       + (-0.5)*n*t^(-1.5)

<a name="ch07a036">
 dφ(t)/dt = 0.5*t^(-1.5)*[m*t-n] ---eqn.AV038
set dφ(t)/dt to zero, find
 m*t0-n=0
or
  t0=n/m 
or
  t0=[√(2)*√(B)]/[√(2)*√(A)]
or
  t0=[√(B)/√(A)] ---eqn.AV039
<a name="ch07a037">
We find optimal  t0, put t0
to φ(t) eqn.AV034 get
  φ(t0) = √(2)*√[√(B)/√(A)]*√(A)
      + √(B)*√(2)/√[√(B)/√(A)] ---eqn.AV040

  φ(t0) = √(2)*(B)1/4*√(A)-1/4*(A)1/2
      + (B)1/2*√(2)*(B)-1/4*(A)1/4 ---eqn.AV041

<a name="ch07a038">
  φ(t0) = √(2)*(B)1/4*(A)1/4
       + √(2)*(B)1/4*(A)1/4 ---eqn.AV042
2010-01-31-18-11 here
  φ(t0) = 2*√(2)*(B)1/4*(A)1/4
  φ(t0) = √(8)*(B)1/4*(A)1/4 ---eqn.AV043

<a name="ch07a039">
From eqn.AV032 and eqn.AV033, 
eqn.AV034 φ(t) bound eqn.7.2. Any 
t all bound. eqn.AV043 φ(t0) is
smallest bound Put A (eqn.AV032) 
and B (eqn.AV033) to eqn.AV043

Result is eqn.7.1 greater than
side exactly. Problem solved.
2010-01-31-18-26 stop






<a name="ch07a040">　Index begin　Index this file
2010-02-01-19-01 start
■　Integration by parts
Textbook next topic use integration 
by parts. LiuHH read calculus book
and write the following notes.

<a name="ch07a041">
Integration by parts in short is
  ∫u*dv = u*v - ∫v*du ---eqn.AV044
both 
  u=u(x) ---eqn.AV045
and
  v=v(x) ---eqn.AV046
are function of x. 

<a name="ch07a042">
A real example is
  ∫x*exp(x)*dx ---eqn.AV047

First use method one (better)
  choose
  u(x)=x   ---eqn.AV048
  dv=exp(x)*dx ---eqn.AV049
<a name="ch07a043">
From eqn.AV049 we know
  dv=exp(x)*dx=d[exp(x)]
then
  v(x)=exp(x) ---eqn.AV050
eqn.AV044 become
  ∫u*[dv] = ∫x*[exp(x)*dx]
  = x*exp(x) - ∫exp(x)*dx ---eqn.AV051
  = u*  v    - ∫  v   *du //side by side
<a name="ch07a044">
whole expression is
  ∫x*exp(x)*dx  ---eqn.AV052
  =x*exp(x)-∫exp(x)*dx
Integral part change from harder
∫x*exp(x)*dx to easier -∫exp(x)*dx
the final result is
  ∫x*exp(x)*dx  ---eqn.AV053
  =x*exp(x)-exp(x) + C

<a name="ch07a045">
Second use method two (worse)
  choose
  u(x)=exp(x) ---eqn.AV054
  dv=x*dx     ---eqn.AV055
From eqn.AV055 we know
  dv=d[x*x/2]
then
  v(x)=x*x/2 ---eqn.AV056
<a name="ch07a046">
eqn.AV044 become
  ∫u*[dv] = ∫exp(x)*[x*dx] ---eqn.AV057
  = exp(x)*(x*x/2) - ∫(x*x/2)*[exp(x)*dx]
  =    u  *   v    - ∫   v   *   [du]
Integral part change 
from hard ∫exp(x)*[x*dx] to 
harder -∫(x*x/2)*exp(x)*dx

<a name="ch07a047">
Integration by parts can relief our
work, if we do the right choice. In
"u*v - ∫v*du", "u*v" is determined 
by ends x=a and x=b. Not involve
x in [a,b] "u*v" is boundary term.

<a name="ch07a048">
Why
  ∫u*dv = u*v - ∫v*du ---eqn.AV044
is true? Let us re-write eqn.AV044
as following
  ∫u*dv + ∫v*du = u*v ---eqn.AV058
or
  ∫u*dv + ∫v*du = ∫d[u*v] ---eqn.AV059
Its differential version is
   u*dv +  v*du =  d[u*v] ---eqn.AV060
I remember! That is product rule!

