Young’s theorem
Let f,g:ℝn→ℝ. Recall that the convolution of f and g at x is
(f∗g)(x)=∫ℝnf(x-y)g(y)𝑑y |
provided the integral is defined.
The following result is due to William Henry Young.
Theorem 1
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 cp,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∈Lp, then y↦f(x-y) belongs to Lp as well, and its Lp-norm is the same as f’s.
-
•
For any y, if f∈Lp, then x↦f(x-y) belongs to Lp as well, and its Lp-norm is the same as f’s.
Proof of case 1.
Suppose f∈Lp(ℝn), g∈Lq(ℝn) with 1/p+1/q=1. Then
|∫f(x-y)g(y)𝑑y|≤∫|f(x-y)g(y)|𝑑y≤∥f∥p∥g∥q. |
This holds for all x∈ℝn, 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)|𝑑y)𝑑x=∫(∫|f(x-y)|𝑑x)|g(y)|𝑑y=∥f∥1∥g∥1, |
thus the function (x,y)↦f(x-y)g(y) belongs to L1(ℝn×ℝn). By Fubini’s theorem, the function y↦f(x-y)g(y) belongs to L1(ℝn) for almost all x, and x↦(f∗g)(x) belongs to L1(ℝn); plus,
∥f∗g∥1≤∫∫|f(x-y)g(y)|𝑑y𝑑x=∥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 L1 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∈Lq′ and v∈Lq with 1/q′+1/q=1,
so uv∈L1 and
∥uv∥1≤∥u∥q′∥v∥q:
but uv=|f(x-y)g(y)|, so point 1 of the theorem is proved.
By Hölder’s inequality
,
|∫f(x-y)g(y)𝑑y|≤∫|f(x-y)g(y)|𝑑y≤∥f∥1/q′1(∫|f(x-y)|⋅|g(y)|q𝑑y)1/q: |
but we know that |f|∗|g|q∈L1, so f∗g∈Lq and point 2 is also proved. Finally,
∥f∗g∥qq≤∥f∥q/q′1∥|f|∗|g|q∥1≤∥f∥q/q′1∥f∥1∥g∥qq=∥f∥1+q/q′1∥g∥qq: |
but 1/q+1/q′=1 means q+q′=qq′ 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 |