1. Hw 7 Part II: The Holder Inequality
Proposition 1.1. Let a, b be positive real numbers and p, q be real numbers such that and
p, q > 1 and 1/p + 1/q = 1. Then
ap bq
ab ≤
+ .
p
q
Proof. Let f : R → R>0 be the exponential function f (x) = ex for x ∈ R. Since f 00 (x) =
ex > 0, f is a convex function, i.e. for any t ∈ [0, 1],
f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y), for any x, y ∈ R.
Let x = p ln a, y = q ln b, and t = 1/p. Then 1 − t = 1/q and hence
ab = eln(ab) = eln a+ln b
1
= ep
=
1
1
≤ ep ln a + eq ln b
p
q
1 ln bq
ap bq
+ e
=
+ .
q
p
q
·p ln a+ 1q ·q ln b
1 ln ap
e
p
Theorem 1.1. Let a1 , · · · , an and b1 , · · · , bn be real numbers. Then
!1
!1
n
n
n
p
q
X
X
X
p
q
ak bk ≤
|ak |
|bk |
.
k=1
k=1
k=1
Proof. By triangle inequality,
n
n
X
X
ak bk ≤
|ak bk | .
k=1
k=1
1
1
P
P
Let A = ( nk=1 |ak |p ) p and B = ( nk=1 |bk |q ) q . By proposition 2.1, we find
1 |ak | p 1 |bk | q
|ak | |bk |
·
≤
+
.
A
B
p
A
q B
Therefore
n
n n X
1 X |ak | p 1 X |bk | q
|ak | |bk |
·
≤ ·
+ ·
A
B
p
A
q
B
k=1
k=1
k=1
Pn
P
1 k=1 |ak |p 1 nk=1 |bk |q
+
=
p
Ap
q
Bq
1 1
= + = 1,
p q
which implies
Pn
k=1 |ak bk |
≤ AB.
Proposition 1.2. (Minkowski inequality). Let a1 , · · · , an and b1 , · · · , bn be real numbers.
Let p ≥ 1.
!1/p
!1/p
!1/p
n
n
n
X
X
X
p
p
p
|ai + bi |
≤
|ai |
+
|bi |
.
i=1
i=1
1
i=1
2
Proof. If all of ai , bj are zero, the inequality is trivial. Assume that ai , bj are not all zero.
Let p = 1. Then |ai + bi | ≤ |ai | + |bi | by triangle inequality. Hence
n
n
n
n
X
X
X
X
|ai + bi | ≤
(|ai | + |bi |) =
|ai | +
|bi |.
i=1
i=1
i=1
i=1
The inequality holds when p = 1. Now assume that p > 1.
For each 1 ≤ i ≤ n,
|ai + bi |p = |ai + bi |p−1 |ai + bi | ≤ |ai + bi |p−1 |ai | + |ai + bi |p−1 |bi |.
Let xi = |ai + bi |p−1 for 1 ≤ i ≤ n. By Holder’s inequality,
!1/p n
!1/q
n
n
X
X
X q
xi |ai | ≤
|ai |p
xi
i=1
i=1
i=1
where q > 1 such that 1/p + 1/q = 1. Since 1/p + 1/q = 1, (p − 1)q = pq − q = p and hence
n
n
n
X
X
X
q
(p−1)q
xi =
|ai + bi |
=
|ai + bi |p .
i=1
i=1
i=1
This shows that
n
X
xi |ai | ≤
n
X
i=1
i=1
n
X
n
X
!1/p
p
|ai |
n
X
!1/q
|ai + bi |
p
.
i=1
Similarly,
xi |bi | ≤
i=1
!1/p
|bi |p
i=1
n
X
!1/q
|ai + bi |p
.
i=1
Combining all of these results, we find
n
X
|ai + bi |p ≤
i=1
n
X
!1/q 
|ai + bi |p
n
X

i=1
!1/p
|ai |p
+
i=1
n
X
!1/p 
|bi |p
.
i=1
Dividing the inequality on both side by ( i=1 |ai + bi |p )1/q , we obtain that
!1−1/q
!1/p
!1/p
n
n
n
X
X
X
p
p
p
≤
+
.
|ai + bi |
|ai |
|bi |
Pn
i=1
i=1
i=1
Since 1/p + 1/q = 1, 1 − 1/q = 1/p. We proved the Minkowski inequality.
