<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="ch09b001">　Index begin　Index this file
2010-03-16-11-14 start
Minkowski's inequality tells us 
that the function h:Realn to Real
defuned by
  h(a)=∥a∥p ---eqn.BB001
is subadditive in the sense 
that one has the bound
  h(a+b)≦h(a)+h(b) ---eqn.BB002
  for all a,b in Realn 
<a name="ch09b002">
Subadditive relations are typically
much more obvious than Riesz's
proof, and one may wonder if there 
is some way to see Minkowski's 
inequality at a glance. The next 
Challenge problem confirms this
suspicion and throws added 
precision into the bargain.
2010-03-16-11-26 here

<a name="ch09b003">　Index begin　Index this file
■　Problem 9.4 (Quasilinearization
　　of the lp Norm) //Textbook p.143
Show that for all 1≦p≦∞ one has
the identity
  ∥a∥p ---eqn.9.16
  =max{∑[k=1,n]akxk:∥x∥q=1}
where //red is given constraint
  a={a1,a2,...,an} ---eqn.BB003
and where p and q are conjugate
<a name="ch09b004">
[so one has
  q=p/(p-1) ---eqn.BB004
 when p＞1, but q=∞ when p=1
 and q=1 when p=∞ ]
Finally, explain why this identity
yields Minkowski's inequality
without any further computation.
2010-03-16-11-35 stop

<a name="ch09b005">
2010-03-16-11-59 start
First do a simple logic analysis.
There are two teams A and B. They 
receive jobs as following.

<a name="ch09b006">
Job1: Team A and team B merge as 
      one team search for maximum 
      value jointly. This result
      is answer1.
Job2: Team A and team B search for
      maximum value independently.
      Add their results as answer2.

<a name="ch09b007">
Now the question is 
answer1 and answer2 which one is
greater? Is it always greater?
 or sometimes answer1 greater
and sometimes answer2 greater?

<a name="ch09b008">　Index begin　Index this file
Assume team A and team B have
equal probability to find maximum 
value, then two chances is better
than one chance. At least not
worse. So the answer is 
answer1 ≦ answer2

<a name="ch09b009">
This analysis give us textbook
page 143, line 16/17 equation
  h(a+b) ---eqn.BB005 begin
  =max[w∈W]{L(a+b,w)}
  =max[w∈W]{L(a,w)+L(b,w)}
  ≦max[w0∈W]{L(a,w0)}
   +max[w1∈W]{L(b,w1)}
  =h(a)+h(b) ---eqn.BB005 end
<a name="ch09b010">
Net result is
  h(a+b)≦h(a)+h(b) ---eqn.BB002
Here L is a function transform
  L:V×W→Real ---eqn.BB006
L is additive in its first variable
  L(a+b,w)=L(a,w)+L(b,w) ---eqn.BB007
Define 
  h(a)=max[w∈W]L(a,w) ---eqn.9.17
2010-03-16-12-32 stop

<a name="ch09b011">
2010-03-16-12-44 start
Please compare the difference
between maximum equation and
minimum equation.

Two team vs. one joint team
For maximum, we have eqn.BB005 
  max[w∈W]{L(a+b,w)} ---eqn.BB005
  ≦max[w0∈W]{L(a,w0)}
   +max[w1∈W]{L(b,w1)}

<a name="ch09b012">
Two team vs. one joint team
For minimum, we have eqn.AZ027 

since two choices are better than 
one. (here minimum is better)
2010-03-16-12-51 here



<a name="ch09b013">　Index begin　Index this file
2010-03-16-12-53
Wonder, textbook page 143, line 16/17
whether eqn.BB005 is better stated as
  max[w∈W]{L(a+b,w)} ---eqn.BB008
  ≦OUTER_max{inner_max[w0∈W][L(a,w0)],
             inner_max[w1∈W][L(b,w1)]}
<a name="ch09b014">
Wonder, textbook page 259, Line 11
whether eqn.AZ027 is better stated as
  OUTER_min{inner_min[x∈D](∑[k=1,n]ak*xk)/n,
            inner_min[x∈D](∑[k=1,n]bk*xk)/n}
  ≦min[x∈D]{∑[k=1,n](ak+bk)*xk}/n ---eqn.BB009

<a name="ch09b015">
eqn.AZ027 two team two min add
together has risk greater than
joint team one min !!
2010-03-16-13-06 stop



<a name="ch09b016">
2010-03-16-15-30 start
Now back to Problem 9.4 
In
  ∥a∥p ---eqn.9.16
  =max{∑[k=1,n]akxk:∥x∥q=1}
question is maximum over what?
Many equation write max[k=1,n]
indicate for value at k=1, k=2
..., k=n, in n values, select
a maximum value. But eqn.9.16
did not specify what varies.
<a name="ch09b017">
LiuHH guessed that for q-norm
  ∥x∥q=1 ---eqn.BB010
sequence x={x1,x2,...,xn} is 
variable. For those different
x seq., evaluate ∑[k=1,n]akxk
and choose the maximum value.

<a name="ch09b018">　Index begin　Index this file
To prove equality eqn.9.16,
we prove twice,
 left≧right and
 left≦right 
to get left=right.
 left is ∥a∥p
right is max{∑[k=1,n]akxk:∥x∥q=1}
2010-03-16-15-47 here

<a name="ch09b019">
First prove eqn.9.16 left≧right, 
that is to show in
  ∥a∥p ---eqn.9.16
  =max{∑[k=1,n]akxk:∥x∥q=1}
we have
  ∥a∥p ---eqn.BB011
  ≧max{∑[k=1,n]akxk:∥x∥q=1}
First consider the set

2010-03-16-16-04 here

<a name="ch09b021">
2010-03-16-16-12 start
After think, suggest change 
from eqn.BB012 to eqn.BB013

<a name="ch09b022">
We compare eqn.BB013 (above) with 
Holder Inequality (below)

