<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="ch11b001">　Index begin　Index this file
2010-04-26-12-30 start
■　Problem 11.3 (Carleson's
　　Convexity Inequality)
Textbook page 173

<a name="ch11b002">
Show that if φ:[0,∞)→Real is
convex and
  φ(0)=0 ---eqn.BI001
//Reason for φ(0)=0 (a), (b)
then for all -1＜α＜∞ one has 
the integral bound

2010-04-26-12-53 stop

<a name="ch11b004"> skip drawing
2010-04-27-15-40 start 
Please access the following file
for Carleson's Inequality curve
http://freeman2.com/carleson.htm
local (drop 'http://freeman2.com/')
carleson.htm draw textbook 
Page 174, fig.11.1. 
<a name="ch11b005">　Index begin　Index this file
There are three run buttons
[Shift curve prob1103]
[Draw A User Data]
[Draw B Default]
First button [Shift curve] is 
initial code. Later find this
button not really help. But
first button [Shift curve] has
code to change function variable 
from 'x' to '(x+(-1.2))'. It
has programming reference value.
<a name="ch11b006">
Second button [Draw A] use user
data.
Third button [Draw B] use default
data and erase user data.
Both [Draw A] and [Draw B] draw
Page 174, fig.11.1. Program let
user define f(x). 

<a name="ch11b007">
You should define convex function 
for Carleson's Inequality curve.
Convex function is function 
shape like a bowl ╰╯ 
In convex section f''(x)＞0

<a name="ch11b008">
Concave function is function 
shape like a hat ╭╮.
In concave section f''(x)＜0

<a name="ch11b009">
Sine curve is partial convex
partial concave. You must choose
domain such that curve look like
a bowl ╰╯ .
2010-04-27-15-57 stop

<a name="ch11b010">　Index begin　Index this file
2010-04-27-16-24 start
Integrand of eqn.11.15 is a
composite function.
For -1＜α＜∞, xα is linear 
 at α=0 : f(x)=x0≡1 ---eqn.BI002
and α=1 : f(x)=x1=x ---eqn.BI003
For -1＜α＜∞, xα has convex 
and concave sections which is 
non-linear. (draw f(x)=x^α)

<a name="ch11b011">
Integrand of eqn.11.15 use exp(x).
  f(x)=exp(x) ---eqn.BI004
which is convex ╰╯ for all x 
in real.

<a name="ch11b012">
Integrand of eqn.11.15 use φ(x)
which is not specified, just
require φ(x) be convex.
If φ(x) is convex, 
then -φ(x) is concave. (sure)
 and -φ(x)/x is
 convex    ╰╯  -(x*log(x))/x
or linear  ／  -pow(x,2)/x
or concave ╭╮ -(exp(0.8*x)-1)/x

<a name="ch11b013">
Convexity of integrand of (11.15)
is complicated. But problem 11.3
concentrate at convexity of φ(x).
Convexity of φ(x) alone is enough
to solve problem 11.3.

<a name="ch11b014">
Convexity always gives us some-
thing useful; in particular, 
convexity provides an estimate of
the shift difference
  φ(x+t)-φ(x) ---eqn.BI005
<a name="ch11b015">
Unfortunately this estimate 
does not seem to help us much 
here. The way Carleson cut the 
Gordian knot was to consider 
instead the scale shift 
difference
  φ(px)-φ(x) ---eqn.BI006
where p＞1 is a parameter that
we can optimize later.
Reason for p＞1 (a), (b)

<a name="ch11b016">　Index begin　Index this file
We change variable as following
  x→py ---eqn.BI007
x, p are variables. y is constant
[y,φ(y)] is a given tangent point.
When p=1, py sit on tangent point.
Horizontal axis change from
x-axis to py-axis.
Vertical function value axis
change from f(x) to f(py) axis.

<a name="ch11b017">
From convexity of φ(x) we know
that at convexity section, a 
tangent line of φ(x) is always
below curve. Tangent and curve
only equal at tangent point one
point. Please see figure 11.1
local Click [Draw B] button.
Red curve is convex. Red tangent
is below red curve in convex
section. Tangent point is 'A'.
2010-04-27-17-02 stop
<a name="ch11b018">
2010-04-27-18-04 start
For any given tangent point,
horizontal axis y value is
given, constant, not change.
'p' in φ(py) is a variable.
(p＞1) When p varies, point
[py, φ(py)] move on curve.
(not move on tangent)

<a name="ch11b019">
Tangent line is specified as
  t(p)=φ(y)+(py-y)*φ'(y) ---eqn.BI008
't' in 't(p)' means tangent.
'p' in 't(p)' means variable.
Worth attention is that 'y'
in eqn.BI008 is not a variable.
'y' is constant in eqn.BI008.
Because we freeze the tangent
line, then y can not change.
//eqn.BI011 reverse y and p

<a name="ch11b020">　Index begin　Index this file
eqn.BI008 say that tangent and
curve touch at the point [y,φ(y)]
Here y measured on horizontal
py-axis and φ(y) measured on
vertical f(py) axis.
eqn.BI008 whole equation is
measured on vertical f(py) axis.
<a name="ch11b021">
(py-y) say that change (p) on
horizontal axis in amount of
(py-y), the slope φ'(y) of
tangent line will change in
vertical f(py) axis in the
amount of (py-y)*φ'(y).
<a name="ch11b022">
p=1 is a sign of no change.
Because if p=1, then 
  (py-y)=y-y=0 ---eqn.BI009
that is no change.

<a name="ch11b023">
Clearly eqn.BI008 is a linear
equation, because variable p
has first power. A linear 
equation is a straight line.

<a name="ch11b024">
Convexity of φ(x) tell us that
tangent line of φ(x) is always
below curve. In other words,
point on curve is above or
equal point on tangent.

<a name="ch11b025">　Index begin　Index this file
point on curve is φ(py)
point on tangent is
 φ(y)+(py-y)*φ'(y)
Convexity condition in math.
form is
  φ(py)≧φ(y)+(py-y)*φ'(y) ---eqn.11.16
<a name="ch11b026">
Equality is true only at one
point where two tangent, and
p=1 at tangent point.

