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I am currently reading the book Spectral Methods in Automorphic Forms, and Iwaniec defines the trace operator in a different way than I am accustomed to. Throughout, assume that everything converges spectacularly - that's not important here.

In particular, if $K: F \times F \longrightarrow \mathbb{C}$ is a $C_0^\infty$ (that is, smooth and bounded) function and $L$ is the integral operator having $K$ as its kernel, i.e. $$ (Lf)(z) = \int_F K(z,w)f(w) d w,$$ then Iwaniec defines the trace of $L$ as the integral across the diagonal, $$ \text{Tr} L = \int_F K(z,z)dz. \tag{1}$$

I'm familiar with the trace of a more generic (linear operator $A$ over a Hilbert space by

$$ \text{Tr} A = \sum_j \langle Ae_j, e_j \rangle,\tag{2}$$

where the $e_j$ form an orthonormal basis of functions. Do these definitions agree? If we suppose in addition that the $e_j$ are eigenfunctions with eigenvalues $\lambda_j$, then I can see the equivalence in the following "wrong" way. Taking the spectral decomposition for $K(z,w)$, $$K(z,w) = \sum_j \lambda_j e_j(z) \overline{e_j(w)},$$ then since the $e_j$ are orthonormal, we have that $$\int_F K(z,z)dz = \sum_j \lambda_j \int_F e_j(z)\overline{e_j(z)}dz = \sum_j \lambda_j.$$ And Lidskii's Theorem says that $$\text{Tr} A = \sum_j \lambda_j,$$ where $\text{Tr} A$ is as in $(2)$. So I can conclude that $(1)$ and $(2)$ should agree, but I would like to see in a more fundamental, less roundabout way that they do actually agree.

davidlowryduda
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  • Question: What if $K$ is defined on the unit square $[0,1]\times[0,1]$ and the integral is respect to Lebesgue measure? If you set $K(x,x)=0$ for $0 \le x \le 1$, then how have you changed the integral operator $Lf=\int_{0}^{1}K(x,y)f(y),dy$ on $L^{2}[0,1]$? And how have you changed $\int_{0}^{1}K(x,x),dx$? – Disintegrating By Parts Oct 21 '14 at 20:49
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    @T.A.E.: That is a great question. It feels like we can define $K$ however we like on a set of measure zero. I can add that to my list of unanswered concerns over this definition of trace. – davidlowryduda Oct 21 '14 at 20:59
  • That's where you need some smoothness in order to make sense of the result. – Disintegrating By Parts Oct 21 '14 at 21:50
  • @T.A.E. Let us suppose that $K$ is smooth and bounded, very reasonable assumptions I think. – davidlowryduda Oct 21 '14 at 21:57
  • Then you can expand in a double Fourier series of smooth functions and use your "wrong" solution. :) However, you need a basis of eigenfunctions to do what you want, and those will inherit smoothness. Maybe some symmetry on $K$ is needed, too? – Disintegrating By Parts Oct 21 '14 at 22:03
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    I asked a similar question some time ago that might be relevant to this question (see here). – Cm7F7Bb Apr 23 '18 at 09:25
  • I don't know what you mean by a more fundamental way: you have a spectral and a geometric expansion of the kernel function, and integrating them in two different ways gives you the equality of these definitions of traces, which in finite dimensions boils down to the trace of a matrix is the sum of the diagonal entries and also the sum of the eigenvalues. – Kimball Apr 27 '18 at 19:08
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    I think what you want is Mercer's theorem; it is mentioned in Kowalski, Spectral theory in Hilbert spaces, https://people.math.ethz.ch/~kowalski/spectral-theory.pdf – Bart Michels Jun 15 '18 at 17:48
  • @barto That seems like a very promising lead. Thank you. I will look into this. – davidlowryduda Jun 16 '18 at 21:19

2 Answers2

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We have the following:

Theorem. (Mercer) Let $X$ be a locally compact sequential topological space, $\mu$ a strictly positive finite measure on $X$ and $k : X \times X \to \mathbb C$ continuous, bounded and with $k(y,x) = \overline{k(x, y)}$. Then the associated bounded convolution operator $K : L^2(X) \to L^2(X)$ is self-adjoint and compact. Suppose that it is nonnegative. Let $(e_n)_{n \geq 1}$ be a $L^2$-orthonormal basis of eigenfunctions for $K$ with eigenvalues $(\lambda_n)_{n \geq 1}$, so that $e_i \in C_b^0(X, \mathbb C)$ when $\lambda_i > 0$. Then $$k(x, y) = \sum_{n}\lambda_n e_n(x) \overline{e_n(y)}$$ uniformly on sets of the form $L \times X$ and $X \times L$ with $L$ compact, and absolute for fixed $(x,y)$.

Mercer proved this for $X = [0,1]$: Functions of positive and negative type, and their connection the theory of integral equations, Phil. Trans. Roy. Soc. London (A) 209 (1909) 415–446. http://rsta.royalsocietypublishing.org/content/209/441-458/415 The proof is also in Werner, Funktionalanalysis, 8th edition, 2018, Satz VI.4.2. The proof in the general case is the same.

In particular:

Corollary. Suppose in addition that $X$ is compact. Then $K$ is trace class and $\DeclareMathOperator{\Tr}{Tr}$ $$\Tr K = \int_X k(x, x) d\mu(x)$$

Proof. By integrating the above equality over the diagonal: $$\begin{align*} \infty > \int_{X} k(x, x) d \mu(x) &= \int_X \sum_{n}\lambda_n e_n(x) \overline{e_n(x)} d\mu(x) \\ &= \sum_{n} \lambda_n \int_X e_n(x) \overline{e_n(x)} d\mu(x) \\ &= \sum_n \lambda_n \\ &= \Tr K \end{align*}$$ because the convergence of the series is uniform, hence $L^1$. $\square$

If $X$ is not compact, the corollary need not hold, c.f. Selberg's trace formula.

Bart Michels
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Let $\{e_j\}$ be an orthonormal basis. Then $$ \int_F \Big(\sum_je_j(x)e(y)\Big)f(y)dy=\sum_je_j(x)\int_Ff(y)e_j(y) dy=f(x), $$ which implies that $$ \sum_je_j(x)e(y)=\delta(x-y). $$ Like you said, we are assuming that everything converges spectacularly. Now we compute the trace $$\begin{split} \mathrm{Tr}\,L&=\sum_j\langle Le_j,e_j\rangle=\sum_j\int_F\Big(\int_F K(x,y)e_j(y)dy\Big)e_j(x)dx\\ &=\int_F\int_F K(x,y)\Big(\sum_je_j(x)e_j(y)\Big)dydx\\ &=\int_F\int_F K(x,y)\delta(x-y)dydx=\int_FK(x,x)dx. \end{split}$$

timur
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    The sum $\sum_je_j(x)e(y)$ worries me, because it doesn't converge. We probably want to work with its partial sums, but that may need some condition on $K$, since for general $K\in L^2$ the integral on the diagonal is ill-defined. – Bart Michels Apr 30 '18 at 05:13
  • @barto: Nothing to worry, because the question assumed everything is fine. – timur Apr 30 '18 at 05:23