The strategy is to split the [convolution](/page/Convolution) kernel at a tunable scale $R \in (0, 1]$ into three spatial regions, and estimate each using a different aspect of $f$'s regularity: the cancellation condition and Hölder [continuity](/page/Continuity) for the inner ball $|y| < R$, the $L^\infty$ bound for the annulus $R \le |y| \le 1$, and the $L^2$ bound with the [Cauchy-Schwarz Inequality](/theorems/432) for the far field $|y| > 1$. The three estimates (Claims 1–3) are combined in Step 5, and $R$ is optimised (Claim 4) to balance the Hölder contribution $C[f]_{C^{0,\alpha}}R^\alpha$ against the logarithmic $L^\infty$ contribution $C\|f\|_{L^\infty}\log(1/R)$, producing the $\log^+$ in the final bound.
Write $K(y) := \Omega(y/|y|)/|y|^n$ for the kernel, so that $Tf(x) = \mathrm{p.v.}\int_{\mathbb{R}^n} K(y)\,f(x-y)\,d\mathcal{L}^n(y)$.
**Step 1: Decomposition.** For a parameter $R \in (0, 1]$ and each $x \in \mathbb{R}^n$, decompose the principal value integral as
\begin{align*}
Tf(x) &= \lim_{\varepsilon \to 0^+} \int_{\varepsilon < |y| < R} K(y)\,f(x-y)\,d\mathcal{L}^n(y) + \int_{R \le |y| \le 1} K(y)\,f(x-y)\,d\mathcal{L}^n(y) + \int_{|y| > 1} K(y)\,f(x-y)\,d\mathcal{L}^n(y).
\end{align*}
The second and third integrals are absolutely convergent (the kernel is bounded on $\{R \le |y| \le 1\}$ and square-[integrable](/page/Integral) on $\{|y| > 1\}$), so only the first requires a principal value [limit](/page/Limit). We will show this limit exists absolutely and then estimate each piece.
**Step 2: Inner estimate.**
[claim:Inner Estimate]
For every $R \in (0,1]$ and $x \in \mathbb{R}^n$,
\begin{align*}
\left|\lim_{\varepsilon \to 0^+} \int_{\varepsilon < |y| < R} K(y)\,f(x-y)\,d\mathcal{L}^n(y)\right| &\le \frac{\|\Omega\|_{L^\infty(\mathbb{S}^{n-1})}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})}{\alpha}\,[f]_{C^{0,\alpha}}\,R^\alpha.
\end{align*}
[/claim]
[proof]
The cancellation condition $\int_{\mathbb{S}^{n-1}} \Omega\,d\mathcal{H}^{n-1} = 0$ implies that $\int_{\varepsilon < |y| < R} K(y)\,d\mathcal{L}^n(y) = 0$ for every $0 < \varepsilon < R$, since in polar coordinates $y = r\theta$ with $r \in (\varepsilon, R)$ and $\theta \in \mathbb{S}^{n-1}$:
\begin{align*}
\int_{\varepsilon < |y| < R} K(y)\,d\mathcal{L}^n(y) &= \int_\varepsilon^R \int_{\mathbb{S}^{n-1}} \frac{\Omega(\theta)}{r^n}\,r^{n-1}\,d\mathcal{H}^{n-1}(\theta)\,d\mathcal{L}^1(r) = \left(\int_\varepsilon^R \frac{d\mathcal{L}^1(r)}{r}\right)\int_{\mathbb{S}^{n-1}} \Omega(\theta)\,d\mathcal{H}^{n-1}(\theta) = 0.
\end{align*}
Subtracting $f(x) \cdot 0$ from the integral allows the replacement $f(x-y) \mapsto f(x-y) - f(x)$:
\begin{align*}
\int_{\varepsilon < |y| < R} K(y)\,f(x-y)\,d\mathcal{L}^n(y) &= \int_{\varepsilon < |y| < R} K(y)\bigl[f(x-y) - f(x)\bigr]\,d\mathcal{L}^n(y).
\end{align*}
Since $f \in C^{0,\alpha}$, the difference satisfies $|f(x-y) - f(x)| \le [f]_{C^{0,\alpha}}\,|y|^\alpha$. The integrand is therefore bounded by $\|\Omega\|_{L^\infty}\,[f]_{C^{0,\alpha}}\,|y|^{\alpha - n}$, which is locally integrable near the origin (since $\alpha > 0$). In particular, the limit $\varepsilon \to 0^+$ is dominated by the absolutely convergent integral over $|y| < R$. Switching to polar coordinates:
\begin{align*}
\int_{|y| < R} |K(y)|\,|y|^\alpha\,d\mathcal{L}^n(y) &= \int_0^R \int_{\mathbb{S}^{n-1}} \frac{|\Omega(\theta)|}{r^n}\,r^\alpha\,r^{n-1}\,d\mathcal{H}^{n-1}(\theta)\,d\mathcal{L}^1(r) \\
&= \|\Omega\|_{L^\infty}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\int_0^R r^{\alpha - 1}\,d\mathcal{L}^1(r) \\
&= \|\Omega\|_{L^\infty}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\,\frac{R^\alpha}{\alpha}.
