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Given an appropriate function $K: \mathbb{R}^2 \to \mathbb{C}$, say continuous of compact support, we obtain a compact operator $T$ on the Hilbert space $L^2(\mathbb{R})$ by the formula $$ (T h)(t) = \int K(s,t) h(t-s) \ ds.$$

Suppose $T$ is trace class and we want its trace. The standard formula is $$ \mathrm{trace}(T) = \sum \langle e_r \vert T e_r \rangle $$ where the sum is taken over an arbitrary orthonormal basis $(e_r)$ for $L^2(\mathbb{R})$ and inner products are conjugate-linear in the 1st slot. Now, correct me if I'm wrong, but I don't think there is any particularly descript basis for $L^2(\mathbb{R})$. However, $\mathbb{R}$ does have nice characters $\epsilon_s(t) = e^{its}$, $s \in \mathbb{R}$. Despite the fact these are not square-integrable, we might attempt to plow on formally, replacing summation by integration, as follows: \begin{align*} \mathrm{trace}(T) &= \int \langle \epsilon_r \vert T \epsilon_r \rangle \ dr \\ &= \int \int \overline{\epsilon_r(t)} (T \epsilon_r)(t) \ dt \ dr \\ &= \int \int e^{-irt} \int K(s,t) \epsilon_r(t-s) \ ds \ dt \ dr \\ &= \int \int e^{-irt} \int K(s,t) e^{irt} e^{-irs} \ ds \ dt \ dr \\ &= \int \int e^{-irs} \int K(s,t) \ dt \ ds \ dr \\ &= \int \int e^{-irs} k(s) \ ds \ dr \\ &= \int \hat k(r) \ dr \\ \end{align*} where we have defined $k$ by $k(s) = \int K(s,t) \ dt$ and written $\hat k$ for the Fourier transform of $k$. My question is:

Does this actually work? That is, does the compact operator $T$ on $L^2(\mathbb{R})$ given by the integral kernel $K$ have trace equal to $\int \hat k$ where $k(s) = \int K(s,\cdot)$?

I'm pretty sure this should be true. For example, I can write down the $K$ which makes $T$ the rank-1 projection onto some compactly supported unit vector $h \in L^2(\mathbb{R})$. Specifically, putting $K(s,t) = \overline{h(t-s)} h(t)$ does the job. In this case, it turns out $k = h^* * h$ where $h^*(t) = \overline{ h(-t)}$ and $*$ is the convolution product. So, $\hat k = \widehat{ h^* * h} = \overline h h = |h|^2$ and $\int \hat k = \|h\|_2^2 = 1$ which equals the trace of $h$ so the formula holds in this case.

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    I don't think convolution on the line against a compactly-supported function can _ever_ be a compact operator, because on the Fourier transform side it's a multiplication operator against the Fourier transform of the compactly-supported function, which is a holomorphic function. In particular, unless identically 0, it is nowhere locally constant, so has no discrete spectrum whatsoever.2012-08-22
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    @paul garrett: I'm afraid I don't understand your comment. Let me give a reference for compactness of $T$ - first adjusting notation. If my $K$ is compactly supported and I define $H(x,y) = K(y-x,y)$ then $H$ is also compactly supported, hence $H \in L^2(\mathbb{R}^2$). The defining formula for $T$ becomes $(Th)(y) = \int K(x,y)h(y-x) \ dx = \int H(y,x) h(x)$. Theorem VI.23 on pp. 211 of Reed and Simon's *Functional Analysis* gives a proof that such operators are Hilbert-Schmidt operators, hence compact.2012-08-23
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    I may have messed up that variable change, but you get the idea...2012-08-23
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    Sorry, @Mike, I misunderstood the intention: so it's not meant to be a convolution operator, really? Rather, really an integral kernel with $L^2$ kernel in two variables, so perhaps written more helpfully as $\int_{\mathbb R} K(x,y)\,f(y)\,dy$? (My earlier comment was logically correct, but irrelevant to the case at hand.)2012-08-23
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    @paul: Right it's not a convolution. I mean, given an $f$ one could use $(x,y) \mapsto f(x-y)$ or similar to get a kernel which does the convolution, but then the kernel is not compactly supported. I admit my convention for defining $T$ may be a little odd. I chose it to make the formula for compositions work out in a particular way. You are of course free to use another convention if you'd rather.2012-08-23
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    About conventions, @Mike: apart from _my_ own hasty mis-reading, for which I accept blame... :) ... other hasty people may have a similar reflex (mistaken though it is), if you're writing for public consumption. Not pretending to make a virtue of my failing, but to note that many others may share the same. :)2012-08-23

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