<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="ch11a001">　Index begin　Index this file
2010-04-21-16-27 start
■■Chapter 11: Hardy's Inequality
　　and the Flop
Textbook page 166

Hardy's Inequality work with 
integration of a function f(x).
Chapter title say "Hardy's 
Inequality　and the Flop". 
<a name="ch11a002">
What is Flop? //here do flop
Textbook page 176 top say:
"If we have a factor on both 
side of an equation and if it
appears to a smaller power on
the 'right' than on the 'left'
then we can clear the factor 
to the left to obtain a new--
and potentially useful--bound."

<a name="ch11a003">
Textbook target eqn.11.1 has
two terms [∫f(u)du]/x and f(x)
They are (finally) separated.
During proving process, the
step of integration by parts
generate [∫f(u)du]/x * f(x)
two terms are multiplied. Until
eqn.11.7 two are separatable.
This separation process is 
called "Flop".

<a name="ch11a004">　Index begin　Index this file
2010-04-21-16-47 here
■　Problem 11.1 
(Hardy's Inequality)

Show that every integrable 
function f:(0,T)→Real 
satisfies the inequality

<a name="ch11a006">
and show, moreover, that the
constant 4 cannot be replaced
with any smaller value.
2010-04-21-17-00 stop


<a name="ch11a007">
2010-04-21-17-01 start
Following is a reading notes
parallel to textbook page 166
to 168 with some change.
Online paper use Hardy's 
Inequality which is slightly
generalized as following

<a name="ch11a009">　Index begin　Index this file
2010-04-21-17-27 here
If p=2, eqn.BH001 become eqn.11.1
eqn.11.1 use p=2 (squared), then
if f(x)≦0, f(x)_squared≧0.
eqn.11.1 do not use absolute
sign for f(x).
<a name="ch11a010">
If choose eqn.BH001 to start,
if p=3, negative^3 is negative.
Some paper require that f(x)≧0
Some paper use absolute sign 
for |f(x)|. To simplify equation
the following use eqn.BH001 and
require f(x)≧0.

<a name="ch11a011">
The term {∫[u=0,x]f(u)du}/x is
an integration average of f(u) 
over u=0 to u=x.
If f(u) is force, and if x is 
distance, variable u upper
bound is u=x. Then u and x 
have same physics dimension
distance, du/x is pure number.
{f(u)du}/x has same dimension 
as f(u) force.
{∫[u=0,x]f(u)du}/x is average 
force between distance=0 to x.

<a name="ch11a012">
eqn.11.1 tell us that the
integral of average_squared
is not too far from
integral of f(x)_squared

<a name="ch11a013">
eqn.11.1 greater than side has
coefficient 4. Whether this
number 4 is critical? The 
following is stress testing
(textbook page 167 line 3)

<a name="ch11a014">　Index begin　Index this file
Textbook set
  f(x)=xα ---eqn.BH002
to test constant C. To set f(x) 
to specific function xα, it is
for stress testing only. Target
at constant C=4. To prove eqn.11.1
f(x) is general. Start at here.
For stress testing, we have next 
equation.

<a name="ch11a017">
u integral is 
  ∫[u=0,x]uαdu
 =∫[u=0,x]d(uα+1)/(α+1)
 =(xα+1-0α+1)/(α+1)
 =(xα+1)/(α+1) ---eqn.BH004
//Power integration is simple
Integration average become
  {∫[u=0,x]uαdu}/x
 =(xα+1)/(α+1)/x
 =(xα)/(α+1) ---eqn.BH005
<a name="ch11a018">
eqn.11.4 is {term} in eqn.11.2
{term} shall be squared and
integrated from x=0 to x=T.
The result is
  ∫[x=0,T]dx*[(xα)/(α+1)]2
 ={∫[x=0,T]x2αdx}/[(α+1)2]
 ={∫[x=0,T]d(x2α+1)/(2α+1)}/[(α+1)2]
 ={T2α+1}/(2α+1)/[(α+1)2] ---eqn.BH006