Figure 11.1 main point is to
illustrate eqn.11.16. local
2010-04-27-18-36 here
<a name="ch11b027">
2010-04-27-18-49 start
View from curve and tangent
relation, p≦1 or p≧1, curve
is always above tangent. Even
if p＜1 still true. But text-
book require p＞1
Reason for p＞1 (a), (b)

<a name="ch11b028">
The reason require p＞1 is
that Holder inequality asks
  1/p+1/q = 1  ---eqn.BI010
and p＞1 and q＞1.

<a name="ch11b029">
To prove problem 11.3, we start
from eqn.11.15 less than side
with small change. 
eqn.11.15 less than side integral
upper bound is infinity. Now
change to finite at x=A＜∞.
//otherwise second point blurred
Later we let A approach to ∞
to get answer. Define IA first

<a name="ch11b031">
2010-04-27-19-15 here
eqn.BI011 first line is problem
given equation (but upper bound
change from infinity to A).
From first line to second line
is change variable
  x→py ---eqn.BI007
xα contribute to pα
dx contribute to p+1. 
Together get pα+1

<a name="ch11b032">
Here, we apply eqn.11.16 to red
term －φ(py) in eqn.BI011 
eqn.11.16 is
  φ(py)≧φ(y)+(py-y)*φ'(y) ---eqn.11.16
We need negative of it
 -φ(py)≦-φ(y)-(p-1)*y*φ'(y) ---eqn.BI012
Reverse sign, reverse inequality.
<a name="ch11b033">
In eqn.BI011, -φ(py) is a 
parameter of exp() function.
exp() is a monotone increase
function. That means
if m＜n, then exp(m)＜exp(n).
if m＞n, then exp(m)＞exp(n).
eqn.BI012 is a given inequality.
[problem require convex φ(x)]
Denominator py is positive, 
not change inequality direction.
//p＞1 and y domain [0,inf)
<a name="ch11b034">
eqn.BI011 change from equality 
to inequality as the following

<a name="ch11b035">　Index begin　Index this file
Inequality from eqn.BI011 
to eqn.BI013 has two causes.
First eqn.BI012 is inequal,
monotone increase function exp() 
follow the same pattern and
cause inequality in eqn.BI013.

<a name="ch11b036">
Second, integration upper bound
change from y=A/p to y=A.
Problem require p＞1
Assume p=2, assume A=10, from
y=A/p=10/2=5 to y=A=10 is an
extension of integration domain.
eqn.BI013 integrand is positive.
[ exp(negative) still ＞0 ]
extend domain increase integral
value. If use ∞ as upper bound,
we can not present second point.
(second point blurred)

<a name="ch11b037">
Here, about to apply Holder's
inequality. Holder ask
  1/p+1/q = 1  ---eqn.BI010
From eqn.BI010 find q to be
  q=p/(p-1)  ---eqn.BI014

<a name="ch11b038">
We know that
  exp(m+n)=exp(m)*exp(n) ---eqn.BI015
eqn.BI013 change to the following

2010-04-27-21-48 stop
<a name="ch11b046">
2010-04-28-14-11 start
eqn.BI020 is still different 
from our goal: greater than 
side of eqn.11.15.

<a name="ch11b047">
eqn.BI020 integration upper 
bound is A
eqn.11.15 integration upper 
bound is ∞
and
eqn.BI020 coef. p(α+1)p/(p-1)
eqn.11.15 coef. e(α+1)

<a name="ch11b048">
We let upper bound A approach
to infinity, solve the first
difference.
For second difference, we let
p approach to one 
p(α+1)p/(p-1) = [pp/(p-1)](α+1)
<a name="ch11b049">
The limit value of pp/(p-1) is
  lim[p→1]pp/(p-1)=e ---eqn.BI021
  =2.71828182845904...
Then eqn.BI020 match eqn.11.15
greater than side exactly.
2010-04-28-14-27 stop

<a name="ch11b050">　Index begin　Index this file
2010-04-28-15-03 start
Now show that
  lim[p→1]pp/(p-1)=e ---eqn.BI021
Let
  m=pp/(p-1) ---eqn.BI022

<a name="ch11b051">
exp() and log() are reverse to
each other. That is
  exp(log(c))=c ---eqn.AX085
therefore
  m=exp{log(pp/(p-1))} ---eqn.BI023
  m=exp{[p/(p-1)]*log(p)} ---eqn.BI024
<a name="ch11b052">
Define
  g(p)=p*log(p)/(p-1) ---eqn.BI025
To find 
  lim[p→1]pp/(p-1)
need find
  lim[p→1]g(p) ---eqn.BI026
 =lim[p→1]{p*log(p)/(p-1)}
<a name="ch11b053">
Direct substitution get
  g(1)={1*log(1)/(1-1)}=0/0 ---eqn.BI027
We need L'Hopital's Rule
  lim[p=1]g(p)={d[p*log(p)]/dp}
              / {d(p-1)/dp} ---eqn.BI028
<a name="ch11b054">
Numerator differentiation is
  d[p*log(p)]/dp=1*log(p)+p/p ---eqn.BI029
Denominator differentiation is
  d(p-1)/dp = 1-0 = 1 ---eqn.BI030

<a name="ch11b055">　Index begin　Index this file
then
  lim[p→1]{p*log(p)/(p-1)}
 =lim[p→1]{[1*log(p)+p/p]/1}
 = [1*log(1)+1/1]/1 = 1 ---eqn.BI031

<a name="ch11b056">
eqn.BI031 is inside of eqn.BI026
The answer of eqn.BI021 is
  lim[p→1]pp/(p-1) ---eqn.BI032
 =lim[p→1]exp{p*log(p)/(p-1)}
 =exp{1}
 =e=2.71828182845904...

Problem 11.3 is done.
2010-04-26-12-30 start
2010-04-28-15-27 stop

<a name="ch11b057">
2010-04-28-17-49 start
Problem 11.3 require φ(y)=φ(0)=0
where use this condition?
why require //second reason
  φ(0)=0 ---eqn.BI001

<a name="ch11b058">
require
  φ(0)=0 ---eqn.BI001
so that 
eqn.11.15 'I' definition 
term 'φ(x)/x' has a finite 
value. 

