<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="ch09c001">　Index begin　Index this file
2010-03-21-12-13
Exercise 9.4 need Exercise 9.7
equation. Solve Exercise 9.7 first


<a name="ch09c002">　Index begin　Index this file
2010-03-21-12-18 start
■　Exercise 9.7 problem statement
　　textbook page 150
（Holder's Inequality for Integrals）

Naturally there are integral 
version of Holder's Inequality 
and, in keeping with the more 
modern custom, there is no 
cause for a name change when
one switch from sum to integrals.

<a name="ch09c003">
Let w:D→[0,∞) be given, and 
reinforce your mastery of
Holder's Inequality by checking
that our earlier argument (page
137) also shows that for all
suitable functions f and g from
D to Real

<a name="ch09c005">
where, as usual, 
1＜p＜∞ and
  p-1+q-1=1 ---eqn.BC002
2010-03-21-12-38 stop





<a name="ch09c006">　Index begin　Index this file
2010-03-21-12-39 start
■　Exercise 9.7 hint
　　textbook page 263

Without loss of generality, one can
assume that the integrals of the 
upper bound do not vanish. Call 
these intrgrals I1 and I2. 
<a name="ch09c007">
Apply Young's inequality (9.6) to
  u=|f(x)|/I11/p ---eqn.BC003
and
  v=|g(x)|/I21/q ---eqn.BC004
multiply by w(x) and integrate.
Holder's Inequality then follows 
by arithmetic. For a thorough job, 
one may want to retrace this 
argument to sort out the case of 
equality.
2010-03-21-12-45 stop


<a name="ch09c008">
2010-03-21-12-51 start
■　Exercise 9.7 solution


First clearly state the definition 
of I1 and I2 as following

<a name="ch09c010">
Please pay attention to the
difference between w(x) and 
f(x),g(x).
w(x)   is  D→[0,∞)  be given,
f(x),g(x): D→(-∞,∞) be suitable.
w(x)   is  non-negative but
f(x),g(x) is negative/zero/positive.
D is integration domain, x∈[-1,1]
or x∈[1,∞) whatever it is, name
it as D.

<a name="ch09c011">　Index begin　Index this file
Please pay attention to the
difference between 
summation Holder's Inequality 
and
 integral Holder's Inequality
summation Holder do not use weight
and
 integral Holder use weight w(x)
2010-03-21-13-14 stop
<a name="ch09c012">
2010-03-21-14-52 start
If I1=0, then either f(x)≡0 or 
w(x)≡0 either case cause eqn.BC001 
both side be zero and equality is
true. Now assume
  I1≠0 ---eqn.BC007
and
  I2≠0 ---eqn.BC008
we can use I1,I2 as denominator.
Define u and v as following

<a name="ch09c017">
2010-03-21-15-39 here
In eqn.BC012, red denominator 
already integrate over D, Red
integral apply to numerator only.
The integration of red numerator
has same value as red denominator,
they cancel to one.
Same reason apply to blue term.
Both red and blue are one, what 
left is 1/p + 1/q which we know
it has value one (eqn.BC002).

<a name="ch09c018">
Then eqn.BC011 whole integration
less-equal to one. Its denominator
[∫D|f(x)|pw(x)dx]1/p*[∫D|g(x)|qw(x)dx]1/q
already integrated, the integration
sign in eqn.BC011 do not apply to
denominator. Denominator is 
integration of positive term,
denominator is positive. We move 
denominator to greater than side 
of inequality where value is one
and inequality direction do not 
change. The result is our target 
equation eqn.BC001

<a name="ch09c019">
Small difference is that eqn.BC011 
numerator applied absolute value 
to f(x) and g(x), but eqn.BC001 
less than side do not use absolute 
value. This is OK, because 
if |-5|＜8 remove absolute sign 
get -5 ＜8 it is still true.

Target equation eqn.BC001 is proved.
2010-03-21-16-05 stop

<a name="ch09c020">
2010-03-21-16-13 start
Integral Holder equality condition 
rely on Young's equality condition.
In eqn.9.6, if
  u=m1/p ---eqn.BC013
  v=m1/q ---eqn.BC014
then Young's inequality eqn.9.6
become
  uv=m1/pm1/q=m1 ---eqn.BC015
<a name="ch09c021">
and
  up=mp/p=m
  vq=mq/q=m
  up/p+vq/q=m*(1/p+1/q)=m*1 ---eqn.BC016
eqn.BC015 and eqn.BC016 indicate
Young's equality.

<a name="ch09c022">
Back to eqn.BC009, if
  f(x)=m(x)1/p
and
  g(x)=m(x)1/q
then
<a name="ch09c023">
  u(x)=m(x)1/p/[∫D|m(x)|w(x)dx]1/p
and
  v(x)=m(x)1/q/[∫D|m(x)|w(x)dx]1/q
eqn.BC011 become
  ∫D|m(x)|w(x)dx = ∫D|m(x)|w(x)dx
which is an equality.