<a name="ch07a049">　Index begin　Index this file
2010-02-01-19-30 here
■　Product rule of differentiation
How to prove eqn.AV060 ?
   u*dv +  v*du =  d[u*v] ---eqn.AV060
differentiation basic rule is
  f'(x) = 
  lim[h→0][f(x+h)-f(x)]/[(x+h)-(x)]
or
<a name="ch07a050">
  f'(x) =
  lim[h→0][f(x+h)-f(x)]/h ---eqn.AV061
Now
  f(x)=u(x)*v(x) ---eqn.AV062
Let us put u(x)*v(x) to eqn.AV061, get
  f'(x) = lim[h→0]
  [u(x+h)*v(x+h)-u(x)*v(x)]/h ---eqn.AV063
Since
  u(x+h) = u(x+h)-u(x)+u(x) ---eqn.AV064
<a name="ch07a051">
write u(x+h)-u(x) as Δu
  u(x+h) = Δu + u(x) ---eqn.AV065
or
  u(x+h) - u(x) = Δu ---eqn.AV066
In the limit h→0, Δu approach to
differential du(x)
Same reason apply to v(x+h) and v(x)
<a name="ch07a052">
eqn.AV063 become
  f'(x) = lim[h→0]
  [(u+Δu)*(v+Δv)-u*v]/h ---eqn.AV067
  =
  [(u*v+Δu*v+Δv*u+Δu*Δv)-u*v]/h
u*v cancel out, get
  f'(x) =  ---eqn.AV068
  lim[h→0] [Δu*v+Δv*u+Δu*Δv]/h
<a name="ch07a053">
In the limit case h→0,
Δu and Δv and h all approach to zero.
  lim[h→0] Δu/h = u'(x) ---eqn.AV069
  lim[h→0] Δv/h = v'(x) ---eqn.AV070
eqn.AV068 become
  f'(x) =  ---eqn.AV071
  u'(x)*v+v'(x)*u + lim[h→0] [Δu*Δv]/h
<a name="ch07a054">
Last term "lim[h→0] [Δu*Δv]/h"
If group as lim[h→0] [Δu/h]*Δv
[Δu/h] may not be zero, but Δv→0
If group as lim[h→0] [Δv/h]*Δu
[Δv/h] may not be zero, but Δu→0
In either case lim[h→0] [Δu*Δv]/h
disappear.
<a name="ch07a055">
In the limit h→0 eqn.AV071 become
  f'(x)=u'(x)*v+v'(x)*u ---eqn.AV072
eqn.AV072 is the product rule.
2010-02-01-20-00 stop

<a name="ch07a056">　Index begin　Index this file
2010-02-02-15-31 start
■　Thank you for your shoulder, Schwarz
Problem 7.2 (Textbook page 107)
Show that if f:[0,∞)→[0,∞) is a
continuous, non-increasing function
which is differentiable on (0,∞),
then for any pair of parameters
0＜α,β＜∞ the integral

<a name="ch07a059">　Index begin　Index this file
2010-02-02-15-55 here
■　Problem 7.2 discussion
Let us look at Schwarz shoulder first,
Hello! Schwarz!

We re-write eqn.7.3 as next
  ∫xα+βf(x)dx = ---eqn.AV073
  ∫{xα√[f(x)]*xβ√[f(x)]}dx
Take xα√[f(x)] as f(x) in eqn.1.19
Take xβ√[f(x)] as g(x) in eqn.1.19
<a name="ch07a060">
Schwarz Inequality eqn.1.19 give us
result similar to eqn.7.4 but change
{1-[(α－β)/(α＋β＋1)]2} to one
Since [(α－β)/(α＋β＋1)]2 ≧ 0
{1-[(α－β)/(α＋β＋1)]2} is tighter
and one is looser.
If we can prove eqn.7.4 is true, then
we solve problem 7.2
2010-02-02-16-15 stop

<a name="ch07a061">
2010-02-02-17-25 start
Problem 7.2 do not apply to any f(x).
Problem 7.2 limit to f(x):[0,∞)→[0,∞)
It says the function f(x) 
domain is [0,∞), x is not negative.
range  is [0,∞), f(x) is not negative.
Problem 7.2 also limit f(x) to be
continuous, non-increasing function
<a name="ch07a062">
We can see for x in [0,∞) start
from x=0 move toward x→∞, we must 
have f(x) value approach to zero
(because continuous, non-increasing)
and f(x)≧0  {because red [0,∞) part
in "f(x):[0,∞)→[0,∞)"}
<a name="ch07a063">
For such function, it must have
f'(x)≦0 for x in [0,∞). 
Whole working space is positive. We
need to use POSITIVE f'(x) to get
positive answer. This is the reason
textbook use |f'(x)| not use f'(x).

