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The properties of convex functions, focusing on holder's and jensen's inequalities. Convex functions are essential in mathematics and have applications in various fields, including physics, engineering, and economics. Holder's inequality is a mathematical result that relates the sum of the products of two sequences to the product of the sums of the sequences. Jensen's inequality, on the other hand, states that the expected value of a convex function of a random variable is less than or equal to the function's value at the expected value of the random variable. Proofs, exercises, and examples to help understand these concepts.
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PROF. MARIA GIRARDI
Notation 1.1. Throughout this presentation:
′ ∈ (1, ∞) is defined by
1 p
1 p′^
Definition 1.2. ϕ : I → R is convex ⇐⇒
x, y ∈ I , t ∈ (0, 1) =⇒ ϕ (tx + (1 − t) y) ≤ tϕ(x) + (1 − t) ϕ(y). (1.1)
Picture.
a x tx + (1 − t) y y b
Date: 7 June 2001.
Lemma 2.1. Let ϕ : I → R be convex. Then:
xi ∈ I , ti ∈ (0, 1) ,
n ∑
i=
ti = 1 =⇒ ϕ
n ∑
i=
tixi
n ∑
i=
ti ϕ(xi) (2.1)
xi ∈ I , ti ∈ (0, 1) =⇒ ϕ
(∑n
i=
tixi ∑n
i=1 ti
∑n
i=
ti ϕ(xi) ∑n
i=1 ti
Proof. (2.1): Use (1.1) and induction. (2.2): Let ˜ti = [
n j=
tj ]
− 1 ti and apply (2.1).
Recall. Geometric-Arithmetic Mean Inequality: GM ≤ AM
xi ≥ 0 , n ∈ N =⇒
n ∏
i=
xi
) 1 /n
n
n ∑
i=
xi.
Using convex functions, we can generalize the GM-AM inequality.
Proposition 2.2. Generalized GM-AM inequality:
xi ≥ 0 , ti ∈ (0, 1) ,
∑^ n
i=
ti = 1 =⇒
∏^ n
i=
x
ti i ≤
∑^ n
i=
ti xi. (2.3)
Proof. Let ϕ : (0, ∞) → R be ϕ(x) = − ln x. Then
ϕ is convex =⇒ − ln
n ∑
i=
ti xi
n ∑
i=
ti (− ln xi)
∑^ n
i=
ln
x
ti i
≤ ln
∑n
i=
ti xi
Now exponentiate both sides of (2.4).
An immediate corollary follows now.
Corollary 2.3. Young’s Inequality:
xi ≥ 0 , 1 < p < ∞ =⇒ x 1 · x 2 ≤
x
p 1
p
x
p′ 2
p
′
Exercise. For sequences {xi}
n i=1 and^ {yi}
n i=1 from^ R^ and^
1 p 1
1 p 2
1 p 3
‖{xi · yi}
n i=1‖`p 3 ≤^ ‖{xi}
n i=1‖`p 1 ·^ ‖{yi}
n i=1‖`p 2.
Hint:
1 p 1 /p 3
1 p 2 /p 3
Recall. For a nice function f : I → R,
‖f ‖ Lp
I
|f (x)|
p dx
] 1 /p
Theorem 2.5. Holder’s Inequality in Lp: For nice functions f, g : I → R,
‖f · g‖ L 1
≤ ‖f ‖ Lp
· ‖g‖ Lp′
that is,
I
|f (x)g(x)| dx ≤
I
|f (x)|
p
] 1 /p
I
|g(x)|
p′
] 1 /p′
Proof. WLOG: Neither ‖f ‖L p
nor ‖g‖L p′^
is 0 or ∞. WLOG: ‖f ‖L p
= 1 = ‖g‖L p′^
for if
not so:
f :=
f
‖f ‖L p
f
Lp
= 1 , ˜g :=
g
‖g‖L p′
=⇒ ‖˜g‖L p′^
By Young’s Inequality
∫
I
|f (x)g(x)| dx ≤
I
|f (x)|
p
p
|g(x)|
p′
p′
p
‖f ‖
p Lp
p′^
‖g‖
p′ Lp′
Here are two exercises of Generalized Holder’s Inequalities in Lp.
Exercise. For nice functions fj : I → R and
∑k
j=
1 pj
k ∏
j=
fj
L 1
k ∏
j=
‖fj ‖ Lpj
Exercise. For nice functions f, g : I → R and
1 p 1
1 p 2
1 p 3
‖f · g‖ Lp 3
≤ ‖f ‖ Lp 1
· ‖g‖ Lp 2
Lemma 3.1. Let ϕ : I → R be convex. Let x, y, z ∈ I with x < y < z. Then
ϕ(y) − ϕ(x)
y − x
ϕ(z) − ϕ(x)
z − x
ϕ(z) − ϕ(y)
z − y
Picture.
a x y z b
Proof. Key idea:
y = tx + (1 − t)z =
z − y
z − x
x +
y − x
z − x
z =⇒ ϕ(y) ≤ tϕ(x) + (1 − t)ϕ(z).
Think of what this says in the picture; the needed algebra then follows easily.
Observation 3.3. Let ϕ : I → R be convex and x 0 ∈ I. Let
ϕ
′ −(x^0 )^ ≤^ m^ ≤^ ϕ
′ +(x^0 )^.
Consider the line
l(x) = m(x − x 0 ) + ϕ(x 0 )
through the point (x 0 , ϕ (x 0 )). Then
l(x) ≤ ϕ(x) ∀x ∈ I. (3.3)
A line y = l(x) through the point (x 0 , ϕ (x 0 )) that satisfies (3.3) is called a supporting line
of y = ϕ(x) at x 0. Draw yourself a picture to see the choice of terminolgy here. Thus
Observation 3.3 says that convex functions always have supporting lines.
Picture.
a xl x 0 xr b
Proof. By (3.2)
ϕ(x 0 ) − ϕ(xl)
x 0 − xl
≤ m ≤
ϕ(xr) − ϕ(x 0 )
xr − x 0
Thus
m(x − x 0 ) ≤ ϕ(x) − ϕ(x 0 ) ∀x ∈ I \ {x 0 }.
Theorem 3.4. Jensen’s Inequality: Let ϕ : R → R be convex and f : I → R be integrable.
If I = (0, 1), then
ϕ
I
f (x) dx
I
ϕ (f (x)) dx (3.4)
If I = (a, b) has finite length, then
ϕ
I
f (x) dx
b − a
I
ϕ (f (x)) dx
b − a
Remark. Theorem 3.4 may be thought of as a continuous version of Lemma 2.1.
Proof. (3.4): Let
I
f (x) dx = x 0 and ϕ
′ −(x^0 )^ ≤^ m^ ≤^ ϕ
′ +(x^0 ). Then by Observation 3.
m(x − x 0 ) + ϕ(x 0 ) ≤ ϕ(x) ∀x ∈ R.
Thus
m(f (x) − x 0 ) + ϕ(x 0 ) ≤ ϕ (f (x)) ∀x ∈ I. (3.6)
Now integrate both sides of (3.6) over I := (0, 1).
(3.5) follows applying (3.4) to
g(x) := f (a + x (b − a)) : I → R.
