Editing Hilbert transform (section)

==Domain of definition==
It is by no means obvious that the Hilbert transform is well-defined at all, as the improper integral defining it must converge in a suitable sense. However, the Hilbert transform is well-defined for a broad class of functions, namely those in <math>L^p(\mathbb{R})</math> for {{math|1 < ''p'' < ∞}}.

More precisely, if {{mvar|u}} is in <math>L^p(\mathbb{R})</math> for {{math|1 < ''p'' < ∞}}, then the limit defining the improper integral

<math display="block">\operatorname{H}(u)(t) = \frac{2}{\pi} \lim_{\varepsilon \to 0} \int_\varepsilon^\infty \frac{u(t - \tau) - u(t + \tau)}{2\tau}\,d\tau</math>

exists for [[almost every]] {{mvar|t}}. The limit function is also in <math>L^p(\mathbb{R})</math> and is in fact the limit in the mean of the improper integral as well. That is,

<math display="block">\frac{2}{\pi} \int_\varepsilon^\infty \frac{u(t - \tau) - u(t + \tau)}{2\tau}\,\mathrm{d}\tau \to \operatorname{H}(u)(t)</math>

as {{math|''ε'' → 0}} in the {{mvar|L<sup>p</sup>}} norm, as well as pointwise almost everywhere, by the [[#Titchmarsh.27s theorem|Titchmarsh theorem]].{{sfn|Titchmarsh|1948|loc=Chapter 5}}

In the case {{math|1=''p'' = 1}}, the Hilbert transform still converges pointwise almost everywhere, but may itself fail to be integrable, even locally.{{sfn|Titchmarsh|1948|loc=§5.14}} In particular, convergence in the mean does not in general happen in this case. The Hilbert transform of an {{math|''L''<sup>1</sup>}} function does converge, however, in {{math|''L''<sup>1</sup>}}-weak, and the Hilbert transform is a bounded operator from {{math|''L''<sup>1</sup>}} to {{math|''L''<sup>1,w</sup>}}.{{sfn|Stein|Weiss|1971|loc=Lemma V.2.8}} (In particular, since the Hilbert transform is also a multiplier operator on {{math|''L''<sup>2</sup>}}, [[Marcinkiewicz interpolation]] and a duality argument furnishes an alternative proof that {{mvar|H}} is bounded on {{math|''L''<sup>''p''</sup>}}.)