In a Hilbert space of bounded integrable functions, let $P$ be an operator such that
$$P(f(x)) = \frac{f(x)+|f(x)|}{2}$$
The complement of $P$ can be written as $Q = I - P$, hence
$$Q(f(x)) = \frac{f(x)-|f(x)|}{2}$$
Both $P$ and $Q$ are idempotent. However, in contrary to many textbook examples, $P$ and $Q$ are not linear, nor are they Hermitian. Still
$$\langle P(f(x))|Q(f(x))\rangle = 0$$
holds. Are we allowed to call $P$ an orthogonal projection and $Q$ its orthogonal complement?
(Sorry for my inaccurate wording, I am a physicist, not a mathematician.)
After a lot of Googling, I am able to come up with an answer myself. There is a well-known definition of non-linear orthogonal projections. Roughly, the procedure is as follows:
To point it out explicitly, the properties of $P$ totally depend on the norm $||\cdot||$ and on the choice of $M$. As a simple example, let $M$ be the unit circle in the complex plain. Then $P(z)=z/|z|$.
If, as a special case, $M$ is a closed linear subspace of $H$, then $P$ turns out to be a linear orthogonal projection.
For more details, see e.g. http://matwbn.icm.edu.pl/ksiazki/apm/apm59/apm5911.pdf.