Young's Inequality for Convolutions

Theorem
Let $p, q, r \in \R_{\ge 1}$.

Let $L^p \left({\R^n}\right)$ and $L^q \left({\R^n}\right)$ be Lebesgue spaces.

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

Let $\left\Vert{f}\right\Vert_p$ denote the $p$-seminorm of $f$.

Then the convolution $f * g$ of $f$ and $g$ satisfies:


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

where:
 * $f * g \in L^r \left({\R^n}\right)$

such that:
 * $1 + \dfrac 1 r = \dfrac 1 p + \dfrac 1 q$

and hence is also Lebesgue $p$-integrable.

Proof
We begin by seeking to simplify $|(f*g)(x)|$: \begin{align*} (f*g)(x) &= \int f(x-y) g(y) ~\mathrm{d}y \\ |(f*g)(x)| & \leq \int |f(x-y)| \cdot |g(y)| ~\mathrm{d}y \\ &= \int |f(x-y)|^{1+p/r - p/r} \cdot |g(y)|^{1+q/r - q/r} ~\mathrm{d}y \\ &= \int |f(x-y)|^{p/r} \cdot |g(y)|^{q/r} \cdot |f(x-y)|^{1-p/r} \cdot  |g(y)|^{1-q/r} ~\mathrm{d}y \\ &= \int \big[ |f(x-y)|^p \cdot |g(y)|^q \big]^{1/r} \cdot  |f(x-y)|^{ (r-p)/r} \cdot  |g(y)|^{ (r-q)/r} ~\mathrm{d}y \\ & \leq \underset{I}{\underbrace{\|\big[ |f(x-y)|^p \cdot |g(y)|^q \big]^{1/r}\|_{ r}}} \cdot \underset{II}{\underbrace{\| |f(x-y)|^{ (r-p)/r} \|_{\textstyle \frac{pr}{r-p} }}} \cdot  \| \underset{III}{\underbrace{|g(y)|^{(r-q)/r} \|_{\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: \begin{equation*} \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 \end{equation*} using the hypotheses on $p$, $q$, and $r$. We now analyze terms $I$, $II$, and $III$ individually: \begin{align*} I: &&\| \big[ |f(x-y)|^p \cdot  |g(y)|^q \big]^{1/r}\|_r &= \left(\int \big[ |f(x-y)|^p \cdot  |g(y)|^q \big]^{\textstyle \frac{1}{r}*r} ~\mathrm{d}y \right)^{1/r} \\ &&	&= \left(\int \big[ |f(x-y)|^p \cdot |g(y)|^q \big] ~\mathrm{d}y \right)^{1/r} \\ II: && \| |f(x-y)|^{\textstyle (r-p)/r} \|_{\textstyle\frac{pr}{r-p}} &= \left( \int |f(x-y)|^{\textstyle \frac{r-p}{r}*\frac{pr}{r-p}} ~\mathrm{d}y \right)^{\textstyle\frac{r-p}{pr}} \\ &&	&= \left( \int |f(x-y)|^{p} ~\mathrm{d}y \right)^{\textstyle\frac{1}{p}*\frac{r-p}{r}} \\ &&	&= \|f\|_p^{(r-p)/r} \\ III:  &&  |g(y)|^{(r-q)/r} \|_{\textstyle\frac{qr}{r-q}}    &= \left( \int |g(y)|^{\textstyle \frac{r-q}{r}* \frac{qr}{r-q}}  ~\mathrm{d}y \right)^{\textstyle \frac{r-q}{qr}} \\ &&	&=\left( \int |g(y)|^{\textstyle\frac{r-q}{r}* \frac{qr}{r-q}} ~\mathrm{d}y \right)^{\textstyle \frac{r-q}{qr}} \\ &&	&=\left( \int |g(y)|^q ~\mathrm{d}y \right)^{\textstyle \frac{1}{q}*\frac{r-q}{r}}\\ && &= \|g\|_q^{(r-q)/r}. \end{align*} With these preliminary calculations out of the way, we turn to the main proof: \begin{align*} \|f*g\|_{r}^{r} &= \int |(f*g)(x)|^r ~\mathrm{d}x \\ & \leq \int \left[ \int \big[ |f(x-y)|^p \cdot  |g(y)|^q \big]^{1/r} \cdot |f(x-y)|^{ (r-p)/r} \cdot  |g(y)|^{(r-q)/q} ~\mathrm{d}y \right]^r \mathrm{d}x \\ &= \int \left[ \left(\int \big[ |f(x-y)|^p \cdot |g(y)|^q \big] ~\mathrm{d}y \right)^{1/r} \cdot \|f\|_p^{(r-p)/r} \cdot \|g\|_q^{(r-q)/r}   \right]^r ~\mathrm{d}x  \\ &= \int \left(\int \big[ |f(x-y)|^p \cdot  |g(y)|^q \big] ~\mathrm{d}y \right) \cdot \|f\|_p^{r-p} \cdot \|g\|_q^{r-q}    ~\mathrm{d}x  \\ &= \|f\|_p^{r-p} ~ \|g\|_q^{r-q} \iint |g(y)|^q | f(x-y)|^p ~\mathrm{d}y ~ \mathrm{d}x \\ &= \|f\|_p^{r-p} ~ \|g\|_q^{r-q} \int |g(y)|^q \left(\int | f(x-y)|^p ~\mathrm{d}x \right)~ \mathrm{d}y \\ &= \|f\|_p^{r-p} ~ \|g\|_q^{r-q} \int |g(y)|^q ~\mathrm{d}y \int | f(x)|^p ~ \mathrm{d}x          \\ &= \|f\|_p^{r-p} ~\|g\|_q^{r-q} ~\|g\|_q^q ~ \|f\|_p^p\\ &= \|f\|_p^r ~ \|g\|_q^r \\ \implies \| f*g \|_r &\leq \|f\|_p \|g\|_q \end{align*}