example of Dirac sequence

example of Dirac sequence

We can construct a

Dirac sequence

{

δ

n

}

n

∈

ℕ

+

by choosing

δ

n

⁢

(

x

)

=

n

π

⁢

(

1

+

n

2

⁢

x

2

)

.

To show that conditions 1 and 3 in the definition of a Dirac sequence are satisfied is trivial and condition 2 is also fulfilled since

∫

-

∞

∞

δ

n

(

x

)

d

x

=

1

π

⋅

∫

-

∞

∞

n

1

+

n

2

⁢

x

2

d

x

=

[

y

=

n

x

d

y

=

n

⋅

d

x

]

=

1

π

⋅

∫

-

∞

∞

1

1

+

y

2

d

y

=

1

π

⋅

arctan

y

|

y

=

-

∞

∞

=

1

π

⋅

π

=

1

for all

n

∈

ℕ

+

, hence

{

δ

n

}

n

∈

ℕ

+

is a Dirac sequence.

To prove that it actually

converges

in

𝒟

′

⁢

(

ℝ

)

(the space of all

distributions

on

𝒟

⁢

(

ℝ

)

) to the

Dirac delta distribution

δ

, we must show that

lim

n

→

∞

⁡

∫

ℝ

δ

n

⁢

(

x

)

⁢

φ

⁢

(

x

)

⁢

𝑑

x

=

φ

⁢

(

0

)

for any test function

φ

∈

𝒟

⁢

(

ℝ

)

(a

topological vector space

of

smooth functions with compact support

). Let us take an arbitrary test function

φ

∈

𝒟

⁢

(

ℝ

)

and assume that the closed and

compact set

supp

⁡

(

φ

)

is contained in some

open interval

(

a

,

b

)

⊂

ℝ

(

a

<

0

and

b

>

0

). Using the

triangle inequality

and the fact that

∫

ℝ

δ

n

⁢

(

x

)

⁢

𝑑

x

=

1

for all

n

∈

ℕ

+

we can write

|

∫

-

∞

∞

δ

n

⁢

(

x

)

⁢

φ

⁢

(

x

)

⁢

𝑑

x

-

φ

⁢

(

0

)

|

=

|

∫

-

∞

∞

δ

n

⁢

(

x

)

⁢

(

φ

⁢

(

x

)

-

φ

⁢

(

0

)

)

⁢

𝑑

x

|

≤

≤

φ

⁢

(

0

)

⁢

∫

-

∞

a

|

δ

n

⁢

(

x

)

|

⁢

𝑑

x

⏟

I

1

+

∫

a

b

|

δ

n

⁢

(

x

)

⁢

(

φ

⁢

(

x

)

-

φ

⁢

(

0

)

)

|

⁢

𝑑

x

⏟

I

2

+

φ

⁢

(

0

)

⁢

∫

b

∞

|

δ

n

⁢

(

x

)

|

⁢

𝑑

x

⏟

I

3

It is easy to see that

lim

n

→

∞

⁡

δ

n

⁢

(

x

)

=

0

,

∀

x

∈

(

-

∞

,

a

]

∪

[

b

,

∞

)

and therefore

lim

n

→

∞

⁡

I

1

=

0

and

lim

n

→

∞

⁡

I

3

=

0

.
Finally we want to estimate

I

2

when

n

→

∞

.

I

2

=

∫

a

b

|

δ

n

⁢

(

x

)

|

⁢

|

(

φ

⁢

(

x

)

-

φ

⁢

(

0

)

)

|

⏟

≤

|

x

|

⋅

sup

⁡

|

φ

′

⁢

(

x

)

|

⁢

𝑑

x

≤

sup

⁡

|

φ

′

⁢

(

x

)

|

⋅

∫

a

b

|

δ

n

⁢

(

x

)

⁢

x

|

⁢

𝑑

x

=

=

sup

⁡

|

φ

′

⁢

(

x

)

|

⋅

1

π

⁢

∫

a

b

|

n

⁢

x

1

+

(

n

⁢

x

)

2

|

⁢

𝑑

x

=

sup

⁡

|

φ

′

⁢

(

x

)

|

⋅

1

π

⁢

(

-

∫

a

0

n

⁢

x

1

+

(

n

⁢

x

)

2

⁢

𝑑

x

+

∫

0

b

n

⁢

x

1

+

(

n

⁢

x

)

2

⁢

𝑑

x

)

=

=

sup

|

φ

′

(

x

)

|

⋅

1

π

(

-

(

1

2

⁢

n

⋅

ln

|

1

+

(

n

x

)

2

|

|

x

=

a

0

)

+

(

1

2

⁢

n

⋅

ln

|

1

+

(

n

x

)

2

|

|

x

=

0

b

)

)

=

=

sup

⁡

|

φ

′

⁢

(

x

)

|

⋅

1

π

⁢

(

1

2

⁢

n

⋅

ln

⁢

|

1

+

(

n

⁢

a

)

2

|

+

1

2

⁢

n

⋅

ln

⁢

|

1

+

(

n

⁢

b

)

2

|

)

We now conclude that

lim

n

→

∞

⁡

I

2

=

0

. This means that

lim

n

→

∞

⁡

I

1

+

I

2

+

I

3

=

0

which shows that

{

δ

n

}

n

∈

ℕ

+

converges to the Dirac delta distribution

δ

.

Title

example of Dirac sequence

Canonical name

ExampleOfDiracSequence

Date of creation

2013-03-22 14:13:10

Last modified on

2013-03-22 14:13:10

Owner

Johan (1032)

Last modified by

Johan (1032)

Numerical id

8

Author

Johan (1032)

Entry type

Example

Classification

msc 46F05

Related topic

Distribution4

Related topic

DeltaDistribution

Related topic

DiracDeltaFunction

Related topic

ConstructionOfDiracDeltaFunction

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