#### Read Bartle_8_beta.pdf text version

Solution of 8. of Bartle, The Elements of Real Analysis, 2/e

Written by Men-Gen Tsai email: [email protected] 8.. In this project, let {a1 , a2 , ..., an }, and so forth, be sets of n positive real numbers. (a) In can be proved (for example, by using the Mean Value Theorem) that if a and b are positive and 0 < < 1, then a b1- a + (1 - )b and that the equality holds if and only if a = b. Assume this, let r > 1 and let s satisfy 1 1 + = 1, r s (so that s > 1 and r + s = rs). Show that if A and B are positive, then Ar B s + , r s and that the equality holds if and only if Ar = B s . AB (b) Let {a1 , ..., an } and {b1 , ..., bn } be positive real numbers. If r, s > 1 and (1/r) + (1/s) = 1, establish H¨lder's Inequality o

n n 1/r n 1/r

aj b j

j=1 j=1

ar j

br j

j=1

(Hint: Let A = ( bj /B.)

ar )1/r and B = ( j

br )1/r and apply part (a) to aj /A and j

(c) Using H¨lder's Inequality, establish Minkowski Inequality o

n

(aj + bj )r

j=1

1/r

n

j=1

ar j 1

1/r

n

+

j=1

br j

1/r

(Hint: (a + b)r = (a + b)(a + b)r/s = a(a + b)r/s + b(a + b)r/s .) (d) Using H¨lder's Inequality, prove that o 1 n 1 n r aj a n j=1 n j=1 j

1/r

(e) If a1 a2 and b1 b2 , then (a1 - a2 )(b1 - b2 ) 0 and hence a1 b 1 + a2 b 2 a1 b 2 + a2 b 1 . Show that if a1 a2 ... an and b1 b2 ... bn , then

n n n

n

j=1

aj b j

j=1

aj

j=1

bj

(f) Suppose that 0 a1 a2 ... an and 0 b1 b2 ... bn and r 1. Establish the Chebyshev Inequality 1 n r a n j=1 j

1/r

1 n r b n j=1 j

1/r

1 n (aj bj )r n j=1

1/r

Show that this inequality must be reversed of {aj } is increasing and {bj } is decreasing. Proof of (a): Let f (x) = x + (1 - ) - x , where x = a/b. Thus f (x) = - x-1 = (1 - x-1 ). Hence f (x) attains its minimum at x = 1 (, a = b,) and f (1) = 0. Hence f (x) = x + (1 - ) - x 0. Put x = a/b I have a a ( ) - (1 - ) - ( ) 0 b b 2

or a b1- a + (1 - )b. Next, let = 1/r, a = A1/ , and b = B 1/(1-) . Hence AB Ar B s + . r s

Proof of (b): Let A = ( aj /A and bj /B:

ar )1/r , B = ( j

r

br )1/r and apply part (a) to j

s

aj b j 1 aj AB r A for all 1 j n. Sum them up: aj b j 1 n aj r j=1 A j=1 A B 1 AB Thus,

n n n n

+

1 bj s B

r

+

1 n bj s j=1 B

s

aj bj 1.

j=1 1/r n 1/r

aj b j

j=1 j=1

ar j

br j

j=1

Proof of (c): By using H¨lder's Inequality: o

r/s

aj aj + b j

r/s

ar j br j

1/r

(aj + bj )r (aj + bj )r

1/s

1/r

1/s

b j aj + b j Thus

r

.

aj + b j (aj + bj )r

1/r

(aj + bj )r ar j

1/r

1/s

ar j br j

1/r

1/r

+

br j

1/r

+ 3

.

Proof of (d): Let bj = 1/n for all j between 1 and n. Thus 1 aj n j=1 Hence

n n

ar j

j=1

1/r

1 ( )r j=1 n

1/r

n

1/s

.

1 n 1 n r aj a n j=1 n j=1 j

Proof of (e): Note that (ai - aj )(bi - bj ) 0 for all i and j. Hence ai b i + aj b j ai b j + aj b i for all i and j. Sum them up as following:

n n n n

ai bi + aj bj

j=1 i=1 n n

ai b j j=1 i=1 n n (bj

j=1 n i=1

+ aj b i

n

(naj bj +

j=1 n i=1 n

ai b i ) ai bi )

i=1 n

ai + aj

i=1 n

bi )

n n

n

j=1

(aj bj + n

ai

i=1 n j=1

bj +

j=1 n

aj

i=1

bi

Hence n

aj b j

j=1 j=1

aj

j=1

bj

Note: If a1 a2 ... an and b1 b2 ... bn (or a1 a2 ... an and b1 b2 ... bn ), then

n n n

n

j=1

aj b j

j=1

aj

j=1

bj

4

It will be used in part (f). Proof of (f ): Note that 0 ar ar ... ar and 0 br br ... br 1 2 n 1 2 n if r 1. Hence by part (e) I have

n n n

ar j

j=1 j=1

br j

n

j=1

(ar br ). j j

Hence

1 n r a n j=1 j

1/r

1 n r b n j=1 j

1/r

1 n (aj bj )r n j=1

1/r

Also, by the note of part (e), this inequality must be reversed of {aj } is increasing and {bj } is decreasing.

5

#### Information

5 pages

#### Report File (DMCA)

Our content is added by our users. **We aim to remove reported files within 1 working day.** Please use this link to notify us:

Report this file as copyright or inappropriate

53744