Euclidean space

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

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

Generalizations

Young's convolution inequality has a natural generalization in which we replace ${\displaystyle \mathbb {R} ^{d}}$ by a 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]

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

Proof by Hölder's inequality

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 unimodular group endowed with a bi-invariant 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.

Proof by interpolation

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