<a name="ch07a064">　Index begin　Index this file
What if lim[x→∞]{f(x)}→ε and ε>0
for example tiny ε=1.0e-12 > 0
f(x) from f(x=0)=big_number
      to  f(x→∞)=ε=1.0e-12 > 0
Every given condition are satisfied.
This is best case for lazy bones.
Because if lim[x→∞]{f(x)}→ε ε>0
then eqn.7.4 read
  I2≦{coef.}*infinity*infinity
It is always true, problem is done!
<a name="ch07a065">
We ruled out trivial case. Then
  lim[x→∞]{f(x)}=0 ---eqn.AV074
or textbook equation

is an important condition in our 
hand. Integration by parts use it.
2010-02-02-17-55 stop

<a name="ch07a066">
2010-02-02-19-43 start
Integration by parts also use
  lim[x→0]{f(x)}=what ---eqn.AV075
We need to find out the answer.

<a name="ch07a067">
Function f(x) is continuous at x=c 
if and only if
1. f(c) is defined. x=c is in domain.
2. limit[x→c]f(x) exist.
3. f(c) = limit[x→c]f(x)
If every point in domain are continuous
point, then function f(x) is called
continuous.

<a name="ch07a068">
Our given condition is
[[
if f:[0,∞)→[0,∞) is a
continuous, non-increasing function
which is differentiable on (0,∞)
]]
f(x) is continuous and domain 
x in [0,∞)
(red term is for range, function output
blue term is for domain function input.)
where x=0 is in domain (important)
<a name="ch07a069">　Index begin　Index this file
We must have 
  lim[x→0]{f(x)}=finite number ---eqn.AV076
and
  lim[x→0]{f(x)}=f(0) ---eqn.AV077

Since f(x) is decrease (non-increase)
for x in [0,∞), f(0) must be highest 
and finite value. In the following
we will meet xα+β+1*f(x)
<a name="ch07a070">
although
  lim[x→0]{f(x)}=f(0)=highest ＞ 0 
but actual term in work is
  lim[x→0]{xα+β+1*f(x)}=what
It is easy to explain that
  lim[x→0]{xα+β+1*f(x)}=zero ---eqn.AV078
Because given condition 0＜α,β＜∞
then xα+β+1 is xgreater_than_1. 
<a name="ch07a071">
Condition xnegative never happen.
  lim[x→0]{xα+β+1*f(x)}=what
is same as
  lim[x→0]{xgreater_than_1*f(x)}=what
is same as
  lim[x→0]{xgreater_than_1}*lim[x→0]{f(x)}=what
is same as
  zero * finite_number = what? = zero !
<a name="ch07a072">
We get another important condition
  lim[x→0]{xα+β+1*f(x)}=0 ---eqn.AV079
Integration by parts use it.
2010-02-02-20-37 here

<a name="ch07a073">　Index begin　Index this file
■　Apply integration by parts
We stand on Schwarz shoulder, the 
main tool is integration by parts.
Our start point is eqn.7.3
  I = ∫[x=0,∞]{xα+β*f(x)}dx ---eqn.AV080
Integration by parts in short is
  ∫u*dv = u*v - ∫v*du ---eqn.AV044
eqn.AV080 is just ∫u*dv part. 

<a name="ch07a074">
How to select u and v?
First try is next (worse)
If we group {xα+β*f(x)}dx as
{xα+β}*{f(x)dx} then
u is xα+β and dv is f(x)dx
eqn.AV044 right side 
need v which is ∫f(x)dx and 
need du which is (α+β)*xα+β-1dx

<a name="ch07a075">
eqn.AV044 u*v become
  xα+β*∫f(x)dx ---eqn.AV081
we can not use eqn.AV074
we can not use eqn.AV079
we can not evaluate eqn.AV081

<a name="ch07a076">
eqn.AV044 -∫v*du become
  -∫[∫f(x)dx]*(α+β)*xα+β-1dx ---eqn.AV082
It has double integral, become 
more complicated.

