another proof of Jensen’s inequality

another proof of Jensen’s inequality

First of all, it’s clear that defining

λ

k

=

μ

k

∑

k

=

1

n

μ

k

we have

∑

k

=

1

n

λ

k

=

1

so it will we enough to prove only the simplified version.

Let’s proceed by

induction

.

1)

n

=

2

; we have to show that, for any

x

1

and

x

2

in

[

a

,

b

]

,

f

⁢

(

λ

1

⁢

x

1

+

λ

2

⁢

x

2

)

≤

λ

1

⁢

f

⁢

(

x

1

)

+

λ

2

⁢

f

⁢

(

x

2

)

.

But, since

λ

1

+

λ

2

must be equal to 1, we can put

λ

2

=

1

-

λ

1

, so that the thesis becomes

f

⁢

(

λ

1

⁢

x

1

+

(

1

-

λ

1

)

⁢

x

2

)

≤

λ

1

⁢

f

⁢

(

x

1

)

+

(

1

-

λ

1

)

⁢

f

⁢

(

x

2

)

,

which is true by

definition

of a

convex function

.

2) Taking as true that

f

⁢

(

∑

k

=

1

n

-

1

μ

k

⁢

x

k

)

≤

∑

k

=

1

n

-

1

μ

k

⁢

f

⁢

(

x

k

)

, where

∑

k

=

1

n

-

1

μ

k

=

1

, we have to prove that

f

⁢

(

∑

k

=

1

n

λ

k

⁢

x

k

)

≤

∑

k

=

1

n

λ

k

⁢

f

⁢

(

x

k

)

,

where

∑

k

=

1

n

λ

k

=

1

.

First of all, let’s observe that

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

=

(

∑

k

=

1

n

λ

k

)

-

λ

n

1

-

λ

n

=

1

-

λ

n

1

-

λ

n

=

1

and that if all

x

k

∈

[

a

,

b

]

,

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

belongs to

[

a

,

b

]

as well. In fact,

λ

k

1

-

λ

n

being non-negative,

a

≤

x

k

≤

b

⇒

λ

k

1

-

λ

n

⁢

a

≤

λ

k

1

-

λ

n

⁢

x

k

≤

λ

k

1

-

λ

n

⁢

b

,

and,

summing

over

k

,

a

⁢

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

≤

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

≤

b

⁢

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

,

that is

a

≤

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

≤

b

.

We have, by definition of a convex function:

f

⁢

(

∑

k

=

1

n

λ

k

⁢

x

k

)

=

f

⁢

(

∑

k

=

1

n

-

1

λ

k

⁢

x

k

+

λ

n

⁢

x

n

)

=

f

⁢

(

(

1

-

λ

n

)

⁢

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

+

λ

n

⁢

x

n

)

≤

(

1

-

λ

n

)

⁢

f

⁢

(

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

)

+

λ

n

⁢

f

⁢

(

x

n

)

.

But, by

inductive hypothesis

, since

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

=

1

, we have:

f

⁢

(

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

)

≤

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

f

⁢

(

x

k

)

,

so that

f

⁢

(

∑

k

=

1

n

λ

k

⁢

x

k

)

≤

(

1

-

λ

n

)

⁢

f

⁢

(

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

x

k

)

+

λ

n

⁢

f

⁢

(

x

n

)

≤

(

1

-

λ

n

)

⁢

∑

k

=

1

n

-

1

λ

k

1

-

λ

n

⁢

f

⁢

(

x

k

)

+

λ

n

⁢

f

⁢

(

x

n

)

=

∑

k

=

1

n

-

1

λ

k

⁢

f

⁢

(

x

k

)

+

λ

n

⁢

f

⁢

(

x

n

)

=

∑

k

=

1

n

λ

k

⁢

f

⁢

(

x

k

)

which is the thesis.

Title

another proof of Jensen’s

inequality

Canonical

name

AnotherProofOfJensensInequality

Date of creation

2013-03-22 15:52:53

Last modified on

2013-03-22 15:52:53

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

12

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 26D15

Classification

msc 39B62

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