For $q:\mathbb{R}\rightarrow\mathbb{R}$ let $\delta$ be a differential operator of the form
$$
\delta=\frac{d}{dx}+q(x),
$$
defined on some dense subspace of $L^2((0,1),d\mu)$, where $\mu$ is a finite measure. Denote by $\delta^*$ a formal adjoint of $\delta$ with respect to $L^2((0,1),d\mu)$. In other words $\delta^*$ satisfies the relation
$$
\langle \delta \phi, \psi\rangle_{d\mu}=\langle \phi, \delta^* \psi\rangle_{d\mu},
$$
for $\phi,\psi\in C_c^\infty(0,1)$, where $\langle ,\rangle_{d\mu}$ stands for a scalar product on $L^2((0,1),d\mu)$.
Moreover, assume that $f\in L^2((0,1),d\mu)$ is an eigenfunction of the second order differential operator $\delta^* \delta$ with the eigenvalue $\lambda$, i.e.
$$
\delta^* \delta f=\lambda f.
$$
$\textbf{Question:}$ Is that true that necessarily $\lambda\ge 0$?
If $\delta$ was a bounded operator $T$ the question would be trivial, since then a formal adjoint would be a usual adjoint and one would have
$$
\lambda\langle f, f\rangle_{d\mu}=\langle\delta^* \delta f, f\rangle_{d\mu}=\langle \delta f, \delta f\rangle_{d\mu}\ge 0,
$$
and consequently $\lambda\ge 0.$
I don't know well the theory of unbounded operators, so I don't know whether one can transfer the above simple argument to that setting. Any hints will be highly appreciated!
Best Answer
I think there are some parts of this question that need further specification, for example, what are the assumptions on $q$. Let me, however, give my best attempt at answering anyway:
For $f\in C^\infty_c(0,1)$ you clearly have positivity of the eigenvalues by the definition of the formal adjoint. On the other hand restricting $\delta^*\delta$ to $C^\infty_c(0,1)$ makes it a positive symmetric operator. Such an operator might have symmetric extensions. One way to check whether this is the case is to consider the deficiency indices. See for example https://en.wikipedia.org/wiki/Extensions_of_symmetric_operators. Such extensions then might not be positive.
On the other hand, assuming e.g. that $q$ is $C^1(0,1)$ and in the special case in which $\mu$ is the Lebesgue measure, we might show that the quadratic form $Q(v)=\langle \delta^*\delta v,v\rangle+\langle v,v\rangle$ is closable, meaning that if $v_n\to 0$ and $Q(v_n-v_m)\to0$ as $n,m\to\infty$ we have $Q(v_n)\to 0$. In that case, there exist a unique self-adjoint operator, $A$, such that $\overline{Q}(v)=\langle Av,v\rangle$ on $\text{dom}(A)$ called the Friedrich extension of $\delta^*\delta+I$ (here $\overline{Q}$ is the closure of $Q$). It then follows from the Min-Max principle that $A-I$, which is a self-adjoint extension of $\delta^*\delta$ only has positive eigenvalues.