<a name="ch07a077">
Now back to
How to select u and v?
Second try is next (better)
If we group {xα+β*f(x)}dx as
{f(x)}*{xα+βdx} then
u is f(x) and dv is xα+βdx
eqn.AV044 right side 
need v which is xα+β+1/(α+β+1) and 
need du which is d[f(x)]/dx=f'(x)

<a name="ch07a078">　Index begin　Index this file
Second group look much better.
because we know f'(x) property
f'(x)≦0 for x in [0,∞)
We apply integration by parts to
  I = ∫[x=0,∞]{xα+β*f(x)}dx ---eqn.AV080
get
<a name="ch07a079">

<a name="ch07a081">
2010-02-02-21-42 here
eqn.7.5a [ square ] term is u*v in
 " u*v - ∫v*du " 
We need evaluate its value at two
points x=0 and x=∞

For x=0, f(0) is finite and xα+β+1
go to zero. Then f(0)*0α+β+1 is zero

<a name="ch07a082">
For x=∞, f(∞) is zero and xα+β+1
go to infinity. Then f(∞)*∞α+β+1 is 
uncertain.
Later part, Professor Steele will
prove that limit[x→∞]f(x)*xα+β+1=0
2010-02-02-21-50 stop

<a name="ch07a083">　Index begin　Index this file
2010-02-03-12-10 start
eqn.7.5a and b are integration by
parts equation. Square bracketed
[term] in eqn.7.5a is boundary.
For x=0, f(0)*0α+β+1 is zero
For x→∞, f(x)*xα+β+1 is zero 
Then eqn.7.5a and b simplify to

<a name="ch07a085">
2010-02-03-12-36 here
Given 0＜α,β＜∞, then α＋β＋1＞1＞0
In eqn.7.5, no 1/0 condition. 
Now in eqn.7.5 move "α＋β＋1" from 
right denominator to left numerator
get
  (α＋β＋1)*I =  ---eqn.AV083
  ∫[x=0,∞]xα+β+1|f'(x)|dx
<a name="ch07a086">
Re-write integrand term
  xα+β+1|f'(x)| =  ---eqn.AV084
  xα+1/2√|f'(x)| * xβ+1/2√|f'(x)|
We can apply Schwarz inequality
to eqn.AV083 right side.
Schwarz inequality use f(x)g(x)
<a name="ch07a087">
f(x) is xα+1/2√|f'(x)| in eqn.AV084
g(x) is xβ+1/2√|f'(x)| in eqn.AV084
Schwarz inequality give us the 
result

<a name="ch07a090">
2010-02-03-13-31 here
To simplify eqn.AV085 left side,
use I definition and eqn.AV083,
then eqn.AV085 left side become
(α＋β＋1)*I. Take square for
 "(α＋β＋1)*I≦eqn.AV086" get 
textbook page 108 line 16 
equation
  (α＋β＋1)2*I2 ≦

<a name="ch07a092">　Index begin　Index this file
2010-02-03-13-50 here
Our target equation eqn.7.4
do not involve f'(x). But eqn.AV087
has f'(x). What to do ?
Integration by parts allow us go
either direction. At first we go 
forward change f(x) to f'(x). Now
we go backward, change f'(x) to 
f(x).

<a name="ch07a093">
Look at eqn.AV087 left integral
  ∫[x2α+1|f'(x)|]dx ---eqn.AV088
Integration by parts in short is
  ∫u*dv = u*v - ∫v*du ---eqn.AV044
We want to recover f(x) from f'(x)
we must choose |f'(x)|dx as dv 
What left is choose x2α+1 as u
<a name="ch07a094">
eqn.AV044 give us the result
  ∫[x2α+1|f'(x)|]dx  ---eqn.AV089
 =x2α+1*f(x) - ∫f(x)*d[x2α+1]
 =x2α+1*f(x) - ∫f(x)*(2α+1)*[x2α]dx
<a name="ch07a095">
Similarly
  ∫[x2β+1|f'(x)|]dx  ---eqn.AV090
 =x2β+1*f(x) - ∫f(x)*d[x2β+1]
 =x2β+1*f(x) - ∫f(x)*(2β+1)*[x2β]dx
Both eqn.AV089 and eqn.AV090 integrate
from x=0 to x=∞
Boundary term (red term) are zero

