7-14

Question:

Prove that

ϕ

{\displaystyle \phi}

 is convex on

(
a
,
b
)

{\displaystyle (a,b)}

 if and only if it is continuous and

ϕ

(

x
1

x
2

2

)

≤

ϕ
(

x
1

)
ϕ
(

x
2

)

2

{\displaystyle \phi \left(\frac{x\_1+x\_2}{2}\right) \leq \frac{\phi(x\_1)+\phi(x\_2)}{2}}

for

x
1

,

x
2

∈
(
a
,
b
)

{\displaystyle x\_1,x\_2 \in (a,b) }

.

Answer:

⇒

{\displaystyle \Rightarrow}

 If

ϕ

{\displaystyle \phi}

 is convex, by definition,

ϕ

(

p
1

x
1

p
2

x
2

)

≤

p
1

ϕ
(

x
1

)

p
2

ϕ
(

x
2

)

{\displaystyle \phi \left(p\_1x\_1+p\_2x\_2\right) \leq p\_1\phi(x\_1)+p\_2\phi(x\_2) }

where

p
1

p
2

=
1

{\displaystyle p\_1+p\_2=1}

.

Choose

p
1

=

p
2

=

1
2

{\displaystyle p\_1=p\_2=\frac{1}{2} }

, then

ϕ

(

x
1

x
2

2

)

=
ϕ

(

1
2

x
1

1
2

x
2

)

≤

1
2

ϕ
(

x
1

)

1
2

ϕ
(

x
2

# )

ϕ
(

x
1

)
ϕ
(

x
2

)

2

.

{\displaystyle \phi \left(\frac{x\_1+x\_2}{2}\right) = \phi \left(\frac{1}{2}x\_1+\frac{1}{2}x\_2\right) \leq \frac{1}{2}\phi(x\_1)+\frac{1}{2}\phi(x\_2) = \frac{\phi(x\_1)+\phi(x\_2)}{2}. }

Since

ϕ

{\displaystyle \phi}

 is convex, the slope

ϕ
(
x
h
)
−
ϕ
(
x
)

h

,

h
>
0

{\displaystyle \frac{\phi(x+h)-\phi(x)}{h}, \quad h>0 }

 decreases when

h

{\displaystyle h}

 decreases.

Also for any

h
,
k
≥
0

{\displaystyle h, k \geq 0 }

, we have

ϕ
(
x
h
)
−
ϕ
(
x
)

h

≥

ϕ
(
x
)
−
ϕ
(
x
−
k
)

k

.

{\displaystyle \frac{\phi(x+h)-\phi(x)}{h}\geq\frac{\phi(x)-\phi(x-k)}{k}. }

This means that

D

ϕ
(
x
# )

lim

h
→

0

ϕ
(
x
h
)
−
ϕ
(
x
)

h

{\displaystyle D^+ \phi(x)= \lim\_{h\to 0^+} \frac{\phi(x+h)-\phi(x)}{h} }

 exists and finite by monotone convergence theorem.

By similar approach, we can show that

D
−

ϕ
(
x
# )

lim

k
→

0
−

ϕ
(
x
)
−
ϕ
(
x
−
k
)

k

{\displaystyle D^- \phi(x) = \lim\_{k\to0^-} \frac{\phi(x)-\phi(x-k)}{k} }

 exists and finite. Then

−
∞
<

D
−

ϕ
(
x
)
≤

D

ϕ
(
x
)
<
∞
.

{\displaystyle -\infty < D^- \phi(x) \leq D^+ \phi(x)<\infty. }

So

lim

h
→

0

ϕ
(
x
h
)
−
ϕ
(
x
# )
0

{\displaystyle \lim\_{h\to0^+} \phi(x+h) - \phi(x) =0 }

, and

lim

k
→

0
−

ϕ
(
x
)
−
ϕ
(
x
−
k
# )
0

{\displaystyle \lim\_{k\to0^-} \phi(x) - \phi(x-k) =0 }

. This guarantees that

ϕ

{\displaystyle \phi}

 is continuous.

⇐

{\displaystyle \Leftarrow}

 If

ϕ

{\displaystyle \phi}

 is continuous, and

ϕ

(

x
1

x
2

2

)

≤

ϕ
(

x
1

)
ϕ
(

x
2

)

2

{\displaystyle \phi \left(\frac{x\_1+x\_2}{2}\right) \leq \frac{\phi(x\_1)+\phi(x\_2)}{2}}

 for

x
1

,

x
2

∈
(
a
,
b
)

{\displaystyle x\_1,x\_2 \in (a,b) }

,

So

ϕ

(

x
1

x
2

⋯

x

2
n

2
n

)

≤

ϕ
(

x
1

)
ϕ
(

x
2

)
⋯
ϕ
(

x

2
n

)

2
n

{\displaystyle \phi \left( \frac{x\_1+x\_2+\dots+x\_{2^n}}{2^n} \right) \leq \frac{\phi(x\_1)+\phi(x\_2)+\dots+\phi(x\_{2^n})}{2^n}}

. Let

# f

x
i

{\displaystyle f = x\_i}

 for

1
≤
i
≤
m

{\displaystyle 1 \le i \le m}

, and

# g

x
i

{\displaystyle g = x\_i }

 for

m
1
≤
i
≤

2
n

{\displaystyle m+1 \leq i \leq 2^n }

 where

1
≤
m
≤

2
n

{\displaystyle 1\leq m \leq 2^n }

 is a integer.

Then we obtain

ϕ

(

m

2
n

f

(

1
−

m

2
n

)

g

)

≤

m

2
n

ϕ
(
f
)

(

1
−

m

2
n

)

ϕ
(
g
)
.

{\displaystyle \phi \left( \frac{m}{2^n}f + \left(1-\frac{m}{2^n}\right)g \right) \leq \frac{m}{2^n}\phi(f) + \left(1-\frac{m}{2^n}\right) \phi(g). }

Since the number

m

2
n

{\displaystyle \frac{m}{2^n} }

 with integers

m
,
n

{\displaystyle m,n}

 and

1
≤
m
≤

2
n

{\displaystyle 1\leq m \leq 2^n }

 is dense in

[
0
,
1
]

{\displaystyle [0,1]}

. And by the continuity, we prove that

ϕ

{\displaystyle \phi}

 is convex.