<a name="ch09b023">　Index begin　Index this file
2010-03-16-16-21 here
Write in p-norm, q-norm shorter
equation, eqn.BB013 is
  S={∑[k=1,n]akxk:∥x∥q=1} ---eqn.BB013a
and Holder Inequality
  ∑[k=1,n]akxk≦∥a∥p∥x∥q ---eqn.9.1a
<a name="ch09b024">
For any element s in S,
  s=∑[k=1,n]akxk ---eqn.BB014
 s must have
  ∥x∥q=1 ---eqn.BB015
because this is required by 
set S specification.
<a name="ch09b025">
then Holder Inequality eqn.9.1a 
tell us that this element s
  ∑[k=1,n]akxk≦∥a∥p*1
that is  s≦∥a∥p
or in eqn.9.16 order
  ∥a∥p≧s ---eqn.BB016
<a name="ch09b026">
eqn.BB016 is valid for any s∈S
include the maximum in set S.
So we have
  ∥a∥p ---eqn.BB011
  ≧max{∑[k=1,n]akxk:∥x∥q=1}
2010-03-16-16-57 here

<a name="ch09b027">
To prove equality eqn.9.16,
we prove twice,
 left≧right and
 left≦right 
to get left=right.
Above is
 left≧right and
<a name="ch09b028">　Index begin　Index this file
Below is
 left≦right 
That is to show that
  ∥a∥p ---eqn.BB017
  ≦max{∑[k=1,n]akxk:∥x∥q=1}
2010-03-16-17-00 stop

<a name="ch09b029">
2010-03-16-17-48 start
We need to use Holder converse 
to prove eqn.BB017.
Holder converse given condition 
is eqn.9.8
Holder converse conclusion is
eqn.9.9
<a name="ch09b030">
Here in Problem 9.4, we use Holder 
converse with slight modification.

Define C=max{s∈S} ---eqn.BB018
where
  s=∑[k=1,n]akxk ---eqn.BB014
 s must have //see eqn.BB013
  ∥x∥q=1 ---eqn.BB015
<a name="ch09b031">
Modified Holder converse is next

<a name="ch09b032">
for all yk in Realn 1≦k≦n,
yk is dummy, when we come to 
Holder converse conclusion, 
yk drop out. Then for
  q=p/(p-1) ---eqn.BA093
one has the bound

<a name="ch09b033">　Index begin　Index this file
eqn.BB018 tell us that
  C=max{s∈S} ---eqn.BB018
eqn.9.9 left side is p-norm ∥a∥p
Holder converse eqn.9.9 is same as
  ∥a∥p≦max{s∈S} ---eqn.BB019
eqn.BB019 is left≦right part

<a name="ch09b034">
Next two relations
  ∥a∥p≦max{s∈S} ---eqn.BB019
and
  ∥a∥p ---eqn.BB011
  ≧max{∑[k=1,n]akxk:∥x∥q=1}
tell us that
  ∥a∥p  ---eqn.BB020
  =max{∑[k=1,n]akxk:∥x∥q=1}
eqn.BB020 is our target eqn.9.16.
Problem 9.4 solved.
2010-03-16-18-25 stop

<a name="ch09b035">
2010-03-17-04-50 start
Recent few days made several 
changes. Whether these changes
are correct? or wrong? review 
as following.

<a name="ch09b036">
First change is 
Young's inequality alert
  uv ≦ up/p + vq/q ---eqn.9.6
If p=3 and q=1.5, then
  1/p + 1/q = 1/3 + 1/1.5 = 1 ---eqn.BA055
satisfy requirement.
<a name="ch09b037">
eqn.9.6 is designed 
for u be length^(1/p) L^1/3
for v be length^(1/q) L^2/3
If we plug length data for u and v,
then eqn.9.6 read as
  length^2 ≦ length^3 + length^1.5
area L^2 compare with volume L^3? 
that is not reasonable. We can 
plug normalized (pure number) to
Young's inequality and safe.
eqn.9.7 has physics dimension
not consistent problem.

<a name="ch09b038">　Index begin　Index this file
Second change is eqn.BB005 
change to eqn.BB008
If team A find maximum be 1
If team B find maximum be 3
If AB joint team find maximum be 2
we need     max{1,3}=3＞2
if use max1+max2=1+3=4＞2
3＞2 is sharper than 4＞2

<a name="ch09b039">
Similar change is eqn.AZ027 
change to eqn.BB009. Here is 
to find minimum, not maximum.
For maximum problem we add
two result to get even higher
greater side, that will not 
reverse inequality direction.

<a name="ch09b040">
But for minimum problem we add
two result to get even higher
less than side, that DO
reverse inequality direction.

<a name="ch09b041">
Third change is eqn.BB012 
change to eqn.BB013 
that is change from constraint
  ∑[k=1,n]|xk|q≦1  ---eqn.BB021
to constraint
  ∥x∥q=1 ---eqn.BB022
<a name="ch09b042">
eqn.BB012 constraint eqn.BB021
is not compatible with problem
statement eqn.9.16 (if change 
constraint then change problem !)
eqn.9.16 use constraint eqn.BB022
eqn.BB013 also use eqn.BB022
which is compatible with given
eqn.9.16

<a name="ch09b043">
After second thought, third thought
above changes look reasonable to
Liu,Hsinhan. But still could be
wrong change. Because LiuHH's 
capability is limited.
2010-03-17-05-33 stop



<a name="ch09b044">　Index begin　Index this file
2010-03-17-10-46 start
■　Stability is a different
　　problem

Please goto drawing board and 
click "12" in a line marked
"f(x) 11 12 p1  p2  ; x(t),y(t)"
<a name="ch09b045">
Please pay attention to red line.
Red line equation is
9*pow(x,4/3)/4-9*pow(x,1/3)
Please open next page (local)
http://freeman2.com/complex2.htm
in "Box3, input JS command" input
[[
x=4
9*pow(x,4/3)/4-9*pow(x,1/3)
]]
<a name="ch09b046">
then click 
"text box3 command output to box4"
box4 output next value
[[
9*pow(x,4/3)/4-9*pow(x,1/3)
0
]]