<a name="ch11b059">
In other words, 
if φ(0) NOT=0, then
exp(－φ(x)/x) has divide by
zero trouble at x=0.
2010-04-28-18-01 stop

<a name="ch11b060">　Index begin　Index this file
2010-04-29-11-06 start
Second reason //first reason
why Problem 11.3 ask φ(0)=0 ?
Please see eqn.11.15
Compare less than side and 
greater than side two {term}
Less side {term} is －φ(x)/x,
greater side {term} is －φ'(x)
<a name="ch11b061">
We know that 
  φ'(x)= ---eqn.BI033
  lim[h→0][φ(x+h)-φ(x)]/[(x+h)-x]
or
  φ'(x)= ---eqn.BI034
  lim[y→x][φ(y)-φ(x)]/(y-x)
<a name="ch11b062">
When we consider φ'(x) at x=0
we have
  φ'(0)= ---eqn.BI035
  lim[x→0][φ(x)-φ(0)]/(x-0)
If φ(0) were NOT =0, then
eqn.11.15 less than side
should write as
  {-[φ(x)-φ(0)]/x}
Otherwise eqn.11.15 has jump
at x=0 which we want to avoid.

<a name="ch11b063">
Carleson choose to simplify
his inequality, 
change {-[φ(x)-φ(0)]/x}
 to    {-[φ(x)     ]/x}
by requiring φ(0)=0
2010-04-29-11-25 stop

2010-04-29-15-24 done  first proofread
2010-04-29-16-40 done second proofread
2010-04-29-17-04 done spelling check



<a name="ch11b064">　Index begin　Index this file
2010-05-02-11-29 start
The following is study notes
from multiple sources. They are
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele

<a name="ch11b065">
2010-04-29-20-41 access (835 KB)
http://epubl.luth.se/1402-1544/2007/53/LTU-DT-0753-SE.pdf
thesis_epubl.luth.se_Carleman_ineq.pdf
from page 19/145 to 21/145 (page mark)
or
from page 35/164 to 37/164 (pdf index)
page mark skip intro. 
pdf index include intro.

<a name="ch11b066">
2010-04-29-21-08 access (443KB)
http://www.maths.soton.ac.uk/EMIS/journals/JIPAM/v4n3/135_02_www.pdf
Carleman_Ineq_History_Proofs_maths.soton.ac.uk.pdf
from page 10/42 to 11/42 (only pdf index)
Above two pdf files come from same
author, same contents about "from 
Carleson to Carleman". Thesis has
more material and controllable pdf
interface. JIPAM has smaller file size.

<a name="ch11b067">
2010-04-25-15-24 access (95 KB)
http://www.math.uconn.edu/~kconrad/math121/121sumbypartsproj.pdf
math.uconn.edu_121sumbypartsproj.pdf

2010-04-25-16-04 access (25 KB)
http://fractal.math.unr.edu/~ejolson/311-06/handout/sumpart.pdf
fractal.math.unr.edu_sumpart.pdf

<a name="ch11b068">
2010-05-01-17-12 access (68 KB)
http://www.math.osu.edu/~fowler/teaching/handouts/math153/difference-calculus.pdf
math.osu.edu_difference-calculus.pdf

<a name="ch11b069">
2010-05-02-11-32 here
Textbook page 173 has 
problem 11.3
The next topic is 
"An Informative Choice of φ"
which start 
from Carleson's inequality
 to  Carleman's inequality.
Although textbook not mark as
problem 11.4. This study note
mark it as problem 11.4.

<a name="ch11b070">　Index begin　Index this file
2010-05-02-11-40 start
■　Problem 11.4 
(from Carleson to Carleman)
Textbook page 174, 175.

Part of the charm of Carleson's
inequality is that it provides
a sly generalization of the 
famous Carleman's inequality,
which we have met twice before
(page 27 and 128). In fact, one
only needs to make a wise 
choice of φ. //What is φ(x)?

<a name="ch11b071">
Given the hint of this possibility 
and a little time for experimenta-
tion, one is quite likely to hit 
on the candidate suggested by 
Figure 11.2. For the function φ
defined there, we have identity
  ∫[x=n-1,n]exp(-φ'(x))dx=an ---eqn.11.17
and since φ(x)/x is nondecreasing
we also have the bound

<a name="ch11b073">
2010-05-02-12-15 here
When we sum the relations (11.17)
and (11.18), we then find by
invoking Carleson's inequality 
(11.15) with α=0 that

<a name="ch11b075">　Index begin　Index this file
2010-05-02-12-28 here
Thus we recover Carleman's 
inequality under the added 
assumption that a1≧a2≧a3≧...
Moreover, this assumption incurs 
no loss of generality as one 
easily confirm in Exercise 11.7.
2010-05-02-12-31 stop

<a name="ch11b076">
2010-05-02-13-52 start
Problem 11.4 is from Carleson to 
Carleman. Now put two inequality
side by side for comparison.

<a name="ch11b079">
There is one point in common, 
that is both greater than
side has a coefficient e. For
compatibility between eqn.2.15 
and eqn.11.15, let us set
  α=0 ---eqn.BI037

<a name="ch11b080">
Carleson's Convexity Inequality 
already proved in Problem 11.3
What we need to do now is to
identify 
Carleson left  to Carleman left
and
Carleson right to Carleman right
2010-05-02-14-08 here

<a name="ch11b081">　Index begin　Index this file
■　What is φ(x)? //answer
Without knowing φ(x) definition
we can not move forward.
What is φ(x)? Textbook say
[[
the candidate suggested by 
Figure 11.2. For the function φ
defined there...
]]
To see Figure 11.2, please goto
http://freeman2.com/carleman.htm 
Please click [D] button. local

<a name="ch11b082">
Figure 11.2 say the following
A simple, but useful, observation:
the slope φ(x)/x of the chord
increase with x.
If y=φ(x) is the curve given by
the linear interpolation of the
points (n,s(n)) 
<a name="ch11b083">
where
  s(n)=log(1/a1)+log(1/a2)
       +...+log(1/an) ---eqn.BI038
then on the interval (n-1,n) we
have 
  φ'(x)=log(1/an) ---eqn.BI039
[textbook put (x), sub-n together
 in one equation.]
<a name="ch11b084">
If we assume that
  an＞an+1 ---eqn.BI040
then φ'(x) is non-decreasing and
φ(x) is convex. Also, since 
  φ(0)=0 ---eqn.BI041
the chord slope φ(x)/x is 
monotone increasing.