Possibly you can do better job
than LiuHH.
2010-03-21-16-30 stop



<a name="ch09c024">　Index begin　Index this file
2010-03-22-08-06 start
■　Exercise 9.4 problem statement
　　textbook page 149
（Interpolation Bound for Moment 
 Sequences）

If φ:[0,∞)→[0,∞) is an integrable
function and t∈(0,∞), then the
integral

<a name="ch09c026">
is called the t-th moment of φ.
Show that if t∈(t0, t1) then
  μt≦μt01-α*μt1α ---eqn.BC018
where
  t=(1-α)t0+αt1 ---eqn.BC019
and 0＜α＜1
<a name="ch09c027">
In other words, the linearly 
interpolated moments is bounded
by the geometric interpolation 
of two extreme moments.
2010-03-22-08-24 stop





<a name="ch09c028">　Index begin　Index this file
2010-03-22-08-30 start
■　Exercise 9.4 hint
　　textbook page 262

Apply the Holder inequality given 
by Exercise 9.7 with D=[0,∞) and
  w(x)=φ(x) ---eqn.BC020
with the natural choice
<a name="ch09c029">
  f(x)=x(1-α)t0 ---eqn.BC021
  g(x)=xαt1 ---eqn.BC022
  p=1/(1-α) ---eqn.BC023
  q=1/α ---eqn.BC024
//given 0＜α＜1 //2010-03-25-14-30
//check 1/p+1/q=(1-α)+α=1 OK
<a name="ch09c030">
One consequence of this bound 
is that if the t-th moment is
infinity, then either t0-th
or t1-th moment must be
infinity.
2010-03-22-08-37 stop



<a name="ch09c031">
2010-03-22-08-46 start
■　Exercise 9.4 solution


Integral version of Holder 
inequality is next

<a name="ch09c033">　Index begin　Index this file
Follow Exercise 9.4 hint
change eqn.BC001 to next

<a name="ch09c035">
2010-03-22-09-13 here
In eqn.BC025, 
  x(1-α)t0*xαt1 = x(1-α)t0+αt1=xt ---eqn.BC027
Second equality in eqn.BC027 
used given condition eqn.BC019
After convert, eqn.BC025 is μt
See eqn.BC017.

<a name="ch09c036">
Refer to eqn.BC018, we need
find μt01-α and μt1α
From eqn.BC023, eqn.BC024 find
  (1-α)p=1 ---eqn.BC028
and 
  αq=1 ---eqn.BC029

<a name="ch09c037">
eqn.BC026 left bracket term 
|x(1-α)t0|p become |xt0| //(1-α)p=1
Because x in D=[0,∞) which is 
non-negative, |xt0| is xt0

eqn.BC026 left bracket become 

<a name="ch09c039">
Above term is eqn.BC026 left
bracket integral, need take power
of 1/p. eqn.BC023 tell us that
  1/p = 1-α ---eqn.BC031
so, eqn.BC026 left bracket term 
is μt01-α which is eqn.BC018 
greater than side first term.

<a name="ch09c040">
Similar reason apply to eqn.BC026
right bracket term, get μt1α which 
is eqn.BC018 greater than side 
second term.

We converted eqn.BC025 + eqn.BC026
to eqn.BC018 and Exercise 9.4 is 
solved.
2010-03-22-10-29 stop



<a name="ch09c041">　Index begin　Index this file
2010-03-22-12-20 start
■　Exercise 9.5 problem statement
　　textbook page 149
（Complex Holder -- and the Case 
　　of Equality）

Holder's inequality for real numbers
implies that for complex numbers
a1,a2,...,an and b1,b2,...,bn one 
has the bound

<a name="ch09c043">
2010-03-22-12-36 here
when p>1 and q>1 satisfy
  1/p + 1/q = 1 ---eqn.BC032
What conditions on the complex 
numbers a1,a2,...,an and 
b1,b2,...,bn are necessary and
sufficient equality to hold in 
the bound (9.30)? Although this
exercise is easy, it nevertheless
offers one useful morsel of
insight that should not be missed.
2010-03-22-12-41 stop





<a name="ch09c044">　Index begin　Index this file
2010-03-22-14-03 start
■　Exercise 9.5 hint
　　textbook page 262

Equality in the bound(9.30) gives
us

<a name="ch09c045">
Now if |a1|,|a2|,...,|an|
is a nonzero sequence, then the
real variable characterization
on page 136 tells us that the
second equality holds if and
only if there exists a constant 
λ＞0 such that 
  λ|ak|1/p=|bk|1/q ---eqn.BC033
for all 1≦k≦n.

<a name="ch09c046">
The novel issue here is to discover
when the first equality holds. If
we set
  akbk=ρkeiθ ---eqn.BC034
where ρk≧0 and θk∈[0,2π)
and if we further set
  pk=ρk/(ρ1+ρ2+...+ρn) ---eqn.BC035
<a name="ch09c047">
then the first equality holds
exactly when the average 
  p1eiθ1+p2eiθ2+...+pneiθn ---eqn.BC036
is on the boundary of the unit
disk, and this is possible if
and only if there exists a θ
such that
  θ=θk ---eqn.BC037
for all k such that pk≠0
<a name="ch09c048">
In other words, the first 
equality holds if and only if 
the values arg{akbk} are equal 
for all k for which arg{akbk} 
is well defined.
2010-03-22-14-30 stop



<a name="ch09c049">　Index begin　Index this file
2010-03-22-14-50 start
■　Exercise 9.5 discussion

Exercise 9.5 title is Complex Holder
But Complex Holder equation eqn.9.30
is GIVEN, not to be proved.

<a name="ch09c050">
Exercise 9.5 hint start at eqn.14.58
In this equation, there are two
equality. What condition ak,bk
must satisfy to get equality?

<a name="ch09c051">
eqn.14.58 second equality must
satisfy eqn.9.3 which reproduced
as eqn.BC033. The difference is 
that eqn.9.3 is for non-negative 
real number sequence.
<a name="ch09c052">
But eqn.BC033 is for complex 
numbers a1,...,an and b1,...,bn.
eqn.BC033 take absolute value for
each elements, output non-negative
real number. Satisfy Holder's
non-negative requirement.

<a name="ch09c053">
eqn.14.58 first equality is
flattened triangle-inequality,
that is "triangle-equality"

<a name="ch09c054">　Index begin　Index this file
In eqn.14.58 assume n=2 for two 
elements easy example
for k=1, set c1=a1b1
for k=2, set c2=a2b2
ak and bk are complex numbers
c1 and c2 are two vectors
(complex number is 2-D vector)

<a name="ch09c055">
If c1 and c2 are not collinear,
then the vector sum |c1+c2| is 
shorter than the sum of two
sides |c1|+|c2|

<a name="ch09c056">
If c1 and c2 are collinear, point to
same direction (flattened triangle)
then the vector sum |c1+c2| is 
same as the sum of two sides 
|c1|+|c2|.

