Decomposition of spectrum of a (unbounded, non-self-adjoint) linear operator in two spatial dimensions

Question

Assume $L$ is unbounded, non-self adjoint operator for functions over two space dimensions $(x,y)\in \mathbb{R}^2$, such that upon fourier transforming w.r.t $y$, one can reduce the operator to (for fixed $y$ frequency $\omega$):

$\tilde{L}_{\omega}:=\partial_{xx}+f(x)\partial_x +(i\omega x)$.

Are there any issues in claiming that $spec(L)=\cup_{\omega\in\mathbb{R}} spec(\tilde{L}_{\omega})$ ?

It seems obvious that if the underlying domain is bounded in $y$ and we work with $y-$periodic functions, then the set of frequencies $\omega$ is just the integers, and above sum is over countable set of frequencies. Hence, I am mostly concerned when we are working with unbounded domain.

Answer 1

What you need are some estimates on the norm of the resolvent, $N(\omega,\lambda) = \| (\tilde{L}_\omega - \lambda)^{-1} \|$, as a function of $\omega$. In general, if $N(\omega,\lambda)$ grows too quickly for large $\omega$ for a fixed $\lambda$, $\lambda$ may fail to be in the spectrum of $L$ even if $\lambda$ is not in the spectrum of $\tilde{L}_\omega$ for any $\omega$, since $(\tilde{L}_\omega - \lambda)^{-1} v(\omega)$ may then fail to be in $L^2(\mathbb{R}^2, dx d\omega)$ even if $v(w) \in L^2(\mathbb{R}^2, dx d\omega)$. Analogously, even if a fixed $\lambda$ is in the spectrum of $\tilde{L}_\omega$ for some value of $\omega$, say $\omega = 0$, the resolvent $(\tilde{L}_\omega - \lambda)^{-1}$ may still end up being bounded on $L^2(\mathbb{R}^2,dx d\omega)$ if $N(\omega,\lambda)$ grows too slowly, e.g., when $N(\omega,\lambda) \le C \log|\omega|$. These are the two pathologies that you need to avoid for your conjecture about the spectrum to hold.