Young's Inequality for Convolutions

Theorem
Let $p, q, r \in \R_{\ge 1}$ satisfy:
 * $1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$

Let $L^p \left({\R^n}\right)$, $L^q \left({\R^n}\right)$, and $L^r \left({\R^n}\right)$ be Lebesgue spaces with seminorms $\left\Vert{\bullet}\right\Vert_p$, $\left\Vert{\bullet}\right\Vert_q$, and $\left\Vert{\bullet}\right\Vert_r$ respectively.

Let $f \in L^p \left({\R^n}\right)$ and $g \in L^q \left({\R^n}\right)$.

Then the convolution $f * g$ is in $L^r \left({\R^n}\right)$ and the following inequality is satisfied:


 * $\left\Vert{f * g}\right\Vert_r \le \left\Vert{f}\right\Vert_p \cdot \left\Vert{g}\right\Vert_q$

Proof
We begin by seeking to bound $\vert(f*g)(x)\vert$: \begin{align*} (f*g)(x) &= \int f(x-y) g(y) ~\mathrm{d}y \\ \vert(f*g)(x)\vert & \leq \int \vert f(x-y)\vert \cdot \vert g(y)\vert ~\mathrm{d}y \\ &= \int \vert f(x-y)\vert ^{1+p/r - p/r} \cdot \vert g(y)\vert ^{1+q/r - q/r} ~\mathrm{d}y \\ &= \int \vert f(x-y)\vert ^{p/r} \cdot \vert g(y)\vert ^{q/r} \cdot \vert f(x-y)\vert ^{1-p/r} \cdot  \vert g(y)\vert ^{1-q/r} ~\mathrm{d}y \\ &= \int \left( \vert f(x-y)\vert ^p \cdot \vert g(y)\vert ^q \right)^{1/r} \cdot  \vert f(x-y)\vert ^{ (r-p)/r} \cdot  \vert g(y)\vert ^{ (r-q)/r} ~\mathrm{d}y \\ & \leq \underset{I}{\underbrace{ \Vert \left( \vert f(x-y)\vert ^p \cdot \vert g(y)\vert ^q \right)^{1/r}\Vert_{ r}}} \cdot \underset{II}{\underbrace{\Vert \vert f(x-y)\vert ^{ (r-p)/r} \Vert_{\textstyle \frac{pr}{r-p} }}} \cdot   \underset{III}{\underbrace{\Vert \vert g(y)\vert ^{(r-q)/r} \Vert_{\textstyle \frac{qr}{r-q}}}} \end{align*} where the last inequality is via the Generalized Hölder Inequality applied to three functions, which we justify by noting that: $$\frac{1}{r} + \frac{r-p}{pr} + \frac{r-q}{qr} = \frac{1}{r} + \frac{1}{p} - \frac{1}{r} + \frac{1}{q} - \frac{1}{r} = \frac{1}{p} + \frac{1}{q} - \frac{1}{r} = 1 $$ using the hypothesis on $p$, $q$, and $r$. We now analyze terms $I$, $II$, and $III$ individually: \begin{align*} I: &&\Vert \left( |f(x-y)|^p \cdot  |g(y)|^q \right)^{1/r}\Vert_r &= \left(\int \left( |f(x-y)|^p \cdot  |g(y)|^q \right)^{\textstyle \frac{1}{r}*r} ~\mathrm{d}y \right)^{1/r} \\ &&	&= \left(\int \vert f(x-y)\vert ^p \cdot  \vert g(y)\vert ^q  ~\mathrm{d}y \right)^{1/r} \\ II: && \| |f(x-y)|^{(r-p)/r} \|_{\textstyle \frac{pr}{r-p}} &= \left( \int \vert f(x-y)\vert ^{\textstyle \frac{r-p}{r}*\frac{pr}{r-p}} ~\mathrm{d}y \right)^{\textstyle\frac{r-p}{pr}} \\ &&	&= \left( \int \vert f(x-y)\vert ^{p} ~\mathrm{d}y \right)^{\textstyle\frac{1}{p}*\frac{r-p}{r}} \\ &&	&= \Vert f\Vert_p^{(r-p)/r} \\ III:  &&  \Vert \vert g(y)\vert^{(r-q)/r} \Vert_{\textstyle\frac{qr}{r-q}}    &= \left( \int \vert g(y)\vert ^{\textstyle \frac{r-q}{r}* \frac{qr}{r-q}}  ~\mathrm{d}y \right)^{\textstyle \frac{r-q}{qr}} \\ &&	&=\left( \int \vert g(y)\vert ^{\textstyle\frac{r-q}{r}* \frac{qr}{r-q}} ~\mathrm{d}y \right)^{\textstyle \frac{r-q}{qr}} \\ &&	&=\left( \int \vert g(y)\vert ^q ~\mathrm{d}y \right)^{\textstyle \frac{1}{q}*\frac{r-q}{r}}\\ && &= \Vert g\Vert_q^{(r-q)/r}. \end{align*} With these preliminary calculations out of the way, we turn to the main proof: \begin{align*} \Vert f*g\Vert_{r}^{r} &= \int \vert (f*g)(x)\vert ^r ~\mathrm{d}x \\ & \leq \int \left[ \int \left( \vert f(x-y)\vert ^p \cdot  \vert g(y)\vert ^q \right)^{1/r} \cdot \vert f(x-y)\vert ^{ (r-p)/r} \cdot  \vert g(y)\vert ^{(r-q)/q} ~\mathrm{d}y \right]^r \mathrm{d}x \\ &= \int \left( \left(\int \left( \vert f(x-y)\vert ^p \cdot \vert g(y)\vert ^q \right) ~\mathrm{d}y \right)^{1/r} \cdot \Vert f\Vert_p^{(r-p)/r} \cdot \Vert g\Vert_q^{(r-q)/r}   \right)^r ~\mathrm{d}x  \\ &= \int \left(\int \left( \vert f(x-y)\vert ^p \cdot  \vert g(y)\vert ^q \right) ~\mathrm{d}y \right) \cdot \Vert f\Vert_p^{r-p} \cdot \Vert g\Vert_q^{r-q}    ~\mathrm{d}x  \\ &= \Vert f\Vert_p^{r-p} ~ \Vert g\Vert_q^{r-q} \iint \vert g(y)\vert ^q \vert f(x-y)\vert ^p ~\mathrm{d}y ~ \mathrm{d}x \\ &= \Vert f\Vert_p^{r-p} ~ \Vert g\Vert_q^{r-q} \int \vert g(y)\vert ^q \left(\int \vert f(x-y)\vert ^p ~\mathrm{d}x \right)~ \mathrm{d}y \\ &= \Vert f\Vert_p^{r-p} ~ \Vert g\Vert_q^{r-q} \int \vert g(y)\vert ^q ~\mathrm{d}y \int \vert f(x)\vert ^p ~ \mathrm{d}x          \\ &= \Vert f\Vert_p^{r-p} ~\Vert g\Vert_q^{r-q} ~\Vert g\Vert_q^q ~ \Vert f\Vert_p^p\\ &= \Vert f\Vert_p^r ~ \Vert g\Vert_q^r \\ \implies \Vert f*g \Vert_r &\leq \Vert f\Vert_p \Vert g\Vert_q \end{align*}