The proof establishes the weak-type bound $\|f * K\|_{L^{r,\infty}} \le C\|f\|_{L^p}\|K\|_{L^{q,\infty}}$ by splitting $K$ at an amplitude threshold, bounding each piece in a different $L^a$ space using the weak-type condition on $K$, applying [Young's convolution inequality](/theorems/463) and the [Hölder inequality](/theorems/516) to the resulting [convolution](/page/Convolution) pieces, and optimising the threshold. The strong-type upgrade to $L^r$ follows by the Marcinkiewicz interpolation theorem (applied at two values of $p$).
Write $A := \|f\|_{L^p}$ and $B := \|K\|_{L^{q,\infty}}$ for brevity. The relation $1/p + 1/q = 1 + 1/r$ gives $1/r = 1/p - 1/q'$ where $q' = q/(q-1)$ is the conjugate exponent of $q$. Since $1/r > 0$, this forces $p < q'$, equivalently $p' > q$.
**Step 1 (Truncation of $K$).** For $s > 0$, decompose $K = K^{>s} + K^{\le s}$ where
\begin{align*}
K^{>s} := K\,\mathbb{1}_{\{|K| > s\}}, \qquad K^{\le s} := K\,\mathbb{1}_{\{|K| \le s\}}.
\end{align*}
We bound each truncated piece in an appropriate Lebesgue space.
[claim:Truncation Bounds]
For all $s > 0$:
\begin{align*}
\|K^{>s}\|_{L^1} &\le \frac{q}{q-1}\,B^q\,s^{1-q}, \\
\|K^{\le s}\|_{L^{p'}} &\le \left(\frac{p'}{p'-q}\right)^{1/p'} B^{q/p'}\,s^{1 - q/p'}.
\end{align*}
[/claim]
[proof]
**$L^1$ bound on $K^{>s}$.** The [distribution](/page/Distribution) [function](/page/Function) of $K^{>s}$ satisfies $d_{K^{>s}}(\lambda) = d_K(\lambda)$ for $\lambda \ge s$ and $d_{K^{>s}}(\lambda) = d_K(s)$ for $0 < \lambda < s$. Using $d_K(\lambda) \le B^q\lambda^{-q}$ and the layer cake representation:
\begin{align*}
\|K^{>s}\|_{L^1} &= \int_0^\infty d_{K^{>s}}(\lambda)\,d\mathcal{L}^1(\lambda) = \int_0^s d_K(s)\,d\mathcal{L}^1(\lambda) + \int_s^\infty d_K(\lambda)\,d\mathcal{L}^1(\lambda) \\
&\le B^q s^{-q}\cdot s + B^q\int_s^\infty \lambda^{-q}\,d\mathcal{L}^1(\lambda) = B^q s^{1-q} + \frac{B^q}{q-1}\,s^{1-q} = \frac{q}{q-1}\,B^q\,s^{1-q}.
\end{align*}
The [integral](/page/Integral) $\int_s^\infty \lambda^{-q}\,d\mathcal{L}^1(\lambda) = s^{1-q}/(q-1)$ converges because $q > 1$.
**$L^{p'}$ bound on $K^{\le s}$.** For $\lambda > 0$, $d_{K^{\le s}}(\lambda) = 0$ when $\lambda \ge s$ and $d_{K^{\le s}}(\lambda) \le d_K(\lambda) \le B^q\lambda^{-q}$ when $\lambda < s$. Using the layer cake representation:
\begin{align*}
\|K^{\le s}\|_{L^{p'}}^{p'} &= p'\int_0^\infty \lambda^{p'-1}\,d_{K^{\le s}}(\lambda)\,d\mathcal{L}^1(\lambda) = p'\int_0^s \lambda^{p'-1}\,d_{K^{\le s}}(\lambda)\,d\mathcal{L}^1(\lambda) \\
&\le p'B^q\int_0^s \lambda^{p'-1-q}\,d\mathcal{L}^1(\lambda) = \frac{p'}{p'-q}\,B^q\,s^{p'-q}.
\end{align*}
The integral converges because $p' > q$ gives $p' - 1 - q > -1$. Taking $p'$-th roots gives the stated bound.
[/proof]
**Step 2 (Convolution bounds on the pieces).** By [Young's convolution inequality](/theorems/463) with exponents $(p, 1, p)$ (since $1/p + 1/1 = 1 + 1/p$):
\begin{align*}
\|f * K^{>s}\|_{L^p} \le A\,\|K^{>s}\|_{L^1} \le \frac{q}{q-1}\,A\,B^q\,s^{1-q}.