<a name="ch11a019">　Index begin　Index this file
Similarly eqn.11.2 right side
integration has the value
  ∫[x=0,T]x2αdx
 =∫[x=0,T]d(x2α+1)/(2α+1)
 =(T2α+1-02α+1)/(2α+1)
 =(T2α+1)/(2α+1) ---eqn.BH007

<a name="ch11a020">
Now put  eqn.BH006 and eqn.BH007
back to eqn.11.2 get

<a name="ch11a021">
2010-04-21-18-29 here
Why it is called "stress testing"?
Because eqn.BH008 both side have
"(2α+1)" when it approach to zero
equation is stressed (divide by 0)
If α=-1/2, we can cancel non-zero 
term T2α+1 in eqn.BH008. But we can 
not cancel zero term (2α+1). We 
require 2α+1＞0, that is require
  α＞-1/2 ---eqn.BH009
<a name="ch11a022">
In the limit case α→-1/2
eqn.BH008 give us 
  lim[α→-1/2]C→1/[(-1/2)+1]2
(non-zero 2α+1 drop out)
[(-1/2)+1]2 has value 4
We get C=4 as the limit value.

<a name="ch11a023">
"limit value" means that C=4 is 
critical or is best. C a little 
bit higher than 4, α＞-1/2 OK
If C=4, α = -1/2, then violate 
the rule α＞-1/2
2010-04-21-18-36 here

Above is stress testing for 
eqn.11.1
<a name="ch11a024">　Index begin　Index this file
Parallel to eqn.11.2, now see
eqn.BH001. (in eqn.BH001
 set p=2 get eqn.11.2)
In eqn.BH001, replace red 
(p/(p-1))p with C
and replace f(x) with xα
calculate as following.

<a name="ch11a026">
Less than side {term} can be
copied from eqn.BH005.
Instead of square eqn.BH005,
now take p-th power to
eqn.BH005 and integrate 
from x=0 to x=T.
<a name="ch11a027">
The result is
  ∫[x=0,T]dx*[(xα)/(α+1)]p
 ={∫[x=0,T]xpαdx}/[(α+1)p]
 ={∫[x=0,T]d(xpα+1)/(pα+1)}/[(α+1)p]
 ={Tpα+1}/(pα+1)/[(α+1)p] ---eqn.BH011

<a name="ch11a028">
Similarly eqn.BH010 right side
integration has the value
  ∫[x=0,T]xpαdx
 =∫[x=0,T]d(xpα+1)/(pα+1)
 =(Tpα+1-0pα+1)/(pα+1)
 =(Tpα+1)/(pα+1) ---eqn.BH012

<a name="ch11a029">　Index begin　Index this file
Now put eqn.BH011 and eqn.BH012
back to eqn.BH010 get

<a name="ch11a030">
2010-04-21-19-29 here
If α=-1/p, we can cancel non-zero 
term Tpα+1 in eqn.BH013.
But we can not cancel zero
term (pα+1). We require pα+1＞0
that is require
  α＞-1/p ---eqn.BH014
<a name="ch11a031">
In the limit case α→-1/p
eqn.BH013 give us 
  lim[α→-1/p]C→1/[(-1/p)+1]p
(non-zero pα+1 drop out)
[(-1/p)+1]p has value (p/(p-1))p
We get C=(p/(p-1))p as the 
limit value.
If p=2, (p/(p-1))p=4.

<a name="ch11a032">
Square in eqn.11.1 and
p-th power in eqn.BH001
are parallel
Coefficient C in eqn.11.1 and
coef. (p/(p-1))p in eqn.BH001
are parallel.
Parallel analysis is success.

<a name="ch11a033">
Above is special case
  f(x)=xα ---eqn.BH002
for stress testing.
The following is general case
f(x) be arbitrary.
2010-04-21-19-48 stop

<a name="ch11a034">　Index begin　Index this file
2010-04-21-20-42 start
LiuHH try to prove eqn.BH001 
instead of eqn.11.1.
In eqn.BH001, if set p=2,
it is same as eqn.11.1.
Prove method still parallel
with textbook.

