Suppose $F(\lambda)~(\lambda\in\mathbb{C})$ is a linear ordinary differential operator (with, say, domain $D$ dense in some Hilbert space), and is also affine-linear in $\lambda$.
Is there a proof that if $u\in D$ satisfies $F(\lambda)u(\cdot,\lambda)=0$ on dom$(u)$ then for each fixed $x\in$ dom$(u)$ we have $u(x,\cdot)$ is entire (i.e., holomorphic in $\mathbb{C}$)?
[To be clear, I have in mind the solution of a differential equation of the form \begin{equation} u^{(n)}+f_1u^{(n-1)}+\ldots+f_nu=\lambda(u^{(m)}+g_1u^{(m-1)}+\ldots+g_mu), \end{equation} where $n>m\geq0$ (an operator pencil), with suitable domain for $u$, and suitably chosen $f_i,g_i$.]
I am specifically interested in the case of $F(\lambda)$ being, in general, not self-adjoint. Any help on this would be hugely appreciated!
Standard theorems on dependence of parameters proves this if we in addition assume, for example, that $\lambda$-independent initial data for $u$ are given in some $\lambda$-independent point. Without some such extra assumption the statement is false, which is easily seen by instead letting initial data be non-analytic in $\lambda$. If $u$ for each $\lambda$ should be in a given Hilbert space the answer is sometimes yes, sometimes no, depending on the properties of the differential operator with respect to the Hilbert space.