Proving Cotlar's Identity of Hilbert transform I am stuck with the following exercise of my textbook:

Cotlar's Identity: For $f\in \mathcal{S}(\mathbb{R})$, we have $(Hf)^2 = f^2 + 2H(f\cdot Hf)$, where $H$ is the Hilbert transform.

The exercise is accompanied by the following hint:

Since $\widehat{Hf}(\xi) = -i\:\text{sgn}(\xi)\hat{f}(\xi)$. Directly compute $$\widehat{Hf}*\widehat{Hf} =\hat{f}*\hat{f} - 2i \: \text{sgn}(\xi)(\hat{f}*\widehat{Hf})$$

But I don't even understand why it's enough to prove the above expression involving convolutions. 
 A: The reason it's enough to prove the statement about convolution is that Fourier transform sends products to convolutions. So, the transform of $f^2$ is $\hat f*\hat f$, the transform of $f\, Hf$ is $\hat f*\widehat{Hf}$, and so on. If two expressions have the same Fourier transform, they are equal. 
To prove the claimed identity, let's write it in symmetric form
$$
\widehat{Hf}*\widehat{Hf} =\hat{f}*\hat{f} - i \, \operatorname{sgn}(\xi)(\hat{f}*\widehat{Hf}) - i \, \operatorname{sgn}(\xi)(\widehat{Hf}*\hat f)
\tag1 $$
By the definition of convolution, 
$$
(\hat{f}*\hat{f})(\xi) = \int_{\mathbb{R}} \hat f(\xi-t)\hat f(t)\,dt
$$
Similarly,
$$
(\widehat{Hf}*\widehat{Hf})(\xi) = \int_{\mathbb{R}} \widehat{Hf}(\xi-t)\widehat{Hf}(t)\,dt = - \int_{\mathbb{R}} \operatorname{sgn}(\xi-t) \operatorname{sgn}(t) \,\hat f(\xi-t)\hat f(t)\,dt
$$
and also 
$$
i\operatorname{sgn} (\xi) (\hat f*\widehat{Hf})(\xi) =  \int_{\mathbb{R}} \operatorname{sgn} (\xi) \operatorname{sgn}(t) \,\hat f(\xi-t)\hat f(t)\,dt
$$
$$
i\operatorname{sgn} (\xi) (\widehat{Hf}*\hat f)(\xi) =  \int_{\mathbb{R}} \operatorname{sgn} (\xi) \operatorname{sgn}(\xi-t) \,\hat f(\xi-t)\hat f(t)\,dt
$$
Comparing these four integrals, we see that (1) boils down to  the algebraic identity
$$
-\operatorname{sgn}(\xi-t) \operatorname{sgn}(t) = 1 - \operatorname{sgn} (\xi) \operatorname{sgn}(t) - \operatorname{sgn} (\xi) \operatorname{sgn}(\xi-t)
\tag2 $$
To verify this, rearrange it as 
$$
(\operatorname{sgn} (\xi)-\operatorname{sgn}(t))
 \operatorname{sgn}(\xi-t)   = 1 - \operatorname{sgn} (\xi) \operatorname{sgn}(t)  
\tag3 $$
If $\xi$ and $t$ have the same sign, both sides of (3) are equal to $0$. If they have opposite signs, both sides of (3) are equal to $2$.

Aside: a much shorter proof comes from complex analysis, according to which $f$ and $Hf$ are  the real and imaginary parts of the boundary values of a holomorphic function $u+iv$ in the upper half-plane. Since $(u+iv)^2 = (u^2-v^2) + 2iuv$ is also holomorphic, the Hilbert transform of $f^2-(Hf)^2$ is $2f Hf$. Hence, $H(2fHf) = (Hf)^2 - f^2$.