<a name="ch11b085">　Index begin　Index this file
Above is Figure 11.2 document.
φ(x) is defined here. We can see
  φ'(x)=log(1/an) ---eqn.BI039
then φ(x) is integration of
φ'(x).

<a name="ch11b086">
log(1/an) is last element of
s(n) and s(n) is defined as
  s(n)=log(1/a1)+log(1/a2)
       +...+log(1/an) ---eqn.BI038
Therefore 
<a name="ch11b087">
φ(n) IS s(n), that is
φ(n) is defined as
  φ(n)=log(1/a1)+log(1/a2)
       +...+log(1/an) ---eqn.BI042
For eqn.BI042 definition, we 
have
  φ'(x)=log(1/an) ---eqn.BI039
Because discrete differentiation
  φ'(n)=φ(n+1)-φ(n) ---eqn.BI043
φ'(n-1) give us eqn.BI039.

<a name="ch11b088">
What is φ(x)?
φ(x) is high degree polynomial
which pass through each (n,φ(n)).
where n=1,2,3,...,N
If we have 101 data points, 
φ(x) is 100th degree polynomial.

ALERT Above explanation
may not be correct !!
2010-05-02-14-37 here 



<a name="ch11b089">　Index begin　Index this file
2010-05-02-17-39 start
For numerical example, use
2, 1.6, 0.9, 0.65, 0.1
assume
a1=2,a2=1.6,a3=0.9,a4=0.65,a5=0.1;
then 
n=1 φ(1)=log(1/a1)=log(1/2)
n=2 φ(2)=log(1/a1)+log(1/a2)
        =log(1/2)+log(1/1.6)
n=3 φ(3)=log(1/2)+log(1/1.6)+log(1/0.9)
n=4 φ(4)=log(1/2)+log(1/1.6)+log(1/0.9)+log(1/0.65)
n=5 φ(5)=log(1/2)+log(1/1.6)+log(1/0.9)+log(1/0.65)+log(1/0.1)

<a name="ch11b090">
φ(0) to φ(5) is next sequence
0     //Reason for φ(0)=0 (a), (b)
-0.693147
-1.163150
-1.057790
-0.627007
1.6755777
Gradually decrease then increase.

<a name="ch11b091">
The derivative of φ(x) is next
Define φ(0)=0
n=1 φ(1)-φ(0)=log(1/2)=-0.693147
n=2 φ(2)-φ(1)=log(1/1.6)=-0.470003
n=3 φ(3)-φ(2)=log(1/0.9)=0.105360
n=4 φ(4)-φ(3)=log(1/0.65)=0.430782
n=5 φ(5)-φ(4)=log(1/0.1)=2.302585
'slope' is gradually increase.

<a name="ch11b092">
Please goto carleman.htm, paste
2, 1.6, 0.9, 0.65, 0.1
to box 1, then click [C] button.
or do nothing, just click [D].
Red line is the one we calculated
above. Red line is also shape like
textbook Figure 11.2 polygon curve.
You can paste your own data to 
box 1 and click [C] button.
You can also click [rand5] and 
click [C] button.
2010-05-02-18-05 stop

<a name="ch11b093">　Index begin　Index this file
2010-05-02-18-25 start
Now convert from //two eqns
Carleson's less than side to
Carleman's less than side 
We set
  α=0 ---eqn.BI037
<a name="ch11b094">
we convert from
  ∫[x=0,∞][exp(-φ(x)/x)]dx ---eqn.BI044
to
  ∑[k=1,∞][a1*a2*...*ak]1/k ---eqn.BI045
and φ(n) is defined as
  φ(n)=log(1/a1)+log(1/a2)
       +...+log(1/an) ---eqn.BI042
<a name="ch11b095">
For the interval x=n-1 to x=n
  ∫[x=n-1,n]exp(-φ(x)/x)dx
 ≧exp(-φ(n)/n) ---eqn.BI046 begin
  //plug in φ(n)
  =exp(-
       [ log(1/a1)+log(1/a2)
        +...+log(1/an)]
      /n)
<a name="ch11b096">
  //let log() absorb '-'
  =exp(
       [-log(1/a1)-log(1/a2)
        -...-log(1/an)]
      /n)
<a name="ch11b097">
  //use -log(1/m)=log([1/m]^[-1])=log(m)
  =exp(
       [+log(a1)+log(a2)
        +...+log(an)]
      /n)
<a name="ch11b098">　Index begin　Index this file
  //use (1/n)*log(m)=log(m^[1/n])
  =exp([+log(a1^[1/n])+log(a2^[1/n])
        +...+log(an^[1/n])]
      )
<a name="ch11b099">
  //use exp(m+n)=exp(m)*exp(n)
  =exp(log(a1^[1/n]))*exp(log(a2^[1/n]))
  *...*exp(log(an^[1/n]))
  //use exp(log(m))=m
  =a1^[1/n]*a2^[1/n]*...*an^[1/n]
  //[1/n] is common power
  =(a1*a2*...*an)^[1/n] ---eqn.BI046 end

<a name="ch11b100">
eqn.BI046 begin is one element in
  ∫[x=0,∞][exp(-φ(x)/x)]dx ---eqn.BI044

eqn.BI046 end is one element in
  ∑[k=1,∞][a1*a2*...*ak]1/k ---eqn.BI045

<a name="ch11b101">
Integrate eqn.BI046 in whole 
domain and sum eqn.BI046 to
infinity, we achieved
convert from
Carleson's less than side to
Carleman's less than side.
2010-05-02-18-56 here


<a name="ch11b102">　Index begin　Index this file
Next convert from //two eqns
Carleson's greater than side to
Carleman's greater than side.
That is convert from
  ∫[x=0,∞] exp{-φ'(x)}dx ---eqn.BI047
to
  ∑[k=1,∞]ak ---eqn.BI048

<a name="ch11b103">
Compare the difference
eqn.BI047 is greater than side  
  ∫[x=0,∞] exp{-φ'(x)}dx ---eqn.BI047
  ∫[x=0,∞][exp(-φ(x)/x)]dx ---eqn.BI044
eqn.BI044 is less than side.
Both are Carleson's which is done 
at Problem 11.3 Through plan A, 
Carleson implies Carleman.