Now change from x=4 to x=4.01
box 4 output
0.03574626267025671

<a name="ch09b047">
In summary
x=4.00 f(x)=0
x=4.01 f(x)=0.03574626267025671
x=0.00 f(x)=0
x=0.01 f(x)=-1.9341437429761236

<a name="ch09b048">
Both x=0 and x=4 have f(x)=0
but x=0.01 f(x)=-1.934
and x=4.01 f(x)=0.0357
We find
at x=0, ∆x=0.01 ∆f=-1.934
at x=4, ∆x=0.01 ∆f=0.0357
same small step ∆x=0.01,
x=0 has great ∆f change -1.934
x=4 has small ∆f change 0.0357
This is stability problem.

<a name="ch09b049">　Index begin　Index this file
Problem solved, root are x=0,4
but ! their stability is a 
different problem.

<a name="ch09b050">
Please goto program integral.htm
input
f(x) =  ---eqn.BB023
9*pow(x,4/3)/4-9*pow(x,1/3)
and
d[f(x)]/dx =  ---eqn.BB024
3*pow(abs(x),1/3)-3*pow(abs(x),-2/3)

<a name="ch09b051">
for dx=0.00001
output
Slope at x=4 is    3.5716553
Analytic Solution: 3.5716523669284483
Slope at x=0 is    -19389.86
Analytic Solution: -Infinity
2010-03-17-11-19 stop

<a name="ch09b052">
2010-03-17-11-46 start
To study Holder Inequality 
stability problem, let us review
Holder Inequality first

<a name="ch09b054">　Index begin　Index this file
Right most (∑[k=1,n]1q)1/q is n1/q
Since 1q is 1 and ∑[k=1,n]1q is n.
Then Holder1409-eqn.9.1 simplify
to
  ∑[k=1,n]ak≦ ---eqn.BB025
    (∑[k=1,n]akp)1/p*n1/q
<a name="ch09b055">
eqn.BB025 take p-th power get
  (∑[k=1,n]ak)p≦ ---eqn.BB026
    (∑[k=1,n]akp)p/p*np/q
From p,q math eqn.BA114
get
  p/q=p-1 ---eqn.BB027
then
  np/q=np-1 ---eqn.BB028
<a name="ch09b056">
eqn.BB026 become
  (∑[k=1,n]ak)p≦ ---eqn.BB029
    (∑[k=1,n]akp)*np-1
Move np-1 to less than side
  (∑[k=1,n]ak)p*n1-p≦
     ∑[k=1,n](akp) ---eqn.BB030
<a name="ch09b057">
Follow eqn.BB030, we write
defect δ(a) as greater than
side minus less than side. 
[then δ(a) is nonnegative]
Define
  δ(a)=∑[k=1,n](akp) ---eqn.BB031
       -(∑[k=1,n]ak)p*n1-p
write in better math equation as

<a name="ch09b059">　Index begin　Index this file
eqn.BB030 inequality direction
determined that defect δ(a)
is a non-negative number.

<a name="ch09b060">
The condition of defect
  δ(a)=0 ---eqn.BB032
worth our attention.
δ(a) come from 
Holder1409-eqn.9.1
<a name="ch09b061">
If Holder1409-eqn.9.1 become 
equality, then defect δ(a)
become zero.
Holder Inequality's equality condition
tell us that for each 1≦k≦n we
<a name="ch09b062">
must have
  ahatkp＝bhatkq　---eqn.BA066
We set b to {1,1,...,1}
then for Holder to be equality 
matching ak must be a constant 
for all k.

<a name="ch09b063">
For defect δ(a) become zero,
we require
  aj=μ for j=1,2,...,n ---eqn.BB033a
The reverse is true,
If given
  a={μ,μ,...,μ} ---eqn.BB033b
then
  δ(a)=0 ---eqn.BB032
2010-03-17-12-46 stop

<a name="ch09b064">　Index begin　Index this file
2010-03-17-13-56 start
■　Problem 9.5 (A Stability Result
　　for Holder's　Inequality)
Show that if
  p≧2 ---eqn.BB034
and if
  aj≧0 for all j=1,2,...,n ---eqn.BB035
<a name="ch09b065">
then there exists a constant
  λ=λ(a,p) ---eqn.BB036
such that
  aj∈[(λ-δ½)2/p, (λ+δ½)2/p] ---eqn.9.20
  for all j=1,2,...,n 
<a name="ch09b066">
In other words, show that if the
difference defect δ=δ(a) is small
then the sequence a1,a2,...,an is
almost constant.
2010-03-17-14-05 here

<a name="ch09b067">
eqn.9.20 tell us we have two
bounds
  (λ-δ½)2/p ≦ aj ---eqn.BB037
and
  aj ≦ (λ+δ½)2/p ---eqn.BB038
<a name="ch09b068">
We assume //justified
  λ-δ½≧0 ---eqn.BB039
otherwise if λ-δ½＜0, then
(λ-δ½)2/p is negative2/p
with p≧2 ---eqn.BB034
negative2/3 is a complex number
which we want to avoid.

