Young's convolution inequality

Template:Short description In mathematics, Young's convolution inequality is a mathematical inequality about the convolution of two functions, named after William Henry Young.

Statement

Euclidean space

In real analysis, the following result is called Young's convolution inequality:

Suppose $f$ is in the Lebesgue space $L^p({ℝ}^d)$ and $g$ is in $L^q({ℝ}^d)$ and $$\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1$$ with $1\le p,q,r\le \mathrm{\infty }.$ Then $$\Vert f*g{\Vert }_r\le \Vert f{\Vert }_p\Vert g{\Vert }_q.$$

Here the star denotes convolution, $L^p$ is Lebesgue space, and $$\Vert f{\Vert }_p=(\underset{{ℝ}^d}{\int }|f(x){|}^{p}dx{)}^{1/p}$$ denotes the usual $L^p$ norm.

Equivalently, if $p,q,r\ge 1$ and $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$ then $$|\underset{{ℝ}^d}{\int }\underset{{ℝ}^d}{\int }f(x)g(x-y)h(y)\mathrm{d}x\mathrm{d}y|\le {(\underset{{ℝ}^d}{\int }|f{|}^p)}^{\frac{1}{p}}{(\underset{{ℝ}^d}{\int }|g{|}^q)}^{\frac{1}{q}}{(\underset{{ℝ}^d}{\int }|h{|}^r)}^{\frac{1}{r}}$$

Generalizations

Young's convolution inequality has a natural generalization in which we replace ${ℝ}^d$ by a unimodular group $G.$ If we let $\mu$ be a bi-invariant Haar measure on $G$ and we let $f,g:G\to ℝ$ or $ℂ$ be integrable functions, then we define $f*g$ by $$f*g(x)=\underset{G}{\int }f(y)g(y^{-1}x)\mathrm{d}\mu (y).$$ Then in this case, Young's inequality states that for $f\in L^p(G,\mu )$ and $g\in L^q(G,\mu )$ and $p,q,r\in [1,\mathrm{\infty }]$ such that $$\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1$$ we have a bound $$\Vert f*g{\Vert }_r\le \Vert f{\Vert }_p\Vert g{\Vert }_q.$$ Equivalently, if $p,q,r\ge 1$ and $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2$ then $$|\underset{G}{\int }\underset{G}{\int }f(x)g(y^{-1}x)h(y)\mathrm{d}\mu (x)\mathrm{d}\mu (y)|\le {(\underset{G}{\int }|f{|}^p)}^{\frac{1}{p}}{(\underset{G}{\int }|g{|}^q)}^{\frac{1}{q}}{(\underset{G}{\int }|h{|}^r)}^{\frac{1}{r}}.$$ Since ${ℝ}^d$ is in fact a locally compact abelian group (and therefore unimodular) with the Lebesgue measure the desired Haar measure, this is in fact a generalization.

This generalization may be refined. Let $G$ and $\mu$ be as before and assume $1<p,q,r<\mathrm{\infty }$ satisfy $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1.$ Then there exists a constant $C$ such that for any $f\in L^p(G,\mu )$ and any measurable function $g$ on $G$ that belongs to the weak $L^q$ space $L^{q,w}(G,\mu ),$ which by definition means that the following supremum $$\Vert g{\Vert }_{q,w}^q:=\underset{t>0}{\sup }{t}^q\mu (|g|>t)$$ is finite, we have $f*g\in L^r(G,\mu )$ andTemplate:Sfn $$\Vert f*g{\Vert }_r\le C\Vert f{\Vert }_p\Vert g{\Vert }_{q,w}.$$

Applications

An example application is that Young's inequality can be used to show that the heat semigroup is a contracting semigroup using the $L^2$ norm (that is, the Weierstrass transform does not enlarge the $L^2$ norm).

Proof

Proof by Hölder's inequality

Young's inequality has an elementary proof with the non-optimal constant 1.

We assume that the functions $f,g,h:G\to ℝ$ are nonnegative and integrable, where $G$ is a unimodular group endowed with a bi-invariant Haar measure $\mu .$ We use the fact that $\mu (S)=\mu (S^{-1})$ for any measurable $S\subseteq G.$ Since $p(2-\frac{1}{q}-\frac{1}{r})=q(2-\frac{1}{p}-\frac{1}{r})=r(2-\frac{1}{p}-\frac{1}{q})=1$ $$\begin{array}{cc}& \underset{G}{\int }\underset{G}{\int }f(x)g(y^{-1}x)h(y)\mathrm{d}\mu (x)\mathrm{d}\mu (y)\\ =& \underset{G}{\int }\underset{G}{\int }{(f(x{)}^{p}g(y^{-1}x{)}^q)}^{1-\frac{1}{r}}{(f(x{)}^{p}h(y{)}^r)}^{1-\frac{1}{q}}{(g(y^{-1}x{)}^{q}h(y{)}^r)}^{1-\frac{1}{p}}\mathrm{d}\mu (x)\mathrm{d}\mu (y)\end{array}$$ By the Hölder inequality for three functions we deduce that $$\begin{array}{cc}& \underset{G}{\int }\underset{G}{\int }f(x)g(y^{-1}x)h(y)\mathrm{d}\mu (x)\mathrm{d}\mu (y)\\ & \le {(\underset{G}{\int }\underset{G}{\int }f(x{)}^{p}g(y^{-1}x{)}^q\mathrm{d}\mu (x)\mathrm{d}\mu (y))}^{1-\frac{1}{r}}{(\underset{G}{\int }\underset{G}{\int }f(x{)}^{p}h(y{)}^r\mathrm{d}\mu (x)\mathrm{d}\mu (y))}^{1-\frac{1}{q}}{(\underset{G}{\int }\underset{G}{\int }g(y^{-1}x{)}^{q}h(y{)}^r\mathrm{d}\mu (x)\mathrm{d}\mu (y))}^{1-\frac{1}{p}}.\end{array}$$ The conclusion follows then by left-invariance of the Haar measure, the fact that integrals are preserved by inversion of the domain, and by Fubini's theorem.

Proof by interpolation

Young's inequality can also be proved by interpolation; see the article on Riesz–Thorin interpolation for a proof.

Sharp constant

In case $p,q>1,$ Young's inequality can be strengthened to a sharp form, via $$\Vert f*g{\Vert }_r\le c_{p,q}\Vert f{\Vert }_p\Vert g{\Vert }_q.$$ where the constant $c_{p,q}<1.$ When this optimal constant is achieved, the function $f$ and $g$ are multidimensional Gaussian functions.

See also

• Template:Annotated link

References

• Template:Bahouri Chemin Danchin Fourier Analysis and Nonlinear Partial Differential Equations 2011

External links

• Young's Inequality for Convolutions at ProofWiki