Advanced High-School Mathematics

SECTION 3.4 Holder Inequality ̈ 159

We may assume that both A, B 6 = 0, else the theorem is clearly
true. Therefore by using Young’s inequality, we see that for each
i= 1, 2 , ..., n, that

|xi|

A

·

|yi|

B

≤

|xi|p

pAp

+

|yi|q

qB

.

Therefore,

1

AB

∑n

i=1

|xiyi| ≤

∑n

i=1

Ñ

|xi|p

pAp

+

|yi|q

qB

é

=

1

p

+

1

q

= 1.

This implies that

∑n

i=1

|xiyi|≤AB=

Ñn

∑

i=1

|xi|p

é 1 /pÑn

∑

j=1

|yj|q

é 1 /q

,

and we’re done.

Note that if we set allyi= 1 then we get

∑n

i=1

|xi|≤n^1 /q

Ñn

∑

i=1

|xi|p

é 1 /p

=n^1 −^1 /p

Ñn

∑

i=1

|xi|p

é 1 /p

,

and so
1
n

∑n

i=1

|xi|≤

Ñ n

∑

i=1

|xi|p/n

é 1 /p

for anyp >1. This proves that

AM(|x 1 |,|x 2 |,...,|xn|)≤pM(|x 1 |,|x 2 |,...,|xn|)

wheneverp >1.
Finally, assume that 0< p < q and assume thatx 1 , x 2 , ..., xn are
non-negative. We shall show that

pM(x 1 ,x 2 ,...,xn)≤qM(x 1 ,x 2 ,...,xn).