$L^2$ boundedness of the Hilbert transform via Cotlar-Stein Lemma Can anyone outline Cotlar's original proof of the $L^2$ boundedness of the Hilbert transform.  I cannot locate the original paper on the web.  I know the Cotlar-Stein lemma but I don't see how to make the estimates needed for the conclusion of the lemma to hold.  A reference would also be helpful if I can access it.
 A: (Sketch) Recall that the Hilbert transform is given (in a principal value sense) by the convolution with the function 1/t. That is the Hilbert transform of $f$ is defined (up to normalization) by
$$Tf(x) = \int_{-\infty}^{\infty} \frac{f(x-t)}{t} dt .$$ 
To apply the Coltar-Stein lemma we wish to decompose T as the infinite sum of "almost orthogonal" operators:
$$Tf = \sum_{i \in \mathbb{Z}} T_i f.$$
The trick is to let each $T_i$ denote the contribution of the kernel $\frac{1}{t}$ at a given dyadic frequency scale. More precisely, $T_i$ should correspond to convolution with the $i$-th component of a Littlewood-Paley decomposition of the function 1/t.
A: Since the paper is hard to find on the web, I sketch the original proof below.
Let $C = \{x \in \mathbb R : 1 \leq |x| \leq 2\}$.
For $j \in \mathbb Z$, let $k_j(x) =1_{C}(2^{-j}x) \dfrac{1}{x}$ and $T_jf(x) = (f*k_j)(x)$.
The kernel $k_j$ satisfies the following properties :

*

*$k_j(-x) = -k_j(x)$

*$k_j(x) = 2^{-j} k_0(2^{-j}x)$

*$\int_{\mathbb R} k_j(x)dx = 0$

*$||k_j||_{L^1} = ||k_0||_{L^1}$

*$\int_{\mathbb R} |k_j(x-t) - k_j(x)|dx \leq M2^{-j}|t|$

*$||k_i * k_j||_{L^1} \leq M2^{-|i-j|}$
for some constant $M > 0$ independent of $i,j,t$.
By Young Convolution Inequality, $T_j : L^2(\mathbb R) \rightarrow L^2(\mathbb R)$, $j \in \mathbb Z$, is a family of bounded operators such that

*

*$||T_j|| \leq ||k_0||_{L^1}$

*$||T_i^* T_j||, ||T_i T_j^*|| \leq ||k_i * k_j||_{L^1} \leq M2^{-|i-j|}$
Now, apply Cotlar-Stein Lemma to conclude that $\sum_{i = -n}^n T_j$ converges in operator norm to some bounded $T : L^2(\mathbb R) \rightarrow L^2(\mathbb R)$ and observe that $T$ must coincide with the Hilbert Transform.
