[proofplan]
We split the integrand $|f(x-y)g(y)|$ into three factors whose exponents are tuned by the relation $1/p + 1/q = 1 + 1/r$, then apply the generalised Hölder inequality with exponents $r$, $pr/(r-p)$, and $qr/(r-q)$ to obtain a pointwise bound on $|(f*g)(x)|$. Raising this bound to the $r$-th power, integrating over $x$, and applying Tonelli's theorem reduces the double integral to $\|f\|_{L^p}^p \|g\|_{L^q}^q$ via translation-invariance of Lebesgue measure. The case $r < \infty$ is treated; endpoint cases involving $r = \infty$ follow by a simpler argument.
[/proofplan]
[step:Rewrite the exponent relation and define the splitting parameters]
Write $p' := p/(p-1)$ for the Hölder conjugate of $p$ (with $1' = \infty$), and similarly $q' := q/(q-1)$.
The exponent condition $1/p + 1/q = 1 + 1/r$ is equivalent to
\begin{align*}
\frac{1}{r} &= \frac{1}{p} + \frac{1}{q} - 1 = \frac{1}{p} - \frac{1}{q'} = \frac{1}{q} - \frac{1}{p'}.
\end{align*}
Define the exponents $\alpha := p/r$ and $\beta := q/r$, so that $\alpha, \beta \ge 0$ and $\alpha + \beta = 1$ when $r < \infty$.
We treat the case $r < \infty$ throughout; the endpoint cases involving $r = \infty$ follow by a simpler argument or by taking limits.
[/step]
[step:Split the integrand and apply the generalised Hölder inequality to bound $|(f*g)(x)|$]
Fix $x \in \mathbb{R}^n$.
The [convolution](/page/Convolution) satisfies $(f*g)(x) = \int_{\mathbb{R}^n} f(x-y)\,g(y) \, d\mathcal{L}^n(y)$.
Factor the integrand as
\begin{align*}
|f(x-y)\,g(y)| &= \bigl[|f(x-y)|^p\,|g(y)|^q\bigr]^{1/r} \cdot |f(x-y)|^{1-p/r} \cdot |g(y)|^{1-q/r}.
\end{align*}
This factorisation is valid since $(p/r) + (1 - p/r) = 1$ accounts for the powers of $|f|$, and similarly for $|g|$.
Apply the generalised [Hölder inequality](/theorems/516) with three exponents $r$, $pr/(r-p)$, and $qr/(r-q)$.
One verifies the exponents sum to $1$:
\begin{align*}
\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{align*}
Applying Hölder to the three factors yields
\begin{align*}
|(f*g)(x)| &\le \left(\int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y)\right)^{1/r} \cdot \left(\int_{\mathbb{R}^n} |f(x-y)|^p \, d\mathcal{L}^n(y)\right)^{(r-p)/(pr)} \\
&\qquad \cdot \left(\int_{\mathbb{R}^n} |g(y)|^q \, d\mathcal{L}^n(y)\right)^{(r-q)/(qr)}.
\end{align*}
By translation-invariance of the [Lebesgue integral](/page/Lebesgue%20Integral), the second factor equals $\|f\|_{L^p}^{p \cdot (r-p)/(pr)} = \|f\|_{L^p}^{1-p/r}$, and the third factor equals $\|g\|_{L^q}^{1-q/r}$.
This gives the pointwise bound: for $\mathcal{L}^n$-a.e. $x \in \mathbb{R}^n$,
\begin{align*}
|(f*g)(x)| &\le \|f\|_{L^p}^{1-p/r}\,\|g\|_{L^q}^{1-q/r} \left(\int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y)\right)^{1/r}.
\end{align*}
[guided]
The goal is to obtain a pointwise estimate on $|(f*g)(x)|$ in which the mixed integral $\int |f(x-y)|^p |g(y)|^q \, d\mathcal{L}^n(y)$ appears with exponent $1/r$, and the remaining factors are pure norms of $f$ and $g$.
The idea is to write $|f(x-y)\,g(y)|$ as a product of three pieces, each tailored to one of the three Hölder exponents.
Factor the integrand as
\begin{align*}
|f(x-y)\,g(y)| &= \underbrace{\bigl[|f(x-y)|^p\,|g(y)|^q\bigr]^{1/r}}_{\text{paired with exponent } r} \cdot \underbrace{|f(x-y)|^{1-p/r}}_{\text{paired with exponent } pr/(r-p)} \cdot \underbrace{|g(y)|^{1-q/r}}_{\text{paired with exponent } qr/(r-q)}.
\end{align*}
Why these exponents?
The three Hölder exponents must satisfy $1/r + (r-p)/(pr) + (r-q)/(qr) = 1$.
Expanding:
\begin{align*}
\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{align*}
where the last equality is precisely the exponent condition $1/p + 1/q = 1 + 1/r$.
So the splitting is designed to make the generalised [Hölder inequality](/theorems/516) applicable.
