Hilbert transform pair

Question

We know the Hilbert transform is defined as: $$ y(t)=\mathcal{H}(x(t))=\operatorname{P.V.}\ \left\{x(t)*\frac{1}{\pi t}\right\} $$ In the frequency domain this is equivalent to: $$ Y(f)=X(f)\times (-j\operatorname{sgn}(f)) $$ Using this twice, we get : $$ Z(f)=Y(f)(-j\operatorname{sgn}(f))=X(f)(-j\operatorname{sgn}(f))^2=-X(f)\operatorname{sgn}^2(f) $$ Many textbooks say $$\operatorname{sgn}^2(f)=1 $$ So $$Z(f)=-X(f) $$ and thus $$\hat{\hat{x}}(t)=-x(t)$$ But in my mind,$$ \operatorname{sgn}(f)|_{f=0}=0$$ and if$$X(0)\neq0,$$ $$Z(0)\neq-X(0)$$ So there is a single point where $$Z(f)\neq X(f).$$ Does this matter? Or Hilbert transform is defined in almost everywhere sense?

Answer 1

Any time you speak of integral operators, equality is almost always (hah) meant in the sense of almost everywhere. This makes sense since integrals don't see points; so as long as two functions only differ on some "small" set (of measure zero), we should expect to get the same output regardless.

As a note for future posts: sometimes in-line math is a good thing. Not every line of LaTeX code needs to be in an equation environment.