This geometric description has same 
conclusion as Exercise 9.5 hint
complex argument.
2010-03-22-15-33 stop


<a name="ch09c057">
2010-03-22-15-35 start
■　Exercise 9.5 solution


Exercise 9.5 hint and
Exercise 9.5 discussion
already did the job.
2010-03-22-15-36 stop



<a name="ch09c058">　Index begin　Index this file
2010-03-22-16-20 start
■　Exercise 9.6 problem statement
　　textbook page 150
（Jensen Implies Minkowski）

By Jensen's inequality, we know 
that for a convex φ and positive
weights w1,w2,...,wn, one has

<a name="ch09c060"> Draw eqn.BC038
Consider the concave function 
  φ(x)=(1+x1/p)p ---eqn.BC038
on [0,∞], and show that by 
making the right choice of the 
weights wk and the values xk in 
Jensen's inequality (9.31) one 
obtain Minkowski's inequality.
2010-03-22-16-41 stop





<a name="ch09c061">　Index begin　Index this file
2010-03-22-16-45 start
■　Exercise 9.6 hint
　　textbook page 262/263

One checks by taking derivative
that 
<a name="ch09c062">
  φ''(x)=(1-p)x-2+1/p[(1+x1/p)-2+p]/p ---eqn.BC039
and this is negative since p＞1
and x≧0. One then applies 
Jensen's inequality (for concave 
function) to 
  wk=|ak|p ---eqn.BC040
and
  xk=|bk|p/|ak|p ---eqn.BC041
<a name="ch09c063">
the rest is arithmetic. This
modestly miraculous proof is
just one more example of how
much one can achieve with
Jensen's inequality, given 
the wisdom to chose the "right"
function.
2010-03-22-16-54 stop



<a name="ch09c064">
2010-03-22-17-57 start
■　Exercise 9.6 discussion

Start from given equation
  φ(x)=(1+x1/p)p ---eqn.BC038
<a name="ch09c065">
First derivative is 
  φ'(x)=d[(1+x1/p)p]/dx
  φ'(x)=p(1+x1/p)p-1 * d(1+x1/p)/dx
  //red to red, blue to blue
  φ'(x)=p(1+x1/p)p-1 * x(-1+1/p)/p
  // p/p cancel to 1/1
<a name="ch09c066">
  φ'(x)=1(1+x1/p)p-1 * x(-1+1/p)/1
  φ'(x)=(1+x1/p)p-1 * x-(p-1)/p
  φ'(x)=(1+x1/p)p-1 * [x-1/p](p-1)
  φ'(x)=(x-1/p+x1/p*x-1/p)p-1
  φ'(x)=(x-1/p+1)p-1 ---eqn.BC042

<a name="ch09c067">　Index begin　Index this file
Second derivative is 
  φ''(x)=d[(x-1/p+1)p-1]/dx
  φ''(x)=(p-1)(x-1/p+1)p-2 *d[x-1/p+1]/dx
  φ''(x)=(p-1)(x-1/p+1)p-2 *[(-1/p)*x-1-1/p+0]
  φ''(x)=[(-p+1)/p](x-1/p+1)p-2*x-1-1/p ---eqn.BC043
2010-03-22-18-20 stop

<a name="ch09c068"> [DrawEx0906] key
2010-03-22-19-27 start
Here is a simple drawing program
It marked next line
＜a name="sideDraw"＞ Exercise 9.6 p:[ 1.75] [DrawEx0906]
Fill a p number in p:[ 1.75] p＞1
then click [DrawEx0906]
If you modify data, please click
[Draw f0(x) to f3(x)] button.

<a name="ch09c069">
Four equations in Javascript code
f0(x)=pow(1+pow(x,1/p),p)
f1(x)=pow(1+pow(x,-1/p),p-1)
f2(x)=((-p+1)/p)*pow(x,-1-1/p)*pow(1+pow(x,-1/p),p-2)
f3(x)= ((1-p)/p)*pow(x,-2+1/p)*pow(1+pow(x,+1/p),p-2)
If you fill 1.75 to  "p:[ 1.75]" box
program replace all p in equation
with 1.75
<a name="ch09c070"> [DrawEx0906]
Four equations in math form are
f0(x)=eqn.BC038=φ(x)   by textbook
f1(x)=eqn.BC042=φ'(x)  by LiuHH
f2(x)=eqn.BC043=φ''(x) by LiuHH
f3(x)=eqn.BC039=φ''(x) by textbook

<a name="ch09c071"> [DrawEx0906]
Key point to watch is that
red line is concave, 
shape like ╭╮ , not ╰╯
f2(x) (blue) and f3(x) (dark brown)
coincide. They are negative.

<a name="ch09c072">
If you input p＜1, red is convex╰╯
f2(x) and f3(x) are positive.
Exercise 9.6 can not use them.
2010-03-22-19-51 stop [DrawEx0906]

<a name="ch09c073">　Index begin　Index this file
2010-03-22-21-16 start
■　Exercise 9.6 solution


We are given Jensen's inequality
for a convex function φ, eqn.9.31
The following eqn.BC044 is 
Jensen's inequality for a concave
function φ. (Inequality reverse 
direction for convex. The 
following is for concave.)