Applying Hölder to the three factors:
\begin{align*}
|(f*g)(x)| &\le \left(\int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y)\right)^{1/r} \cdot \left(\int_{\mathbb{R}^n} |f(x-y)|^p \, d\mathcal{L}^n(y)\right)^{(r-p)/(pr)} \\
&\qquad \cdot \left(\int_{\mathbb{R}^n} |g(y)|^q \, d\mathcal{L}^n(y)\right)^{(r-q)/(qr)}.
\end{align*}
The second factor simplifies by translation-invariance of the [Lebesgue integral](/page/Lebesgue%20Integral): substituting $z = x - y$ gives $\int |f(x-y)|^p \, d\mathcal{L}^n(y) = \int |f(z)|^p \, d\mathcal{L}^n(z) = \|f\|_{L^p}^p$.
Raising to the power $(r-p)/(pr)$ yields $\|f\|_{L^p}^{p \cdot (r-p)/(pr)} = \|f\|_{L^p}^{(r-p)/r} = \|f\|_{L^p}^{1 - p/r}$.
Similarly, the third factor equals $\|g\|_{L^q}^{1 - q/r}$.
This produces the pointwise bound:
\begin{align*}
|(f*g)(x)| &\le \|f\|_{L^p}^{1-p/r}\,\|g\|_{L^q}^{1-q/r} \left(\int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y)\right)^{1/r}.
\end{align*}
[/guided]
[/step]
[step:Raise to the $r$-th power, integrate over $x$, and apply Tonelli's theorem]
Raise the pointwise bound to the $r$-th power and integrate over $x \in \mathbb{R}^n$:
\begin{align*}
\int_{\mathbb{R}^n} |(f*g)(x)|^r \, d\mathcal{L}^n(x) &\le \|f\|_{L^p}^{r-p}\,\|g\|_{L^q}^{r-q} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y) \, d\mathcal{L}^n(x).
\end{align*}
By Tonelli's theorem (the integrand is non-negative) and translation-invariance of $\mathcal{L}^n$:
\begin{align*}
\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y)\,d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n} |g(y)|^q \left(\int_{\mathbb{R}^n} |f(x-y)|^p \, d\mathcal{L}^n(x)\right) d\mathcal{L}^n(y) \\
&= \|f\|_{L^p}^p\,\|g\|_{L^q}^q.
\end{align*}
Substituting back:
\begin{align*}
\|f*g\|_{L^r}^r &\le \|f\|_{L^p}^{r-p}\,\|g\|_{L^q}^{r-q} \cdot \|f\|_{L^p}^p\,\|g\|_{L^q}^q = \|f\|_{L^p}^r\,\|g\|_{L^q}^r.
\end{align*}
Taking $r$-th roots gives $\|f*g\|_{L^r} \le \|f\|_{L^p}\,\|g\|_{L^q}$.
[guided]
We now upgrade the pointwise bound to a global $L^r$ estimate.
Raising both sides of the pointwise inequality to the $r$-th power:
\begin{align*}
|(f*g)(x)|^r &\le \|f\|_{L^p}^{r-p}\,\|g\|_{L^q}^{r-q} \int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y).
\end{align*}
Integrating both sides over $x \in \mathbb{R}^n$:
\begin{align*}
\|f*g\|_{L^r}^r &\le \|f\|_{L^p}^{r-p}\,\|g\|_{L^q}^{r-q} \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(x-y)|^p\,|g(y)|^q \, d\mathcal{L}^n(y)\,d\mathcal{L}^n(x).
\end{align*}
The double integral can be evaluated by swapping the order of integration.
The integrand $|f(x-y)|^p |g(y)|^q$ is non-negative and measurable, so Tonelli's theorem applies without any integrability hypothesis.
After swapping:
\begin{align*}
\int_{\mathbb{R}^n} |g(y)|^q \left(\int_{\mathbb{R}^n} |f(x-y)|^p \, d\mathcal{L}^n(x)\right) d\mathcal{L}^n(y).
\end{align*}
The inner integral $\int_{\mathbb{R}^n} |f(x-y)|^p \, d\mathcal{L}^n(x)$ equals $\|f\|_{L^p}^p$ by translation-invariance of $\mathcal{L}^n$ (substitute $z = x - y$, then $d\mathcal{L}^n(x) = d\mathcal{L}^n(z)$ and the domain remains $\mathbb{R}^n$).
So the double integral collapses to $\|f\|_{L^p}^p \int_{\mathbb{R}^n} |g(y)|^q \, d\mathcal{L}^n(y) = \|f\|_{L^p}^p \|g\|_{L^q}^q$.
Substituting:
\begin{align*}
\|f*g\|_{L^r}^r &\le \|f\|_{L^p}^{r-p} \cdot \|g\|_{L^q}^{r-q} \cdot \|f\|_{L^p}^p \cdot \|g\|_{L^q}^q = \|f\|_{L^p}^r\,\|g\|_{L^q}^r.
\end{align*}
Taking $r$-th roots yields $\|f*g\|_{L^r} \le \|f\|_{L^p}\,\|g\|_{L^q}$.
[/guided]
[/step]