2. Passing to infinity
Lemma 2.1. Let (an ) and (bn ) be sequence of real numbers such that an ≤ bn for any
n ≥ 1. If limn→∞ an = a and limn→∞ bn = b, then a ≤ b.
Proof. The proof is left to the reader.
Proposition 2.1. (Holder inequality for infinite series) Let p, qPbe positive real
Pnumbers
∞
p and
q
such that p, q > 1 and 1/p + 1/q = 1. Assume that Suppose that ∞
|a
|
n=1 n
n=1 |bn |
are both convergent infinite series. Then
∞
!1
!1 ∞
∞
p
q
X
X
X
q
p
|b
|
.
a
b
≤
|a
|
n
n n
n
n=1
n=1
k=1
3
Pn
P
Proof. Let us prove that ∞
i=1 |ai bi | for
n=1 an bn is absolutely convergent. Define sn =
n ≥ 1. Then (sn ) is a nondecreasing sequence of nonnegative real numbers. If we can show
that (sn ) is bounded above, by monotone sequence property, (sn ) is convergent. For each
n ≥ 1, Holder inequality implies that
!1
!1
n
n
p
q
X
X
sn ≤
.
|ai |p
|bi |q
i=1
Let un =
Pn
p
i=1 |ai | and tn =
i=1
Pn
q
i=1 |bi | for n ≥ 1. Then
1/q
0 ≤ sn ≤ u1/p
n tn for any n ≥ 1.
(2.1)
By assumption, (un ) and (tn ) are both convergent in R and hence they are bounded. Choose
M > 0 so that 0 ≤ un , tn ≤ M for any n ≥ 1. Hence
1/q
1/p+1/q
0 ≤ u1/p
=M
n tn ≤ M
for any n ≥ 1. This implies that 0 ≤ sn ≤ M for any n ≥ 1. Therefore (sn ) is convergent in
R. Let s = limn→∞ sn and u = limn→∞ un and v = limn→∞ vn . By (3.1) and Lemma 3.1,
0 ≤ s ≤ u1/p v 1/q which is the desired inequality.
P∞
Proposition
2.2. (Minkowski inequality for infinite series) Let p ≥ 1. Suppose n=1 |an |p
P∞
p
and n=1 |bn | are both convergent infinite series in R. Then
!1/p
!1/p
!1/p
∞
∞
∞
X
X
X
.
|bn |p
+
≤
|an |p
|an + bn |p
n=1
n=1
n=1
Pn
Pn
Proof. Let un = i=1 |ai |p and tn = i=1 |bi |p for any n ≥ 1. By assumption. (un ) and (tn )
are convergent
P in R. Hence they are bounded in R. Choose M > 0 so that 0 ≤ un , tn ≤ M.
Let sn = ni=1 |ai + bi |p for each n ≥ 1. By Minkowski inequality,
0 ≤ sn ≤ M 1/p + M 1/q for any n ≥ 1.
Hence (sn ) is a bounded sequence in R. Since (sn ) is nondecreasing and bounded, by monotone sequence property, (sn ) is convergent. For n ≥ 1,
1/p
0 ≤ sn ≤ u1/p
n + tn .
Let s = limn→∞ sn , u = limn→∞ un and t = limn→∞ tn . By Lemma 3.1,
1/p
1/p
1/p
1/p
s ≤ lim (u1/p
+ t1/p .
n + tn ) = lim un + lim tn = u
n→∞
The inequality s ≤
u1/p
+
n→∞
t1/p
n→∞
is exactly the Minkowski inequality for infinite series.