<a name="ch09c075">
We also have concave function 
  φ(x)=(1+x1/p)p ---eqn.BC038
on [0,∞],
Hint suggest choose
  wk=|ak|p ---eqn.BC040
and
  xk=|bk|p/|ak|p ---eqn.BC041
Target is Minkowski's Inequality

<a name="ch09c076">
eqn.BC044 greater than side 
φ(x_AM) need x_AM. Find x_AM
as following
  x_AM=(w1x1+w2x2+...+wnxn)
      /(w1+w2+...+wn) ---eqn.BC045
Refer to eqn.BC040 and eqn.BC041
  x_AM=(|b1|p+|b2|p+...+|bn|p)
      /(|a1|p+|a2|p+...+|an|p)
  x_AM=∑[k=1,n](|bk|p)
      /∑[k=1,n](|ak|p) ---eqn.BC046
<a name="ch09c077">
Apply φ(x) find
  φ(x_AM)=(1+x_AM1/p)p
  φ(x_AM)=[1+(∑|bk|p/∑|ak|p)1/p]p
  φ(x_AM)={[(∑|ak|p)1/p+(∑|bk|p)1/p]
          /(∑|ak|p)1/p}p ---eqn.BC047
Better view: eqn.BC049 left side.
2010-03-22-21-52 here
// x_AM≦AM_RANGE WRONG! 2010-03-25-16-03
// f(x_AM)≦AM_RANGE CORRECT!
<a name="ch09c078">　Index begin　Index this file
Above is eqn.BC044 greater than
side φ(x_AM)
Below is eqn.BC044 less than
side AM in range.
  wkφ(xk)=wk(1+xk1/p)p
<a name="ch09c079">
Apply eqn.BC040 and eqn.BC041
  wkφ(xk)=|ak|p(1+[|bk|p/|ak|p]1/p)p
       =(|ak|+|ak|*[|bk|p/|ak|p]1/p)p
       =(|ak|+[|ak|p|bk|p/|ak|p]1/p)p
       =(|ak|+[|bk|p|ak|p/|ak|p]1/p)p
       =(|ak|+[|bk|p]1/p)p
       =(|ak|+|bk|)p

<a name="ch09c080">
Above is eqn.BC044 numerator one 
term. ∑[k=1,n]wkφ(xk) is numerator.
eqn.BC044 less than side 
numerator/denominator is
  ∑(|ak|+|bk|)p / ∑|ak|p ---eqn.BC048

Put eqn.BC044 greater than side
eqn.BC047 and less than side
eqn.BC048 together as following

<a name="ch09c082">
eqn.BC049 whole equation take 
p-th root, cancel denominator.
Two side numerator is exactly
Minkowski's Inequality eqn.9.13
2010-03-22-22-38 stop



<a name="ch09c083">　Index begin　Index this file
2010-03-23-10-58 start
■　Exercise 9.8 problem statement
　　textbook page 150
（Legendre Transform and
 Young's Inequality）

<a name="ch09c084">
If f:(a,b)→Real then the function
g:Real→Real defined by
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
is called the Legendre Transform
of f. It is used widely in the
theory of inequalities, and part
<a name="ch09c085">
of its charm is that it helps us
relate products to sums. For
example, the definition (9.32)
gives us the immediate bound
  xy≦f(x)+g(y) ---eqn.9.33
//xy is product, f+g is sum
for all (x,y)∈(a,b)×Real
<a name="ch09c086">
(a) Find the Legendre Transform
    of
    f(x)=xp/p for p＞1 ---eqn.BC050
    and compare the general bound
    (9.33) to Young's Inequality
    (9.6) //uv≦up/p+vq/q
<a name="ch09c087">
(b) Find the Legendre Transform
    of
    f(x)=ex ---eqn.BC051
    and
    φ(x)=x*log(x)-x ---eqn.BC052
<a name="ch09c088">
(c) Show that for any function f
    the Legendre Transform g is
    convex.
2010-03-23-11-12 stop





<a name="ch09c089">　Index begin　Index this file
2010-03-23-11-27 start
■　Exercise 9.8 hint
　　textbook page 263

The natural calculus exercise shows
the Legendre Transform of
  f(x)=xp/p for p＞1 ---eqn.BC050
is 
<a name="ch09c090">
  g(y)=yq/q  ---eqn.BC053
where
  q=p/(p-1) ---eqn.BC054
Thus, the bound (9.33) simply puts
Young's Inequality (9.6) into a
larger context. Similarly, one
<a name="ch09c091">
finds the Legendre Transform pair
  f(x)=ex ---eqn.BC051
  →
  g(y)=y*log(y)-y ---eqn.BC055
and
  φ(x)=x*log(x)-x ---eqn.BC052
  →
  γ(y)=ey ---eqn.BC056
<a name="ch09c092">
This example suggests the conjecture
that for a convex function, the
Legendre Transform of its Legendre 
is the original function. This
conjecture is indeed true. 
/** <a name="ch09c092Alert">
2010-03-25-16-24 Alert start
Alert: problem (c) say:
"for any function f the Legendre 
Transform g is convex." Hint say:
"for a convex function, the
Legendre Transform of its Legendre 
is the original function." then,
for a concave function, the
Legendre Transform of its Legendre 
is NOT the original function.
/* 2010-03-25-16-30 Alert stop */
<a name="ch09c093">
Finally, for part (c), we take 
0≦p≦1 and note that
  g(py1+(1-p)y2)= ---eqn.BC057
   sup[x∈D]{x(py1+(1-p)y2)-f(x)}
also equals //next line is ---eqn.BC058
  sup[x∈D](p{xy1-f(x)}+(1-p){xy2-f(x)})
//sup[x∈D](p{(xy1-f(x)} ... ●●change
//"(" in textbook page 263 line 20
//which is not listed in three 
//errata pages 2010-03-23-11-52
<a name="ch09c094"> //"bounded" = "≦"
Since this is bounded by
  sup[x∈D]p{xy1-f(x)}
 +sup[x∈D](1-p){xy2-f(x)} ---eqn.BC059
which equals
  pg(y1)+(1-p)g(y2) ---eqn.BC060
we see that g is convex.
2010-03-23-11-59 stop



