# Adjoint of a non-linear operator

I am a retired aerospace engineer, embarking on a self-study of QM. In reading S. Weinberg's book Lectures on QM (second ed.) I found the following definition (pag.65):

"The adjoint $$A^\dagger$$ of any operator (linear or not) is defined as the operator (if there is one) for which $$(\psi',A^\dagger\psi)=(A\psi',\psi)$$ for any two vectors $$\psi$$ and $$\psi'$$.

What puzzles me are the words "linear or not". All other sources that I have seen specify that $$A$$ is a linear operator before introducing the adjoint. To make things more confusing, Weinberg himself in speaking about symmetries (pag 76) says:

"For anti-unitary operators the adjoint is changed to $$(U^\dagger\phi,\psi)=(\phi,U\psi)^*$$, etc..".

Note the complex conjugation. Now, an anti-unitary operator is non-linear, alright, so he is consistent with his previous "linear or not", but why does he need to change the definition then? There must be something subtle that I am not able to grasp.

From the identity $$(Ax, y) = (x, A^\dagger y)$$ valid for every pair $$x,y$$ and assuming that the scalar product is linear in one entry and antilinear in the other one, you can easily prove that if $$A^\dagger$$ exists, then
(a) $$A^\dagger$$ is linear,
(b) $$A$$ is linear as well.
Indeed, concerning (a), assuming linearity in the right entry (the other case is analogous) $$(x, A^\dagger(ay+bz)) = (Ax, ay + bz) = a(Ax, y) + b(Ax, z) = (x, aA^\dagger y+ bA^\dagger z).$$ In summary $$(x, A^\dagger(ay+bz) - (aA^\dagger y+ bA^\dagger z))=0$$ for every vector $$x$$ in the Hilbert space. The only vector orthogonal to all other vectors is $$0$$ itself, so that, $$A^\dagger(ay+bz) = aA^\dagger y+ bA^\dagger z$$ namely, the adjoint is linear proving (a). With the same argument swapping the role of the two operators the initial identity entails that $$A$$ is also linear establishing (b).
Hence the former comment by Weinberg is actually a bit misleading, since $$A$$ must be linear if it admits adjoint satisfying the initial identity. If $$A$$ is antiliner, the initial identity is therefore untenable. A possible change to make the definition consistent is the one pointed out by Weinberg in his second comment. This assumption also agrees with the requirement that the antiunitary operators preserve the norm.