<a name="ch11a035">
We start from eqn.BH001 less
than side. Textbook suggest
use integration by parts.
//Previous notes: ch07a040

Calculus product rule tell us
that
  d[u(x)*v(x)]/dx
 = u(x)*d[v(x)]/dx
 + v(x)*d[u(x)]/dx ---eqn.BH015

<a name="ch11a036">
Let u'(x)=d[u(x)]/dx ---eqn.BH016
let v'(x)=d[v(x)]/dx ---eqn.BH017
then 
  d[u(x)*v(x)] ---eqn.BH018
 =[u(x)*v'(x)+u'(x)*v(x)]dx
<a name="ch11a037">
Integration get
  ∫d[u(x)*v(x)]
 =∫u(x)*v'(x)dx
 +∫u'(x)*v(x)dx ---eqn.BH019
or
<a name="ch11a038">
   ∫u(x)*v'(x)dx
 =-∫u'(x)*v(x)dx
  +∫d[u(x)*v(x)] ---eqn.BH020
eqn.BH020 is common used form.
∫d[u(x)*v(x)] is boundary term

<a name="ch11a039">　Index begin　Index this file
re-write eqn.BH001 less than 
side as next (not in textbook)

<a name="ch11a041">　Index begin　Index this file
2010-04-21-21-38 here
eqn.BH022 right side is boundary 
term. It evaluate at two points
x=T and x=0.
boundary term evaluation are

<a name="ch11a043">
2010-04-21-22-03 here
Upper side x=T has no problem.
Lower side x=0 is 0/0 which is
undetermined.
We need step back and see what
is the limit value of this 0/0

<a name="ch11a045">
We need L'Hopital's Rule
If g(a)=h(a)=0 and g'(a), h'(a)
exist, assume h'(a)≠0, then
  lim[x→a]g(x)/h(x)
         =g'(x)/h'(x) ---eqn.BH027

<a name="ch11a046">　Index begin　Index this file
Now in eqn.BH026
g(x) is {∫[u=0,x]f(u)du}p
h(x) is xp-1
point a is a=0
g(0)=0 because integration domain
       ∫[u=0,0] is zero.
h(0)=0 because p-1＞0, then 0p-1=0
<a name="ch11a047">
L'Hopital's Rule tell us to
differentiate both g(x) and h(x)
and evaluate g'(a)/h'(a) 
( here []' represent multiple
  differentiation)
until not see 0/0. The last 
g'(a)/h'(a) is answer.

<a name="ch11a048">
Look at eqn.BH026,
{∫[u=0,x]f(u)du}p has power p,
differentiate p times to constant
xp-1 has power p-1,
differentiate p-1 times to 
constant.
<a name="ch11a049">
The competition result is that
after differentiate p-1 times
{∫[u=0,x]f(u)du}p still has 
power one, ∫[u=0,0] is zero.
xp-1 become constant (p-1)!
We get 0/(p-1)! which is zero.

<a name="ch11a050">
Lower boundary term eqn.BH025 is 
zero !!
Upper boundary term eqn.BH024≦0

Because p＞1, then －p+1＜0. 
eqn.BH024 integration≧0

<a name="ch11a051">　Index begin　Index this file
eqn.BH022 is equality with a
negative boundary term. 
What to do? 
This is inequality class.
Drop negative boundary term. 
eqn.BH022 change from equality
to inequality 
<a name="ch11a052">
eqn.BH022 remaining non-boundary
term will be eqn.BH001 greater 
than side. 
But still need flop operation.
2010-04-21-22-46 stop

<a name="ch11a053">
2010-04-22-11-36 start
We start at eqn.BH021=eqn.BH022
Integration by parts tell us that
boundary terms are non-positive,
we dropped boundary term from
eqn.BH022.
 then  eqn.BH021=eqn.BH022 
become eqn.BH021≦eqn.BH022
<a name="ch11a054">
The whole inequality is next

2010-04-22-11-51 here
<a name="ch11a055">
2010-04-22-12-16 start
eqn.BH028 greater than side 
has a differentiation to an 
integration. Variable x located 
at integration upper bound. We 
need Leibniz's rule
eqn.BH028 become next