<a name="ch09c095"> (a) (b1) (b2) (c)
2010-03-23-14-30　Index begin　Index this file
■　Exercise 9.8 solution


"for all (x,y)∈(a,b)×Real"
means that 
x is defined in the domain (a,b)
y is defined in the domain (-∞,+∞)
where -∞＜a＜b＜+∞

<a name="ch09c096"> (a) (b1) (b2) (c)
Legendre Transform is defined by
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
Do transform for f(x) defined by
  f(x)=xp/p for p＞1 ---eqn.BC050

<a name="ch09c097">
"natural calculus exercise" is next
  g(y)=sup[x∈(a,b)]{xy-xp/p} ---eqn.BC061
Here g(y) take y as variable,
sup[x∈(a,b)] take x as variable.
To find sup (max) function value, 
take x as variable. Define 
  h(x)=xy-xp/p ---eqn.BC062
where p＞1. Please goto sideDraw
click "Ex.9.8A". Red curve is 
eqn.BC062, make sure h'(x) has 
maximum, not minimum.
<a name="ch09c098">
  h'(x)=d[xy-xp/p]/dx
  h'(x)=y-p*xp-1/p
  h'(x)=y-xp-1 ---eqn.BC063
For sup value, set h'(x)=0.
After this setting h'(x)=0, 
x is not arbitrary value any
more. x has a specific value
<a name="ch09c099">
At x=x0 maximum function
value occurs. To simplify the
following work, write x0 as 
x, with understanding that x 
is not a variable. Get
  y-xp-1=0 ---eqn.BC064
<a name="ch09c100">
Solve for x=k(y) to eliminate
x from eqn.BC061.
  y=xp-1 ---eqn.BC065
Take (p-1)th root get
  x=y1/(p-1) ---eqn.BC066
Substitute eqn.BC066 to eqn.BC061
  g(y)=sup[x∈(a,b)]{xy-xp/p} ---eqn.BC061

<a name="ch09c101">　Index begin　Index this file
Because x in eqn.BC066 is already
sup value (since set h'(x)=0). We 
drop "sup[x∈(a,b)]" in eqn.BC061
get
  g(y)=y1/(p-1)*y-[y1/(p-1)]p/p
<a name="ch09c102">
  g(y)=y1+[1/(p-1)]-[yp/(p-1)]/p
  g(y)=y(p-1+1)/(p-1)-[yp/(p-1)]/p
  g(y)=yp/(p-1)-[yp/(p-1)]/p
  g(y)=yp/(p-1)*(1-1/p)
  g(y)=yp/(p-1)*(p-1)/p
// here use  q=p/(p-1) ---eqn.BC054
  g(y)=yq/q ---eqn.BC067 (eqn.BC053)
Please goto sideDraw
click "Ex.9.8B". Red curve is 
eqn.BC067, make sure Legendre 
Transform result is convex╰╯

<a name="ch09c103">
Young's inequality eqn.9.6 is
  uv≦up/p+vq/q ---eqn.9.6 simplify
It is same as
  xy≦xp/p+yq/q ---eqn.BC068 (eqn.9.6)
We started from Legendre Transform
  g(y)=sup[x∈(a,b)]{xy-xp/p} ---eqn.BC061
<a name="ch09c104">
End up with
  g(y)=yq/q=sup[x∈(a,b)]{xy-xp/p} ---eqn.BC069
//Left side yq/q contains x0 which 
//sit on max. point. Right side 
//x is variable.
Replace "=sup[x∈(a,b)]" by "≧"
The red part is
  yq/q≧xy-xp/p
or
<a name="ch09c105">
  xy≦xp/p+yq/q ---eqn.BC070
compare
  xy≦f(x)+g(y) ---eqn.9.33
eqn.BC070 is Young's inequality 
eqn.9.6!!

<a name="ch09c106">
f(x) and g(y) in eqn.9.33 are
not arbitrary, they are related
by
[[
If f:(a,b)→Real then the function
g:Real→Real defined by
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
]]
2010-03-23-15-35 stop

<a name="ch09c107"> (a) (b1) (b2) (c)
2010-03-23-17-51　Index begin　Index this file
Next solve problem (b)
[[
(b) Find the Legendre Transform
    of
    f(x)=ex ---eqn.BC051
    and
    φ(x)=x*log(x)-x ---eqn.BC052
]]
<a name="ch09c108">
Legendre Transform is defined by
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
Do transform for f(x) defined by
  f(x)=ex ---eqn.BC051
We find
  g(y)=sup[x∈(a,b)]{xy-ex} ---eqn.9.32
<a name="ch09c109">
To find sup value, take x as 
variable. Define 
  h(x)=xy-ex ---eqn.BC071
Please goto sideDraw click 
"Ex.9.8A". Black curve is 
eqn.BC071, make sure h'(x) has 
maximum, not minimum.
  h'(x)=d[xy-ex]/dx
  h'(x)=y-ex ---eqn.BC072
<a name="ch09c110">
For maximum value, set h'(x)=0
we get
  y-ex=0
  y=ex ---eqn.BC073
Solve for x
  x=log(y) ---eqn.BC074
<a name="ch09c111">
Legendre Transform general equation
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
consider problem (b) given
  f(x)=ex ---eqn.BC051
eqn.9.32 and eqn.BC051 become
  g(y)=sup[x∈(a,b)]{xy-ex}
which is same as
  g(y)=sup[x∈(a,b)]{xy-exp(x)} ---eqn.BC075