<a name="ch07a096">
Put eqn.AV089 and eqn.AV090 into
eqn.AV087 get

<a name="ch07a097">
2010-02-03-14-24 here
Our target equation eqn.7.4 has
coefficient
  1-[(α－β)/(α+β+1)]2 ---eqn.AV092
eqn.AV091 has coefficient
  (2α+1)*(2β+1)/[(α+β+1)]2 ---eqn.AV093
<a name="ch07a098">
They are identical, since
  [α+β+1]2 - [α－β]2
 =[α+β+1+α-β]*[α+β+1-α+β]
 =[2α+1]*[2β+1] ---eqn.AV094
problem 7.2 solved. Except 
We need to prove that
 limit[x→∞]f(x)*xα+β+1=0 ---eqn.AV095
2010-02-03-14-31 stop

<a name="ch07a099">　Index begin　Index this file
2010-02-03-19-00 start
■　Verify infinity boundary value
　　be zero
To prove
 limit[x→∞]f(x)*xα+β+1=0 ---eqn.AV095
we need two conditions
eqn.7.7 is necessary
  ∫[x=0,∞]x2αf(x)dx ＜ ∞ ---eqn.AV096=eqn.7.7
if our problem is non-trivial
Second necessary condition is
f(x) is monotone decreasing.
Problem statement use
[[
continuous, non-increasing function
]]
it is same as monotone decreasing

<a name="ch07a100">
Our goal is to prove a condition of
limit[x→∞]. We set x in [0,t] and
let t approach to infinity. 
In 2α, 2β, α+β, assume 2α is largest
one. if largest power is bounded,
other smaller power term must be
bounded. Now use integral by parts

<a name="ch07a103">
2010-02-03-20-01 here
We have given condition
0＜α,β＜∞, then α＞0, 2α+1＞1＞0
if f:[0,∞)→[0,∞) says
x, t ≧ 0 {blue [0,∞)} and
f(x) ≧ 0 {red [0,∞) }
Under above conditions
eqn.7.8b middle non-integral term
has every elements positive. Drop 
a positive term f(t)*t2α+1/(2α+1)
from eqn.7.8b, equality become 
next inequality.

<a name="ch07a105">
2010-02-03-20-27 here
We do not want a trivial problem.
We found a necessary bound eqn.7.7
eqn.7.8c tell us that eqn.7.7 bound 
right side integral when t→∞.
Back to equality eqn.7.8b, both 
integral are bounded when t→∞.
The difference of two integral
  f(t)*t2α+1/(2α+1)
must be bounded when t→∞. We can 
drop (2α+1), since (2α+1) do not
vary with t→∞.

<a name="ch07a106">
Here is the key point. 
  lim[t→∞]{f(t)*t2α+1}=c ---eqn.AV097
When t approach to infinity, 
f(t)*t2α+1 approach to a constant 
c. What is this c value ?

First from eqn.AV097 we find the
physics dimension relation
  f(t)*t2α = c/t ---eqn.AV098

<a name="ch07a107">
Non-zero c is the limit value of
  lim[t→∞]{f(t)*t2α+1}
any positive number smaller than c
(for example 0.9*c or 0.5*c etc.)
will never be approached by
  lim[t→∞]{f(t)*t2α+1}
<a name="ch07a108">
Assume this number is c/2 (range)
There is a border point T (domain)
on t axis, such that for all t＞T
function {f(t)*t2α+1} value
always greater than c/2. Then we 
find the following integral
  ∫[x=0,∞]{f(x)*x2α}dx ≧
  ∫[x=T,∞]{f(x)*x2α}dx ＝
  ∫[x=T,∞]{(c/2)/x}dx ＝ 
  (c/2)*∫[x=T,∞]{dx/x} ＝ 
  (c/2)*[log(∞)-log(T)] ---eqn.AV099

<a name="ch07a109">　Index begin　Index this file
eqn.AV099 first line '≧' is due to
positive value integral from 0 to 10
is greater than
positive value integral from 9 to 10
Simple reasom. Here 9 play 'T'
10 play '∞'.

eqn.AV099 second line '＝' is due to
eqn.AV098.

<a name="ch07a110">
eqn.AV099 third line '＝' is due to
∫dx/x = log(x).
The result log(∞) explode to infinity.
But we want non-trivial problem
eqn.7.7 is guarding. 

eqn.AV099 violate eqn.7.7. 
Conclusion: constant c=0 !!

