# Formula for trace of compact operators on $L^2(\mathbb{R})$ given by integral kernels?

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. – paul garrett Aug 22 '12 at 23:41
@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. – Mike F Aug 23 '12 at 21:14
I may have messed up that variable change, but you get the idea... – Mike F Aug 23 '12 at 21:21
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.) – paul garrett Aug 23 '12 at 22:07
@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. – Mike F Aug 23 '12 at 22:17

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. – paul garrett Aug 23 '12 at 0:27
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. – Jonathan Aug 23 '12 at 0:33
Verily... ok... and this question merits considerably greater clarification and precisification. – paul garrett Aug 23 '12 at 1:10