<a name="ch09c112">
Substitute eqn.BC074 to eqn.BC075
Since set h'(x)=0 and eqn.BC074 is
a maximum value condition equation.
We can drop "sup[x∈(a,b)]",  get
  g(y)=[log(y)]*y-exp(log(y))
  g(y)=y*log(y)-y ---eqn.BC076
Please goto sideDraw
click "Ex.9.8B". Black curve is 
eqn.BC076, make sure Legendre 
Transform result is convex╰╯
2010-03-23-18-20 here
<a name="ch09c113">
eqn.BC076 is eqn.BC055
Above proved
  f(x)=ex ---eqn.BC051
  →
  g(y)=y*log(y)-y ---eqn.BC055

<a name="ch09c114"> (a) (b1) (b2) (c)
Below prove　Index begin　Index this file
  φ(x)=x*log(x)-x ---eqn.BC052
  →
  γ(y)=ey ---eqn.BC056
Still start from beginning
<a name="ch09c115">
Legendre Transform is defined by
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
Do transform for f(x) defined by
  f(x)=φ(x)=x*log(x)-x ---eqn.BC052
We find
  g(y)=sup[x∈(a,b)]{xy-f(x)} ---eqn.9.32
  g(y)=sup[x∈(a,b)]{xy-[x*log(x)-x]} ---eqn.BC077
<a name="ch09c116">
To find sup value, take x as 
variable. Define 
  h(x)=xy-[x*log(x)-x] ---eqn.BC078
Please goto sideDraw click 
"Ex.9.8A". Blue curve is 
eqn.BC078, make sure h'(x) has 
maximum, not minimum.
<a name="ch09c117">
  h'(x)=d{xy-[x*log(x)-x]}/dx ---eqn.BC079
  h'(x)=y-d[x*log(x)-x]/dx
  h'(x)=y-d[x*log(x)]/dx+dx/dx
  h'(x)=y+1-{[dx/dx]log(x)+x*d[log(x)]/dx}
  h'(x)=y+1-{log(x)+x*1/x}
  h'(x)=y+1-{log(x)+1}
  h'(x)=y-{log(x)} ---eqn.BC080
<a name="ch09c118">
For sup value (maximum), 
set h'(x)=0 ---eqn.BC081
get
  y-log(x)=0 ---eqn.BC082
Solve for x
  y=log(x)
  x=exp(y) ---eqn.BC083
<a name="ch09c119">
Substitute x in eqn.BC083 to
  g(y)=sup[x∈(a,b)]{xy-[x*log(x)-x]} ---eqn.BC077
Drop "sup[x∈(a,b)]", since 
eqn.BC083 is already maximum 
value point equation. get
  g(y)={exp(y)*y-[exp(y)*log(exp(y))-exp(y)]} ---eqn.BC084
<a name="ch09c120">
We know
  log(exp(y))=y ---eqn.BC085
Because log() and exp() are
reverse operation to each other
eqn.BC084 simplify to
  g(y)={exp(y)*y-[exp(y)*y-exp(y)]}
  g(y)={exp(y)*y-exp(y)*y+exp(y)}
<a name="ch09c121">
  g(y)=exp(y) ---eqn.BC086
eqn.BC086 is our answer, which is
same as problem requested
  γ(y)=ey ---eqn.BC056
Please goto sideDraw
click "Ex.9.8B". Blue curve is 
eqn.BC086, make sure Legendre 
Transform result is convex╰╯
2010-03-23-18-42 here

<a name="ch09c122"> (a) (b1) (b2) (c)
2010-03-23-18-52　Index begin　Index this file
Above done part (b)
Below is part (c)
(c) Show that for any function f
    the Legendre Transform g is
    convex. Ex.9.8 hint

<a name="ch09c123"> more
■　Convex: AM_in_domain ≦ AM_in_range
First, review convex function
definition.
[[
<a name="ch09c124">
A function f:[a,b]→Real is said to be
convex function provided that for all
x,y∈[a,b] and all
  0≦p≦1  ---eqn.AQ003
one has　　Do we prove eqn.6.1?
  f(px +(1-p)y)≦
  pf(x)+(1-p)f(y)    ---eqn.6.1
 xe=px +(1−p)y ∈ [a,b] ---eqn.AQ004
]]

<a name="ch09c125"> more
Use AM for Arithmetic Mean.
eqn.6.1 can be described as
AM_in_domain ≦ AM_in_range
 xe=px +(1−p)y is x_AM
f(x_AM) is AM_in_domain
pf(x)+(1-p)f(y) is AM_in_range

<a name="ch09c126">
Now back to Exercise 9.8 part (c)
//the following y is domain, 
//y is not range. 2010-03-25-17-11
Exercise 9.8 Hint said
[[
for part (c), we take 0≦p≦1 and
note that
  g(py1+(1-p)y2)= ---eqn.BC057
   sup[x∈D]{x(py1+(1-p)y2)-f(x)}
also equals //next line is ---eqn.BC058
  sup[x∈D](p{xy1-f(x)}+(1-p){xy2-f(x)})
]]
<a name="ch09c127">　Index begin　Index this file
"py1+(1-p)y2" in g(py1+(1-p)y2) is
  y_AM = py1+(1-p)y2 ---eqn.BC087
y1 and y2 are domain_y begin and 
end points. p and (1-p) are their
weights. Simple Arithmetic Mean
we have p=1/2 and (1-p)=1/2.
p and (1-p) must sum to one.