<a name="ch07a111">
The equation
  lim[t→∞]{f(t)*t2α+1}=c ---eqn.AV097
is actually
  lim[t→∞]{f(t)*t2α+1}=0
Integration by parts assume boundary
terms are zero for example
For x→∞, f(x)*xα+β+1 is zero 
which is true. 
We assume boundary terms are zero 
in three integration by parts 
application all justified.
2010-02-03-21-18 stop

<a name="ch07a112">　Index begin　Index this file
2010-02-04-13-29 start
■　Problem 7.3 (A Pointwise Bound)
Show that if f:[0,∞)→Real satisfies
the two integral bounds

<a name="ch07a116">
2010-02-04-13-51 here
and consequently,
  √x*|f(x)|→0  as x→∞  ---eqn.AV102
2010-02-04-13-55 here

<a name="ch07a117">　Index begin　Index this file
■　Problem 7.3 discussion
Compare Problem 7.1, 7.2, 7.3
P.7.1 give f:(-∞,∞)→(-∞,∞)
P.7.2 give f:[0,∞)→[0,∞)
P.7.3 give f:[0,∞)→(-∞,∞)
<a name="ch07a118">
P.7.1 f(x) domain (-∞,∞)
P.7.2 f(x) domain [0,∞)
P.7.3 f(x) domain [0,∞) same as P.7.2
but
P.7.1 f(x) range  (-∞,∞)
P.7.2 f(x) range  [0,∞)
P.7.3 f(x) range  (-∞,∞) different from P.7.2

<a name="ch07a119">
Given condition is eqn.AV100 and 
eqn.AV101, involve f(x) and f'(x).
Target eqn.7.11 involve f(x) and 
f'(x). In calculus theorem which 
involve both f(x) and f'(x) is the
fundamental theorem of calculus
2010-02-04-14-08 stop
2010-02-04-16-05 start
<a name="ch07a120">
We use fundamental theorem of 
calculus and take absolute value,
find

<a name="ch07a122">　Index begin　Index this file
2010-02-04-16-30 here
where
  (∫[u=x,x+t]{1*1*du})1/2
  = (x+t-x)1/2 = t1/2 ---eqn.AV105
eqn.AV104 left side use absolute 
value, which is non-negative.
eqn.AV104 right side work for
positive term only, result is
also non-negative. Then we see
  -[right term]≦[left term] ---eqn.AV106
               ≦+[right term]
<a name="ch07a123">
left term is bounded from up and 
from below.
Left term is |f(x+t)－f(x)|, compare
with |f(x+t)|－|f(x)| we find
  |f(x+t)|－|f(x)| ≦ |f(x+t)－f(x)| ---eqn.AV107
for example
  |(-6)|-|(+3)|=3 ≦ 9=|(-6) - (+3)|
  |(+6)|-|(+3)|=3 ≦ 3=|(+6) - (+3)|
  |(+6)|-|(-3)|=3 ≦ 9=|(+6) - (-3)|
  |(-6)|-|(-3)|=3 ≦ 3=|(-6) - (-3)|
  |(-3)|-|(-6)|=-3≦ 3=|(-3) - (-6)|

<a name="ch07a124">
Another thing worth attention is
eqn.AV104 integration upper bound
change from u=x+t to u=∞.
The integrand |f ' (u)|2 ＞ 0
from u=x to u=∞, increment du ＞ 0
Then integrate from u=x to u=∞ has
greater value than from u=x to u=x+t
Put everything together, find upper
bound

<a name="ch07a127">　Index begin　Index this file
2010-02-04-17-15 here
Define short symbol for frequently
used integration as following.

Keep in mind that
F(x) relate to f(u) and
D(x) relate to f'(u) .
2010-02-04-17-27 here
<a name="ch07a129">
2010-02-04-17-41 start
Although read textbook at least three
times, but before write reading notes
it was rough, not digest well. Now
find something new.

<a name="ch07a130">
Problem 7.3 give f:[0,∞)→Real
"Real" governing function output.
Problem 7.3 allow f(x) be positive 
and negative like sine wave curve.
Textbook page 111 line seven says
[[
note that for any value of 0≦h 
such that 
  h1/2 ≦ f(x)/D(x) ---eqn.AV110
one has .....
]]
<a name="ch07a131">
Here h is a positive number and
h1/2 is positive. D(x) definition
say D(x) is a positive number. 
But problem 7.3 allow f(x) be 
negative. If f(x) happen to be 
negative, then eqn.AV110 is 
not true. 
<a name="ch07a132">
If change 
Textbook page 111 line seven to
[[
note that for any value of 0≦h 
such that 
  h ≦ [f(x)/D(x)]2 ---eqn.AV111
one has .....
]]
may be better.
2010-02-04-17-56 stop
2010-02-04-20-14 found errata
has correction.

