# Young’s theorem

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

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

provided the integral is defined.

The following result is due to William Henry Young.

###### Theorem 1

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

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

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

1. 1.

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

2. 2.

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

3. 3.

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

 $\|{f\ast g}\|_{r}\leq c\cdot\|f\|_{p}\cdot\|g\|_{q}$

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.

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

2. 2.

$p=1$, $q\in[1,\infty)$, $r=q$

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

 $\left|\int f(x-y)g(y)dy\right|\leq\int|f(x-y)g(y)|dy\leq\|f\|_{p}\|g\|_{q}\;.$

This holds for all $x\in\mathbb{R}^{n}$, therefore $\|f\ast g\|_{\infty}\leq\|f\|_{p}\|g\|_{q}$ 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 $\tilde{f}$ 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,

 $\int\left(\int|f(x-y)g(y)|dy\right)dx=\int\left(\int|f(x-y)|dx\right)|g(y)|dy=% \|f\|_{1}\|g\|_{1}\,,$

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

 $\|f\ast g\|_{1}\leq\int\int|f(x-y)g(y)|dydx=\|f\|_{1}\|g\|_{1}\;.$

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 $\|uv\|_{1}\leq\|u\|_{q^{\prime}}\|v\|_{q}:$ but $uv=|f(x-y)g(y)|$, so point 1 of the theorem is proved. By Hölder’s inequality,

 $\left|\int f(x-y)g(y)dy\right|\leq\int|f(x-y)g(y)|dy\leq\|f\|_{1}^{1/q^{\prime% }}\left(\int|f(x-y)|\cdot|g(y)|^{q}dy\right)^{1/q}\;:$

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,

 $\|f\ast g\|_{q}^{q}\leq\|f\|_{1}^{q/q^{\prime}}\||f|\ast|g|^{q}\|_{1}\leq\|f\|% _{1}^{q/q^{\prime}}\|f\|_{1}\|g\|_{q}^{q}=\|f\|_{1}^{1+q/q^{\prime}}\|g\|_{q}^% {q}\;:$

but $1/q+1/q^{\prime}=1$ means $q+q^{\prime}=qq^{\prime}$ and thus $1+q/q^{\prime}=q$, 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 YoungsTheorem 2013-03-22 18:17:44 2013-03-22 18:17:44 Ziosilvio (18733) Ziosilvio (18733) 15 Ziosilvio (18733) Theorem msc 44A35