<a name="ch09c128">
g(py1+(1-p)y2) is g(y_AM) it is
called as
  AM_in_domain=g(py1+(1-p)y2) ---eqn.BC088
<a name="ch09c129"> Compare above and below
  AM_in_range=p*g(y1)+(1-p)*g(y2) ---eqn.BC089
We need AM_in_range, it is above
line. If we can prove
  AM_in_domain≦AM_in_range
then this function g(y) satisfy
convex definition eqn.6.1, hence
g(y) is convex (when if is true)

<a name="ch09c130">
Above analysis is important. For
derivation, please see hint part
(c).
AM_in_domain  is eqn.BC057
≦            use phrase "bounded"
AM_in_range   is eqn.BC060

Exercise 9.8 is done
2010-03-23-19-39 stop

<a name="ch09c131"> more
2010-03-23-19-48 start
Why for convex function we have
  AM_in_domain≦AM_in_range
?
Please consider this way.
<a name="ch09c132">　Index begin　Index this file
AM_in_domain is g(y_AM), the point
[y_AM, g(y_AM)] must be on curve
AM_in_range is p*g(y1)+(1-p)*g(y2)
Variable is p, two end points are
[y1, g(y1)] and [y2, g(y2)]
then AM_in_range is a straight
line connect two end points.

<a name="ch09c133"> more
Interpolation on straight line
has a function value different 
from a point on curve.
AM_in_range  point move on chord,
AM_in_domain point move on curve.

<a name="ch09c134">
When curve shape like a bowl╰╯
AM_in_domain is lower than AM_in_range
  AM_in_domain≦AM_in_range
Curve shape is defined as convex.

<a name="ch09c135"> more
When curve shape like a hat ╭╮
AM_in_domain is higher than AM_in_range
  AM_in_domain≧AM_in_range
Curve shape is defined as concave.
2010-03-23-20-01 stop



<a name="ch09c136">　Index begin　Index this file
2010-03-24-11-40 start
■　Exercise 9.9 problem statement
　　textbook page 151
（Self-Generalization of Holder's
 Inequality）

Holder's inequality is self-
generalizing in the sense that
it implies several apparently 
more general inequalities. This
exercise address two of the
most pleasing of these
generalizations.

<a name="ch09c137">
(a) Show that for positive p,q 
    bigger than r one has

2010-03-24-12-00 here
<a name="ch09c138">
(b) Given p,q, and r are bigger
    than 1, show that if

2010-03-24-12-13 stop





<a name="ch09c140">　Index begin　Index this file
2010-03-24-12-17 start
■　Exercise 9.9 hint
　　textbook page 263

Part (a) follows by applying 
Holder's inequality for the
conjugate pair (p/r,q/r) to 
the splitting ajr◦bjr.
<a name="ch09c141">
Part (b) can be obtained by two 
similar application of Holder's 
inequality but one saves 
arithmetic and gain insight by 
following Riesz's pattern. By 
the AM-GM inequality one has
  xyz≦xp/p+yq/q+zr/r ---eqn.BC093
<a name="ch09c142">
and, after applying this to the
corresponding normalized values
a_hatj, b_hatj, c_hatj, one can
finish exactly as before.
2010-03-24-12-33 stop



<a name="ch09c143">
2010-03-24-12-37 start
■　Exercise 9.9 solution


Here we require s＞1 and t＞1
and
  1/s + 1/t = 1 ---eqn.BC094
Set
  s=p/r ---eqn.BC095
  t=q/r ---eqn.BC096
<a name="ch09c144">
Put eqn.BC095 and eqn.BC096 into
eqn.BC094 get
  r/p + r/q = 1
or
  1/p + 1/q = 1/r ---eqn.BC097

Holder Inequality is next

<a name="ch09c145">　Index begin　Index this file
Do the following replacement
  cj⇒ajr ---eqn.BC098
  dj⇒bjr ---eqn.BC099
Holder Inequality change 
to next

<a name="ch09c146">
Holder1417 whole equation
take r-th root, //therefore we act
Holder1417 left side become //because 
eqn.BC090 less than side.   //we want
Holder1417 right side has
two parts. First part is next

<a name="ch09c148">
2010-03-24-14-00 here
Because we have
  s=p/r ---eqn.BC095
which give us
  rs=p ---eqn.BC101
We used eqn.BC101 in eqn.BC100
right side equality.

Holder1417 right side second 
part is the same. See next.
(after take r-th root)

<a name="ch09c150">　Index begin　Index this file
In eqn.BC102 we used
  t=q/r ---eqn.BC096

Holder1417 left side take r-th
root is less or equal to
eqn.BC100 right end multiply by
eqn.BC102 right end 
<a name="ch09c151">
This verbal equation is exactly
eqn.BC090 inequality part.
Exercise 9.9 part (a) is done
2010-03-24-14-10 stop

<a name="ch09c152">
2010-03-24-17-04 start
Exercise 9.9 part (b) give us
  1/p+1/q+1/r = 1 ---eqn.BC091
ask us to prove three sequence
Holder's inequality eqn.BC092 
We had two sequence Holder's 
inequality eqn.9.1. That proof 
start from Riesz's version 
Exercise 9.9 hint suggest to
follow Riesz steps to "saves 
arithmetic and gain insight".

<a name="ch09c153">
Riesz's version start from two
variables Young's inequality.
Here we need to derive three
variables Young's inequality.
  xyz≦xp/p+yq/q+zr/r ---eqn.BC093
given p,q,r＞1 and
  1/p+1/q+1/r=1 ---eqn.BC091

Below is copied from tute0033.htm  
and modified.