<a name="ch09b069">　Index begin　Index this file
With λ-δ½≧0 assumption, we take
p/2 power for both eqn.BB037
and eqn.BB038, get
  λ-δ½ ≦ ajp/2 ---eqn.BB040
and
  ajp/2 ≦ λ+δ½ ---eqn.BB041
<a name="ch09b070">
re-write as
  -δ½ ≦ ajp/2-λ ---eqn.BB042
and
  ajp/2-λ ≦ δ½ ---eqn.BB043
where ajp/2-λ is common
we have
  -δ½ ≦ ajp/2-λ ≦ δ½ ---eqn.BB044
<a name="ch09b071">
Square eqn.BB044, we get
  (ajp/2-λ)2 ≦ δ(a) ---eqn.BB045
eqn.BB045 is in textbook page 145
line 12 right end.
eqn.9.20 has n bound equations
eqn.BB045 satisfy all n of them.
What we do next step is
eqn.BB045 left side sum n times,
but right side NOT to do so. 
We get

<a name="ch09b073">
eqn.9.21 left side summed
n times (∑[j=1,n]), but right
side still keep one δ(a).
It is possible that eqn.9.21 asks
for too much. but 
eqn.9.21 is such a nice conjecture 
that it deserve our attention.
2010-03-17-14-46 stop

<a name="ch09b074">　Index begin　Index this file
2010-03-17-15-58 start
Why eqn.9.21 is a nice conjecture?
Is there any trace indicate that
eqn.9.21 is possibly true?
Yes, for one special case, eqn.9.21
is actually an identity.
If we set 
  p=2 ---eqn.BB046
and set
  λ=(a1+a2+...+an)/n ---eqn.BB047
eqn.9.21 become next

<a name="ch09b076">
where "?＝?" mean possibly
an identity, need to verify.
defect δ(a) is defined in eqn.9.19
For p=2, eqn.9.19 is next

<a name="ch09b078">
To simplify work, define
  SumA=a1+a2+...+an ---eqn.BB050
eqn.BB048 left side is our starting
point.
  ∑[j=1,n]{(aj-SumA/n)2}= ---eqn.BB051 begin
  ∑[j=1,n]{aj2-2*aj*SumA/n
       + SumA*SumA/n/n }
  =
    ∑[j=1,n]{aj2}
 -2*∑[j=1,n]{aj*SumA/n}
 +  ∑[j=1,n]{SumA*SumA/n/n}
<a name="ch09b079">　Index begin　Index this file
  =
    ∑[j=1,n]{aj2}
 -{2*SumA/n}*∑[j=1,n]{aj}
 +{SumA*SumA/n/n}*∑[j=1,n]{1}
  =
    ∑[j=1,n]{aj2}
 -{2*SumA/n}*{SumA}
 +{SumA*SumA/n/n}*n
<a name="ch09b080">
  =
    ∑[j=1,n]{aj2}
 -{2*SumA*SumA/n}
 +{  SumA*SumA/n}
  =
  ∑[j=1,n]{aj2}
  -SumA*SumA/n ---eqn.BB051 end

<a name="ch09b081">
eqn.BB051 end is same as eqn.BB049
right side !!
eqn.BB051 begin and eqn.BB051 end 
is same as eqn.BB048 
"?=?" in eqn.BB048 can be replaced 
by "=".

<a name="ch09b082">
Conjecture eqn.BB048 is true for
  p=2 ---eqn.BB046
and 
  λ=(a1+a2+...+an)/n ---eqn.BB047
Then how about general case?
Whether conjecture eqn.BB048 is 
true for arbitrary p and λ ?
2010-03-17-16-45 here

<a name="ch09b083">
eqn.9.21 is in quadratic of λ.
If eqn.9.21 is equality, and
if eqn.9.21 has two distinct
real root for λ, then between
these two roots, inequality 
eqn.9.21 exist.
<a name="ch09b084">　Index begin　Index this file
Now let us write eqn.9.21 as
equality and use eqn.9.19 for
δ(a) to find a quadratic
equation for λ. Start from
equality of eqn.9.21

<a name="ch09b086">
Do algebraic expansion and
simplification as following.
2010-03-17-17-03 here

<a name="ch09b087">
2010-03-17-18-15 start
Terms not contain λ are constant
to this λ-quadratic equation.
  ∑[j=1,n]{ajp/2-λ}2 ---eqn.BB053 begin
 =∑[j=1,n]{aj2p/2-2λajp/2+λ2}
 =∑[j=1,n]ajp
 -∑[j=1,n]2λajp/2
 +∑[j=1,n]λ2
 =∑[j=1,n]ajp
 -2λ∑[j=1,n]ajp/2
 +n*λ2 ---eqn.BB053 end
<a name="ch09b088">
eqn.BB053 end is eqn.BB052 left
side, which equal right side.
eqn.BB053 end ∑[j=1,n]ajp and
eqn.BB052 right side ∑[j=1,n]ajp
cancel out.
Move eqn.BB052 right side blue 
term to left side (to eqn.BB053
end) we get

<a name="ch09b090">
Compare eqn.9.22 with a model
quadratic equation
  Aλ2 + 2Bλ + C =0 ---eqn.BB054
they have real roots if and 
only if
  AC≦B2 ---eqn.BB055
Here
  A=n ---eqn.BB056
  B=-λ∑[j=1,n]ajp/2 ---eqn.BB057
  C=eqn.9.22 blue term

<a name="ch09b091">
Discriminant
  AC≦B2 ---eqn.BB055
expand to next equation

<a name="ch09b093">
In Holder1410-eqn.9.1 we set
  s=p/2  ---eqn.BB058
  p=2s ---eqn.BB059
  q=p/(p-1)=2s/(2s-1)  ---eqn.BB060
verify
  1/p + 1/q = 1/(2s) + (2s-1)/(2s)
   = (2s-1+1)/(2s) = 1  ---eqn.BB061
Requirement 1/p + 1/q =1 is OK

<a name="ch09b094">　Index begin　Index this file
How to link t with p,q?
  1/p+1/q =1 =1/s+1/t ---eqn.BB062
  1/p+1/q-1/s = 1/t 
  1/t = 1/p+1/q-1/s 
  1/t = 1/(2s)+(2s-1)/(2s)-2/(2s)
  1/t = (1+2s-1-2)/(2s) = (s-1)/s ---eqn.BB063
then
  1/t = (s-1)/s = ((p/2)-1)/(p/2)
  1/t = (p-2)/(p) ---eqn.BB064
and
  t = p/(p-2) ---eqn.BB065