<a name="ch11b104">
For the interval x=n-1 to x=n
  exp{-φ'(x)} ---eqn.BI049 begin
 =exp{-φ'(n)} //why equal?
 //use φ'(n)=φ(n+1)-φ(n) ---eqn.BI043
 =exp{-[
        log(1/a1)+log(1/a2)
       +...+log(1/an)+log(1/an+1)
       -log(1/a1)-log(1/a2)
       -...-log(1/an)
      ] }
<a name="ch11b105">
  //cancel all, but one left
 =exp{-[log(1/an+1)] }
  //use (-1)*log(m)=log(m^[-1])
 =exp{log(an+1) }
  //use exp(log(m))=m
 =an+1 ---eqn.BI049 end

<a name="ch11b106">
Integrate eqn.BI049 begin 
and sum eqn.BI049 end, convert
from
  ∫[x=0,∞] exp{-φ'(x)}dx ---eqn.BI047
to
  ∑[k=1,∞]ak ---eqn.BI048

<a name="ch11b107">　Index begin　Index this file
Carleson's Inequality greater than 
side has e=2.718281828459045... 
Carleman's Inequality greater than 
side has same e
Put e back to summed eqn.BI049
we achieved convert from
Carleson's greater than side to
Carleman's greater than side.

<a name="ch11b108">
Here, conclude that 
Carleson's Inequality implies
Carleman's Inequality.
Problem 11.4 is done.



<a name="ch11b109">
Do you have any question? If
you are still not certain,
please read textbook page 174
to 175 and read online paper
http://epubl.luth.se/1402-1544/2007/53/LTU-DT-0753-SE.pdf
from page 19/145 to 21/145 (page mark)
or
from page 35/164 to 37/164 (pdf index)
They may do better job in
explaining 
from Carleson to Carleman.
2010-05-02-19-24 stop

<a name="ch11b110">　Index begin　Index this file
2010-05-02-20-20 start Carleman local
On 2010-04-29-11-25 done
Problem 11.3
When LiuHH read next problem
Feel it may take longer time
to understand. Decide upload
one Problem 11.3 immediately.

<a name="ch11b111">
The section of
"An Informative Choice of φ"
say "For the function φ
defined there" (Figure 11.2)
LiuHH expect "function φ defined"
should give an analytic equation
expression. But LiuHH see only
s(n) and not see φ(x) !
Possible answer is here

<a name="ch11b112"> Carleman local
Second wondering point is
[[
on the interval (n-1,n) we
have 
  φ'(x)=log(1/an) ---eqn.BI039
]]
Why a derivative of observed
data sequence is φ'(x)=log(1/an) ?

Reader can find answer in any
calculus of finite differences
textbooks.

<a name="ch11b113">
Third wondering point is
given 
  a1≧a2≧...≧an ＞0
why "φ(x)/x is nondecreasing"?

//We also require that the ak 
//sequence be all positive
//number. Because there is 
//log(1/ak) operation.

<a name="ch11b114"> Carleman local
Possible answer: because 
given a1≧a2≧...≧an ＞0
then  1/a1≦1/a2≦...≦1/an
log() function is monotone increase
log(1/a1)≦log(1/a2)≦...≦log(1/an)
Textbook say log(1/a1) is slope.
From a1 to an slope increase
φ(x)/x is proportion to slope.
slope increase, φ(x)/x increase too.

<a name="ch11b115">　Index begin　Index this file
LiuHH need to draw φ(x) to believe.
From tute0012.htm cut program 
section and create carleman.htm
Few hours time curve show up.
LiuHH arbitrary set '2' in random 
data range upper bound '10^2'
<a name="ch11b116"> Carleman local
Draw many times, never see a
curve like textbook Figure 11.2
It is always slider shape curve.
But figure 11.2 is a bowl shape
and curve tail go up above
horizontal axis. After think
understand why.

<a name="ch11b117">
Textbook say curve slope is
  φ'(x)=log(1/an) ---eqn.BI039
then zero slope is log(1)
which is an=1
10^2=100, random data range
from 0 to 100. 99% data are
greater than 1, only 1% data
less than one. No wonder it
is a slider shape, almost no
tail. 
<a name="ch11b118"> Carleman local
LiuHH change [10^] box from 2 
to 0.301, because 10^0.301=2
Data range from 0 to 2, half
greater than one, half less.
After this change, Click 'B'
or 'C' get red curve similar
to textbook Figure 11.2. 
<a name="ch11b119">
The curve output does show
that φ(x) is convex (╰╯)
and φ'(x) increase only.
2010-05-02-20-52 stop

2010-05-03-14-09 done proofread

<a name="ch11b120">
2010-05-04-15-33
why equal?
When convert from
Carleson's less than side to
Carleman's less than side 
<a name="ch11b121">
At the step
[[
For the interval x=n-1 to x=n
  ∫[x=n-1,n]exp(-φ(x)/x)dx
 ≧exp(-φ(n)/n) ---eqn.BI046 begin
]]
Link to eqn.11.18 to show that
inequality is correct.
∫[x=n-1,n]exp(-φ(x)/x)dx≧exp(-φ(n)/n)

<a name="ch11b122">
When convert from
Carleson's greater than side to
Carleman's greater than side 
At the step
[[
For the interval x=n-1 to x=n
  exp{-φ'(x)} ---eqn.BI049 begin
 =exp{-φ'(n)} //why equal?
]]
<a name="ch11b123">
If still use "inequality is correct"
then
C'man less≦C'son less //plan A
          ≦C'son great≦C'man great
become
C'man less≦C'son less //plan B
         ?≦?C'man great≦C'son great

<a name="ch11b124">
only plan A can bridge 
from Carleson to Carleman
plan B can NOT.

