The Hilbert transform 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.

I am wondering if there are other operators $\tilde {\mathcal H}:L^2(\mathbb R)\to L^2(\mathbb R)$ with similar properties.

Question: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 $x=y$

(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.

Namely, it satisfies $$ \forall x\in \mathbb R,\qquad\qquad \lim_{t\to \infty}\;\;\;\; t\cdot K(t,x-t) = 0. $$

**Variant:** 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.

Right now, I actually believe that such an integral kernel does not exist,
but this is purely a gut feeling...

If someone has any ideas about how to prove the non-existence of $\tilde {\mathcal H}:L^2(\mathbb R)\to L^2(\mathbb R)$ with the above properties, then I would very interested to hear them.