Let $E$, $F$ be vector bundles with metric over a smooth (not necessarily compact) manifold $X$.
Let $L:C^\infty(E) \rightarrow C^\infty(F)$ be a differential operator.
Let $L^*:C^\infty(F) \rightarrow C^\infty(F)$ be the formal adjoint of $L$ with respect to the $L^2$-inner product on the $L^2$-sections of $E$ and $F$, i.e. $L^*$ satisfies:
$$
\langle L f,g \rangle_{L^2(F)} =
\langle f, L^*g \rangle_{L^2(E)}
$$
for all $f \in C^\infty(E)$, $f \in C^\infty(F)$ with compact support.
Then $L^*$ is again a differential operator and can therefore be applied to all smooth sections of $F$.
Question:
Are all eigenvalues of $L^*L$ acting on smooth sections non-negative?
Here, smooth sections need not be in $L^2$.
$L^* L$ acting on $L^2_2(E)$ has only non-negative eigenvalues, because of
$$
\langle L^* L f, f \rangle
=
\langle Lf, Lf \rangle
=
|L f|^2 \geq 0.
$$
But according to what I know, it could be possible that there exists some section $f \in C^\infty(E)$ that is not in $L^2_2(E)$, for example because its $L^2$-integral is infinite, but satisfies $Lf=-f$.
Context about my application:
I am interested in the case where $P$ is a principal bundle over an asymptotically conical manifold, $A$ is an asymptotically flat connection on $P$, and $L=\nabla_A$ on $\operatorname{Ad} P$.
I am also interested only in sections $f \in C^\infty(E)$ which decay at infinity, but they may not decay fast enough to be in $L^2(E)$.
Best Answer
The answer to the question is: No.
Example: take $X=\mathbb{R}$ and $E$ and $F$ to be the trivial bundle over $\mathbb{R}$. Let $L=\frac{d}{dt}$. Then $L^*=-\frac{d}{dt}$ and $L^*L=-\frac{d^2}{dt^2}$, i.e. the operator $L^*L$ is the Laplacian in one dimension.
Let $f(t)=e^{t}$, then $L^*L (f)=-f$, i.e. $f$ is an eigenfunction of $L^*L$ for the negative eigenvalue $-1$.
(The answer to the refined question from "Context about my application" in the original post may still be positive, I don't know that.)