PlanetMath: Young's theorem

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Young's theorem

(Theorem)

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

$\displaystyle (f\ast g)(x)=\int_{\ensuremath{\mathbb{R}}^n}f(x-y)g(y)dy$

provided the integral is defined.

The following result is due to William Henry Young.

Theorem Let $p,q,r\in[1,\infty]$ satisfy \begin{equation} \label{eq:young-coeff} \frac{1}{p}+\frac{1}{q}-\frac{1}{r}=1 \end{equation}with the convention $1/\infty=0$ . Let $f\in L^p(\ensuremath{\mathbb{R}}^n)$ , $g\in L^q(\ensuremath{\mathbb{R}}^n)$ . Then

the function $y\mapsto f(x-y)g(y)$ belongs to $L^1(\ensuremath{\mathbb{R}}^n)$ for almost all $x$ ,
the function $x\mapsto(f\ast g)(x)$ belongs to $L^r(\ensuremath{\mathbb{R}}^n)$ , and
there exists a constant $c=c_{p,q}\leq 1$ , depending on $p$ and $q$ but not on $f$ or $g$ , such that

$\displaystyle \Vert{f\ast g}\Vert _r \leq c\cdot\Vert f\Vert _p\cdot\Vert g\Vert _q$

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 (

). The cases

$1/p+1/q=1$ , $r=\infty$
$p=q=r=1$

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

$\displaystyle \left\vert\int f(x-y)g(y)dy\right\vert \leq\int\vert f(x-y)g(y)\vert dy \leq\Vert f\Vert _p\Vert g\Vert _q\;.$

This holds for all $x\in\ensuremath{\mathbb{R}}^n$ , therefore $\Vert f\ast g\Vert _\infty\leq\Vert f\Vert _p\Vert g\Vert _q$ as well.

Proof of case 2.

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 fo $f$ 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,

$\displaystyle \int\left(\int\vert f(x-y)g(y)\vert dy\right)dx =\int\left(\int\vert f(x-y)\vert dx\right)\vert g(y)\vert dy =\Vert f\Vert _1\Vert g\Vert _1\,,$

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

$\displaystyle \Vert f\ast g\Vert _1 \leq\int\int\vert f(x-y)g(y)\vert dydx =\Vert f\Vert _1\Vert g\Vert _1\;.$

Bibliography

1: W. Rudin. Real and complex analysis. McGraw-Hill 1987.
2: W. H. Young. On the multiplication of successions of Fourier constants. Proc. Roy. Soc. Lond. Series A 87 (1912) 331-339.

"Young's theorem" is owned by Ziosilvio.

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Cross-references: plus, Fubini's theorem, Tonelli's theorem, theorem, Lebesgue measure, FO, Borel measurable functions, Borel measurable, measurable, proof, product, pointwise, similar, analogy, almost all, belongs, function, integral, convolution

This is version 4 of Young's theorem, born on 2008-08-02, modified 2009-08-05.
Object id is 10912, canonical name is YoungsTheorem.
Accessed 2452 times total.

Classification:

AMS MSC:

44A35 (Integral transforms, operational calculus :: Convolution)

Pending Errata and Addenda

None.

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