<a name="ch11b125">
Why use inequality at the step
C'man less≦C'son less
Why same inequality would lead to
C'man great≦C'son great //trouble
change to equality?

<a name="ch11b126">
Both textbook page 175 line 8
right side use equality and
Maria Johansson thesis
page 36/164 (20/145) last line
use equality.
Why less than side use inequality?
Why greater than side use equality?
Liu,Hsinhan is still thinking.
2010-05-04-15-57 stop



2010-05-04-16-20 done second proofread

2010-05-04-18-15 done spelling check



<a name="ch11b127">　Index begin　Index this file
2010-05-04-22-03 start
■　Exercise 11.1 problem statement
　　textbook page 175
（The Lp Flop and a
 General Principle）

Suppose that 1＜α＜β and suppose 
that the bounded nonnegative 
function φ and ψ satisfies the
inequality

<a name="ch11b130">
The bound (11.20) is just one
example of a general (but vague)
principle. If we have a factor 
on both sides 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.
2010-05-04-22-22 stop





<a name="ch11b131">　Index begin　Index this file
2010-05-04-23-42 start
■　Exercise 11.1 hint
　　textbook page 271

By applying Holder's inequality
with 
  p=β/α ---eqn.BI050
//why set p=β/α? why bgn, why end
and
  q=β/(β-α) ---eqn.BI051
<a name="ch11b132">
to the right side of bound (11.19)
//textbook use (11.20), LiuHH 
//●●change to (11.19). Because
//(11.20) right side has just
//one ψ(x), but (11.19) right 
//side has ψ(x) and φ(x). We
//know that Holder like double.
//2010-05-04-23-52
<a name="ch11b133">
we obtain the inequality

<a name="ch11b134">
There is no loss of generality 
if we assume that the first integral
factor on the right is nonzero, so
we may divide both side by that
factor. If we then raise both sides 
of the resulting bound to the power
β/(β-α) to get our target bound
(11.20)

<a name="ch11b135">
It is only the division step where
we use the condition that φ is
bounded. The inequality for bounded
functions can then be used to prove
a corresponding inequality for
functions that need not be bounded.
<a name="ch11b136">
It is quite common in arguments that
call on the flop for one to first
consider bounded functions so that
one can side step any inappropriate
arithmetic with integrals that 
might be infinite.
2010-05-05-00-17 stop



2010-05-05-11-52 start
<a name="ch11b137">　Index begin　Index this file
■　Exercise 11.1 solution


Our start point is eqn.11.19, 
it has φ(x) at both side.
end point is eqn.11.20. it has 
φ(x) at left side only.
<a name="ch11b138">
Key step is to put φ(x) to one 
side.
eqn.11.19 left side has φβ(x)
eqn.11.19 right side has φα(x)
they are in different power
and can NOT be combined.
But Holder's inequality will
change power. We can use 
Holder's change power function.

<a name="ch11b139">
eqn.11.19 right side has two
terms multiply φα(x)ψ(x).
We apply Holder's to φα(x)ψ(x).
[not to left side single φβ(x)]
eqn.BC001 tell us that
∫φα(x)ψ(x)dx will be less than
[∫φαp(x)dx]1/p multiply by
[∫ψq(x)dx]1/q
<a name="ch11b140">
The key point is αp in φαp(x)
Both α and β are given, we can 
not alter α, β.
p has a simple requirement that
  p＞1 ---eqn.BI053
and
  1/p + 1/q = 1 ---eqn.BI054
<a name="ch11b141">
p is under our control !
We want a p such that
φβ(x) equals φαp(x)
We require
  β=αp ---eqn.BI055
eqn.BI055 is same as 
  p=β/α ---eqn.BI050
<a name="ch11b142">
Because require p＞1, or β/α＞1
It is necessary that β＞α for
this logic to work.
2010-05-05-12-32 stop

<a name="ch11b143">　Index begin　Index this file
2010-05-05-14-44 start
Because 
  1/p + 1/q = 1 ---eqn.BI054
Find q as following
  1/q = 1-1/p = (p-1)/p
then 
  q=p/(p-1)
  q=(β/α)/[(β/α) - 1]
  q=(β/α)/[(β-α)/α]
  q=β/(β-α) ---eqn.BI051
<a name="ch11b144">
Above state why p and q be 
defined as eqn.BI050/eqn.BI051.
With this p, q in hand, we apply
Holder's inequality to eqn.11.19
right side as following

<a name="ch11b146">
2010-05-05-15-17 here
eqn.11.19 greater than side is 
same as eqn.BI056 less than side
C∫[x=0,T]φα(x)ψ(x)dx is intermediate
term where Holder apply directly.
Now put eqn.11.19 less than side
and eqn.BI056 greater than side
into one equation as following

<a name="ch11b148">　Index begin　Index this file
In eqn.BI057, black ∫φβ(x)dx
and red ∫φβ(x)dx are like terms
which is the result of eqn.BI055
requirement.
Let us FLOP, move red ∫φβ(x)dx 
to black ∫φβ(x)dx

<a name="ch11b150">
eqn.BI058 less than side term
has power 1-α/β.
Black '1-' is eqn.11.19 original
less than side term.
Red 'α/β' is Holder converted
for us which is from eqn.11.19
greater than side.

