Young’s theorem

Young’s theorem

Let

f

,

g

:

ℝ

n

→

ℝ

.
Recall that the

convolution

of

f

and

g

at

x

is

(

f

∗

g

)

⁢

(

x

)

=

∫

ℝ

n

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

⁢

𝑑

y

provided the

integral

is defined.

The following result is due to William Henry Young.

Theorem 1

Let

p

,

q

,

r

∈

[

1

,

∞

]

satisfy

1

p

+

1

q

-

1

r

=

1

(1)

with the convention

1

/

∞

=

0

.
Let

f

∈

L

p

⁢

(

R

n

)

,

g

∈

L

q

⁢

(

R

n

)

.
Then:

1.

The

function

y

↦

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

belongs to

L

1

⁢

(

ℝ

n

)

for almost all

x

.

2.

The function

x

↦

(

f

∗

g

)

⁢

(

x

)

belongs to

L

r

⁢

(

ℝ

n

)

.

3.

There exists a

constant

c

=

c

p

,

q

≤

1

,
depending on

p

and

q

but not on

f

or

g

, such that

∥

f

∗

g

∥

r

≤

c

⋅

∥

f

∥

p

⋅

∥

g

∥

q

Observe the

analogy

with the

similar

result
with convolution replaced by ordinary (

pointwise

)

product

,
where the requirement is

1

/

p

+

1

/

q

=

1

/

r

—

i.e.

,

1

/

p

+

1

/

q

-

1

/

r

=

0

—instead of (

1

).
The cases

1.

1

/

p

+

1

/

q

=

1

,

r

=

∞

2.

p

=

1

,

q

∈

[

1

,

∞

)

,

r

=

q

are the most widely known;
for these we provide a proof, supposing

c

p

,

q

=

1

.
We shall use the following facts:

•

If

x

↦

f

⁢

(

x

)

,

x

↦

g

⁢

(

x

)

are

measurable

,
then

(

x

,

y

)

↦

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

is measurable.

•

For any

x

, if

f

∈

L

p

,
then

y

↦

f

⁢

(

x

-

y

)

belongs to

L

p

as well,
and its

L

p

-norm is the same as

f

’s.

•

For any

y

, if

f

∈

L

p

,
then

x

↦

f

⁢

(

x

-

y

)

belongs to

L

p

as well,
and its

L

p

-norm is the same as

f

’s.

Proof of case

1

.

Suppose

f

∈

L

p

⁢

(

ℝ

n

)

,

g

∈

L

q

⁢

(

ℝ

n

)

with

1

/

p

+

1

/

q

=

1

.
Then

|

∫

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

⁢

𝑑

y

|

≤

∫

|

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

|

⁢

𝑑

y

≤

∥

f

∥

p

⁢

∥

g

∥

q

.

This holds for all

x

∈

ℝ

n

,
therefore

∥

f

∗

g

∥

∞

≤

∥

f

∥

p

⁢

∥

g

∥

q

as well.

Proof of case

2

.

First, suppose

q

=

1

.
We may suppose

f

and

g

are

Borel measurable

:
if they are not, we replace them with Borel measurable functions

f

~

and

g

~

which are equal to

f

and

g

, respectively,
outside of a set of

Lebesgue measure

zero;
apply the

theorem

to

f

~

,

g

~

, and

f

~

∗

g

~

;
and deduce the theorem for

f

,

g

, and

f

∗

g

.
By Tonelli’s theorem,

∫

(

∫

|

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

|

⁢

𝑑

y

)

⁢

𝑑

x

=

∫

(

∫

|

f

⁢

(

x

-

y

)

|

⁢

𝑑

x

)

⁢

|

g

⁢

(

y

)

|

⁢

𝑑

y

=

∥

f

∥

1

⁢

∥

g

∥

1

,

thus the function

(

x

,

y

)

↦

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

belongs to

L

1

⁢

(

ℝ

n

×

ℝ

n

)

.
By Fubini’s theorem,
the function

y

↦

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

belongs to

L

1

⁢

(

ℝ

n

)

for almost all

x

,
and

x

↦

(

f

∗

g

)

⁢

(

x

)

belongs to

L

1

⁢

(

ℝ

n

)

;
plus,

∥

f

∗

g

∥

1

≤

∫

∫

|

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

|

⁢

𝑑

y

⁢

𝑑

x

=

∥

f

∥

1

⁢

∥

g

∥

1

.

Suppose now

q

>

1

;
choose

q

′

so that

1

/

q

+

1

/

q

′

=

1

.
By the

argument

above,

y

↦

|

f

⁢

(

x

-

y

)

|

⋅

|

g

⁢

(

y

)

|

q

belongs to

L

1

for almost all

x

:
for those

x

, put

u

⁢

(

y

)

=

|

f

⁢

(

x

-

y

)

|

1

/

q

′

,

v

⁢

(

y

)

=

|

f

⁢

(

x

-

y

)

|

1

/

q

⁢

|

g

⁢

(

y

)

|

.

Then

u

∈

L

q

′

and

v

∈

L

q

with

1

/

q

′

+

1

/

q

=

1

,
so

u

⁢

v

∈

L

1

and

∥

u

⁢

v

∥

1

≤

∥

u

∥

q

′

⁢

∥

v

∥

q

:

but

u

⁢

v

=

|

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

|

, so point

1

of the theorem is proved.
By Hölder’s

inequality

,

|

∫

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

⁢

𝑑

y

|

≤

∫

|

f

⁢

(

x

-

y

)

⁢

g

⁢

(

y

)

|

⁢

𝑑

y

≤

∥

f

∥

1

1

/

q

′

⁢

(

∫

|

f

⁢

(

x

-

y

)

|

⋅

|

g

⁢

(

y

)

|

q

⁢

𝑑

y

)

1

/

q

:

but we know that

|

f

|

∗

|

g

|

q

∈

L

1

,
so

f

∗

g

∈

L

q

and point

2

is also proved.
Finally,

∥

f

∗

g

∥

q

q

≤

∥

f

∥

1

q

/

q

′

⁢

∥

|

f

|

∗

|

g

|

q

∥

1

≤

∥

f

∥

1

q

/

q

′

⁢

∥

f

∥

1

⁢

∥

g

∥

q

q

=

∥

f

∥

1

1

+

q

/

q

′

⁢

∥

g

∥

q

q

:

but

1

/

q

+

1

/

q

′

=

1

means

q

+

q

′

=

q

⁢

q

′

and thus

1

+

q

/

q

′

=

q

,
so that point

3

is also proved.

References

1

G. Gilardi.

Analisi tre.

McGraw-Hill 1994.

2

W. Rudin.

Real and complex analysis.

McGraw-Hill 1987.

3

W. H. Young.
On the

multiplication

of successions of Fourier constants.

Proc. Roy. Soc. Lond. Series A

87

(1912) 331–339.

Title

Young’s theorem

Canonical

name

YoungsTheorem

Date of creation

2013-03-22 18:17:44

Last modified on

2013-03-22 18:17:44

Owner

Ziosilvio (18733)

Last modified by

Ziosilvio (18733)

Numerical id

15

Author

Ziosilvio (18733)

Entry type

Theorem

Classification

msc 44A35

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