\end{align*}
Multiplying by $[f]_{C^{0,\alpha}}$ gives the claim.
[/proof]
**Step 3: Medium estimate.**
[claim:Medium Estimate]
For every $R \in (0,1]$ and $x \in \mathbb{R}^n$,
\begin{align*}
\left|\int_{R \le |y| \le 1} K(y)\,f(x-y)\,d\mathcal{L}^n(y)\right| &\le \|\Omega\|_{L^\infty(\mathbb{S}^{n-1})}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\,\|f\|_{L^\infty}\,\log\frac{1}{R}.
\end{align*}
[/claim]
[proof]
Bounding $|f(x-y)| \le \|f\|_{L^\infty}$ and $|\Omega(y/|y|)| \le \|\Omega\|_{L^\infty}$, then switching to polar coordinates:
\begin{align*}
\left|\int_{R \le |y| \le 1} K(y)\,f(x-y)\,d\mathcal{L}^n(y)\right| &\le \|\Omega\|_{L^\infty}\,\|f\|_{L^\infty}\int_{R \le |y| \le 1} |y|^{-n}\,d\mathcal{L}^n(y) \\
&= \|\Omega\|_{L^\infty}\,\|f\|_{L^\infty}\int_R^1 r^{-n}\,r^{n-1}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\,d\mathcal{L}^1(r) \\
&= \|\Omega\|_{L^\infty}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\,\|f\|_{L^\infty}\int_R^1 \frac{d\mathcal{L}^1(r)}{r} \\
&= \|\Omega\|_{L^\infty}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\,\|f\|_{L^\infty}\,\log\frac{1}{R}.
\end{align*}
[/proof]
**Step 4: Outer estimate.**
[claim:Outer Estimate]
For every $x \in \mathbb{R}^n$,
\begin{align*}
\left|\int_{|y| > 1} K(y)\,f(x-y)\,d\mathcal{L}^n(y)\right| &\le \|\Omega\|_{L^\infty(\mathbb{S}^{n-1})}\,\sqrt{\frac{\mathcal{H}^{n-1}(\mathbb{S}^{n-1})}{n}}\,\|f\|_{L^2}.
\end{align*}
[/claim]
[proof]
Define $h: \mathbb{R}^n \to \mathbb{R}$ by $h(y) := K(y)\,\mathbb{1}_{\{|y|>1\}}(y)$. The integral equals $\int_{\mathbb{R}^n} h(y)\,f(x-y)\,d\mathcal{L}^n(y)$, and since $g \mapsto \int h(y)\,g(x-y)\,d\mathcal{L}^n(y)$ acts by convolution with the reflected kernel $\check{h}$, the [Cauchy-Schwarz Inequality](/theorems/432) gives
\begin{align*}
\left|\int_{\mathbb{R}^n} h(y)\,f(x-y)\,d\mathcal{L}^n(y)\right| &\le \|h\|_{L^2}\,\|f\|_{L^2}.
\end{align*}
Computing $\|h\|_{L^2}^2$ in polar coordinates:
\begin{align*}
\|h\|_{L^2}^2 &= \int_{|y|>1} |K(y)|^2\,d\mathcal{L}^n(y) = \int_1^\infty \int_{\mathbb{S}^{n-1}} \frac{|\Omega(\theta)|^2}{r^{2n}}\,r^{n-1}\,d\mathcal{H}^{n-1}(\theta)\,d\mathcal{L}^1(r) \\
&\le \|\Omega\|_{L^\infty}^2\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})\int_1^\infty r^{-n-1}\,d\mathcal{L}^1(r) \\
&= \|\Omega\|_{L^\infty}^2\,\frac{\mathcal{H}^{n-1}(\mathbb{S}^{n-1})}{n},
\end{align*}
where the tail integral converges since $n \ge 1$: $\int_1^\infty r^{-n-1}\,d\mathcal{L}^1(r) = [-r^{-n}/n]_1^\infty = 1/n$. Taking the square root gives the claim.