<a name="ch09b095">
Now from Holder1410-eqn.9.1 
change s,t,1/s, 1/t to p, get

<a name="ch09b096">
2010-03-17-19-57 here
Compare eqn.BB066 left side with
eqn.9.23 blue term, we need take
p-th power for eqn.BB066. Before
do so, pay attention to eqn.BB066
right end term, it is summation
of ∑[j=1,n]{1} which is n, then
take (p-2)/(p) power.
Next, take p-th power for eqn.BB066

<a name="ch09b098">
From eqn.BB067 to eqn.9.23 
is just one step, move n(p-2)
from greater than side to less
than side and done!!
eqn.9.23 is true, then
conjecture eqn.BB048 is true for
general case.
Problem 9.5 solved.
2010-03-17-20-13 stop

<a name="ch09b099">
2010-03-17-20-30
Problem 9.5 is understand during 
writing. Because during writing
LiuHH did detail calculation.
2010-03-17-20-31

<a name="ch09b100">　Index begin　Index this file
2010-03-17-20-58 start
We made one assumption before.
We assumed
  λ-δ½≧0 ---eqn.BB039
Now done proof. Discriminant
  AC≦B2 ---eqn.BB055
indicate real roots exist.
Then assumed one root δ½
for which λ-δ½≧0
this assumption is justified.
2010-03-17-21-01 stop



<a name="ch09b101">
2010-03-18-12-17 start
Please compare p-norm ∥a∥p with Mt 
Next copied from tute0033.htm and
tute0030.htm.
<a name="ch09a078">
Instead of square and square root
we can use p-power and p-root to
define a p-norm as following

2010-03-18-12-26 stop



<a name="ch09b103">
2010-03-18-13-52 start
Next discuss interpolation of 
lp-norm between l1-norm and l∞-norm 
lp-norm definition is in eqn.9.11 
If change p to 1, we get l1-norm
which is simply summation of all
sequence elements.
  l1=a1+a2+...+an ---eqn.BB068
l∞-norm is the maximum element of 
all sequence elements
  l∞=max{a1,a2,...,an} ---eqn.BB069

<a name="ch09b104">　Index begin　Index this file
2010-03-18-14-19 start
■　Problem 9.6 (An Illustration
　　of l1→l∞ Interpolation)
Let cjk, 1≦j≦m, 1≦k≦n, be an
array of nonnegative real numbers
such that

<a name="ch09b106">
for all xk, 1≦k≦n
If 1＜p＜∞ and 
  q=p/(1-p) ---eqn.BB072
show that one also has the
interpolation bound

for all xk, 1≦k≦n.
2010-03-18-14-52 stop

<a name="ch09b108">
2010-03-18-16-29 start
Please pay attention to that
in eqn.9.24 if set p=1, then
eqn.9.24 become eqn.BB070 
in eqn.9.24 if set p=∞, then
eqn.9.24 become eqn.BB071 

<a name="ch09b109">　Index begin　Index this file
When p=1, q=p/(1-p)=1/0=∞
then B1/q=B1/∞=B0=1
and
  {∑[k=1,n]|xk|p}1/p
          =∑[k=1,n]|xk| ---eqn.BB073
Similarly, when p=∞, A1/p=1
and
  {∑[k=1,n]|xk|p}1/p
          =max[k=1,n]|xk| ---eqn.BB074
Both recover l1 and l∞
inequalities eqn.BB070 and
eqn.BB071.

<a name="ch09b110">
Second attention is that in
eqn.BB070, if drop constant A
then inequality could be the
same if all cjk are small.
But inequality also could be 
reversed if all cjk are great.
Inequality eqn.BB070 exist
mainly because of constant A.
Same reason for eqn.BB071.

<a name="ch09b111">
Textbook page 147 top say that
the p-th root in equation 9.24 
is troublesome and remove p-th 
root is first step. The tool to
remove p-th root is Holder 
converse.

<a name="ch09b112">
Eqn.9.24 is Problem 9.6 wait to 
be proved result equation. Now
assume eqn.9.24 is true and
derive the following step see
whether result is reasonable?

<a name="ch09b113">
Compare eqn.9.24 with Holder 
converse assumption eqn.9.8

<a name="ch09b116">　Index begin　Index this file
If set
  C=A1/p*B1/q ---eqn.BB075
then eqn.9.8 and  eqn.9.24 both
greater than side are the same.

Textbook page 147 upper half say
apply Holder converse conclusion 
to eqn.9.24 we get
  ∑[j=1,m]∑[k=1,n]cjkxkyj
  ≦A1/p*B1/q ---eqn.9.25

<a name="ch09b117">
2010-03-18-17-28 continue think
how to convert eqn.9.24 left
side to eqn.9.25 left side ?
What is the role of yj?
2010-03-18-17-30 continue think
<a name="ch09b118">
Textbook say:
"The reformulation (9.25) offers
signs of real progress, in
particular, the pth roots are 
gone."
2010-03-18-17-33 continue think

2010-03-18-18-36 
Problem 9.6 NOT DONE



<a name="ch09b119">　Index begin　Index this file
2010-03-18-18-41 start
■　Exercise 9.1 problem statement
　　textbook page 148
（Doing the Sum for Holder）

In Exercise 1.8 we saw that the
effective use of Cauchy's
inequality may depend on having 
an estimate for one of bounding
sums and, in this respect, Holder's
inequality is a natural heir. 
<a name="ch09b120">
As a warm-up, check that for real
aj j=1,2,..., one has

2010-03-18-19-18 here




<a name="ch09b121">
2010-03-18-19-20 start
■　Exercise 9.1 hint
　　textbook page 261
For the first bound one apply
Holder's inequality with p=5/4
and q=5 
//verify: 1/p+1/q=4/5+1/5=1 OK 
and finishes with the
telescoping identity

<a name="ch09b123">
For the second bound one uses
p=4/3 and q=4 
//verify: 1/p+1/q=3/4+1/4=1 OK 
and finishes with
Euler's classic sum

