{{Proofread}} We begin by seeking to bound $\left\vert{\left({f * g}\right) \left({x}\right)}\right\vert$: {{begin-eqn}} {{eqn |l = \left({f * g}\right) \left({x}\right)        |r = \int f \left({x - y}\right) g \left({y}\right) \ \mathrm d y }} {{eqn |l = \left\vert{ \left({f * g}\right) \left({x}\right) }\right\vert        |o = \le       |r = \int \left\vert{ f \left({x - y}\right)}\right\vert \cdot \left\vert{g \left({y}\right)}\right\vert \ \mathrm d y }} {{eqn |r = \int \left\vert{ f \left({x - y}\right) }\right\vert^{1 + p/r - p/r} \cdot \left\vert{ g \left({y}\right) }\right\vert^{1 + q/r - q/r} ~\mathrm d y }} {{eqn |r = \int \left\vert{ f \left({x - y}\right)}\right\vert ^{p/r} \cdot \left\vert{g \left({y}\right)}\right\vert ^{q/r} \cdot \left\vert{ f \left({x - y}\right)}\right\vert ^{1-p/r} \cdot \left\vert{g \left({y}\right)}\right\vert ^{1-q/r} ~\mathrm d y }} {{eqn |r =  \int \left({\left\vert{ f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right)^{1/r} \cdot  \left\vert{ f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} \cdot \left\vert{ g \left({y}\right)}\right\vert^{ (r-q)/r} ~\mathrm d y }} {{eqn |o = \le       |r = \underset{I} {\underbrace{ \left\Vert{ \left({ \left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q }\right)^{1/r} }\right\Vert_r} } \cdot \underset{II}{\underbrace{\left\Vert{\left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} }\right\Vert_{\textstyle \frac{pr}{r-p} } } } \cdot \underset{III}{\underbrace{\left\Vert{\left\vert{ g \left({y}\right)}\right\vert^{(r-q)/r} }\right\Vert_{\textstyle \frac{qr}{r-q} } } } }}   {{end-eqn}} where the last inequality is via the Generalized Hölder Inequality applied to three functions. Note that the relation of conjugate exponents in the Generalized Hölder Inequality is satisfied:   {{begin-eqn}} {{eqn |l = \frac 1 r + \frac {r-p} {p r} + \frac {r-q} {q r}        |r = \frac 1 r + \frac 1 p - \frac 1 r + \frac 1 q - \frac 1 r  }} {{eqn |r = \frac 1 p + \frac 1 q - \frac 1 r }} {{eqn |r = 1       |c = by hypothesis on $p$, $q$, $r$ }} {{end-eqn}} We now analyze terms $I$, $II$, and $III$ individually: {{begin-eqn}} {{eqn |n = I       |l = \left\Vert{ \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert ^q}\right)^{1/r} }\right\Vert_r        |r = \left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert ^q}\right)^{\textstyle \frac 1 r \times r} \ \mathrm d y}\right)^{1/r}  }} {{eqn |r = \left({\int \left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q \ \mathrm d y}\right)^{1/r} }} {{eqn |n = II       |l = \left\Vert{\left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} }\right\Vert_{\textstyle \frac {p r} {r - p} }       |r = \left({\int \left\vert{f \left({x - y}\right)}\right\vert^{\textstyle \frac {r - p} r \times \frac {p r} {r - p} } \ \mathrm d y}\right)^{\textstyle\frac {r - p} {p r} }  }} {{eqn |r = \left({\int \left\vert{f \left({x - y}\right)}\right\vert^p \ \mathrm d y}\right)^{\textstyle\frac 1 p \times \frac {r-p} r} }} {{eqn |r = {\left\Vert{f}\right\Vert_p}^{\left({r - p}\right) /r} }} {{eqn |n = III       |l = \left\Vert{\left\vert{g \left({y}\right)}\right\vert^{(r-q)/r} }\right\Vert_{\textstyle\frac{qr}{r-q} }       |r = \left({\int \left\vert{g \left({y}\right)}\right\vert^{\textstyle \frac{r-q} r \times \frac{q r}{r-q} } \ \mathrm d y}\right)^{\textstyle \frac{r-q}{qr} } }} {{eqn |r = \left({\int \left\vert{g \left({y}\right)}\right\vert^{\textstyle \frac{r-q} r \times \frac{q r}{r-q} } \ \mathrm d y}\right)^{\textstyle \frac{r-q}{qr} } }} {{eqn |r = \left({\int \left\vert{g \left({y}\right)}\right\vert^q \ \mathrm d y}\right)^{\textstyle \frac 1 q \times \frac{r-q} r} }} {{eqn |r = {\left\Vert{g}\right\Vert_q}^{(r-q)/r} }} {{end-eqn}} With these preliminary calculations out of the way, we turn to the main proof: {{begin-eqn}} {{eqn |l = {\left\Vert{f * g}\right\Vert_r}^r        |r = \int \vert \left({f * g}\right) \left({x}\right)\vert ^r ~\mathrm d x }} {{eqn |o = \le       |r = \int \left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right)^{1/r} \cdot \left\vert{f \left({x - y}\right)}\right\vert^{\left({r - p}\right) /r} \cdot \left\vert{g \left({y}\right)}\right\vert^{(r-q)/r} \ \mathrm d y}\right)^r \mathrm d x }} {{eqn |r = \int \left({\left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right)}\right\vert^q}\right) \ \mathrm d y}\right)^{1/r} \cdot {\left\Vert{f}\right\Vert_p}^{\left({r - p}\right) /r} \cdot {\left\Vert{g}\right\Vert_q}^{(r-q)/r} }\right)^r \ \mathrm d x         |c = See $I$, $II$, $III$ }} {{eqn |r = \int \left({\int \left({\left\vert{f \left({x - y}\right)}\right\vert^p \cdot \left\vert{g \left({y}\right) }\right\vert^q}\right) \ \mathrm d y}\right) \cdot {\left\Vert{f}\right\Vert_p}^{r-p} \cdot {\left\Vert{g}\right\Vert_q}^{r-q} \ \mathrm d x }} {{eqn |r = {\left\Vert{f}\right\Vert_p}^{r-p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \iint \left\vert{g \left({y}\right)}\right\vert^q \left\vert{f \left({x - y}\right)}\right\vert^p \ \mathrm d y \ \mathrm d x }} {{eqn |r = {\left\Vert{f}\right\Vert_p}^{r-p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \int \vert g \left({y}\right)\vert^q \left({\int \vert f \left({x - y}\right)\vert^p \ \mathrm d x}\right) \ \mathrm d y        |c = Fubini's Theorem }} {{eqn |r = {\left\Vert{f}\right\Vert_p}^{r-p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \int \left\vert{ g \left({y}\right) }\right\vert^q \ \mathrm d y \int \left\vert{f \left({x}\right)}\right\vert^p \ \mathrm d x }} {{eqn |r = {\left\Vert{f}\right\Vert_p}^{r - p} \ {\left\Vert{g}\right\Vert_q}^{r-q} \ {\left\Vert{g}\right\Vert_q}^q \ {\left\Vert{f}\right\Vert_p}^p }} {{eqn |r = {\left\Vert{f}\right\Vert_p}^r \ {\left\Vert{g}\right\Vert_q}^r }} {{eqn |ll= \implies       |l = \left\Vert{f * g}\right\Vert_r        |o = \le       |r = \left\Vert{f}\right\Vert_p \ \left\Vert{g}\right\Vert_q }} {{end-eqn}} {{qed}} {{Namedfor|William Henry Young|cat=Young}} Category:Measure Theory Category:Inequalities ci4x471dcaaeni7hjt0i6zzo9l9do1p