<a name="ch07a133">　Index begin　Index this file
2010-02-04-18-03 start
Textbook page 111 line 7 use
  h1/2 ≦ f(x)/D(x)
Whether this relation correct 
physics dimensionally?

<a name="ch07a134">
Next find h, f(x) and D(x) relation.
x, t, u, h four variables are all
domain elements reside on x-axis.
In terms of physics dimension, they
are equivalent. Let us look at D(x)
definition eqn.AV109
Here only consider physics dimension

<a name="ch07a135">
We can write
  dim(D*D)=dim(f'*f')*dim(du) ---eqn.AV112
x, t, u, h are all domain elements
replace 'u' by 'h'
  dim(D*D)=dim(f'*f')*dim(h) ---eqn.AV113

<a name="ch07a136">
Integral sign '∫' and differential
sign 'd' both not carry physics 
dimension. Both '∫' and 'd' can be
dropped out of dimension equation.
On the other hand, f' = df(x)/dx
  dim(f') = f/x = f/h ---eqn.AV114
Put eqn.AV114 to eqn.AV113 find
  dim(D*D)=dim(f*f/h/h)*dim(h) ---eqn.AV115
<a name="ch07a137">
Now move h to left side and move D
to right side, get
  h*h/h = h = f*f/(D*D) ---eqn.AV116
Take squre root get
  h1/2 = f(x)/D(x) ---eqn.AV117
Finally verified textbook page 111
line 7 is correct dimensionally.
<a name="ch07a138">　Index begin　Index this file
Since h1/2 and f(x)/D(x) have same
dimension, comparison
  h1/2 ≦ f(x)/D(x) ---eqn.AV118
is possible.
2010-02-04-18-19 here
<a name="ch07a139">
eqn.AV118 dimensionally reasonable.
Whether numerically possible.
Do we worry h never ≦ f*f/(D*D) ?

<a name="ch07a140">
Textbook page 111 line seven says
[[
note that for any value of 0≦h 
such that 
  h1/2 ≦ f(x)/D(x) ---eqn.AV110
one has .....
]]
<a name="ch07a141">
Textbook say we are free to choose
h value, therefore 
  h ≦ f*f/(D*D)
is numerically possible.
2010-02-04-18-28 here

<a name="ch07a142">
2010-02-04-19-17 start
We found upper bound eqn.7.12 and
we found lower bound eqn.7.13
Textbook suggest lower bound will
help us solve problem 7.3. Problem
pioneer must spend time think. We
follower read just one line and
ride on right track.

<a name="ch07a143">　Index begin　Index this file
Since lower bound is useful, now 
rewrite lower bound in terms of D(x)
eqn.AV109, we will use it later.
  |f(x+t)|≧|f(x)|－t1/2*D(x) ---eqn.AV119

<a name="ch07a144">
Wrok start from F2(x) eqn.AV108
eqn.AV108 integrand is positive, 
then F2(x) integrate value for 
u∈[x,∞) is greater than same 
integrand integrate value for 
u∈[x,x+h].
Here 0≦x＜x+h＜∞ {f:[0,∞)→Real}
We find the following inequality

<a name="ch07a146">
Next change variable, let
  t = u - x  ---eqn.AV121
or
  u = t + x  ---eqn.AV122
variable u change to variable t.
Here x is constant, du = dt
Integration lower bound u=x
change to t=0
Integration upper bound u=x+h
change to t=h
eqn.AV120 change to next

<a name="ch07a149">
2010-02-04-20-25 here
LiuHH added absolute sign for 
"┃f(x)┃" in eqn.AV124.

eqn.AV124 is integration for t from
 t=0 to t=h. Integrand involve t.
How to integrate eqn.AV124?

