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8/8/2019 Bartle 8 Beta
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Solution of 8.β of Bartle, TheElements of Real Analysis, 2/eWritten by Men-Gen Tsai
email: [email protected]
8.β . 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 < α < 1, then
aαb1−α ≤ αa + (1 − α)b
and that the equality holds if and only if a = b. Assume this, let r > 1 and
let s satisfy1
r+
1
s= 1,
(so that s > 1 and r + s = rs). Show that if A and B are positive, then
AB ≤Ar
r
+Bs
s
,
and that the equality holds if and only if Ar = Bs.
(b) Let {a1,...,an} and {b1,...,bn} be positive real numbers. If r, s > 1
and (1/r) + (1/s) = 1, establish Holder’s Inequality
n j=1
a jb j ≤ n
j=1
ar j
1/r n j=1
br j
1/r
(Hint: Let A = (
ar j)1/r and B = (
br j)1/r and apply part (a) to a j/A and
b j/B.)
(c) Using Holder’s Inequality, establish Minkowski Inequality n
j=1
(a j + b j)r1/r
≤ n j=1
ar j
1/r+ n j=1
br j
1/r
1
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(Hint: (a + b)r = (a + b)(a + b)r/s = a(a + b)r/s + b(a + b)r/s.)
(d) Using Holder’s Inequality, prove that
1
n
n j=1
a j ≤
1
n
n j=1
ar j
1/r
(e) If a1 ≤ a2 and b1 ≤ b2, then (a1 − a2)(b1 − b2) ≥ 0 and hence
a1b1 + a2b2 ≥ a1b2 + a2b1.
Show that if a1 ≤ a2 ≤ ... ≤ an and b1 ≤ b2 ≤ ... ≤ bn, then
nn
j=1
a jb j ≥ n
j=1
a j
n j=1
b j
(f) Suppose that 0 ≤ a1 ≤ a2 ≤ ... ≤ an and 0 ≤ b1 ≤ b2 ≤ ... ≤ bn and
r ≥ 1. Establish the Chebyshev Inequality1
n
n j=1
ar j
1/r1
n
n j=1
br j
1/r≤
1
n
n j=1
(a jb j)r1/r
Show that this inequality must be reversed of {a j} is increasing and {b j} is
decreasing.
Proof of (a): Let f (x) = αx + (1 − α) − xα, where x = a/b. Thus
f
(x) = α − αxα−1
= α(1 − xα−1
).
Hence f (x) attains its minimum at x = 1 (, a = b,) and f (1) = 0. Hence
f (x) = αx + (1 − α)− xα ≥ 0. Put x = a/b I have
α(a
b) − (1 − α)− (
a
b)α ≥ 0
2
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or
aαb1−α ≤ αa + (1 − α)b.
Next, let α = 1/r, a = A1/α, and b = B1/(1−α). Hence
AB ≤Ar
r+
Bs
s.
Proof of (b): Let A = (
ar j)1/r, B = (
br j)1/r and apply part (a) to
a j/A and b j/B:a jA
b jB≤
1r
a jA
r
+1s
b jB
s
for all 1 ≤ j ≤ n. Sum them up:
n j=1
a jA
b jB≤
1
r
n j=1
a jA
r
+1
s
n j=1
b jB
s
1
AB
n j=1
a jb j ≤ 1.
Thus,n
j=1
a jb j ≤
n j=1
ar j
1/r
n j=1
br j
1/r
Proof of (c): By using Holder’s Inequality:
a j
a j + b j
r/s
≤
ar j
1/r(a j + b j)r
1/s
b j
a j + b j
r/s
≤
br j
1/r
(a j + b j)r1/s
.
Thus
a j + b j
r
≤
(a j + b j)r1/s
ar j
1/r+
br j
1/r
(a j + b j)r
1/r≤
ar j
1/r+
br j
1/r.
3
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Proof of (d): Let b j = 1/n for all j between 1 and n. Thus
n j=1
a j1
n≤ n j=1
ar j
1/r n j=1
(1
n)r1/s
.
Hence1
n
n j=1
a j ≤
1
n
n j=1
ar j
1/r
Proof of (e): Note that
(ai − a j)(bi − b j) ≥ 0
for all i and j. Hence
aibi + a jb j ≥ aib j + a jbi
for all i and j. Sum them up as following:
n
j=1
n
i=1
aibi + a jb j ≥n
j=1
n
i=1
aib j + a jbi
n j=1
(na jb j +n
i=1
aibi) ≥n
j=1
(b jn
i=1
ai + a jn
i=1
bi)
nn
j=1
(a jb j + nn
i=1
aibi) ≥ n
i=1
ai
n j=1
b j
+ n j=1
a j
ni=1
bi
Hence
nn
j=1
a jb j ≥ n
j=1
a j
n j=1
b j
Note: If a1 ≤ a2 ≤ ... ≤ an and b1 ≥ b2 ≥ ... ≥ bn (or a1 ≥ a2 ≥ ... ≥ an
and b1 ≤ b2 ≤ ... ≤ bn), then
nn
j=1
a jb j ≤ n
j=1
a j
n j=1
b j
4
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It will be used in part (f).
Proof of (f ): Note that 0 ≤ ar1 ≤ ar
2 ≤ ... ≤ arn and 0 ≤ br1 ≤ br2 ≤ ... ≤ brn
if r ≥ 1. Hence by part (e) I have
n j=1
ar j
n j=1
br j
≤ n
n j=1
(ar jbr j).
Hence 1
n
n j=1
ar j
1/r1
n
n j=1
br j
1/r≤
1
n
n j=1
(a jb j)r1/r
Also, by the note of part (e), this inequality must be reversed of {a j} is
increasing and {b j} is decreasing.
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