# Examining the Hilbert transform of functions over the positive real line

$$\DeclareMathOperator\supp{supp}$$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.

• 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 $\mathbb{R}_{-}$. Mar 14 at 21:21
• No. $\{ f: \textrm{supp }f\subseteq [0,\infty) \}$ is a proper closed subspace of $L^2$ and $H$ is unitary. Mar 14 at 21:21
• thanks to both for the comments Mar 14 at 21:25
• @ChristianRemling the no is for the first question I underestand. Any idea for the rest? Mar 14 at 21:41

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.