Suppose that $T$ is abounded self adjoint operator on $L^2(\mathbb R)$ and $K$ be the integral kernel of $T$. Define a function $T_K(x):=K(x,x), x \in \mathbb R.$
My question: Can the following commutator make sense:
$$[-\Delta+ (w\ast T_K), K]$$?
Here, $w:\mathbb R\to \mathbb R$ nice function, and $[X, Y]=XY-YX$ is commutator and $-\Delta=-\frac{d^2}{dx^2}$ laplacian, $*$ denote the convolution.
So, the usual convention is that functions are identified with multiplication operators. In this case for example the function $V = w * T_K$ acts as an operator on a function $\psi$ sufficiently regular as $(V \psi)(x) = V(x)\,\psi(x)$. Hence the action of the commutator on a function $\psi$ will be $$ [-\Delta + V,K]\psi(x) = ((-\Delta + V)\,K - K\,(-\Delta + V))\psi(x) \\ = -\Delta(K\psi)(x) + V(x)\, (K\psi)(x) + K(\Delta\psi)(x) + V(x)\,\psi(x). $$
Now when are all these objects well-defined? One possibility is to take $\psi\in H^2(\Bbb R)$ so that $\Delta\psi\in L^2$. If the function $w$ is nice and $K$ is a trace class operator, then the function $V$ should formally be $$ V(x) = \int_{\Bbb R} w(x-y) \,T(y,y)\,\mathrm d y = \mathrm{Tr}(w_x\,T) $$ where $w_x$ is the operator of multiplication by $y\mapsto w(x-y)$. In particular, if $w$ is a bounded function, then $V(x) \leq \|w\|_{L^\infty}\,\|T\|_1$ where $\|T\|_1$ denotes the trace norm of $T$.