<a name="ch09c154">　Index begin　Index this file
■　Young's inequality 3 variables

Start from power mean inequality
with m,n,o＞0, we can write
  m^(α)*n^(β)*o^(γ) //GM≦AM
≦m*(α)+n*(β)+o*(γ) ---eqn.BC103
and require
  α+β+γ=1 ---eqn.BC104
(why eqn.BC103 is true? You goto
 document for eqn.BA044. Same
 thing. 2010-03-24-17-23)

<a name="ch09c155">
We can write eqn.BC103 as
  m^(α)*n^(β)*o^(γ)≦ ---eqn.BC105
  (α/1)*m^1+(β/1)*n^1+(γ/1)*o^1

which is
  m^(α)*n^(β)*o^(γ)≦ ---eqn.BC106
  (α/(α+β+γ))*m^(α+β+γ)
 +(β/(α+β+γ))*n^(α+β+γ)
 +(γ/(α+β+γ))*o^(α+β+γ)
<a name="ch09c156">
and one more twist, multiply
with 1=α/α and 1=β/β and 1=γ/γ, get
  m^(α)*n^(β)*o^(γ)≦ ---eqn.BC107
  (α/(α+β+γ))*m^[α(α+β+γ)/α]
 +(β/(α+β+γ))*n^[β(α+β+γ)/β]
 +(γ/(α+β+γ))*o^[γ(α+β+γ)/γ]
Define //Textbook page 136, line -7
  u=m^(α) ---eqn.BC108
  v=n^(β) ---eqn.BC109
  w=o^(γ) ---eqn.BC110
<a name="ch09c157">
Attention: if m,n,o are all length
and if α=1/7, β=2/7, γ=4/7 (alert why)
eqn.BC108 require u be length^1/7
eqn.BC109 require v be length^2/7
eqn.BC110 require w be length^4/7
<a name="ch09c158">
eqn.BC107 become
  u*v*w≦ ---eqn.BC111
  (α/(α+β+γ))*u^[(α+β+γ)/α]
 +(β/(α+β+γ))*v^[(α+β+γ)/β]
 +(γ/(α+β+γ))*w^[(α+β+γ)/γ]
again, define
  p=(α+β+γ)/α ---eqn.BC112
  q=(α+β+γ)/β ---eqn.BC113
  r=(α+β+γ)/γ ---eqn.BC114
<a name="ch09c159">　Index begin　Index this file
then eqn.BC111 become
  u*v*w≦ ---eqn.BC115
  (1/p)*u^p + (1/q)*v^q + (1/r)*w^r
for all u,v,w in [0,infinity)
Better math equation is next.

<a name="ch09c160">
eqn.BC112 to eqn.BC114 definition
automatic satisfy
  1/p + 1/q + 1/r = 1 ---eqn.BC091
since //why require sum to one?
  1/p + 1/q + 1/r ---eqn.BC116
= α/(α+β+γ) + β/(α+β+γ) + γ/(α+β+γ)
  1/p + 1/q + 1/r 
= (α+β+γ)/(α+β+γ) = 1 ---eqn.BC117
<a name="ch09c161">
eqn.BC115 is generalized humble 
bound. eqn.BC115 is named as
Young's inequality.

Above copied from tute0033.htm 
and modified.
2010-03-24-17-52 stop

<a name="ch09c162">
2010-03-24-19-18 start
The following is still copied
from tute0033.htm and modified.

<a name="ch09c163">　Index begin　Index this file
■　Addition to summation magic
//drop eqn.9.7 from the analysis.
The key point
is to use normalization. Define
normalized a_hat and b_hat and 
c_hat as below.

<a name="ch09c167">
ahatk is normalized ak,
bhatk is normalized bk,
chatk is normalized ck.
They are reasonable sequences,
they are pure numbers (ratio)
and satisfy Young's inequality. 
<a name="ch09c168">　Index begin　Index this file
Put ahatk, bhatk, chatk into 
Young's inequality eqn.BC115
get next equation.

<a name="ch09c169">
2010-03-24-20-08 here
'hat' means normalized quantity.
For example,
length to unit length ratio
time   to unit time   ratio
mass   to unit mass   ratio
are all normalized quantity,
they are pure numbers and fit
Young's inequality perfect. why

eqn.BC121 not sum over k. We
will sum over k at later time.

<a name="ch09c170">
If  ahatk take p-th power
and bhatk take q-th power
and chatk take r-th power
they are

<a name="ch09c174">　Index begin　Index this file
ahatk, bhatk, chatk must satisfy
Young's inequality eqn.BC121. 
When plug into eqn.BC121,
carry out summation over k. 
eqn.BC115 less than side 
use eqn.BC118 for ahatk
use eqn.BC119 for bhatk
use eqn.BC120 for chatk
<a name="ch09c175">
eqn.BC121 greater than side 
use eqn.BC122 for ahatkp
use eqn.BC123 for bhatkq
use eqn.BC124 for chatkr
(do not get confused !! can not
 all use eqn.BC122 to eqn.BC124
 2010-03-24-20-26
)

<a name="ch09c176">
eqn.BC122 to eqn.BC124 denominator
are not function of k (already
done sum k). eqn.BC122 to eqn.BC124
numerator will sum over k. The 
result is equal numerator and 
denominator, they become one. 
Next step is

<a name="ch09c178">
2010-03-24-20-45 here
All three red terms are ONE. 
Ignore red one, eqn.BC126 is
  1/p + 1/q + 1/r = 1 ---eqn.BC091
Now put eqn.BC118 for ahatk
and put eqn.BC119 for bhatk
and put eqn.BC120 for chatk
into eqn.BC125. They become

<a name="ch09c180">
2010-03-24-20-55 here
eqn.BC127 left side sum over k
from k=1 to k=n. Its denominator
are already summed over k and is
not a function of k. 
eqn.BC127 left side denominator
can be moved out of k-summation.
denominator is a positive term.
<a name="ch09c181">
Move denominator to greater than 
side will not reverse inequality
direction. The result is our goal
eqn.BC092
Holder Inequality for three 
variables is proved.
Exercise 9.9 is done
2010-03-24-20-57 stop

2010-03-25-18-21 done first  proofread
2010-03-25-20-06 done second proofread
2010-03-25-20-25 done spelling check



<a name="docB001">
2010-03-24-11-51 add arrow
to file end inside <--⇒-->
2010-04-19-14-01 show up
arrows in tute0035.htm
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If not install complete font,
some arrow appears as a square.
Liu,Hsinhan's MSIE 6.0 show most
as square. But in notepad, all
arrow show up.
2010-04-19-14-10 stop

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

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

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