Given the operator known as Hilbert transform, that is $Hf(x)= \left(p.v.\left(\frac{1}{x-y}\right), f(y) \right)$, it can be shown that it is bounded from $L^1$ to $L^1 _w$ and from $L^2$ to $L^2$. Now, this is an example of what is called a "Calderon-Zygmund operator". Basically the same proof used for Hilbert transform works for every operator of this kind, but the thesis is that a Calderon-Zygmund operator is bounded from $L^p$ to $L^p$ with $p \in (1,\infty)$.
The problem is that the proof only provides boundedness for $p\in (1,2]$, and doesn't say anything if $p \in [2, \infty)$. So I think the only way this could make sense is to define the Hilbert transform on $L^p$ with $p >2$ as the adjoint of the Hilbert transform from $L^q$ to $L^q$ with $q \in (1,2]$.
The question is: is the adjoint of the Hilbert transform still itself? I mean, in general I don't know what the adjoint of an operator is, it is kinda abstract, but in this case does it preserve its structure? can I still describe it the same way I've done at the beginning?
And what happens with a generic Calderon-Zygmund operator? Can its adjoint still be represented with the same convolution?
A rough reasoning is the following:
The Hilbert transform is given by $K\ast f$, where $K$ is a distribution kernel, here we assume that $f$ is a Schwartz function, the kernel satisfies the Calderon-Zygmund conditions or the like.
And you will see that for Schwartz functions $f,g$, we have \begin{align*} \left<K\ast f,g\right>&=\int K\ast f(x)g(x)dx\\ &=\int\widehat{K\ast f}(\xi)g^{\vee}(\xi)d\xi\\ &=\int\widehat{K}(\xi)\widehat{f}(\xi)g^{\vee}(\xi)d\xi\\ &=\int\widehat{f}(\xi)(K')^{\vee}(\xi)g^{\vee}(\xi)d\xi\\ &=\int \widehat{f}(\xi)(K'\ast g)^{\vee}(\xi)d\xi\\ &=\int f(x)K'\ast g(x)dx\\ &=\left<f,K'\ast g\right> \end{align*} where we have used Plancherel Theorem, and $K'(x)=K(-x)$, the kernel $K'$ still satisfies the Caleron-Zygmund conditions, so we are in the position to use the boundedness just proved for the case that $p\in(1,2]$.