<a name="ch11a059">
2010-04-22-12-58 here
Let us pay attention to the
blue term {∫[u=0,x]f(u)du}/x
and red term f(x)
Blue term show up at both less
than side and greater than side. 
<a name="ch11a060">
Red term show up at greater
than side only. Can we make 
blue at one side and red at
the other side of inequality?
Flop operation is to separate
blue from red.
2010-04-22-13-15 stop

<a name="ch11a061">　Index begin　Index this file
2010-04-22-14-31 start
To simplify work, textbook 
define blue term as φ(x) and
define red term as ψ(x) 

<a name="ch11a065">
where C=2.
The critical feature of this
inequality (11.6) is that the
function φ(x) is raised to a 
higher power on the left side
of the equation than on the
right. This is far from a 
minor detail; it open up the
possibility of a maneuver 
which has featured in thousands
of investigations.

<a name="ch11a066">　Index begin　Index this file
Now apply Schwarz's inequality to 
the right hand side of eqn.BH031.
Get the following [ignore p/(p-1)]

<a name="ch11a072">
Red terms are common at both side.
Cancel a square root of them, only
less than side left a square root 
of red term. We have

<a name="ch11a074">
Refer to eqn.BH033,
p/(p-1) become C in eqn.11.8
with p=2 , C=2/(2-1)=2
eqn.11.8 take square, recover
φ(x) and ψ(x) we have target 
equation eqn.11.1.

Problem 11.1 is solved.
2010-04-22-17-11 stop

<a name="ch11a075">
2010-04-22-19-25 start
Now try solve eqn.BH001
Get trouble at this step
[[
Now apply Schwarz's inequality to 
the right hand side of eqn.BH031.
]]
<a name="ch11a076">　Index begin　Index this file
Because Schwarz's inequality
use square root operation.
That is in
  1/p + 1/q = 1 ---eqn.BH034
Schwarz's inequality insist
p=q=2.
<a name="ch11a077">
Exercise 9.7
Holder's Inequality for Integrals
eqn.BC001 can be used.
Re-write eqn.BC001 as following
but change p,q to m,n. Because
eqn.BH031 use p and can not be 
mixed.

<a name="ch11a079">
where, as usual, 1＜m＜∞ and
  m-1+n-1=1 ---eqn.BC002

Now apply Integral Holder to 
the right hand side of eqn.BH031
Consider
  eqn.BC001 f(x) = φ(x)p-1 ---eqn.BH035
  eqn.BC001 g(x) = ψ(x)  ---eqn.BH036
  eqn.BC001 w(x) ≡ 1 ---eqn.BH037
get the following

<a name="ch11a081">　Index begin　Index this file
eqn.BH031 less than side is 
          our concern
eqn.BH038 less than side is
          intermediate step
eqn.BH038 greater than side 
          is our concern
Now put eqn.BH031 less than 
side and eqn.BH038 greater 
than side into one equation
as following

<a name="ch11a083">
Pay attention to φ(x) power.
Less than side φ(x) power is p
Greater than side φ(x) power is
(p-1)*m where m is variable.
We want 
  p=(p-1)*m ---eqn.BH040
so that eqn.BH039 two sides have
similar term which can be merged.
<a name="ch11a084">
Solve eqn.BH040 for m get
  m=p/(p-1) ---eqn.BH041
Because
  1/m + 1/n = 1 ---eqn.BH042
solve for n get
  1/n = 1-1/m = 1-(p-1)/p = 1/p
get
  n = p ---eqn.BH043

<a name="ch11a085">
Now put m and n into eqn.BH039
get

<a name="ch11a089">
Recover φ(x) and ψ(x) from
eqn.11.5A and eqn.11.5B.
we find that
eqn.BH047 is the solution of
generalized problem eqn.BH001
2010-04-22-20-43 stop


<a name="ch11a090">　Index begin　Index this file
2010-04-23-12-33 start
■　Problem 11.2 
The Discrete Hardy's Inequality

Show that for any sequence of 
nonnegative real numbers a1,a2,
...,an one has the inequality

2010-04-23-12-59 stop
<a name="ch11a093">
2010-04-23-14-11 start
eqn.BH048 tell us that 
Problem 11.2 is a two
sequence comparison problem.
First sequence is  a1,a2,...,an
Second sequence is 
gradually AMed first sequence.
(see eqn.BH048 less than side)
<a name="ch11a094">
eqn.BH048 compare the sum
of element squared result.
If without factor 4, second 
sequence has greater value.
Second sequence square sum is
not greater than four times
of first sequence square sum.