[/proof]
**Step 5: Combination and optimisation.**
[claim:Optimisation Of The Scale Parameter]
Combining the Inner, Medium, and Outer Estimates and optimising over $R \in (0, 1]$ yields
\begin{align*}
\|Tf\|_{L^\infty} &\le C\Bigl(\|f\|_{L^2} + \|f\|_{L^\infty}\Bigl[1 + \log^+\frac{[f]_{C^{0,\alpha}}}{\|f\|_{L^\infty}}\Bigr]\Bigr),
\end{align*}
where $C = C(n, \alpha, \Omega) > 0$.
[/claim]
[proof]
Set $A := \|\Omega\|_{L^\infty}\,\mathcal{H}^{n-1}(\mathbb{S}^{n-1})$. For any $R \in (0, 1]$, the triangle inequality and Claims 1–3 give
\begin{align*}
|Tf(x)| &\le \frac{A}{\alpha}\,[f]_{C^{0,\alpha}}\,R^\alpha + A\,\|f\|_{L^\infty}\,\log\frac{1}{R} + C_1\|f\|_{L^2},
\end{align*}
where $C_1 := \|\Omega\|_{L^\infty}\sqrt{\mathcal{H}^{n-1}(\mathbb{S}^{n-1})/n}$. Two cases arise.
**Case 1:** $[f]_{C^{0,\alpha}} \le \|f\|_{L^\infty}$. Take $R = 1$. Then $\log(1/R) = 0$ and the inner term satisfies
\begin{align*}
\frac{A}{\alpha}\,[f]_{C^{0,\alpha}} \cdot 1 &\le \frac{A}{\alpha}\,\|f\|_{L^\infty},
\end{align*}
giving $|Tf(x)| \le (A/\alpha)\,\|f\|_{L^\infty} + C_1\|f\|_{L^2}$. Since $\log^+([f]_{C^{0,\alpha}}/\|f\|_{L^\infty}) = 0$ in this case, this is bounded by $C(\|f\|_{L^2} + \|f\|_{L^\infty})$.
**Case 2:** $[f]_{C^{0,\alpha}} > \|f\|_{L^\infty}$. Choose $R$ so that the inner and medium terms are comparable. Set
\begin{align*}
R &:= \left(\frac{\|f\|_{L^\infty}}{[f]_{C^{0,\alpha}}}\right)^{1/\alpha}.
\end{align*}
Since $[f]_{C^{0,\alpha}} > \|f\|_{L^\infty} > 0$, we have $R \in (0, 1)$, so $R \in (0, 1]$ as required. Substituting $R^\alpha = \|f\|_{L^\infty}/[f]_{C^{0,\alpha}}$ into the inner term:
\begin{align*}
\frac{A}{\alpha}\,[f]_{C^{0,\alpha}}\,R^\alpha &= \frac{A}{\alpha}\,[f]_{C^{0,\alpha}} \cdot \frac{\|f\|_{L^\infty}}{[f]_{C^{0,\alpha}}} = \frac{A}{\alpha}\,\|f\|_{L^\infty}.
\end{align*}
For the medium term, $\log(1/R) = \frac{1}{\alpha}\log(1/R^\alpha) = \frac{1}{\alpha}\log([f]_{C^{0,\alpha}}/\|f\|_{L^\infty})$. Since $[f]_{C^{0,\alpha}} > \|f\|_{L^\infty}$, this logarithm is positive, so $\log([f]_{C^{0,\alpha}}/\|f\|_{L^\infty}) = \log^+([f]_{C^{0,\alpha}}/\|f\|_{L^\infty})$. The medium term becomes
\begin{align*}
A\,\|f\|_{L^\infty}\,\log\frac{1}{R} &= \frac{A}{\alpha}\,\|f\|_{L^\infty}\,\log^+\frac{[f]_{C^{0,\alpha}}}{\|f\|_{L^\infty}}.
\end{align*}
Combining:
\begin{align*}
|Tf(x)| &\le \frac{A}{\alpha}\,\|f\|_{L^\infty} + \frac{A}{\alpha}\,\|f\|_{L^\infty}\,\log^+\frac{[f]_{C^{0,\alpha}}}{\|f\|_{L^\infty}} + C_1\|f\|_{L^2} \\
&= \frac{A}{\alpha}\,\|f\|_{L^\infty}\Bigl[1 + \log^+\frac{[f]_{C^{0,\alpha}}}{\|f\|_{L^\infty}}\Bigr] + C_1\|f\|_{L^2}.
\end{align*}
In both cases, $|Tf(x)| \le C(\|f\|_{L^2} + \|f\|_{L^\infty}[1 + \log^+([f]_{C^{0,\alpha}}/\|f\|_{L^\infty})])$ with $C = \max\{A/\alpha, C_1\}$. Since the bound is uniform in $x$, taking the supremum gives the claimed $L^\infty$ estimate.
[/proof]
