Let $H$ be a Hilbert space, and let $A:U \rightarrow L(H)$ be a smooth$^{1}$ family of unbounded closed operators with common dense domain. Suppose furthermore that $A_{t}$ is normal for each $t \in U$. Finally, suppose that for each $t \in U$ the operator $A_{t}$ has discrete real spectrum, that never contains $0$.
Define $P_{t}:H \rightarrow H$ to be the projection onto the positive eigenspace $\text{Eig}_{>0}(A)$.
Is the family $P: U \rightarrow B(H)$ smooth?
It is well known that there are examples of smooth families of operators (even in finite dimensions) such that there does not exist a smooth diagonalization. This has to do with the crossing of eigenvalues, for example if the family $A$ has only multiplicity free eigenvalues, then it is smoothly diagonalizable. This tells us that it might be important to demand that $0$ is never contained in the spectrum.
Here is a stab at a proof, but I am not familiar enough with unbounded operators to tell if it holds any water.
Formally one may verify the identity \begin{equation} P = \frac{1}{2}A^{-1} \left( \sqrt{A^{2}} + A \right). \end{equation} Here $A^{-1}$ should be read as a left inverse of $A$, which should exist since $\ker(A)$ is empty ($0$ is never an eigenvalue). Furthermore, since $A$ only has real eigenvalues, $A^{2}$ is positive and has a unique positive square root. Assuming everything exists, the equation holds, since $\sqrt{A^{2}} + A = A$ on $\text{Eig}_{>0}(A)$ and $\sqrt{A^{2}} + A = 0$ on $\text{Eig}_{<0}(A)$. Now assuming that the mappings $A \mapsto A^{-1}$, $A \mapsto A^{2}$ and $A \mapsto \sqrt{A}$ are smooth, then $P:t \mapsto P_{t}$ is smooth.
(1) Smooth means that the mapping $t \mapsto \langle A(t)v, u \rangle$ is smooth for each $v \in \text{Dom}(A)$ and each $u \in H$.