Are singular integral operators bounded on $L\log L$?

Question

My question is regarding singular integrals of Calderon Zygmund type. It is known that the maximal function is bounded on $L\log L \mapsto L^1$ (but not on $L^1$) and satisfies the same operator bounds as singular integral operators. Are singular integral operators therefore generally bounded from $L\log L \mapsto L^1$?

I know about the endpoint embedding $L^\infty\mapsto BMO$ but this question is of great interest to me.

Answer 1

The answer is no. Certain singular integrals can be used to characterize function spaces, such as the Hardy space $H^{1}(\mathbb{R}^{d})$. For example, the Hilbert transform is bounded $H^{1}(\mathbb{R})\rightarrow L^{1}(\mathbb{R})$ and in fact, this boundedness completely characterizes $H^{1}(\mathbb{R})$.

Theorem. A function $f\in L^{1}(\mathbb{R})$ belongs to the Hardy space $H^{1}(\mathbb{R})$ if and only if its Hilbert transform $H(f)$ is integrable.

For a proof of this result, see L. Grafakos, Modern Fourier Analysis Cor. 2.4.7.

Since functions in $H^{1}(\mathbb{R}^{d})$ necessarily have mean zero--this is a trivial consequence of the atomic decomposition--it suffices to exhibit a function $f\in L\log L(\mathbb{R})$ such that $\int f\neq 0$. I'll leave this to you.