<a name="ch07a150">
The textbook is inequality,
we can increase greater than side
value and keep valid inequality.
we can decrease less than side
value and keep valid inequality.
The following we decrease less
than side value in two places.
eqn.AV124 less than side has t. 
t is positive from t=0 to t=h＞0

<a name="ch07a151">
We do the following two changes.
First,
In (x+t)2 drop positive t,
change to (x)2
t disappear and
  (x+t)2≧(x)2 ---eqn.AV125

<a name="ch07a152">
Second,
In －t1/2*D(x) change t to its 
maximum value h. t disappear 
and
  －t1/2*D(x)≧－h1/2*D(x) ---eqn.AV126

<a name="ch07a153">　Index begin　Index this file
Variable t disappear from integrand,
x and h are constants, both move out
of integration sign. Integration of
eqn.AV124 simplified to ∫dt=h-0=h
( Thank God, it is inequality ! )
Bold 'h' goto eqn.AV127.
<a name="ch07a154">
We stretch eqn.AV124 to wider gap 
inequality as next.
  F2(x)≧ ---eqn.AV127
  hx2{|f(x)|-h1/2D(x)}2
eqn.AV125 and eqn.AV126 twice 
reduce eqn.AV124 less than side 
value, result is eqn.AV127.
Too many squares, take square 
root get
  F(x)≧ ---eqn.AV128
  h1/2x{|f(x)|-h1/2D(x)}
<a name="ch07a155">
In eqn.AV128, h1/2 is what we 
can free to choose. (h1/2 is
variable now) Write h1/2 as y
Write eqn.AV128 less than side
as
  g(y)=y*a{b-y*c}
or
  g(y)=a*{y*b-y*y*c} ---eqn.AV129
here a=x, b=|f(x)|, c=D(x)
<a name="ch07a156">
  g'(y)=a*{b-2*y*c} ---eqn.AV130
set g'(y)=0 find extremum value point
  y0=b/(2*c) ---eqn.AV131
Recover their original symbol
  h1/20=|f(x)|/(2*D(x)) ---eqn.AV132
This y0 has maximum value? or
minimum value? (together extremum)
Exam eqn.AV129, g(y) is quadratic
in y, y=h1/2＞0, when y→∞
-y*y*c bring g(y) to -infinity.
Then y0 has maximum value.

<a name="ch07a157">
Bring h1/20 in eqn.AV132 to
eqn.AV128 find
  F(x)≧ ---eqn.AV133
  h1/2x{|f(x)|-h1/2D(x)}
  = |f(x)|/(2*D(x))x{|f(x)|-|f(x)|/(2*D(x))*D(x)}
  = |f(x)|/(2*D(x))x{|f(x)|-|f(x)|/2}
  = |f(x)|/(2*D(x))x{|f(x)|/2}
  = x|f(x)|2/[4*D(x)]
<a name="ch07a158">
or
  x|f(x)|2 ≦ 4*F(x)*D(x)
or
  |f(x)|2 ≦ (4/x)*F(x)*D(x) ---eqn.AV134
recover F(x) and D(x) terms.
eqn.AV134 is same as the target
eqn.7.11 
<a name="ch07a159">
Re-write eqn.AV134 as
  x*|f(x)|2 ≦ (4)*F(x)*D(x)
take square root
  √x*|f(x)| ≦ 2*[F(x)*D(x)]1/2 ---eqn.AV135
when x→∞, eqn.AV135 greater than 
side is bounded. Refer to eqn.AV108
Less than side must approach to 
zero.
Problem 7.3 solved.
2010-02-04-21-18 stop

2010-02-06-12-10 done proofread
2010-02-06-12-48 done spelling check






<a name="docB001">　Index begin　Index this file
2010-02-02-11-50 start
This file contain the following
external javascript files
jsgraph2.js
tutelink.js
ineqinde.js
mathinc2.js
complex2.js
drawcod2.js
rocsitee.js
jslist1e.js
<a name="docB002">
Where 
graph code are jsgraph2.js, drawcod2.js
math  code are mathinc2.js, complex2.js
link  list  is tutelink.js
Inequality contents list ineqinde.js
Javascript program  list jslist1e.js
Freeman site banner file rocsitee.js

<a name="docB003">
If you want the drawing program work
must include both jsgraph2.js, drawcod2.js
They are at
http://freeman2.com/jsgraph2.js
http://freeman2.com/drawcod2.js
Other *.js files use same domain 
http://freeman2.com/filename.js

<a name="docB004">
mathinc2.js define more math functions
asinh(x) change to rasinh(x) // 'r' = real
acosh(x) change to racosh(x)
atanh(x) change to ratanh(x)

complex2.js define complex functions
asinh(x) and
acosh(x) and
atanh(x) have real and complex capability

<a name="docB005">
jsgraph2.js is XYGraph v2.3 code written by
Mr. J. Gebelein. You can find code at
http://www.structura.info/XYGraph/XYGraph.zip

mathinc2.js contain code written by other
authors.
2010-02-02-12-00 stop



<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56