Schur test and it's relation to representation theory I was told by analyst who doesn't know about such things that Schur's test relating to boundedness of integral operators is somehow a version of Schur's lemma on irreducible representations, the group being $\mathbb R$. The two statements look very different to me (although there seem to be many things named after Schur and I have little experience in rep theory). Does anyone know the details of this connection?
 A: It is true that there are many things named after I. Schur, and they are not necessarily otherwise related! :)
I can't quite guess what your colleague had in mind, but it is true that in many fortuitous circumstances relevant operators are given as integral operators, and if the two-variable "kernel" $K(x,y)$ of such an operator is in $L^2$ of the product space $X\times X$, then it is not only "bounded" (really, continuous, which is more explanatory), but is a "Hilbert-Schmidt" operator, which is an easily-testable case of "compact operator". Self-adjoint operators are our best friends, because their spectral theory is as close to finite-dimensional spectral theory of self-adjoint operators as one could reasonable hope: orthogonal basis of eigenvectors. 
Thus, compact topological groups (not $\mathbb R$...) turn out to have the nice property that all Hilbert-space repns decompose as direct sums of irreducibles.
Already the real line $G=\mathbb R$ does not behave this well: $L^2(G)$ does not decompose as a direct sum of irreducible repns of $G$. Rather, Fourier inversion shows that it is a "direct integral" of irreducibles, each generated by the (not $L^2$!!!) functions $x\to e^{i\xi x}$.
