I'm trying to verify the following estimate, which appears in a paper I'm reading. It seems I'm missing something easy, I just can't figure this out.
$\textbf{Background}:$
For a function $f \in \mathcal{S}(\mathbb{R}^2)$ (Schwarz class), define the (sometimes called double Riesz) transform:
$$R_{11}(f)(x) = \int_{\mathbb{R}^2} \frac{\xi_1^2}{\|\xi\|^2} \hat{f}(\xi)~ e^{i\langle \xi, x\rangle} d\xi.$$
Define also the frequency projection operators $P_N$ by
$$\widehat{(P_Nf)}(\xi) = [\varphi(\xi/N) - \varphi(\xi/2N)]\hat{f}(\xi),$$
where $\varphi$ is a smooth function supported inside $B_2(0)$ and $\varphi \equiv 1$ on $B_1(0)$.
$\textbf{Estimate I want}:$
$$\|R_{11}(f)\|_{\infty} \leq \mathrm{const}. \sum_N \|P_Nf\|_{\infty},$$
where the above sum is understood to be over dyadic integers.
$\textbf{What I tried}:$
It seems that the double Riesz transform is not necessarily bounded on $L^{\infty}$, which is why this is a little tricky. Looking at the integral that defines $R_{11}(f)(x)$, we can write
$$R_{11}(f)(x) = \sum_N \int_{\mathbb{R}^2} \frac{\xi_1^2}{\|\xi\|^2} \widehat{P_Nf}(\xi)~ e^{i\langle \xi, x\rangle} d\xi.$$
Now, if we could just use the inequality $\xi_1^2/\|\xi\|^2 \leq 1$ directly above, the desired estimate would follow. However, this is nonsense, since the above integral is complex-valued. Taking absolute values destroys the term $e^{i\langle \xi, x\rangle}$, so I don't know what else to do.
Any help is really appreciated!