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 $L^2((0,1))$?



A density question for the Hilbert transform
Yes, it is dense.


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 $g\equiv 0$.


