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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.

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Could you explain the third condition? E.g. when $K=1/(x-y)$ one gets $t \cdot K(t,x-t) = t/(2t-x) $... –  Piero D'Ancona Apr 29 '11 at 21:53
@:Piero. That's right. And $lim_{t\to\infty}t/(2t-x)$ is not zero. But if you have something like $K=1/(x-y)^\alpha$ with $\alpha>1$, then you get $lim_{t\to\infty}t/(2t-x)^\alpha=0$. –  André Henriques Apr 30 '11 at 2:23
Well, if K=K(x-y) then the operator is translation invariant, then it is a constant coefficient pseudodifferential operator with a symbol $a(\xi)$, and by your second condition $a(\xi)$ must take only the values $\pm i$. Thus you have jump singularities in the symbol which should always produce a decay of order $\sim t^{-1}$. I guess. –  Piero D'Ancona Apr 30 '11 at 13:54
@Piero: I do not require translation invariance. But I agree with you that it looks like it's not possible to achieve my decay condition. –  André Henriques Apr 30 '11 at 14:31
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2 Answers

EDIT: This solution does not satisfy the third condition, which rules out the Hilbert transform itself. So, this is an answer to different question. I do not delete it in hope it may be useful for someone.

Let $\phi(x)$ be a smooth monotone function such that $x-\phi(x)$ has compact support. This is a diffeomorphism of the real line and the pullback $\phi^*$ is a linear operator acting on $L^2(\mathbb R)$. The singular integral operator $$(\phi^*)^{-1}{\mathcal H}\phi^*$$ has all the properties you need.

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It is possible also to reason thus. Too fast decay could lead to compactness of the operator if not in $L_2$ then in some suitably defined spaces. And this cannot be since its degree is identity and the last operator is compact only in finite dimension spaces.

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