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$\DeclareMathOperator\supp{supp}\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Bigabs[1]{\Bigl\lvert#1\Bigr\rvert}$Given a Schwartz function $f \in \mathcal{S}(\mathbb{R})$ with $\supp(f) \subseteq [-A,A]$, I want to show the following integral estimate when $\abs x \geq 2A$: $$\frac{1}{\pi}\Bigabs{\int_{-\infty}^\infty \frac{f(x-t)}{t+iy}dt} \leq \frac{C_f}{1+\abs x +\abs y}$$ for a constant $C_f>0$ that only depends on $f$.

This estimate is part of a longer proof when showing sharp constants for the $L^p$ boundedness of the Hilbert transform. The case $\abs x \leq 2A$ is shown, see: Grafakos - Best bounds for the Hilbert transform on $L^p(\mathbb{R}^1)$: A corrigendum.

However, for $\abs x \geq 2A$ it is not obvious to me why this is case. I am trying to see how to use the Schwartz property together with the support, but with no luck so far. Would be greatly appreciated if anyone can see the way to showing this integral estimate.

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$\DeclareMathOperator\supp{supp}\newcommand\abs[1]{\lvert#1\rvert}\newcommand\Bigabs[1]{\Bigl\lvert#1\Bigr\rvert}$

We have $$ \Bigabs{\int_{-\infty}^\infty \frac{f(x-t)}{t+iy}dt}= \Bigabs{\int_{x-A}^{x+A} \frac{f(x-t)}{t+iy}dt}=:I$$ When $t\in (x-A, x+A)$ and $|x|\geqslant 2A$ we get $3|t|\geqslant 3|x|-3A\geqslant |x|+A$ and $$ I \leqslant \sqrt{2}{\int_{x-A}^{x+A} \frac{|f(x-t)|}{|t|+|y|}dt}\leqslant \frac{\sqrt{2}\int |f|}{|x|/3+A/3+\abs y}\leqslant \frac{C_f}{1+|x|+|y|} $$ if $C_f\cdot A/3\geqslant \sqrt{2} \int|f|$ and $C_f/3\geqslant \sqrt{2} \int|f|$.

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