If the operator A has only real eigenvalues and Y contains all eigenfunctions associated with positive eigenvalues, and Z contains all eigenfunctions associated with negative eigenvalues, then yes, the operator A can not have eigenfunctions with a non-trivial projection onto Y and a non-trivial projection onto Z.

Because every eigensubspace associated with a positive eigenvalue is included in subspace Y, every eigensubspace associated with a negative eigenvalue is included in subspace Z and other eigensubspaces do not exist.

Unfortunately, I don't know that.

Let a,b reals, a>0 and b0.

In this case the sum Y+Z is clearly a direct sum and Y, Z are orthogonal subspaces of X. Moreover, Y is the subspace generated by the union of all eigensubspaces corresponding to postive eigenvalues, Z is the subspace generated by the union of all eigensubspaces corresponding to negative eigenvalues. Easy to prove, because the linear operator A being self adjoint, is continuous.

Correct, but I don't know whether Y is generated by the union of all eigenspaces associated with positive eigenvalues (nor do I know the analogous for Z and negative eigenvalues). Thanks anyway!

If we suppose A self-adjoint, then to every two different eigenvalues, the corresponding eigensubspaces are orthogonal, and due to the continuity of A, each eigensubspace is closed.

Now we can take Y as the subspace of X generated by the union of eigensubspaces corresponding to positive eigenvalues, and analogous Z for negative eigenvalues.

As I remember, Y is identical to the subspace of X generated by the direct sum of all eigensubspaces corresponding to positive eigenvalues and analogous for Z.

Y and Z are orthogonal, thus X is the direct sum Y+Z.

I'm not sure, but I think, due to the Projection Theorem( which ensures uniqueness of decomposition) in Hilbert spaces, that every other decomposition of X as direct sum of two orthogonal subspaces satisfying the above conditions is identical to the one presented above.

I see. Thanks!