Tue, 10/18/2011 - 06:54 — wick

We next turn to the study of singular integral operators. The canonical example to keep in mind is the Hilbert transform, which is given by

$$

Hf(x)=p.v.\,\int_{R}\frac{f(y)}{x-y}dy.

$$

Using the Fourier Transform it is easy to see that the Hilbert transform is bounded on $L^2(R)$. Then using connections to complex analysis it follows relatively easily that it is in fact bounded on all $L^p(R)$.

In this lecture we will study certain generalizations of the Hilbert transform. The kernel associated to the Hilbert transform, $\frac{1}{x}$, is just barely non square integrable on $R$, but instead possesses certain size and smoothness properties that are suitable replacements. We will consider relatively general convolution-type singular integral operators and show that the general behavior as indicated by the Hilbert transform in fact persists. This lecture will introduce fundamental tools in harmonic analysis: interpolation theorems and Calderon-Zygmund Decompositions.

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## Sorry, this was a little

Sorry, this was a little confusing.

If we assume that the kernel K satisfies the conditions in Theorem 1.1, namely size, smoothness and cancellation, then we can show that the truncations of the kernel have uniformly bounded Fourier transform, independent of the truncation $\epsilon$. The idea is carried out now in Lecture 4 where we study a simple case of multi-parameter singular integral operators.

Next, if we assume that the kernel $K$ satisfies the conditions of Theorem 1.1, then it is possible to show that $K_\epsilon$ also satisfies the same estimates as the original kernel $K$, except we may have a slightly worse constant depending on the dimension. To see this, one should first consider the kernel $K_1$ and show that one has the same estimates as the original kernel. Then one should recognize that the kernels in question have a certain dilation invariance that allows one to reduce the case of general $K_\epsilon$.

Here, unfortunately, I had a reference to the wrong theorem, and I think this is where the confusion may have arisen. I have corrected the lecture, and posted an update. Please let me know if this doesn't clarify things for you.

Regards,

Brett

## cut-off kernel

Hi Brett,

Can you please explain what one needs to prove about the cut-off kernel K_epsilon? So initially K satisfies the condition from Theorem 1.1, but I don't understand exactly what we have to show. And then you say that K_epsilon satisfies the kernel conditions from Theorem 1.2. For this one, you assume that the initial kernel K satisfies the conditions from Theorem 1.1 or 1.2?

Thank you,

Andrei