# A density question for the Hilbert transform

Let $$\mathscr Hf$$ denote the Hilbert transform of a function $$f$$ defined on the real-line $$\mathbb R$$. Are the set of functions $$\{(f+\mathscr Hf)_{|_{(0,1)}}\,:\, f \in C^{\infty}(\mathbb R)\quad \text{and}\quad \textrm{supp} f \Subset (0,\infty)\}$$ dense in $$L^2((0,1))$$?

• Just curious what $\Subset$ means? (in comparison to $\subset$) – TheSimpliFire Oct 13 at 15:57
• It just means that its sitting inside some compact subset of $(0,\infty)$. – Ali Oct 13 at 19:51

Indeed, if $$g$$ is an $$L^2$$ function supported on $$[0,1]$$ such that $$g$$ is orthogonal to every $$f+\mathscr Hf$$ with $$f$$ compactly supported on $$(0,+\infty)$$, then $$g-\mathscr Hg=0$$ on $$(0,+\infty)$$. However, $$\mathscr H$$ is an isometry in $$L^2(\mathbb R)$$, so this would imply that $$\mathscr Hg=g$$ on $$(0,1)$$ and, hence, $$\mathscr Hg=0$$ a.e. outside $$[0,1]$$, i.e., that $$\mathscr Hg=g$$ in $$L^2(\mathbb R)$$, which is impossible unless $$g\equiv 0$$.
• @PietroMajer : $\mathcal H^* = - \mathcal H$ (it is $i \mathcal H$ that is self-adjoint). – Mikael de la Salle Oct 13 at 8:23