Hilbert transform pair 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?
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.
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.