<a name="ch11a095">　Index begin　Index this file
gradually AMed sequence is
not new. We saw gradually GMed 
sequence eqn.2.15 before.

<a name="ch11a096">
Integral Hardy's inequality
start from target equation
eqn.11.1 less than side 
using integration by parts
After we drop boundary term,
we get "preflop" inequality
eqn.11.4

<a name="ch11a097">
Discrete Hardy's Inequality
start from target equation
eqn.11.9 less than side via 
summation by parts, we get 
the next eqn.11.10

<a name="ch11a099">
Let us define new {b} sequence 
Its element bn is defined as
  bn=(a1+a2+...+an)/n ---eqn.BH049
//bn and An eqn.BH072 are identical
{b} sequence is 
gradually AMed {a} sequence.
// AM=Arithmetic Mean 

2010-04-23-15-09 stop
<a name="ch11a102">
2010-04-23-16-29 start
Summation by parts is the
natural analog of integration 
by parts, although it is a bit 
less mechanical. Here, for 
example, we must decide how to
represent 1/n2 as a difference.
<a name="ch11a103">
There are two choices.
First define sn as sum of 1/k2 
from k=n to k=∞
Second define s~n as sum of 1/k2 
from k=1 to k=n
Mathematics expression is next

<a name="ch11a105">　Index begin　Index this file
Above two summation are very
different.
 Red sn sum from k=n to k=∞
Blue s~n sum from k=1 to k=n
Red sn sum better? or
Blue s~n sum better?
Textbook suggest experimentation
and choose red sn sum.

<a name="ch11a106">
We define TN=eqn.11.9 less than 
side, use sn to express TN as 
following

<a name="ch11a108">
  sn sum 1/k2 from k=n to k=∞
sn+1 sum 1/k2 from k=n+1 to k=∞
sn－sn+1 remain only 1/n2
Expand eqn.BH054 as following.

<a name="ch11a111">　Index begin　Index this file
eqn.BH055 positive to eqn.BH056 
positive (black) not change.
eqn.BH055 negative (blue) to 
eqn.BH056 negative (red) changed.

<a name="ch11a112">
Verify.
Blue start from n=1, for n=1
sn+1*(a1+a2+...+an)2
become s2*(a1)2 ---eqn.BH057
Red  start from n=2, for n=2
sn*(a1+a2+...+an-1)2
become s2*(a1)2 ---eqn.BH058
eqn.BH057 and eqn.BH058 are 
the same.

<a name="ch11a113">
Blue end at n=N, for n=N
sn+1*(a1+a2+...+an)2
become sN+1*(a1+a2+...+aN)2 ---eqn.BH059
Red  end at n=N+1, for n=N+1
sn*(a1+a2+...+an-1)2
become sN+1*(a1+a2+...+aN)2 ---eqn.BH060
eqn.BH059 and eqn.BH060 are 
the same.

<a name="ch11a114">
Shift index of negative term
from eqn.BH055 to eqn.BH056
is correct.

<a name="ch11a115">
Now let us target at eqn.BH056
sn unified for positive and
negative terms. But their sum
range are different. 
We keep common (between positive
and negative in eqn.BH056) and 
isolate not paired terms.
2010-04-23-17-55 here

<a name="ch11a116">　Index begin　Index this file
In our focus eqn.BH056
positive sum from n=1 to n=N
negative sum from n=2 to n=N+1

positive n=1 is single
negative n=N+1 is single
from n=2 to n=N are paired

<a name="ch11a117">
TN=  ---eqn.BH061 begin
 +s1*a12
 +∑[n=2,N]sn*(a1+a2+...+an)2
 -∑[n=2,N]sn*(a1+a2+...+an-1)2
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH061 end

