Is the Hilbert transform computable "losslessly"? The Hilbert transform is a procedure that can be used to perform a 90° phase shift on a signal.

Is the computation of (a discretized) Hilbert transform a "lossless" process? I.e. it merely phase shifts the signal, but does not alter it in any other way.

The reason for asking is that the following picture here looks like it introduces slight variation in the Hilbert transformed signal.


http://www.originlab.com/doc/Origin-Help/Hilbert-Transform
The loss you are seeing is due to the use of a finite length rectangular window.  If your signal is long enough that you can ignore the Hilbert filter transients at the both ends of the window (e.g. if it is infinite in length or so long that you don't care about the ends below some noise floor), then the process won't have the particular visible losses that your example points out.