The [Hilbert transform][1] on the real Hilbert space $L^2(\mathbb R)$ is the singular integral operator
$$
\mathcal H(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty \frac{1}{x-y} f(y) dy.
$$

It satisfies $\mathcal H^2=-Id_{L^2(\mathbb R)}$, and in that sense, it is a complex structure on the Hilbert space $L^2(\mathbb R)$ of real-valued, square integrable functions on the real line.
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I am wondering if there are other operators $\tilde {\mathcal H}:L^2(\mathbb R)\to L^2(\mathbb R)$ with similar properties.

> <b>Question:</b> Does there exists a function $K:\mathbb R^2\to \mathbb R$ with the following properties:
>
> - The function $K(x,y)$ looks like $\frac{1}{x-y}$ in a neighborhood of the diagonal
$x=y$<br> (here, by "looks like", I mean for instance as "$K(x,y) = \frac{1}{x-y} +$ smooth function").
>
> - The singular integral operator
> $$
> \tilde {\mathcal H}(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty K(x,y) f(y) dy.
> $$
> satisfies $\tilde {\mathcal H}^2=-Id_{L^2(\mathbb R)}$, and thus defines a complex structure on $L^2(\mathbb R)$.
> 
> - The function $K$ goes to zero faster than $\frac{1}{x-y}$ along the antidiagonals.<br>
> Namely, it satisfies
> $$
> \forall x\in \mathbb R,\qquad\qquad \lim_{t\to \infty}\;\;\;\; t\cdot K(t,x-t) = 0.
> $$

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<b>Variant:</b> In case it turns out difficult to produce an example of an operator $\tilde {\mathcal H}:L^2(\mathbb R)\to L^2(\mathbb R)$ as above, I would be happy to replace $L^2(\mathbb R)$ by $L^2(\mathbb R;\mathbb R^n)$, the Hilbert space of $\mathbb R^n$-valued $L^2$ functions on the real line.

In that case, I would be looking for an integral kernel
$$
K:\mathbb R^2\to \mathit{Mat}_{n\times n}(\mathbb R)
$$
with all the properties listed above.

  [1]: http://en.wikipedia.org/wiki/Hilbert_transform