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It is well-known that Cotlar-Stein's Lemma can be used to prove the $L^2$ boundedness of the Hilbert transform. See e.g. $L^2$ boundedness of the Hilbert transform via Cotlar-Stein Lemma. Then using the Euler's formula $e^{ix}=\cos x+i\sin x$, we can easily show the uniform $L^2$ boundedness of the Dirichlet kernel: $||D_\lambda||_{L^2\to L^2}\le C$, where $$D_\lambda f(t)=\int \frac{\sin(\lambda(t-s))}{t-s}f(s)ds.$$

My question is: Can we prove the uniform $L^2$ boundedness of the Dirichlet kernel directly from the Cotlar-Stein Lemma, without using the $L^2$ boundedness of Hilbert transform? It suffices to show $$\int\Big| \int \frac{\sin(\lambda(t-s))\sin(\lambda s)}{(t-s)s}1_{2^{-j}\le |s|\le 2^{1-j}}\ 1_{2^{-k}\le |t-s|\le 2^{1-k}}\ ds\Big|dt\le C 2^{-|j-k|},$$ for $j,k\in \mathbb{Z}$. To get the bound on the RHS, it seems that one need to consider the cancellation in the integral with $ds$ and get a delicate estimate.

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    $\begingroup$ In both cases, it's much easier to work with the Fourier transforms; this makes these claims immediately obvious. In fact, this way you see that $H$ is unitary and $\|D_{\lambda}\|_{L^2\to L^2}$ is independent of $\lambda$. $\endgroup$
    – Christian Remling
    Commented Nov 21, 2016 at 17:20
  • $\begingroup$ @ChristianRemling Sure. But I am not asking for a different (or easier) proof for it. I want to know how to apply the Cotlar-Stein Lemma to this problem directly. $\endgroup$
    – Mr.right
    Commented Nov 21, 2016 at 19:29

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