While analyzing an operator corresponding to a term in a PDE, I ran into a curiosity that I'm not totally sure about. Let $\Omega \subset \mathbb{R}$ be a bounded subset and the operator $A : L^2(\Omega) \to L^2(\Omega)$ be defined by $f \mapsto xf$ for any $x \in \Omega$. This would be the case if we have the term $xf(x)$ appear in the PDE for all $x \in \Omega$. How would you describe the kernel of $A$, and does the coefficient $x$ matter? Is this operator naively defined in that we would need to fix the coefficient $x \in \Omega$ first, in which case we would have an uncountable amount of these operators?
My attempt is to suppose $f \in \text{dom}(A)$ is such that $Af = xf = 0$. Then this holds when $f = 0$, but also when $x=0$. This originally made me think that the kernel of $A$ is something like $\{f=0 ~ \text{a.e.}\} \cap \{x=0\}$. However, since only $f \in L^2(\Omega)$ is being mapped by $A$, I suppose the kernel of $A$ is just $\{f=0 ~ \text{a.e.}\}$.
$x$ is not a constant here, it is meant to denote the function $x\mapsto x$, and $xf$ is the function $x\mapsto xf(x)$. You're mapping functions to functions (technically, equivalence classes of functions to equivalence classes of functions). The question you need to answer is: Which functions are mapped to the zero function $x\mapsto 0$?
So the question becomes: for which functions $f:\Omega\to\mathbb R$ is the function $xf$ equivalent to the zero function (equivalence meaning almost everywhere equality). That is, which functions satisfy $x f(x)=0$ for almost all $x\in\Omega$? You will find that they need to satisfy $f(x)=0$ for almost all $x\in \Omega$. Which makes $f$ equivalent to the zero function. And since we only care about equivalence classes, the kernel is $\{0\}$ ($0$ denoting the equivalence class of the zero function).