<a name="ch09b124">　Index begin　Index this file
While for the third bound
one use p=3/2 and q=3 
//verify: 1/p+1/q=2/3+1/3=1 OK 
and 
finishes with the geometric
sum // for 0≦x＜1
  1+x3+x3+...=1/(1-x3) ---eqn.BB081
2010-03-18-19-35 stop

<a name="ch09b125">
2010-03-18-19-41 start
Textbook original statement for
Exercise 9.1 (b) is
[[ 
For the second bound one uses
p=3/4 and q=4 and .....
]] textbook page 261 line 15
<a name="ch09b126">
Liu,Hsinhan change to
[[
For the second bound one uses
p=4/3 and q=4 and .....
]]
This change is not found in three
errata pages.
2010-03-18-19-46 stop




<a name="ch09b127">　Index begin　Index this file
2010-03-18-19-55 start
■　Exercise 9.1 solution (a)(b)(c)

For eqn.BB076 consider Holder's
inequality [Here is Exercise 9.1 (a)]

<a name="ch09b128"> eqn.BB076
2010-03-18-19-59 think

bk=[k(k+1)]-1/5
bk5=[k(k+1)]-1
∑[k=1,n]bk5=∑[k=1,n][k(k+1)]-1
=1 - 1/(n+1) < 1

<a name="ch09b129">
2010-03-18-20-14 start
Compare Holder1411-eqn.9.1 with 
eqn.BB076, {∑[k=1,n]|ak|5/4}4/5
is the same. Need to identify 
bk. From eqn.BB076 less than side
we see bk should be
  bk=[k(k+1)]-1/5 ---eqn.BB082
to match eqn.BB076 less than side.
<a name="ch09b130">
With bk defined, goto 
Holder1411-eqn.9.1 right most
term, find
  ∑[k=1,n]bk5=∑[k=1,n][k(k+1)]-1
  =1 - 1/(n+1) < 1 ---eqn.BB083
For expanded equation, please see 
eqn.2.19. eqn.2.19 sum from j=k
to j=infinity.
<a name="ch09b131">
Here eqn.BB083 sum from k=1 to k=n
result is 1 - 1/(n+1). let us drop
- 1/(n+1), what left is 1
Holder1411-eqn.9.1 tell us we are
at greater than side. Assume
5*(1-0.1)=4.5 is greater than side
After drop -0.1, get
5*(1)=5
<a name="ch09b132">
greater than side become even 
greater. Inequality look like
less than side ≦ 4.5 ＜ 5
This is inequality class, the
drop operation is correct. 
After drop -1/(n+1),
1 - 1/(n+1) become 1,
Holder1411-eqn.9.1 right most 
term has outer power 1/5. 
1^(1/5) is still one. 
We proved eqn.BB076.
Exercise 9.1 (a) is done.
2010-03-18-20-35 stop

<a name="ch09b133"> (a)(b)(c)
2010-03-18-20-38　Index begin　Index this file
For eqn.BB077 consider Holder's
inequality [Here is Exercise 9.1 (b)]

<a name="ch09b134"> eqn.BB077
2010-03-18-20-39 here
Compare Holder1412-eqn.9.1 with 
eqn.BB077, {∑[k=1,n]|ak|4/3}3/4
is the same. Need to identify 
bk. From eqn.BB077 less than side
we see bk should be
  bk=1/√k=k-1/2 ---eqn.BB084
to match eqn.BB077 less than side.
<a name="ch09b135">
With bk defined, goto 
Holder1412-eqn.9.1 right most
term, find
  ∑[k=1,n]bk4=∑[k=1,n]k-4/2
    =∑[k=1,n]{1/k2} ---eqn.BB085
eqn.BB085 is finite sum,
eqn.BB080 is infinite sum.
<a name="ch09b136">
Both sum same function 1/k2
which is positive. So infinite sum
is greater than finite sum. Again
this occur at greater than side.
<a name="ch09b137">
Let us push greater than side
even greater, use Euler's classic 
sum which has value PI*PI/6.
Holder1412-eqn.9.1 right most
term need take 1/4 power.
After set (PI*PI/6) to
(PI*PI/6)^(1/4)
Exercise 9.1 (b) is done.
2010-03-18-21-00 stop

<a name="ch09b138"> (a)(b)(c)
2010-03-18-21-03　Index begin　Index this file
For eqn.BB078 consider Holder's
inequality [Here is Exercise 9.1 (c)]

<a name="ch09b139"> eqn.BB078
2010-03-18-21-05 here
Compare Holder1413-eqn.9.1 with 
eqn.BB078, {∑[k=1,n]|ak|3/2}2/3
is the same. Need to identify 
bk. From eqn.BB078 less than side
<a name="ch09b140">
we see bk should be
  bk=xk ---eqn.BB086
to match eqn.BB078 less than side.
Attention problem given for 0≦x＜1

<a name="ch09b141">
With bk defined, goto 
Holder1413-eqn.9.1 right most
term, find
  ∑[k=1,∞]bk3=∑[k=1,∞]xk ---eqn.BB087
Geometric sum tell us //for 0≦x＜1
  1+x3+x3+...=1/(1-x3) ---eqn.BB081
Holder1413-eqn.9.1 right most
term need take 1/3 power.
After set 1/(1-x3) to
1/(1-x3)^(1/3)
Exercise 9.1 (c) is done.

