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.

  • $\begingroup$ 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 $\int_{0}^{1}K(x,x)\,dx$? $\endgroup$ Oct 21, 2014 at 20:49
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    $\begingroup$ @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. $\endgroup$
    – davidlowryduda
    Oct 21, 2014 at 20:59
  • $\begingroup$ That's where you need some smoothness in order to make sense of the result. $\endgroup$ Oct 21, 2014 at 21:50
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    $\begingroup$ I asked a similar question some time ago that might be relevant to this question (see here). $\endgroup$
    – Cm7F7Bb
    Apr 23, 2018 at 9:25
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    $\begingroup$ I think what you want is Mercer's theorem; it is mentioned in Kowalski, Spectral theory in Hilbert spaces, people.math.ethz.ch/~kowalski/spectral-theory.pdf $\endgroup$ Jun 15, 2018 at 17:48

2 Answers 2


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.


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 $$\begin{split} \mathrm{Tr}\,L&=\sum_j\langle Le_j,e_j\rangle=\sum_j\int_F\Big(\int_F K(x,y)e_j(y)dy\Big)e_j(x)dx\\ &=\int_F\int_F K(x,y)\Big(\sum_je_j(x)e_j(y)\Big)dydx\\ &=\int_F\int_F K(x,y)\delta(x-y)dydx=\int_FK(x,x)dx. \end{split}$$

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    $\begingroup$ 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. $\endgroup$ Apr 30, 2018 at 5:13
  • $\begingroup$ @barto: Nothing to worry, because the question assumed everything is fine. $\endgroup$
    – timur
    Apr 30, 2018 at 5:23

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