Quadratically modulated Bilinear Hilbert transform

Here is a simple but surprising fact.

Recall that the Hilbert transform Hf(x) := p.v. \int f(x-t) \frac{dt}{t} is L^p \to L^p bounded for all {1 < p < \infty} (and even L^1 \to L^{1,\infty} bounded, of course). The quadratically modulated Hilbert transform is the operator

\displaystyle H_q f(x) := p.v. \int \int f(x-t) e^{-i t^2} \frac{dt}{t};

this operator is also known to be L^p \to L^p bounded for all {1 < p < \infty} , but the proof is no corollary of that for H , it's a different beast requiring oscillatory integral techniques and almost orthogonality and is due to Ricci and Stein (interestingly though, H_q is also L^1 \to L^{1,\infty} bounded, and this can indeed be obtained by a clever adaptation of Calderón-Zygmund theory due to Chanillo and Christ).

The bilinear Hilbert transform instead is the operator

\displaystyle BHT(f,g)(x) := p.v. \int \int f(x-t)g(x+t)\frac{dt}{t}

and it is known, thanks to foundational work of Lacey and Thiele, to be L^p \times L^q \to L^r bounded at least in the range given by p,q>1, r>2/3 with the exponents satisfying the Hölder condition 1/p + 1/q = 1/r (this condition is strictly necessary to have boundedness, due to the scaling invariance of the operator). This operator has an interesting modulation invariance (corresponding to the fact that its bilinear multiplier is \mathrm{sgn}(\xi - \eta), which is invariant with respect to translations along the diagonal): indeed, if \mathrm{Mod}_{\theta} denotes the modulation operator \mathrm{Mod}_{\theta} f(x) := e^{- i \theta x} f(x), we have

\displaystyle BHT(\mathrm{Mod}_{\theta}f,\mathrm{Mod}_{\theta}g) = \mathrm{Mod}_{2 \theta} BHT(f,g);

it is this fact that suggests one should use time-frequency analysis to deal with this operator.
Now, analogously to the linear case, one can consider the quadratically modulated bilinear Hilbert transform, given simply by

\displaystyle BHT_q(f,g)(x) := p.v. \int \int f(x-t)g(x+t) e^{-i t^2} \frac{dt}{t}.

One might be tempted to think, by analogy, that this operator is harder to bound than BHT – at least, I would naively think so at first sight. However, due to the particular structure of the bilinear Hilbert transform, the boundedness of BHT_q is a trivial corollary of that of BHT ! Indeed, this is due to the trivial polynomial identity

\displaystyle (x+t)^2 + (x-t)^2 = 2x^2 + 2t^2;

thus if \mathrm{QMod}_{\theta} denotes the quadratic modulation operator \mathrm{QMod}_{\theta}f(x) = e^{-i \theta x^2} f(x) we have

\displaystyle \begin{aligned} BHT_q(f,g)(x) = & \int f(x-t)g(x+t) e^{-it^2} \frac{dt}{t} \\ = & \int f(x-t)g(x+t) e^{ix^2}e^{-i(x+t)^2/2}e^{-i(x-t)^2/2} \frac{dt}{t} \\ = & e^{ix^2}\int e^{-i(x-t)^2/2}f(x-t)e^{-i(x+t)^2/2}g(x+t)  \frac{dt}{t} \\ = & \big[ \mathrm{QMod}_{-1} BHT( \mathrm{QMod}_{1/2} f, \mathrm{QMod}_{1/2} g )\big](x). \end{aligned}

Of course this trick is limited to quadratic modulations, so for example already the cubic modulation of BHT

is non-trivial to bound (but the boundedness of the cubic modulation of the trilinear Hilbert transform would again be a trivial consequence of the boundedness of the trilinear Hilbert transform itself… too bad we don’t know if it is bounded at all!). Polynomial modulations of bilinear singular integrals (thus a bilinear analogue of the Ricci-Stein work) have been shown to be bounded by Christ, Li, Tao and Thiele in “On multilinear oscillatory integrals, nonsingular and singular“.

UPDATE: Interesting synchronicity just happened: today Dong, Maldague and Villano have uploaded on ArXiv their paper “Special cases of power decay in multilinear oscillatory integrals” in which they extend the work of Christ, Li, Tao and Thiele to some special cases that were left out. Maybe I should check my email for the arXiv digest before posting next time.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s