Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $U\in HS$, Hilbert--Schmidt operators on $V = L^2(\mathbb{R}^n)$. There is a natural isomorphism between HS operators and elements in $W = L^2(\mathbb{R}^n\times\mathbb{R}^n)$, in particular, for every $U\in HS$ there is a $u\in W$ such that $$ [U\varphi](x) = \int_{\mathbb{R}^n} u(x,y)\varphi(y) dy. $$

Suppose now that $U$ is also trace-class. Informally (i.e., in physics), $$ \operatorname{Tr} U = \int_{\mathbb{R}^3} u(x,x)dx. $$

Question: To what extent is the latter statement rigorous? What confuses me is that $u$ is defined only up to a set of measure zero and the integral is taken over such a set. Is there some notion of trace operator that can be used here, which gives the "correct diagonal"? Taking the limit of successively narrow "strips" along the diagonal?

I can "prove" it as follows: Let $\{f_\mu\}$ be an orthonormal basis for $V$, say Hermite functions, such that $\{f_\mu\otimes f_\nu\}$ is an orthonormal basis for $W$, and we have an expansion $$ u = \sum_{\mu\nu} u_{\mu\nu} f_\mu\otimes f_\nu.$$ Now we have $$ \int u(x,x) dx = \sum_{\mu\nu} u_{\mu\nu} \int f_\mu(x) f_{\nu}(x) = \sum_{\mu\nu} u_{\mu\nu} \langle{f_\mu,f_\nu}\rangle = \sum_\mu u_{\mu\mu} = \operatorname{Tr} U. $$

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.