<a name="ch11a118">
First two lines come from positive.
Last two lines come from negative.
Middle two lines ∑[n=2,N] has the
form c*c-d*d, which is (c+d)(c-d)

<a name="ch11a119">
TN=  ---eqn.BH062 begin
 +{ ∑[n=2,N]sn*
  [(a1+a2+...+an)+(a1+a2+...+an-1)]
 *[(a1+a2+...+an)-(a1+a2+...+an-1)]
  }
 +s1*a12
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH062 end

<a name="ch11a120">
TN=  ---eqn.BH063 begin
 +{ ∑[n=2,N]sn*
  [2*(a1+a2+...+an-1) + an]
 *[an]
  }
 +s1*a12
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH063 end

<a name="ch11a121">　Index begin　Index this file
TN=  ---eqn.BH064 begin
 +{∑[n=2,N]sn*[2*an(a1+a2+...+an-1)
               + an*an]
  }
 +s1*a12
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH064 end

<a name="ch11a122">
Here, we insert two zeros.
TN=  ---eqn.BH065 begin
 +{∑[n=2,N]sn*[2*an(a1+a2+...+an-1)
               + an*an]
  } +s1*a12
 +(s1*a12-s1*a12) //red is zero
 +∑{n=2,N}(sn*an2-sn*an2) //blue is zero
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH065 end

<a name="ch11a123">
TN=  ---eqn.BH066 begin
 +{∑[n=2,N]sn*[2*an(a1+a2+...+an-1)
               + an*an]
  } +s1*a12
 +(s1*a12-s1*a12)
 +∑{n=2,N}(sn*an2-sn*an2)
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH066 end

<a name="ch11a124">
Purple s1*a12 contribute to change 
from ∑[n=2,N] to ∑[n=1,N]
Green sn*an2 contribute to extend
from (a1+a2+...+an-1)
 to  (a1+a2+...+an)
After they move to right place

<a name="ch11a125">
TN=  ---eqn.BH067 begin
 +{∑[n=1,N]sn*2*an(a1+a2+...+an)}
 -s1*a12
 -∑{n=2,N}sn*an2
 -sN+1*(a1+a2+...+aN)2 ---eqn.BH067 end

<a name="ch11a126">　Index begin　Index this file
Up to here eqn.BH067 is equality.
We drop three negative terms.
TN become less than or equal to
the positive term. 
// 5=7-2, drop '-2', get 5＜7
Recover TN original face, 
we have

<a name="ch11a128">
eqn.11.11 is very close to our 
goal eqn.11.10.
If we want flop eqn.11.11, that 
is to move (a1+a2+...+an) from
right side to left side. We need
the red sn change to 1/n, to meet 
left side (a1+a2+...+an)/n
2010-04-23-19-10 stop

<a name="ch11a129">
2010-04-23-20-06 start
sn is defined in eqn.BH052 
Next find sn and 1/n relation.

<a name="ch11a132">
eqn.BH069 red equality is 
telescoping result.
Expand eqn.BH069 summation term,
except first 1/(n-1), all other
terms are cancelled by the next
twin terms. Result is 1/(n-1)
which is less than equal to 2/n
for n greater than one.

<a name="ch11a133">
The case n=1, s1 is
  s1= 1/1 + s2---eqn.BH070
eqn.BH068 and eqn.BH069 tell
us that s2≦2/2, so
  s1≦ 1/1 + 2/2 = 2
or
  s1≦ 2/n = 2/1---eqn.BH071
For all n≧1, we have

<a name="ch11a135">　Index begin　Index this file
eqn.11.12 let us change the 
bound from sn to 2/n
With this replacement, eqn.11.11 
become

<a name="ch11a137">
Busy few hours, change from
sn to 2/n and get the 1/n.
Our goal is eqn.11.10
Compare eqn.11.10 with eqn.11.13
there is one point different.
eqn.11.10 require coefficient 2
eqn.11.13 give us coefficient 4.
Integral version eqn.11.4 use 2
2010-04-23-21-02 here
2010-04-23-21-17 start
If we flop eqn.11.13, we get 8
but target eqn.11.9 require 4.

