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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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Disclaimer (in response to Paul Garret): The following assumes that the integral kernel $K(x,y)$ is nice enough that all integrals and manipulations below make sense. Determining sufficient hypotheses is a standard exercise (say at the level of Folland's Real Analysis), so I leave the details to you. I make no claims about necessary hypotheses.

Take the operator $T$ to be defined by $(Tf)(x) = \int K(x, y) f(y) dy$. We would like to compute the trace of $T$ (provided it exists) in terms of $K$. Let $e_n(x)$ be an orthonormal basis of $L^2(\mathbb{R})$.

\begin{align} \mathrm{tr}(T) &= \sum_n \langle e_n|T| e_n \rangle \\ &= \sum_n \int e_n(x) (T e_n)(x) dx \\ &= \sum_n \int \int e_n(x) K(x,y) e_n(y) dy dx \end{align}

Since the $e_n$ form a complete basis, we have a resolution of the identity $$ 1 = \sum_n | e_n \rangle \langle e_n | $$ Now, \begin{align} f(x) = (1 \cdot f)(x) &= \sum_n \langle e_n | f \rangle | e_n \rangle \\ &= \sum_n e_n(x) \int e_n(y) f(y) dy \end{align} Hence $$ \sum_n e_n(x) \int K(x, y) e_n(y) dy = K(x,x) $$ So we find $$ \mathrm{tr}(T) = \int K(x,x) dx $$

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    Yes, indeed, in circumstances that an operator given by a kernel is trace-class, its trace is given by that integral. These heuristics suggest the correct thing, truly, but are not legit without considerable hypotheses. Not hard to make fallacious arguments if one isn't careful.2012-08-23
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    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*.2012-08-23
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    Verily... ok... and this question merits considerably greater clarification and precisification.2012-08-23
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    Thanks for the answer, permit me to think about this for a little while...2012-08-23
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    Could you please explain to me where that 2nd from final equation comes from?2012-08-23
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    And, as commented above to the question, I misread the question itself as being a convolution operator... which here would not be compact... but with suitable kernel and $\int K(x,y)\,f(y)\,dy$, it certainly can be compact! Sorry for being confused. :)2012-08-23
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    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.2012-08-24
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    Ah of course, thank you. So I guess the appeal of your convention $(Tf)(x) = \int K(x, y) f(y) dy$ to get operators from integral kernels is that it resembles the convention for matrices and vectors viz. $(Ax)_i = \sum_j a_{ij} x_j$?2012-08-24
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    And we get $T_K \circ T_H = T_L$ where $L(x,y) = \int K(x,z) H(z,y) \ dz$ which resembles matrix multiplication, how lovely! And your trace formula carries the analogy one step further.2012-08-24