<a name="ch11b151">
eqn.BI058 whole equation take 
power of 1/(1-α/β), that is 
take power of β/(β-α), let less
than side keep power one. Result 
is

eqn.BI059 is same as our target
eqn.11.20
Exercise 11.1 is done.
2010-05-05-15-53 stop



<a name="ch11b153">　Index begin　Index this file
2010-05-05-17-38 start
■　Exercise 11.2 problem statement
　　textbook page 176
（Rudimentary Example of a General
 Principle）

The principle of Exercise 11.1 can
be illustrated with the simplest of
tools. For example, 
<a name="ch11b154">
show for non-negative x and y that
  2x3≦y3+y2x+yx2 ---eqn.BI060
implies
  x3≦2y3 ---eqn.BI061





<a name="ch11b155">
2010-05-05-17-42 here
■　Exercise 11.2 hint
　　textbook page 271

By the AM-GM inequality we have
  2x3≦y3+y2x+yx2 
 ≦y3+2y3/3+x3/3+y3/3+2x3/3
 =2y3 +x3 ---eqn.BI062
<a name="ch11b156">
In this example, the higher power 
on the left made the transformation
possible, but for the transformation
to be nontrivial one also needed
cooperation from the constant 
factors. 
<a name="ch11b157">
If we replace 2 by 1/2 in 
the original problem, we only obtain
the trivial bound 
 -x3≦4y3 ---eqn.BI063
2010-05-05-17-52 stop



<a name="ch11b158">　Index begin　Index this file
2010-05-05-17-53 start
■　Exercise 11.2 solution


Start from given
  2x3≦y3+y2x+yx2 ---eqn.BI060
Do in three steps.
<a name="ch11b159">
First step
  2x3≦y3+etc1 ---eqn.BI064
Here y3 is on focus, and
'y2x+yx2' goto 'etc1'
We can not apply Holder's to y3
because Holder need two function
multiply together.

<a name="ch11b160">
Second step
  2x3≦y2x+etc2 ---eqn.BI065
Here y2x is on focus, 
y3+yx2 goto 'etc2'
Greater than side x has power 1,
less than side x has power 3. 
<a name="ch11b161">
Let us apply Holder's to y2x
Here set p=3, q=3/2
  x*y2=(x3)1/3*[(y2)3/2]2/3
2010-05-05-18-10 here wondering
  x*y2=(x3)1/3*[(y3)]2/3 ---eqn.BI066

<a name="ch11b162">
Third step
  2x3≦yx2 + etc3 ---eqn.BI067
  yx2=y(x3)2/3
2010-05-05-18-16 here

<a name="ch11b163">　Index begin　Index this file
Change direction.
Use humble bound to
  2x3≦y3+y2x+yx2 ---eqn.BI060

Greater than side term y2x
  x*y2≦(1/3)x3 + (2/3)y3 ---eqn.BI068
Multiply become add, power 1&2 
become power 3. Below same.

Greater than side term x2y
  x2y≦(2/3)x3 + (1/3)y3 ---eqn.BI069

<a name="ch11b164">
eqn.BI060 become next
  2x3≦y3+[(1/3)x3 + (2/3)y3]
        +[(2/3)x3 + (1/3)y3] ---eqn.BI070
that is
  2x3≦2y3+x3 ---eqn.BI071
or
   x3≦2y3 ---eqn.BI072
eqn.BI072 is answer eqn.BI061
2010-05-05-18-27

<a name="ch11b165">
Change problem. Change 
from  2*x3
 to  (1/2)*x3
Use humble bound to
  (1/2)*x3≦y3+y2x+yx2 ---eqn.BI073
Follow first problem, get
  (1/2)*x3≦2y3+x3 ---eqn.BI074
that is trivial bound 
 -x3≦4y3 ---eqn.BI063
2010-05-05-18-33 stop



<a name="ch11b166">　Index begin　Index this file
2010-05-05-19-16 start
■　Exercise 11.3 problem statement
　　textbook page 175
（An Exam-Time Discovery of F. Riesz）

<a name="ch11b167">
Show that there is a constant A 
(not depending on u and v) such
that for each pair of functions 
u and v on [-π,π] for which one
has

<a name="ch11b170">
According to J.E. Littlewood 
(1988, p.194) F. Riesz was trying
to set an examination problem when 
he observed almost by accident 
that the bound (11.21) holds for 
the real u and imaginary v parts
of f(eiθ) when f(z) is a continuous
function that is analytic in the
unit disk. This observation and 
the inference (11.22) subsequently
put Riesz on the trail of some of
his most important discoveries.
2010-05-05-19-35 stop





<a name="ch11b171">　Index begin　Index this file
2010-05-05-19-38 start
■　Exercise 11.3 hint
　　textbook page 271

The hypothesis (11.21) and 
Schwarz's inequality give us

<a name="ch11b173">
which is 
  x2≦c2+6cx ---eqn.BI076
with the natural identification.
If we solve this for the case of
equality, we find
  x=c(6±√40)/2 ---eqn.BI077
so the hypothesis (11.21) implies
the bound if we take
  A={(6+√40)/2}2 ---eqn.BI078
2010-05-05-19-56 stop




<a name="ch11b174">　Index begin　Index this file
2010-05-05-20-09 start
■　Exercise 11.3 solution


We apply Schwarz's inequality to
the term 6*∫[θ=-π,π]u2(θ)v2(θ)dθ
in eqn.11.21 the result is
  ∫[θ=-π,π]u2(θ)v2(θ)dθ
 ≦{∫[θ=-π,π]u4(θ)dθ}1/2
  *{∫[θ=-π,π]v4(θ)dθ}1/2 ---eqn.BI079
Substitute eqn.BI079 into eqn.11.21 
get eqn.BI075
<a name="ch11b175">
Now define
  c={∫[θ=-π,π]u4(θ)dθ}1/2 ---eqn.BI080
  x={∫[θ=-π,π]v4(θ)dθ}1/2 ---eqn.BI081
then eqn.BI075 is
  x2≦c2+6cx ---eqn.BI076
We solve equality
  x2=c2+6cx ---eqn.BI082
for x in terms of c
  x2-6cx-c2=0 ---eqn.BI083
<a name="ch11b176">
let
  x2+Bx+D=0 ---eqn.BI084
then
  x={-B±√(B*B-4*D)}/2 ---eqn.BI085
  x={-(-6c)±√[(-6c)*(-6c)-4*(-c2)]}/2
  x={6c±√[40c2]}/2 
  x=c(6±√40)/2 ---eqn.BI077

<a name="ch11b177">
c and x are defined at eqn.BI080 
and eqn.BI081, recover c and x 
get
  {∫[θ=-π,π]v4(θ)dθ}1/2 ---eqn.BI086
 ={∫[θ=-π,π]u4(θ)dθ}1/2*[(6±√40)/2]
Square eqn.BI086 get
  {∫[θ=-π,π]v4(θ)dθ} ---eqn.BI087
 ={∫[θ=-π,π]u4(θ)dθ}*[(6±√40)/2]2
<a name="ch11b178">　Index begin　Index this file
If we define
  A=[(6+√40)/2]2 ---eqn.BI088
then we have
  x^2=A*c^2 ---eqn.BI089
2010-05-05-20-40 here
<a name="ch11b179">
Equality eqn.BI082 help us find
constant A. 
From given inequality eqn.11.21 
and the border equality eqn.BI089
we have the target eqn.11.22.

