# Young's convolution inequality

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 $L^{p}(\mathbb {R} ^{d})$ and $g$ is in $L^{q}(\mathbb {R} ^{d})$ and ${\frac {1}{p}}+{\frac {1}{q}}={\frac {1}{r}}+1$ with $1\leq p,q,r\leq \infty .$ Then $\|f*g\|_{r}\leq \|f\|_{p}\|g\|_{q}.$
Here the star denotes convolution, $L^{p}$ is Lebesgue space, and $\|f\|_{p}={\Bigl\int _{\mathbb {R} ^{d}}|f(x)|^{p}\,dx{\Bigr )}^{1/p}$ denotes the usual $L^{p}$ norm.
Equivalently, if $p,q,r\geq 1$ and 1/p+1/q+1/r=2{\textstyle {\frac {1}{p}}+{\frac {1}{q}}+{\frac {1}{r}}=2} then $\left|\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}f(x)g(x-y)h(y)\,\mathrm {d} x\,\mathrm {d} y\right|\leq \left(\int _{\mathbb {R} ^{d}}\vert f\vert ^{p}\right)^{\frac {1}{p}}\left(\int _{\mathbb {R} ^{d}}\vert g\vert ^{q}\right)^{\frac {1}{q}}\left(\int _{\mathbb {R} ^{d}}\vert h\vert ^{r}\right)^{\frac {1}{r}}$

### Generalizations

Young's convolution inequality has a natural generalization in which we replace $\mathbb {R} ^{d}$ by a unimodular group $G.$ If we let $\mu$ be a bi-invariant Haar measure on $G$ and we let $f,g:G\to \mathbb {R}$ or $\mathbb {C}$ be integrable functions, then we define $f*g$ by $f*g(x)=\int _{G}f(y)g(y^{-1}x)\,\mathrm {d} \mu (y).$ Then in this case, Young's inequality states that for $f\in L^{p}(G,\mu )$ and $g\in L^{q}(G,\mu )$ and $p,q,r\in [1,\infty ]$ such that ${\frac {1}{p}}+{\frac {1}{q}}={\frac {1}{r}}+1$ we have a bound $\lVert f*g\rVert _{r}\leq \lVert f\rVert _{p}\lVert g\rVert _{q}.$ Equivalently, if $p,q,r\geq 1$ and 1/p+1/q+1/r=2{\textstyle {\frac {1}{p}}+{\frac {1}{q}}+{\frac {1}{r}}=2} then $\left|\int _{G}\int _{G}f(x)g(y^{-1}x)h(y)\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right|\leq \left(\int _{G}\vert f\vert ^{p}\right)^{\frac {1}{p}}\left(\int _{G}\vert g\vert ^{q}\right)^{\frac {1}{q}}\left(\int _{G}\vert h\vert ^{r}\right)^{\frac {1}{r}}.$ Since $\mathbb {R} ^{d}$ 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 $1<p,q,r<\infty$ satisfy 1/p+1/q=1/r+1.{\textstyle {\tfrac {1}{p}}+{\tfrac {1}{q}}={\tfrac {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 $\|g\|_{q,w}^{q}~:=~\sup _{t>0}\,t^{q}\mu (|g|>t)$ is finite, we have $f*g\in L^{r}(G,\mu )$ and $\|f*g\|_{r}~\leq ~C\,\|f\|_{p}\,\|g\|_{q,w}.$

## 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 $f,g,h:G\to \mathbb {R}$ 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(2−1/q−1/r)=q(2−1/p−1/r)=r(2−1/p−1/q)=1{\textstyle p(2-{\tfrac {1}{q}}-{\tfrac {1}{r}})=q(2-{\tfrac {1}{p}}-{\tfrac {1}{r}})=r(2-{\tfrac {1}{p}}-{\tfrac {1}{q}})=1} ${\begin{aligned}&\int _{G}\int _{G}f(x)g(y^{-1}x)h(y)\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\\=&\int _{G}\int _{G}\left(f(x)^{p}g(y^{-1}x)^{q}\right)^{1-{\frac {1}{r}}}\left(f(x)^{p}h(y)^{r}\right)^{1-{\frac {1}{q}}}\left(g(y^{-1}x)^{q}h(y)^{r}\right)^{1-{\frac {1}{p}}}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\end{aligned}}$ By the Hölder inequality for three functions we deduce that ${\begin{aligned}&\int _{G}\int _{G}f(x)g(y^{-1}x)h(y)\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\\&\leq \left(\int _{G}\int _{G}f(x)^{p}g(y^{-1}x)^{q}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right)^{1-{\frac {1}{r}}}\left(\int _{G}\int _{G}f(x)^{p}h(y)^{r}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right)^{1-{\frac {1}{q}}}\left(\int _{G}\int _{G}g(y^{-1}x)^{q}h(y)^{r}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right)^{1-{\frac {1}{p}}}.\end{aligned}}$ 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 $\|f*g\|_{r}\leq c_{p,q}\|f\|_{p}\|g\|_{q}.$ where the constant $c_{p,q}<1.$ When this optimal constant is achieved, the function $f$ and $g$ are multidimensional Gaussian functions.