Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:27:05Z http://mathoverflow.net/feeds/question/52995 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52995/complex-structure-on-l2-mathbb-r-generalizing-the-hilbert-transform Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform. André Henriques 2011-01-23T21:19:13Z 2011-04-30T21:58:53Z <p>The <a href="http://en.wikipedia.org/wiki/Hilbert_transform" rel="nofollow">Hilbert transform</a> on the real Hilbert space $L^2(\mathbb R)$ is the singular integral operator $$\mathcal H(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty \frac{1}{x-y} f(y) dy.$$</p> <p>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.</p> <hr> <p>I am wondering if there are other operators $\tilde {\mathcal H}:L^2(\mathbb R)\to L^2(\mathbb R)$ with similar properties.</p> <blockquote> <p><b>Question:</b> Does there exists a function $K:\mathbb R^2\to \mathbb R$ with the following properties:</p> <ul> <li><p>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").</p></li> <li><p>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 $L^2(\mathbb R)$.</p></li> <li><p>The function $K$ goes to zero faster than $\frac{1}{x-y}$ along the antidiagonals.<br> Namely, it satisfies $$\forall x\in \mathbb R,\qquad\qquad \lim_{t\to \infty}\;\;\;\; t\cdot K(t,x-t) = 0.$$</p></li> </ul> </blockquote> <hr> <p><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.</p> <p>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.</p> <hr> <p>Right now, I actually believe that such an integral kernel does not exist, but this is purely a gut feeling...<br> 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.</p> http://mathoverflow.net/questions/52995/complex-structure-on-l2-mathbb-r-generalizing-the-hilbert-transform/63501#63501 Answer by Alex Gavrilov for Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform. Alex Gavrilov 2011-04-30T06:26:45Z 2011-04-30T07:32:27Z <p>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.</p> <p>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 <code>$\phi^*$</code> is a linear operator acting on $L^2(\mathbb R)$. The singular integral operator <code>$$(\phi^*)^{-1}{\mathcal H}\phi^*$$</code> has all the properties you need.</p> http://mathoverflow.net/questions/52995/complex-structure-on-l2-mathbb-r-generalizing-the-hilbert-transform/63562#63562 Answer by Andrew for Complex structure on $L^2(\mathbb R)$ generalizing the Hilbert transform. Andrew 2011-04-30T21:58:53Z 2011-04-30T21:58:53Z <p>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.</p>