derivations on a ring of continuous functions

derivations on a ring of continuous functions

Let

X

be a

topological space

and denote by

ℝ

the set of reals. Of course the set of all

continuous functions

C

⁢

(

X

,

ℝ

)

is a

ℝ

-algebra. Let

c

∈

ℝ

. By the symbol

c

¯

we will denote

constant function

at

c

, i.e.

c

¯

:

X

→

ℝ

is defined by

c

¯

⁢

(

x

)

=

c

.

Proposition

.

If

D

:

C

⁢

(

X

,

ℝ

)

→

C

⁢

(

X

,

ℝ

)

is a

ℝ

-

derivation

, then

D

⁢

(

x

)

=

0

¯

for any

x

∈

C

⁢

(

X

,

ℝ

)

.

Proof.

Step one. We will prove that

D

⁢

(

c

¯

)

=

0

¯

for any

c

∈

ℝ

. Indeed

D

⁢

(

1

¯

)

=

D

⁢

(

1

¯

⋅

1

¯

)

=

1

¯

⋅

D

⁢

(

1

¯

)

+

1

¯

⋅

D

⁢

(

1

¯

)

=

D

⁢

(

1

¯

)

+

D

⁢

(

1

¯

)

=

2

⋅

D

⁢

(

1

¯

)

and thus

D

⁢

(

1

¯

)

=

0

¯

. Now from linearity of

D

we obtain that

D

⁢

(

c

¯

)

=

D

⁢

(

c

⋅

1

¯

)

=

c

⋅

D

⁢

(

1

¯

)

=

c

⋅

0

¯

=

0

¯

.

Step two. If

f

:

X

→

ℝ

is

continuous

and

c

∈

ℝ

, then

f

+

c

¯

is continuous and obviously

D

⁢

(

f

)

=

D

⁢

(

f

+

c

¯

)

.

Moreover, if

x

∈

X

then

D

⁢

(

f

-

f

⁢

(

x

)

¯

)

=

D

⁢

(

f

)

, but

(

f

-

f

⁢

(

x

)

¯

)

⁢

(

x

)

=

0

. Thus we may assume that

f

⁢

(

x

)

=

0

for fixed

x

∈

X

.

Let

x

0

∈

X

. Now we will restrict only to such maps

f

:

X

→

ℝ

that

f

⁢

(

x

0

)

=

0

.

Step three. We now decompose

f

into sum of two nonnegative

functions

. Indeed, if

f

:

X

→

ℝ

is continuous, then define

f

+

,

f

-

:

X

→

ℝ

by the

formula

:

f

+

⁢

(

x

)

=

max

⁢

(

f

⁢

(

x

)

,

0

)

;

f

-

⁢

(

x

)

=

max

⁢

(

-

f

⁢

(

x

)

,

0

)

.

Of course both

f

-

and

f

+

are continous, nonnegative and

f

=

f

+

-

f

-

. Thus

D

⁢

(

f

)

=

D

⁢

(

f

+

)

-

D

⁢

(

f

-

)

,

so it is enough to show that

D

⁢

(

f

)

=

0

¯

only for nonnegative and continuous functions.

Step four. Assume that

f

:

X

→

ℝ

is nonnegative, continuous and

f

⁢

(

x

0

)

=

0

. Then there exists

g

:

X

→

ℝ

continuous such that

g

2

=

f

(indeed

g

=

f

and it is well defined,

continuous map

, because

f

was nonnegative). Then we have

D

⁢

(

f

)

=

D

⁢

(

g

2

)

=

g

⋅

D

⁢

(

g

)

+

g

⋅

D

⁢

(

g

)

=

2

⋅

g

⋅

D

⁢

(

g

)

.

Now we have

g

⁢

(

x

0

)

=

f

⁢

(

x

0

)

=

0

and thus

D

⁢

(

f

)

⁢

(

x

0

)

=

2

⋅

g

⁢

(

x

0

)

⋅

D

⁢

(

g

)

⁢

(

x

0

)

=

0

.

Now we can take any

x

∈

X

and repeat steps two, three and four to get that for any

x

∈

X

we have

D

⁢

(

f

)

⁢

(

x

)

=

0

and thus

D

⁢

(

f

)

=

0

¯

,

which

completes

the proof.

□

Remark.

Note that this proof cannot be repeated if we (for example) consider the set of all

smooth functions

C

∞

⁢

(

M

,

ℝ

)

on a

smooth manifold

M

, because

f

+

,

f

-

and

f

need not be smooth.

Title

derivations on a

ring of continuous functions

Canonical name

DerivationsOnARingOfContinuousFunctions

Date of creation

2013-03-22 18:37:25

Last modified on

2013-03-22 18:37:25

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

14

Author

joking (16130)

Entry type

Example

Classification

msc 17A36

Classification

msc 16W25

Classification

msc 13N15

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