Jonathan
Reputation
1,614
Next privilege 2,000 Rep.
 Mar15 awarded Good Answer Sep24 awarded Autobiographer Sep1 awarded Enlightened Sep1 awarded Nice Answer Jun24 awarded Yearling Oct14 answered Reference for Gauss-Manin connection Jun24 awarded Yearling Dec15 awarded Revival Aug27 awarded Enlightened Aug27 awarded Nice Answer Aug24 comment Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? If $K$ is nice, then for each fixed $z$, $K(z,x)$ is a well-behaved function of $x$, hence by the 3rd last equation can express $K(z,x)$ as the sum of $n$ of the integral of $K(z,y) e_n(y)$ with respect to $y$. Now plug in $z = x$ to obtain the result. Aug23 revised Help solving differential equation added 222 characters in body Aug23 answered Help solving differential equation Aug23 revised Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? added 356 characters in body Aug23 comment Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? Yes, I am well aware and absolutely agree. I answered as above, assuming $K(x,y)$ to be as nice as necessary, since it didn't seem that Mike was asking for the weakest possible hypotheses under which this holds, but rather whether it can be made rigorous at all. Aug22 answered Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels? Aug21 answered relation between integral and summation Aug21 answered Method of characteristics with constant PDE Aug13 answered Definition of the rank of the isometry group of a manifold. Aug10 comment Are there n-th roots of differential operators? This technique depends sensitively on both the given differential operator, as well as the space of functions on which it acts. In the case of the Dirac operator, it is not the Laplacian on functions that admits an algebraic square root, but rather the Laplacian on spinors. This also works for vector-valued functions, since we can decompose vectors into products of spinors. These kinds of $n$th roots are completely different than what I described in my answer, but both can be useful depending on the situation.