Young's convolution inequality

In real analysis, the following result is called Young's convolution inequality:^[2]

Suppose ${\displaystyle f}$ is in the Lebesgue space ${\displaystyle L^{p}(\mathbb {R} ^{d})}$ and ${\displaystyle g}$ is in ${\displaystyle L^{q}(\mathbb {R} ^{d})}$ and

$${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}={\frac {1}{r}}+1}$$

with ${\displaystyle 1\leq p,q,r\leq \infty .}$ Then

$${\displaystyle \|f*g\|_{r}\leq \|f\|_{p}\|g\|_{q}.}$$

Here the star denotes convolution, ${\displaystyle L^{p}}$ is Lebesgue space, and

$${\displaystyle \|f\|_{p}={\Bigl (}\int _{\mathbb {R} ^{d}}|f(x)|^{p}\,dx{\Bigr )}^{1/p}}$$

denotes the usual ${\displaystyle L^{p}}$ norm.

Equivalently, if ${\displaystyle p,q,r\geq 1}$ and ${\textstyle {\frac {1}{p}}+{\frac {1}{q}}+{\frac {1}{r}}=2}$ then

$${\displaystyle \left|\int _{\mathbb {R} ^{d}}\int _{\mathbb {R} ^{d}}f(x)g(x-y)h(y)\,\mathrm {d} x\,\mathrm {d} y\right|\leq \left(\int _{\mathbb {R} ^{d}}\vert f\vert ^{p}\right)^{\frac {1}{p}}\left(\int _{\mathbb {R} ^{d}}\vert g\vert ^{q}\right)^{\frac {1}{q}}\left(\int _{\mathbb {R} ^{d}}\vert h\vert ^{r}\right)^{\frac {1}{r}}}$$

Young's convolution inequality has a natural generalization in which we replace ${\displaystyle \mathbb {R} ^{d}}$ by a ${\displaystyle \sigma }$-compact unimodular group ${\displaystyle G.}$ If we let ${\displaystyle \mu }$ be a bi-invariant Haar measure on ${\displaystyle G}$ and we let ${\displaystyle f,g:G\to \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$ be integrable functions, then we define ${\displaystyle f*g}$ by

$${\displaystyle f*g(x)=\int _{G}f(y)g(y^{-1}x)\,\mathrm {d} \mu (y).}$$

Then in this case, Young's inequality states that for ${\displaystyle f\in L^{p}(G,\mu )}$ and ${\displaystyle g\in L^{q}(G,\mu )}$ and ${\displaystyle p,q,r\in [1,\infty ]}$ such that

$${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}={\frac {1}{r}}+1}$$

we have a bound

$${\displaystyle \lVert f*g\rVert _{r}\leq \lVert f\rVert _{p}\lVert g\rVert _{q}.}$$

Equivalently, if ${\displaystyle p,q,r\geq 1}$ and ${\textstyle {\frac {1}{p}}+{\frac {1}{q}}+{\frac {1}{r}}=2}$ then

$${\displaystyle \left|\int _{G}\int _{G}f(x)g(y^{-1}x)h(y)\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right|\leq \left(\int _{G}\vert f\vert ^{p}\right)^{\frac {1}{p}}\left(\int _{G}\vert g\vert ^{q}\right)^{\frac {1}{q}}\left(\int _{G}\vert h\vert ^{r}\right)^{\frac {1}{r}}.}$$

Since ${\displaystyle \mathbb {R} ^{d}}$ 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 ${\displaystyle G}$ and ${\displaystyle \mu }$ be as before and assume ${\displaystyle 1<p,q,r<\infty }$ satisfy ${\textstyle {\tfrac {1}{p}}+{\tfrac {1}{q}}={\tfrac {1}{r}}+1.}$ Then there exists a constant ${\displaystyle C}$ such that for any ${\displaystyle f\in L^{p}(G,\mu )}$ and any measurable function ${\displaystyle g}$ on ${\displaystyle G}$ that belongs to the weak ${\displaystyle L^{q}}$ space ${\displaystyle L^{q,w}(G,\mu ),}$ which by definition means that the following supremum

$${\displaystyle \|g\|_{q,w}^{q}~:=~\sup _{t>0}\,t^{q}\mu (|g|>t)}$$

is finite, we have ${\displaystyle f*g\in L^{r}(G,\mu )}$ and^[3]

$${\displaystyle \|f*g\|_{r}~\leq ~C\,\|f\|_{p}\,\|g\|_{q,w}.}$$

Young's inequality has an elementary proof with the non-optimal constant 1.^[4]

We assume that the functions ${\displaystyle f,g,h:G\to \mathbb {R} }$ are nonnegative and integrable, where ${\displaystyle G}$ is a ${\displaystyle \sigma }$-compact unimodular group endowed with a bi-invariant ${\displaystyle \sigma }$-finite Haar measure ${\displaystyle \mu .}$ We use the fact that ${\displaystyle \mu (S)=\mu (S^{-1})}$ for any measurable ${\displaystyle S\subseteq G.}$ Since ${\textstyle p(2-{\tfrac {1}{q}}-{\tfrac {1}{r}})=q(2-{\tfrac {1}{p}}-{\tfrac {1}{r}})=r(2-{\tfrac {1}{p}}-{\tfrac {1}{q}})=1}$

$${\displaystyle {\begin{aligned}&\int _{G}\int _{G}f(x)g(y^{-1}x)h(y)\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\\={}&\int _{G}\int _{G}\left(f(x)^{p}g(y^{-1}x)^{q}\right)^{1-{\frac {1}{r}}}\left(f(x)^{p}h(y)^{r}\right)^{1-{\frac {1}{q}}}\left(g(y^{-1}x)^{q}h(y)^{r}\right)^{1-{\frac {1}{p}}}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\end{aligned}}}$$

By the Hölder inequality for three functions we deduce that

$${\displaystyle {\begin{aligned}&\int _{G}\int _{G}f(x)g(y^{-1}x)h(y)\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\\&\leq \left(\int _{G}\int _{G}f(x)^{p}g(y^{-1}x)^{q}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right)^{1-{\frac {1}{r}}}\left(\int _{G}\int _{G}f(x)^{p}h(y)^{r}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right)^{1-{\frac {1}{q}}}\left(\int _{G}\int _{G}g(y^{-1}x)^{q}h(y)^{r}\,\mathrm {d} \mu (x)\,\mathrm {d} \mu (y)\right)^{1-{\frac {1}{p}}}.\end{aligned}}}$$

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.

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