<a name="ch09b142">
Attention, Previous two problems 
use less than sign, but eqn.BB078 
use less than equal to sign.
Because Exercise 9.1 (c) did not
push greater than side to even
greater. eqn.BB078 sum to infinity.
How can we be greater than infinity?
2010-03-18-21-17 stop



<a name="ch09b143">　Index begin　Index this file
2010-03-19-10-56 start
■　Exercise 9.2 problem statement
　　textbook page 148
　　（An Inclusion Radius Bound）
For a polynomial
  P(z)=zn+an-1zn-1+...+a1z+a0 ---eqn.BB088
with real or complex coefficients,
//LiuHH note: coefficient can be 
//real or complex, but independent 
//variable z IS complex.
<a name="ch09b144">
the smallest value r(P) such all
roots of P are contained in the 
disk
  {z:|z|≦r(P)} ---eqn.BB089
is called the inclusion radius for
P. Show that 
<a name="ch09b145">
for any conjugate pair 
  P＞1 ---eqn.BB090
and
  q=p/(p-1)＞1 ---eqn.BB091
one has the bound
  r(P)＜(1+Apq)1/q ---eqn.9.29
where
  Ap=(∑[j=0,n-1]|aj|p)1/p ---eqn.BB092
2010-03-19-11-09 stop





<a name="ch09b146">　Index begin　Index this file
2010-03-19-11-20 start
■　Exercise 9.2 hint
　　textbook page 261

Consider z such that 
  |z|＞1 ---eqn.BB093
and note by Holder's inequality 
that one has the bound

<a name="ch09b150">
2010-03-19-11-50 here
Thus we have
  |P(z)|＞0 ---eqn.BB097
if
  Ap/(|z|q-1)1/q≦1 ---eqn.BB098
That is we have |P(z)|＞0 for all 
z such that
  |z|＞(1+Apq)1/q ---eqn.BB099

<a name="ch09b151">
The bound (9.29) for the inclusion
radius is due to M. Kuniyeda, and
it provides a useful reminder how
one can benefit from the flexibility
afforded by Holder's inequality.
<a name="ch09b152">
Here, for a given polynomial, a wise
choice of the power p sometimes
leads to an inclusion radius that
is dramatically smaller than the
one given by p=2. This result and
many other bounds for the inclusion 
radius are developed in Mignotte
and Stefanescu (1999).
2010-03-19-12-04 stop




2010-03-19-15-50 start
<a name="ch09b153">　Index begin　Index this file
■　Exercise 9.2 solution


Given a polynomial
  P(z)=zn+an-1zn-1+...+a1z+a0 ---eqn.BB088
If we can solve for roots, then
inclusion radius is simply the
maximum absolute value of all
roots. 
<a name="ch09b154">
Now we use Holder's
inequality to find inclusion 
radius, that is a signal that
solving for complex polynomial
is not available. Then we step 
back, try find an estimation
for inclusion radius.

<a name="ch09b155">
What we have in hand is P(z)'s
coefficients 1, an-1, an-2,
..., a1, a0.
These coefficients from a0 to
an-1 are our starting point.
In Exercise 9.2, 
 we do not have ∑[j=1,n]
instead we have ∑[j=0,n-1] 
to match polynomial P(z).

<a name="ch09b156">
Exercise 9.2 hint tell us to
begin at eqn.BB094 
Equation left side is
  |∑[j=0,n-1]ajzj| ---eqn.BB100
Let us see Holder's inequality

<a name="ch09b157">
Holder1414-eqn.9.1 is eqn.BB094 
in hint. Slight difference is
that eqn.BB094 use Ap eqn.BB092
and eqn.BB094 added absolute 
value to less than side. We 
can do same thing to eqn.9.1
and match eqn.BB094 exactly.

<a name="ch09b158">　Index begin　Index this file
If P(z) is complex polynomial,
we have to take absolute value
to Holder1414-eqn.9.1, because
we can not compare complex 
number. Only real number can.
2010-03-19-16-30 here

<a name="ch09b159">
Polynomial P(z) is
  P(z)=zn+an-1zn-1+...+a1z+a0 ---eqn.BB088
that is
  P(z)=zn+ ∑[j=0,n-1]ajzj ---eqn.BB101
Because zn do not show up in
summation, change eqn.BB101
to next
  P(z)-zn = ∑[j=0,n-1]ajzj ---eqn.BB102
eqn.BB102 right side summation 
is same as eqn.BB094 less than 
side. 

<a name="ch09b160">
Can we push less than side 
even smaller and bring |P(z)| 
into equation? Take absolute
value for eqn.BB102 get
  |P(z)-zn| = |∑[j=0,n-1]ajzj| ---eqn.BB103
Left side has next relation
  |P(z)-zn|≧|zn|-|P(z)|
           =|z|n-|P(z)| ---eqn.BB104
<a name="ch09b161">
Combine eqn.BB103 and eqn.BB104
  |∑[j=0,n-1]ajzj|
     ≧|z|n-|P(z)| ---eqn.BB105
Then we get
  |P(z)|≧|z|n ---eqn.BB106
        -|∑[j=0,n-1]ajzj|
<a name="ch09b161b">
"-|∑[j=0,n-1]ajzj|" part refer 
to eqn.BB094. We need take 
negative for eqn.BB094 and
reverse the inequality order.
The negative of eqn.BB094 is next

Substitute eqn.BB125 to eqn.BB106

<a name="ch09b163">　Index begin　Index this file
In eqn.BB107 in summation
multiply by one = |z|nq/|z|nq
eqn.BB108 summation variable is j
|z|nq/|z|nq is not a function of j
move numerator out of summation,
keep denominator in sum, we get

<a name="ch09b164">
From eqn.BB108 to eqn.BB095 is
a small step.
2010-03-19-17-24 here

2010-03-19-17-40 start
eqn.BB096 less than side is 
summation of 1/|z|kq 
from k=1 to k=n
(k=1 meet j=n-1, k=n meet j=0)

<a name="ch09b165">
eqn.BB096 greater than side is 
summation of 1/|z|kq 
from k=0 to k=∞
(k=0 meet j=0, k=∞ meet j=∞)

<a name="ch09b166">
Because 1/|z|q is positive value
function.  Partial sum k=1 to k=n
is less than total sum k=0 to k=∞
and total sum has a finite value
1/(|z|q - 1)
2010-03-19-17-47 stop and think

<a name="ch09b167">
2010-03-19-18-03 start
Equality in eqn.BB096 should be
right, but LiuHH did not find
related formula from Peirce 
Integral Table.

