8-3

Proof of Theorem 8.12, the Holder's inequality

Case 1:

# p
1

{\displaystyle p = 1}

,

# q
∞

{\displaystyle q = \infty}

,

‖
a
b

‖
1

=

∑
k

a
k

b
k

≤

(

∑
k

a
k

)

sup

k

b

k

=

a

‖
1

‖
b

‖
∞

{\displaystyle \begin{align}
\end{align}}

Case 2:

# p
∞

{\displaystyle p=\infty}

,

# q
1

{\displaystyle q = 1}

 ,

similar proof in case 1

Case 3:

1
<
p
,
q
<
∞

{\displaystyle 1 < p, q < \infty}

If

a

p

=
0

 or

b

p

=
0

, Holder's inequality is obvious holds.

For

0
<

a

p

,

b

p

<
∞

, take

‖
A

‖
p

=
‖
B

‖
q

=
1

By Young's inequality,

‖
A
B

‖
1

=

∑
k

A
k

B
k

≤

∑
k

(

A
k

p

p

B
k

q

q

)

=

‖
A

‖
p
p

p

‖
B

‖
q
q

q

=

1
p

1
q

=
1

{\displaystyle \begin{align}
\end{align}}

Let

# A

a

‖
a

‖
p

,
# B

b

‖
b

‖
p

‖
a
b

‖
1

=

∑
k

a
k

b
k

=
‖
a

‖
p

‖
b

‖
q

∑
k

A
k

B
k

≤
‖
a

‖
p

‖
b
‖

q

{\displaystyle \begin{align}
\end{align}}

Proof of Theorem 8.13, the Minkowski's inequality

Case 1:

# p
1

{\displaystyle p = 1}

‖
a
b

‖
1

=
a
# b
‖
a

‖
1

‖
b

‖
1

{\displaystyle \begin{align}
\end{align}}

Case 2:

# p
∞

{\displaystyle p=\infty}

By triangle inequality,

‖
a
b

‖
∞

=

sup

k

a

k

b

k

≤

sup

k

b

k

sup

k

b

k

=
‖
a

‖
∞

‖
b

‖
∞

{\displaystyle \begin{align}
\|a+b\|\_\infty
&= \sup \_{k}\left|a\_{k} + b\_{k}\right| \\
\end{align}}

Case 3:

1
<
p
<
∞

{\displaystyle 1 < p < \infty}

‖
a
b

‖
p
p

=

∑
k

a
k

b
k

p

≤

∑
k

a
k

b
k

p
−
1

a
k

∑
k

a
k

b
k

p
−
1

b
k

≤

(

∑
k

a
k

b
k

p

)

p
−
1

p

(

∑
k

a
k

p

)

1
p

(

∑
k

a
k

b
k

p

)

p
−
1

p

(

∑
k

b
k

p

)

1
p

=
‖
a
b

‖
p

p
−
1

(

‖
a

‖
p

‖
b

‖
p

)

{\displaystyle \begin{align}
\end{align}}

Divide both sides by

‖
a
b

‖
p

p
−
1

{\displaystyle \|a+b\|\_p^{p-1}}

, we get

‖
a
b

‖

p

≤
‖
a

‖

p

‖
b

‖

p

{\displaystyle
}

A counter example for

p
<
1

{\displaystyle p<1}

Let

# a

{

a

k

}

{\displaystyle a=\left\{a\_{k}\right\}}

,

# b

{

b

k

}

{\displaystyle b=\left\{b\_{k}\right\}}

,Suppose

k

0

∈

Z

{\displaystyle k\_{0} \in \mathbb{Z}^{+}}

,

k

0

>
1

{\displaystyle k\_{0}>1}

a

k

=

{

1
2

,

⩽
1
≤
k
≤

k

0

0
,

k
>

k

0

b

k

=

{

0
,

1
⩽
k
⩽

k

0

 or
k
>
2

k

0

1
2

,

k

0

<
k
⩽
2

k

0

# c
a
# b

{

1
2

}

# k
1

2

k
0

‖
a
b

‖

p

=

(

∑

# k
1

2

k

0

(

1
2

)

p

)

1
p

=

(

2

k

0

)

1
p

⋅

1
2

‖
a

‖

p

‖
b

‖

p

=

(

∑

# k
1

k
0

(

1
2

)

p

)

1
p

(

∑

# k

k
0

1

2

k
0

(

1
2

)

p

)

1
p

=
2

(

∑

# k
1

k

0

(

1
2

)

p

)

1
p

=
2
⋅

k

0

1
p

⋅

1
2

{\displaystyle
\begin{array}{l}
a\_{k}=\left\{\begin{array}{ll}
\frac{1}{2}, & \leqslant 1 \leq k \leq k\_{0} \\
0, & k>k\_{0}
\end{array}
 b\_{k}=\left\{\begin{array}{ll}
0, & 1 \leqslant k \leqslant k\_{0} \text { or } k > 2 k\_{0} \\
\frac{1}{2}, & k\_{0}<k \leqslant 2 k\_{0}
\end{array}\right.\right. \\
c=a+b=\left\{\frac{1}{2}\right\}\_{k=1}^{2k\_0} \\
\|a+b\|\_{p}=\left(\sum\_{k=1}^{2 k\_{0}}\left(\frac{1}{2}\right)^{p}\right)^{\frac{1}{p}}=\left(2 k\_{0}\right)^{\frac{1}{p}} \cdot \frac{1}{2} \\
\end{array}}

Since

p
<
1

{\displaystyle p<1}

,

2

1
p

>
2
⇒

(

2

k

0

)

1
p

>
2

k

0

1
p

{\displaystyle 2^{\frac{1}{p}}>2 \Rightarrow\left(2 k\_{0}\right)^{\frac{1}{p}}>2 k\_{0}^{\frac{1}{p}} }