Let a, b be positive real numbers and p, q be real numbers such that and p, q &gt

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

ab ≤ a^p p +b^q

q .

Proof. Let f : R → R>0 be the exponential function f (x) = e^x for x ∈ R. Since f⁰⁰(x) = e^x> 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 = e^ln(ab)= e^{ln a+ln b}

= e^{1p·p ln a+1q·q ln b}≤ 1

pe^{p ln a}+1 qe^{q ln b}

= 1

pe^{ln ap}+1

qe^{ln bq} = a^p p +b^q

q .

 Theorem 1.1. Let a1, · · · , an and b1, · · · , bn be real numbers. Then

n

X

k=1

akbk

≤

n

X

k=1

|a_k|^p

!¹

p n

X

k=1

|b_k|^q

!¹

q

. Proof. By triangle inequality,

n

X

k=1

akbk

≤

n

X

k=1

|a_kbk| .

Let A = (Pn

k=1|a_k|^p)^p1 and B = (Pn

k=1|b_k|^q)^1q . By proposition 2.1, we find

|a_k| A ·|b_k|

B ≤ 1 p

 |a_k| A

p

+1 q

 |b_k| B

q

. Therefore

n

X

k=1

|a_k| A ·|b_k|

B ≤ 1 p·

n

X

k=1

 |a_k| A

p

+1 q ·

n

X

k=1

 |b_k| B

q

= 1 p

Pn k=1|a_k|^p

A^p + 1 q

Pn k=1|b_k|^q

B^q

= 1 p+ 1

q = 1, which impliesPn

k=1|a_kbk| ≤ AB. 

Proposition 1.2. (Minkowski inequality). Let a₁, · · · , a_n and b₁, · · · , b_n be real numbers.

Let p ≥ 1.

n

X

i=1

|a_i+ b_i|^p

!1/p

≤

n

X

i=1

|a_i|^p

!1/p

+

n

X

i=1

|b_i|^p

!1/p

.

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 |a_i+ b_i| ≤ |a_i| + |b_i| by triangle inequality. Hence

n

X

i=1

|a_i+ bi| ≤

n

X

i=1

(|ai| + |b_i|) =

n

X

i=1

|a_i| +

n

X

i=1

|b_i|.

The inequality holds when p = 1. Now assume that p > 1.

For each 1 ≤ i ≤ n,

|a_i+ b_i|^p = |a_i+ b_i|^p−1|a_i+ b_i| ≤ |a_i+ b_i|^p−1|a_i| + |a_i+ b_i|^p−1|b_i|.

Let xi= |ai+ bi|^p−1 for 1 ≤ i ≤ n. By Holder’s inequality,

n

X

i=1

xi|a_i| ≤

n

X

i=1

|a_i|^p

!1/p n

X

i=1

x^q_i

!1/q

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

X

i=1

x^q_i =

n

X

i=1

|a_i+ b_i|^(p−1)q =

n

X

i=1

|a_i+ b_i|^p. This shows that

n

X

i=1

xi|a_i| ≤

n

X

i=1

|a_i|^p

!1/p n

X

i=1

|a_i+ bi|^p

!1/q

. Similarly,

n

X

i=1

xi|b_i| ≤

n

X

i=1

|b_i|^p

!1/p n

X

i=1

|a_i+ bi|^p

!1/q

. Combining all of these results, we find

n

X

i=1

|a_i+ b_i|^p≤

n

X

i=1

|a_i+ b_i|^p

!1/q



n

X

i=1

|a_i|^p

!1/p

+

n

X

i=1

|b_i|^p

!1/p

. Dividing the inequality on both side by (Pn

i=1|a_i+ bi|^p)^1/q, we obtain that

n

X

i=1

|a_i+ bi|^p

!1−1/q

≤

n

X

i=1

|a_i|^p

!1/p

+

n

X

i=1

|b_i|^p

!1/p

. 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 ≤ b_n 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, q be positive real numbers such that p, q > 1 and 1/p + 1/q = 1. Assume that Suppose thatP∞

n=1|a_n|^p andP∞ n=1|b_n|^q are both convergent infinite series. Then

∞

X

n=1

anbn

≤

∞

X

n=1

|a_n|^p

!¹_{p ∞} X

k=1

|b_n|^q

!¹_q .

3

Proof. Let us prove that P∞

n=1anbn is absolutely convergent. Define sn = Pn

i=1|a_ibi| for n ≥ 1. Then (s_n) 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

s_n≤

n

X

i=1

|a_i|^p

!¹_{p n} X

i=1

|b_i|^q

!¹_q . Let u_n=Pn

i=1|a_i|^p and t_n=Pn

i=1|b_i|^q for n ≥ 1. Then (2.1) 0 ≤ s_n≤ u^1/p_n t^1/q_n for any n ≥ 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

0 ≤ u^1/p_n t^1/q_n ≤ M^1/p+1/q = M

for any n ≥ 1. This implies that 0 ≤ s_n≤ M for any n ≥ 1. Therefore (s_n) 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 ≤ u^1/pv^1/q which is the desired inequality. 

Proposition 2.2. (Minkowski inequality for infinite series) Let p ≥ 1. SupposeP∞ n=1|a_n|^p and P∞

n=1|b_n|^p are both convergent infinite series in R. Then

∞

X

n=1

|a_n+ bn|^p

!1/p

≤

∞

X

n=1

|a_n|^p

!1/p

+

∞

X

n=1

|b_n|^p

!1/p

. Proof. Let un=Pn

i=1|a_i|^pand tn=Pn

i=1|b_i|^pfor any n ≥ 1. By assumption. (un) and (tn) are convergent in R. Hence they are bounded in R. Choose M > 0 so that 0 ≤ un, t_n≤ M.

Let sn=Pn

i=1|a_i+ 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 mono- tone sequence property, (s_n) is convergent. For n ≥ 1,

0 ≤ s_n≤ u^1/p_n + t^1/p_n .

Let s = limn→∞sn, u = limn→∞un and t = limn→∞tn. By Lemma 3.1, s ≤ lim

n→∞(u^1/p_n + t^1/p_n ) = lim

n→∞u^1/p_n + lim

n→∞t^1/p_n = u^1/p+ t^1/p.

The inequality s ≤ u^1/p+ t^1/p is exactly the Minkowski inequality for infinite series.