Now continue.
<a name="ch09b168">　Index begin　Index this file
eqn.BB096 less than side (positive)
is
eqn.BB095 less than side summation
term, which is negative.
Multiply -1 to eqn.BB096, reverse 
its inequality direction. Use 
eqn.BB096 re-write eqn.BB095 as

2010-03-19-18-15 stop and think
<a name="ch09b169b">
2010-03-19-18-39 start
Assume eqn.BB109 square bracket
term ≧0, that is

<a name="ch09b170">
then
  1≧Ap[1/(|z|q-1)]1/q ---eqn.BB111
  1q≧Apq[1/(|z|q-1)]q/q
  1≧Apq[1/(|z|q-1)]
  |z|q-1≧Apq
  |z|q≧1+Apq
  |z|≧(1+Apq)1/q ---eqn.BB112

<a name="ch09b171">
Why need assume that
  |z|＞1 ---eqn.BB093
??

Why need assume eqn.BB109 square 
bracket term ≧0 ??

<a name="ch09b172">
Why eqn.BB112 tell us that 
inclusion radius r(P)
  r(P)＜(1+Apq)1/q ---eqn.9.29
??

If |z|＜r(P) and if z is not a root
do we have |P(z)| not ＞ 0 ??

<a name="ch09b173">
LiuHH still think, and 
Exercise 9.2 is NOT DONE

LiuHH did not take 
complex analysis class.
Learned complex analysis
from engineering math.
2010-03-19-19-16 stop



<a name="ch09b174">　Index begin　Index this file
2010-03-19-20-02 start
■　Exercise 9.3 problem statement
　　textbook page 149
（Cauchy Implies Holder）

Prove that Cauchy's inequality 
implies Holder's inequality.
More specifically, show that
Cauchy's inequality implies
Holder's inequality for 
p in {8/1,8/2,8/3,...,8/6,8/7}
by first showing

<a name="ch09b176">
By the same method, one can prove
Holder's inequality for all
  p=2k/j ---eqn.BB114
1≦j≦2k. One can then call on
continuity to obtain Holder's 
inequality for all 1≦p＜∞.

<a name="ch09b177">
This argument serves as a reminder
that an l2-result may sometimes
be applied iteratively to obtain an
lp-result. The inequalities one
finds this way are often proved 
more elegantly by other methods,
but iteration is still a remarkably
effective tools for the discovery 
of new bounds.
2010-03-19-20-23 stop




<a name="ch09b178">　Index begin　Index this file
2010-03-19-20-28 start
■　Exercise 9.3 hint
　　textbook page 262

This method is worth understanding
but the exercise does not leave 
much to do. First apply Cauchy's
inequality to the sum of αjβj
where 
  αj=ajbjcjdj ---eqn.BB115
  βj=ejfjgjhj ---eqn.BB116
<a name="ch09b179">
then repeat the natural splitting
twice more. It is obvious (but
easy to overlook!) that each
p in [1,∞) can be approximated
as closely as we like by a 
rational number of the form
  p=2k/j ---eqn.BB117
where 1≦j＜2k
2010-03-19-20-35 stop



<a name="ch09b180">
2010-03-19-20-58 start
■　Exercise 9.3 solution


Next is Cauchy's Inequality

<a name="ch09b181">
where 
  αj=ajbjcjdj ---eqn.BB115
  βj=ejfjgjhj ---eqn.BB116
Cauchy142's ∑[j=1,n]αj*αj is next

<a name="ch09b183">　Index begin　Index this file
Take aj2bj2 as first sequence
 and cj2dj2 as second sequence.
Second time apply Cauchy's Inequality

Cauchy143 left side is 
Cauchy142 left square bracket
term without take square root.
If take square root, then 
<a name="ch09b184">
Cauchy143 right side has power
1/4 instead of 1/2.
Cauchy143 right side has aj4,
bj4, cj4, dj4. Repeat same 
procedure one more time.
aj will rise to power 8
same thing happen to bj to hj
<a name="ch09b185">
Also inequality greater than 
side outer power change to 1/8.
Finally, whole equation take
8-th power, result is eqn.BB113.
2010-03-19-21-24 here

<a name="ch09b186">
Now, eqn.BB113 take eighth root
and compare with Holder's 
inequality as following.

<a name="ch09b188">　Index begin　Index this file
2010-03-19-21-38 here
Cauchy's Inequality allow many 
sequences. Holder allow only two.
Merge eight sequences to two.
In eqn.BB119, if set
  aj=bj=cj=dj=mj1/4 ---eqn.BB120
  ej=fj=gj=hj=nj1/4 ---eqn.BB121
simple case

<a name="ch09b189">
In eqn.BB119, if set
  aj=bj=cj=dj=ej=mj1/5 ---eqn.BB122
  fj=gj=hj=nj1/3 ---eqn.BB123

eqn.BB119 become

<a name="ch09b191">
eqn.BB119 is Cauchy's inequality
eqn.BB124 is Holder's inequality
The setting of eqn.BB122 and
eqn.BB123 convert Cauchy to
Holder successfully for rational
number {8/1,8/2,8/3,...,8/6,8/7}
<a name="ch09b192">
Here 8=23, we can use same
method to go to higher k in 2k.
For irrational number, argue that
two rational number will enclose
irrational from both side and
number system is continuous.
We obtained Holder's inequality 
for all 1≦p＜∞.
2010-03-19-22-30 stop

2010-03-20-13-30 done  first proofread
2010-03-20-18-55 done second proofread
2010-03-20-19-56 done spelling check

<a name="ch09b193">
2010-03-25-20-53
"Update 2010-03-22" change the next
From "[SumA}" to "{SumA}"
"can be dropped" to "can be replaced"
2010-03-25-20-58



<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