Young’s theorem

Let $f,g:{ℝ}^n\to ℝ$. Recall that the convolution of $f$ and $g$ at $x$ is

$$(f\ast g)(x)={\int }_{{ℝ}^n}f(x-y)g(y)𝑑y$$

provided the integral is defined.

The following result is due to William Henry Young.

Theorem 1

Let $p\mathrm{,}q\mathrm{,}r\mathrm{\in }\mathrm{[}\mathrm{1}\mathrm{,}\mathrm{\infty }\mathrm{]}$ satisfy

$$\frac{1}{p}+\frac{1}{q}-\frac{1}{r}=1$$ (1)

with the convention $\mathrm{1}\mathrm{/}\mathrm{\infty }\mathrm{=}\mathrm{0}$. Let $f\mathrm{\in }{L}^p\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$, $g\mathrm{\in }{L}^q\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$. Then:

1. The function $y\mapsto f(x-y)g(y)$ belongs to $L^1({ℝ}^n)$ for almost all $x$.

2. The function $x\mapsto (f\ast g)(x)$ belongs to $L^r({ℝ}^n)$.

3. There exists a constant $c=c_{p,q}\le 1$, depending on $p$ and $q$ but not on $f$ or $g$, such that

   $${\parallel f\ast g\parallel }_r\le c\cdot {\parallel f\parallel }_p\cdot {\parallel g\parallel }_q$$

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. $1/p+1/q=1$, $r=\mathrm{\infty }$

2. $p=1$, $q\in [1,\mathrm{\infty })$, $r=q$

are the most widely known; for these we provide a proof, supposing $c_{p,q}=1$. We shall use the following facts:

• If $x\mapsto f(x),x\mapsto g(x)$ are measurable, then $(x,y)\mapsto f(x-y)g(y)$ is measurable.

• For any $x$, if $f\in L^p$, then $y\mapsto f(x-y)$ belongs to $L^p$ as well, and its $L^p$-norm is the same as $f$’s.

• For any $y$, if $f\in L^p$, then $x\mapsto f(x-y)$ belongs to $L^p$ as well, and its $L^p$-norm is the same as $f$’s.

Proof of case 1.

Suppose $f\in L^p({ℝ}^n)$, $g\in L^q({ℝ}^n)$ with $1/p+1/q=1$. Then

$$\left|\int f(x-y)g(y)𝑑y\right|\le \int |f(x-y)g(y)|𝑑y\le {\parallel f\parallel }_p{\parallel g\parallel }_q.$$

This holds for all $x\in {ℝ}^n$, therefore ${\parallel f\ast g\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_p{\parallel g\parallel }_q$ 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 $\tilde{f}$ and $\tilde{g}$ which are equal to $f$ and $g$, respectively, outside of a set of Lebesgue measure zero; apply the theorem to $\tilde{f}$, $\tilde{g}$, and $\tilde{f}\ast \tilde{g}$; and deduce the theorem for $f$, $g$, and $f\ast g$. By Tonelli’s theorem,

$$\int \left(\int |f(x-y)g(y)|𝑑y\right)𝑑x=\int \left(\int |f(x-y)|𝑑x\right)|g(y)|𝑑y={\parallel f\parallel }_1{\parallel g\parallel }_1,$$

thus the function $(x,y)\mapsto f(x-y)g(y)$ belongs to $L^1({ℝ}^n\times {ℝ}^n)$. By Fubini’s theorem, the function $y\mapsto f(x-y)g(y)$ belongs to $L^1({ℝ}^n)$ for almost all $x$, and $x\mapsto (f\ast g)(x)$ belongs to $L^1({ℝ}^n)$; plus,

$${\parallel f\ast g\parallel }_1\le \int \int |f(x-y)g(y)|𝑑y𝑑x={\parallel f\parallel }_1{\parallel g\parallel }_1.$$

Suppose now $q>1$; choose $q^{\prime }$ so that $1/q+1/q^{\prime }=1$. By the argument above, $y\mapsto |f(x-y)|\cdot {|g(y)|}^q$ belongs to $L^1$ for almost all $x$: for those $x$, put $u(y)={|f(x-y)|}^{1/q^{\prime }},v(y)={|f(x-y)|}^{1/q}|g(y)|.$ Then $u\in L^{q^{\prime }}$ and $v\in L^q$ with $1/q^{\prime }+1/q=1$, so $uv\in L^1$ and ${\parallel uv\parallel }_1\le {\parallel u\parallel }_{q^{\prime }}{\parallel v\parallel }_q:$ but $uv=|f(x-y)g(y)|$, so point 1 of the theorem is proved. By Hölder’s inequality,

$$\left|\int f(x-y)g(y)𝑑y\right|\le \int |f(x-y)g(y)|𝑑y\le {\parallel f\parallel }_1^{1/q^{\prime }}{\left(\int |f(x-y)|\cdot {|g(y)|}^q𝑑y\right)}^{1/q}:$$

but we know that $|f|\ast {|g|}^q\in L^1$, so $f\ast g\in L^q$ and point 2 is also proved. Finally,

$${\parallel f\ast g\parallel }_q^q\le {\parallel f\parallel }_1^{q/q^{\prime }}{\parallel |f|\ast {|g|}^q\parallel }_1\le {\parallel f\parallel }_1^{q/q^{\prime }}{\parallel f\parallel }_1{\parallel g\parallel }_q^q={\parallel f\parallel }_1^{1+q/q^{\prime }}{\parallel g\parallel }_q^q:$$

but $1/q+1/q^{\prime }=1$ means $q+q^{\prime }=q{q}^{\prime }$ and thus $1+q/q^{\prime }=q$, so that point 3 is also proved.

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