continuity of convex functions

continuity of convex functions

We will prove below that every

convex function

on an
open (

http://planetmath.org/Open

)

convex subset

A

of a

finite-dimensional

real vector space

is

continuous

. This statement becomes false if we
do not require

A

to be open,
since we can increase the value of

f

at any point of

A

which is
not a

convex combination

of two other points without affecting the
convexity of

f

. An example of this is shown in Figure

1

.

Figure 1:

A convex function on a non-open set need not be continuous.

Let

A

be an open

convex set

in a finite-dimensional

vector space

V

over

ℝ

, and let

f

:

A

→

ℝ

be a convex
function. Let

x

∈

A

be arbitrary, and let

P

be a parallelepiped
centered at

x

and lying completely inside

A

. Here “a
parallelepiped centered at

x

” means a subset of

V

of the form

P

=

{

x

+

∑

i

=

1

n

λ

i

⁢

b

i

:

-

1

≤

λ

i

≤

1

⁢

for

⁢

i

=

1

,

2

,

…

,

n

}

,

where

{

b

1

,

…

,

b

n

}

is some basis of

V

. Furthermore, let

∂

⁡

P

=

{

x

+

∑

i

=

1

n

λ

i

⁢

b

i

:

max

1

≤

i

≤

n

⁡

|

λ

i

|

=

1

}

denote the boundary of

P

. We will show that

f

is continuous at

x

by showing that

f

attains a

maximum

on

∂

⁡

P

and by
estimating

|

f

⁢

(

y

)

-

f

⁢

(

x

)

|

in
of this maximum as

y

→

x

.

The idea is to use the condition of convexity to ‘squeeze’ the graph
of

f

near

x

, as is shown in Figure

2

.

Figure 2:

Given the values of

f

in

x

and on

∂

⁡

P

=

{

y

1

,

y

2

}

, the convexity condition restricts the
graph of

f

to the grey area.

For

λ

∈

[

0

,

1

]

and

y

∈

∂

⁡

P

, the convexity of

f

implies

f

⁢

(

(

1

-

λ

)

⁢

x

+

λ

⁢

y

)

≤

(

1

-

λ

)

⁢

f

⁢

(

x

)

+

λ

⁢

f

⁢

(

y

)

(1)

=

f

⁢

(

x

)

+

λ

⁢

(

f

⁢

(

y

)

-

f

⁢

(

x

)

)

.

On the other hand, for all

μ

∈

[

0

,

1

/

2

]

we have

f

⁢

(

x

)

=

f

⁢

(

(

1

-

μ

)

⁢

[

(

1

-

2

⁢

μ

)

⁢

x

1

-

μ

+

μ

⁢

y

1

-

μ

]

+

μ

⁢

(

2

⁢

x

-

y

)

)

≤

(

1

-

μ

)

⁢

f

⁢

(

(

1

-

2

⁢

μ

)

⁢

x

1

-

μ

+

μ

⁢

y

1

-

μ

)

+

μ

⁢

f

⁢

(

2

⁢

x

-

y

)

.

Dividing by

1

-

μ

and setting

λ

=

μ

1

-

μ

∈

[

0

,

1

]

gives

(

1

+

λ

)

⁢

f

⁢

(

x

)

≤

f

⁢

(

(

1

-

λ

)

⁢

x

+

λ

⁢

y

)

+

λ

⁢

f

⁢

(

2

⁢

x

-

y

)

.

(2)

From the two

inequalities

(

1

) and (

2

) we obtain

-

λ

⁢

(

f

⁢

(

2

⁢

x

-

y

)

-

f

⁢

(

x

)

)

≤

f

⁢

(

x

+

λ

⁢

(

y

-

x

)

)

-

f

⁢

(

x

)

≤

λ

⁢

(

f

⁢

(

y

)

-

f

⁢

(

x

)

)

.

(3)

Note that both

y

and

2

⁢

x

-

y

∂

⁡

P

, and that

f

is

bounded

on

P

(hence in particular on

∂

⁡

P

). Indeed, the
convexity of

f

implies that

f

is bounded by its values at two
faces of

P

, and repeatedly applying
this shows
that

f

attains a maximum at one of the corners of

P

.

Write

P

λ

for the parallelepiped

P

shrunk by a

λ

relative to

x

:

P

λ

=

{

x

+

λ

⁢

(

y

-

x

)

:

y

∈

P

}

.

Now the inequality (

3

) implies that for all

λ

∈

[

0

,

1

]

and all

z

∈

∂

⁡

P

λ

,
we have

|

f

⁢

(

z

)

-

f

⁢

(

x

)

|

≤

λ

⁢

|

max

y

∈

∂

⁡

P

⁡

f

⁢

(

y

)

-

f

⁢

(

x

)

|

.

Consequently, the same inequality holds for all

λ

∈

(

0

,

1

]

and
all

z

in the open neighbourhood

P

λ

∖

∂

⁡

P

λ

of

x

. The right-hand
of this inequality goes to
zero as

λ

→

0

, from which we conclude that

f

is continuous
at

x

.

Title

continuity of convex functions

Canonical name

ContinuityOfConvexFunctions

Date of creation

2013-03-22 15:28:00

Last modified on

2013-03-22 15:28:00

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

5

Author

pbruin (1001)

Entry type

Result

Classification

msc 26A51

Classification

msc 26B25

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