Let $\mathscr Hf$ denote the Hilbert transform of a function $f$ defined on the realline $\mathbb R$. 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 $L^2((0,1))$?
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$\begingroup$ Just curious what $\Subset$ means? (in comparison to $\subset$) $\endgroup$– TheSimpliFireOct 13, 2020 at 15:57

1$\begingroup$ It just means that its sitting inside some compact subset of $(0,\infty)$. $\endgroup$– AliOct 13, 2020 at 19:51
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1 Answer
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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 $g\equiv 0$.

2$\begingroup$ @PietroMajer : $\mathcal H^* =  \mathcal H$ (it is $i \mathcal H$ that is selfadjoint). $\endgroup$ Oct 13, 2020 at 8:23