<a name="ch11a138">
Go back to eqn.11.10 which
textbook reprint as eqn.11.14.

<a name="ch11a139">
Above two failures both start
from eqn.11.9 less than side
and use summation by parts. The
next third method, textbook use
delta calculation. To simplify
expression, define
  An=(a1+a2+...+an)/n ---eqn.BH072
//bn eqn.BH049 and An are identical
Attention: An definition do not 
contain summation (∑[n=1,n=N])

<a name="ch11a140">　Index begin　Index this file
Define
  Δn=An*An-2*An*an ---eqn.BH073
Δn do not contain summation.
Δn is eqn.11.14 less than
side minus greater than side 
and no n-sum.
Less than side minus greater 
than side, that is Δn≦0
Below, we verify Δn≦0
eqn.11.14 is simply summation 
of Δn

<a name="ch11a141">
Δn definition contains an. First 
thing to do is to eliminate an.
Express Δn in terms of An and
An-1

<a name="ch11a142">
  an=(a1+a2+...+an)
    -(a1+a2+...+an-1)  ---eqn.BH074
  an=An*n - An-1*(n-1) ---eqn.BH075
Use eqn.BH075 to eliminate an in
eqn.BH073, get
  Δn=An*An-2*An*[An*n - An-1*(n-1)]
  Δn=(1-2n)*An*An+2*(n-1)*An*An-1 ---eqn.BH076
<a name="ch11a143">
We do not want the term An*An-1
We want An2+An-12
Humble bound will give us
  An*An-1≦(An2+An-12)/2 ---eqn.BH077
Then we have
  Δn≦(1-2n)*An2+(n-1)*(An2+An-12) ---eqn.BH078
or
  Δn≦(n-1)*An-12-n*An2 ---eqn.BH079
<a name="ch11a144">
Now we sum Δn from n=1 to n=N

<a name="ch11a146">
Red cancel red, blue cancel blue,
purple cancel purple, all the way
to the end. Only two terms left 
are first and last
  (1－1)*A1－12－N*AN2
But (1-1)=0, the net result is
  ∑[n=1,N]Δn≦－N*AN2 ---eqn.BH081

<a name="ch11a147">
∑[n=1,N]Δn is short writing of
eqn.11.14 which is same as 
eqn.11.10
Delta calculation let us verified
eqn.11.10.
2010-04-23-22-28 here

<a name="ch11a148">
eqn.11.10 is same as eqn.BH051 
It is easier to see from eqn.BH051.
(eqn.BH051 is verified with Delta
 calculation)
Apply Cauchy's inequality to
 eqn.BH051 right hand side
(eqn.11.10 right hand side)
we get
  ∑[n=1,N]bn2≦2*∑[n=1,N]bnan
 ≦2*√{∑[n=1,N]bnbn}*√{∑[n=1,N]anan} ---eqn.BH082
//bn and An are identical; an not
<a name="ch11a149">
eqn.BH082 first line (black≦red)
is eqn.11.10
eqn.BH082 red part is Cauchy's 
inequality.
Head and tail of eqn.BH082
  ∑[n=1,N]bn2 ---eqn.BH083
 ≦2*√{∑[n=1,N]bn2}*√{∑[n=1,N]an2}
is equation just about flop.
<a name="ch11a150">
We move blue term to one side 
of the equation, this separation
is flop. The result is
  √{∑[n=1,N]bn2} ---eqn.BH084
 ≦2*√{∑[n=1,N]an2}
<a name="ch11a151">
bn is defined at eqn.BH049
Recover bn and square eqn.BH084
we get our goal eqn.11.9

Problem 11.2 solved.
2010-04-23-22-50 stop

2010-04-24-21-40 done first  proofread
2010-04-25-12-02 done second proofread
2010-04-25-12-40 done spelling check

<a name="ch11a152">
2010-04-25-23-42
"Update 2010-04-26" add three "∑{n=2,N}"
You can still find many small bugs.
No one proofread for me.



<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