<a name="ch11b180">
u(θ) and v(θ) are multiplied at 
eqn.11.21, we separate u(θ) from
v(θ) and conclude eqn.11.22 is 
true.
Exercise 11.3 is also a FLOP !!
What is Flop?
2010-05-05-20-59 stop



<a name="ch11b181">　Index begin　Index this file
2010-05-06-10-05 start
■　Exercise 11.4 problem statement
　　textbook page 176
（The Lp Norm of the Average）

Show that if f:(0,∞)→Real+, then
one has

2010-05-06-10-15 stop





<a name="ch11b183">
2010-05-06-10-28 start
■　Exercise 11.4 hint
　　textbook page 272

Only a few obvious changes are 
needed to convert the proof of 
the L2 bound (11.1) to a 
proof for the corresponding Lp
bound (11.23). By that analogy
we first note

<a name="ch11b187">
which is the Lp analog of the
preflop L2 bound(11.4). One
now finishes with the Lp flop
precisely as in Exercise 11.1
provided that one sets
  α=p-1  ---eqn.BI093
and
  β=p  ---eqn.BI094
2010-05-06-11-02 stop



<a name="ch11b188">
2010-05-06-11-05 start
■　Exercise 11.4 solution


Exercise 11.4 is the topic in
problem 11.1
Textbook consider L2 case,
LiuHH solve for Lp case 
from ch11a008 to ch11a088
which is same as Exercise 11.4.

Exercise 11.4 already solved.
2010-05-06-11-08 stop



<a name="ch11b189">　Index begin　Index this file
2010-05-06-13-00 start
■　Exercise 11.5 problem statement
　　textbook page 176
（Hardy and the Qualitative
 Version of Hilbert）

Use the discrete version (11.9)
of Hardy's inequality to prove
that

<a name="ch11b191">
This was the qualitative version 
of Hilbert's inequality that
Hardy had in mind when he first
considered the Problems 11.1
and 11.2.
2010-05-06-13-13 stop

<a name="ch11b192">
2010-05-06-14-09 start
eqn.BI095 in textbook
 was ∑[n=1,∞]∑[n=1,∞]
 now ∑[m=1,∞]∑[n=1,∞] ●●change
2010-05-06-14-12 stop




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2010-05-06-14-14 start
■　Exercise 11.5 hint
　　textbook page 272

Without loss of generality we
assume that an≧0 for all n=1,2,...
<a name="ch11b194">
We then set
  An=a1+a2+...+an ---eqn.BI096
and apply Cauchy's inequality
followed by Hardy's inequality
(11.9) to get

2010-05-06-14-32 stop




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2010-05-06-15-21 start
■　Exercise 11.5 solution


Our starting point is that
given
  S defined=∑[n=1,∞]an2＜∞ ---eqn.BI099
our target point is that
to prove
  ∑[m=1,∞]∑[n=1,∞]
  {anam/(m+n)} converge ---eqn.BI100

<a name="ch11b198">
What is T?
Cauchy's inequality apply to who?

In the text, no where define T,
then equality of eqn.BI097 must
be T definition.

<a name="ch11b199">
T is dot product of two sequences
first seq. is a1,a2,...,an,...,a∞
second seq. is A1/1,A2/2,...,An/n,
           ...,A∞/∞
An is defined at eqn.BI096
T is an intermediate step between
starting point eqn.BI099 and
end point eqn.BI100.

<a name="ch11b200">
Now we find two sequences,
therefore Cauchy has job to do.
Cauchy's inequality apply to above
first sequence and second sequence.
Cauchy's result is

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We are given that
S defined=∑[n=1,∞]an2 (eqn.BI099)
be bounded. The intermediate step
eqn.BI101 An/n term should link
to an. 
<a name="ch11b203">
An is defined to be
  An=a1+a2+...+an ---eqn.BI096
An/n is Arithmetic Mean of first 
n term of a1,a2,...,a∞ sequence.
The Discrete Hardy's Inequality
eqn.11.9 give us exactly what 
we need. It has the following
result

<a name="ch11b205">
2010-05-06-16-28 here
Use eqn.BI102 to replace An/n in 
eqn.BI101, we get eqn.BI097.

<a name="ch11b206">
We are working in the following 
logic
  ∑[m=1,∞]∑[n=1,∞]
  {anam/(m+n)} ≦ T ---eqn.BI103
and 
  T ≦ ∑[n=1,∞]an2 ---eqn.BI104
Blue is given bounded condition.
Red is ask to prove.
T is intermediate term.

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eqn.BI097 finish T ≦ ∑[n=1,∞]an2
We still have work to do, that
is to show
  ∑[m=1,∞]∑[n=1,∞]
  {anam/(m+n)} ≦ T

<a name="ch11b208">
Exercise 11.5 hint tell us that 
the second step is "even simpler"
eqn.BI098 left inequality is next

<a name="ch11b212">　Index begin　Index this file
2010-05-06-17-06 here
eqn.BI107 is clearly true. For
one typical ai, we have
  ai/n ?≧? ai/(i+n) ---eqn.BI108
for i=1,2,...,n. Because 
  i+n＞n  ---eqn.BI109
True! i+n＞n IS simpler!
eqn.BI098 left side inequality
is true.
eqn.BI098 equality is just re-
arrange summation symbol.

<a name="ch11b213">
eqn.BI103 and eqn.BI104 are both
true and the problem target
equation is proved. that is
given
  S defined=∑[n=1,∞]an2＜∞ ---eqn.BI099
we have
  ∑[m=1,∞]∑[n=1,∞]
  {anam/(m+n)} converge ---eqn.BI100
2010-05-06-17-22 stop

2010-05-07-11-51 done first proofread
2010-05-07-16-28 done second proofread
2010-05-07-17-57 done spelling check



<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