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$\DeclareMathOperator\supp{supp}$Let $H:L^{2}(\mathbb{R})\to L^{2}(\mathbb{R})$ be the Hilbert transform. Let suppose we have a compaclty supported function $f \in L^{2}(\mathbb{R})$ such that $\supp(f)\subseteq \mathbb{R}_{+}$, the support of $H(f)$ is probably no longer in $\mathbb{R}_{+},$ on the other hand if we have an arbitrary $g \in L^{2}(\mathbb{R})$ is well know that $g=H(-Hg)$, that is $g$ is the Hilbert transform of someone.

Again, given an arbitrary $g \in L^{2}(\mathbb{R})$, is it possible to find a $f \in L^{2}(\mathbb{R})$ such that $\supp(f)\subseteq \mathbb{R}_{+}$ and $Hf$ is near $g$ ? Or putit in another form is the set $\{H(f)/ f \in L^{2}(\mathbb{R}) \wedge \supp(f)\subseteq \mathbb{R}_{+}\}$ dense in $L^{2}(\mathbb{R})$?

Another related question, given $g \in L^{2}(\mathbb{R})$ such that $\supp(g)\subseteq \mathbb{R}_{-},$ is it possible to find a $f \in L^{2}(\mathbb{R})$ such that $\supp(f)\subseteq \mathbb{R}_{+}$ and $\chi_{\mathbb{R}_{-}}(x)Hf(x)=g(x)$? In this last question I want $Hf$ equals $g$ only over the negatives, so any $f \in L^{2}(\mathbb{R})$ such that $\supp(f)\subseteq \mathbb{R}_{+}$ who satisfies $Hf=\hat{g}$ (where $\hat{g}$ and $g$ coincide over the negatives) will be useful.

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    $\begingroup$ For the first question of course not. If $Hf$ is close to $g$, then (since $H$ is continuous) $f = (-H)Hf$ is close to $-Hg$, so we can approximate $g$ (and indeed be equal to it) if and only if $-Hg$ is $0$ on $\mathbb{R}_{-}$. $\endgroup$ Mar 14 at 21:21
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    $\begingroup$ No. $\{ f: \textrm{supp }f\subseteq [0,\infty) \}$ is a proper closed subspace of $L^2$ and $H$ is unitary. $\endgroup$ Mar 14 at 21:21
  • $\begingroup$ thanks to both for the comments $\endgroup$ Mar 14 at 21:25
  • $\begingroup$ @ChristianRemling the no is for the first question I underestand. Any idea for the rest? $\endgroup$ Mar 14 at 21:41

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The answer to the second question is negative as well. Take for example $g$ supported in $(-\infty,-1)$ and discontinuous in some point. If $f$ is supported in $\mathbb{R}_+$ and $y,z<-1$ it holds that $$|Hf(y)-Hf(z)| \leq |z-y|\int_{\mathbb{R}}\frac{|f(x)|}{|x-y||x-z|}dx|\leq |z-y| \Big( \int_{\mathbb{R}}|f(x)|^2 \Big)^\frac12 \Big( \int_\mathbb{R}\frac{1}{|x-z|^2}dx \Big)^\frac12 \leq M|y-z|. $$So, the Hilbert transform must be continuous in $(-\infty,-1)$ and so it cannon coincide with $g$ there.

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