TITLE: Application of Schwartz Kernel Theorem to Quantum Mechanics
QUESTION [8 upvotes]: I am currently reading Quantum Mechanics for Mathematicians and have a question about a statement made in the book:
Remark. By the Schwartz kernel theorem, the operator B can be represented by an integral operator with distributional kernel $K(q,q')$. Then the commutativity $BQ = QB$ implies that, in the distributional sense, 
$$
(q-q')K(q,q') = 0,
$$
so that $K$ is "proportional" to the Dirac delta-function, i .e., 
$$
K(q,q') = f(q)\delta(q-q'). 
$$
This argument is usually given in physics textbooks.
$B$ and $Q$ are both operators in the coordinate representation. $Q$ is the position operator and $B$ is bounded. 

So, first of all, I can't find much about the kernel theorem online. I've been using this as a reference. According to this document, there is a relation between bilinear forms and distributions. However, I don't know how to view $B$ as a bilinear form and thus apply the theorem. Can someone elucidate how the kernel theorem is applicable to $B$? Secondly, what arguments/notation in physics is the author (of the QM book) referencing?

REPLY [3 votes]: If $T$ is a distribution such that $$\forall g \in C^\infty_c, \langle T, x g  \rangle = 0$$ then pick $\phi \in C^\infty_c$ constant $=1$ on $[-1,1]$, let $$g = \frac1x (f-f(0)\phi)\quad\in C^\infty_c$$ then $$ \langle T,f \rangle = \langle T,f(0)\phi \rangle + \langle T,xg\rangle =f(0) \langle T, \phi \rangle$$