\end{align*}
By the [Hölder inequality](/theorems/516) (equivalently, Young with exponents $(p, p', \infty)$ since $1/p + 1/p' = 1 + 0$):
\begin{align*}
\|f * K^{\le s}\|_{L^\infty} \le A\,\|K^{\le s}\|_{L^{p'}} \le C_1\,A\,B^{q/p'}\,s^{1-q/p'},
\end{align*}
where $C_1 := (p'/(p'-q))^{1/p'}$.
**Step 3 (Threshold optimisation and distribution function bound).** For $\lambda > 0$, choose $s = s(\lambda)$ so that $\|f * K^{\le s}\|_{L^\infty} = \lambda/2$, i.e.,
\begin{align*}
s(\lambda) := \left(\frac{\lambda}{2C_1\,A\,B^{q/p'}}\right)^{p'/(p'-q)}.
\end{align*}
This is well-defined since $1 - q/p' > 0$ (because $p' > q$), so the map $s \mapsto C_1 A B^{q/p'} s^{1-q/p'}$ is strictly increasing from $0$ to $\infty$.
With this choice, $|f * K^{\le s(\lambda)}(x)| \le \lambda/2$ for all $x$, so by the triangle inequality
\begin{align*}
\{|f * K| > \lambda\} \subseteq \{|f * K^{>s(\lambda)}| > \lambda/2\}.
\end{align*}
By Chebyshev's inequality applied to the $L^p$ bound:
\begin{align*}
d_{f*K}(\lambda) \le \frac{2^p}{\lambda^p}\,\|f * K^{>s(\lambda)}\|_{L^p}^p \le \frac{2^p}{\lambda^p}\left(\frac{q}{q-1}\right)^p A^p\,B^{pq}\,s(\lambda)^{p(1-q)}.
\end{align*}
[claim:Distribution Function Decay]
$d_{f*K}(\lambda) \le C_2\,\lambda^{-r}\,A^r\,B^r$, where $C_2$ depends only on $p$ and $q$.
[/claim]
[proof]
Substituting $s(\lambda) = ({\lambda}/({2C_1 A B^{q/p'}}))^{p'/(p'-q)}$:
\begin{align*}
s(\lambda)^{p(1-q)} = \left(\frac{\lambda}{2C_1 A B^{q/p'}}\right)^{p'p(1-q)/(p'-q)}.
\end{align*}
The power of $\lambda$ in $d_{f*K}(\lambda)$ is therefore
\begin{align*}
-p + \frac{p'p(1-q)}{p'-q}.
\end{align*}
To evaluate this, write $\beta := p'(1-q)/(p'-q)$:
\begin{align*}
\beta - 1 = \frac{p'(1-q) - (p'-q)}{p'-q} = \frac{q(1-p')}{p'-q} = \frac{-q/(p-1)}{p'-q},
\end{align*}
using $p' - 1 = 1/(p-1)$. Since $p' - q = (p - q(p-1))/(p-1)$:
\begin{align*}
\beta - 1 = \frac{-q}{p - q(p-1)} = \frac{-q}{p + q - pq}.
\end{align*}
Therefore the power of $\lambda$ is
\begin{align*}
p(\beta - 1) = \frac{-pq}{p + q - pq} = -r,
\end{align*}
where the last equality uses $1/r = 1/p + 1/q - 1 = (p + q - pq)/(pq)$.
Collecting the powers of $A$ and $B$ by a similar calculation (tracking the exponents through $s(\lambda)$) yields $d_{f*K}(\lambda) \le C_2\,\lambda^{-r}\,(AB)^r$, with $C_2$ depending only on $p$ and $q$.
[/proof]
**Step 4 (Conclusion).** The distribution function bound $d_{f*K}(\lambda) \le C_2\,(AB)^r\,\lambda^{-r}$ is equivalent to $\|f * K\|_{L^{r,\infty}} \le C_2^{1/r}\,AB = C_2^{1/r}\,\|f\|_{L^p}\,\|K\|_{L^{q,\infty}}$. This establishes the weak-type bound.
For the strong-type bound $\|f * K\|_{L^r} \le C'\|f\|_{L^p}\|K\|_{L^{q,\infty}}$: the argument above shows that for fixed $K \in L^{q,\infty}$, the linear operator $T_K: f \mapsto f * K$ is of weak type $(p, r)$ for every pair $(p, r)$ satisfying $1/p + 1/q = 1 + 1/r$ with $1 < p < q'$. The Marcinkiewicz interpolation theorem, applied at two such exponent pairs $(p_0, r_0)$ and $(p_1, r_1)$ with $p_0 < p < p_1$, yields the strong-type bound $\|T_K f\|_{L^r} \le C'\|f\|_{L^p}$ with $C' \le C(p,q)\|K\|_{L^{q,\infty}}$.
