The trace of an operator

Question

My question is derived from A. Deitmar's book: A First Course in Harmonic Analysis (second edition), p22, Exercise 1.17. Let me rewrite it again:

Let $k:\mathbb{R}^2 \rightarrow \mathbb{C}$ be smooth (i.e., infinitely differentiable) and invariant under the natural action of $\mathbb{Z}^2$; i.e., $k(x+k,y+l) = k(x,y)$ for all $k,l \in \mathbb{Z}$ and $x,y \in \mathbb{R}$. For $\varphi \in C(\mathbb{R}/\mathbb{Z})$ set $$ K \varphi(x) = \int^1_0 k(x,y)\varphi(y)dy. $$ Show that $K$ satisfies $$ \| K \varphi \|^2_2 \leq \| \varphi \|^2_2 \int^1_0 \int^1_0 |k(x,y)|^2 dxdy. $$ Show that the sum $$ \text{tr}K=\sum_{k \in \mathbb{Z}} \langle K e_k, e_k \rangle $$ converges absolutely and that $$ \text{tr}K = \int^1_0 k(x,x)dx. $$

---

This following inequality $$ \| K \varphi \|^2_2 \leq \| \varphi \|^2_2 \int^1_0 \int^1_0 |k(x,y)|^2 dxdy $$ is easy to verify.

Using the method of integration by parts, I can prove the sum converges absolutely.

My question is: How to prove the following equation $$ \text{tr}K = \int^1_0 k(x,x)dx ~? $$

This problem has been bothering me for a long time. Thanks in advance.

Answer 1

Let $\{ e_{j} \}$ be an orthonormal basis of $L^{2}[0,1]$ consisting of real functions. Define $f_{i,j}(x,y)=e_{i}(x)e_{j}(y)$. Then $\{ f_{i,j} \}_{i,j}$ is a complete orthonormal basis of $L^{2}([0,1]\times[0,1])$; this can be verified by showing that $(f,f_{i,j})=0$ for all $i$, $j$ for a continuous $f(x,y)$ implies $f=0$. Notice that $$ \begin{align} (k,f_{i,j})_{L^{2}([0,1]\times[0,1])} & =\int_{[0,1]\times[0,1]}k(x,y)e_{i}(x)e_{j}(y)\,dx\,dy \\ & =\int_{0}^{1}\left(\int_{0}^{1}k(x,y)e_{j}(y)\,dy\right)e_{i}(x)\,dx \\ & = (Ke_{i},e_{j}) \end{align} $$ Therefore, $$ k(x,y) = \sum_{i,j}(k,f_{j,k})_{L^{2}([0,1]\times[0,1])}f_{j,k}(x,y) \\ k(x,x)=\sum_{j,k}(Ke_{j},e_{k})e_{j}(x)e_{k}(x). $$ Integrating gives $$ \int_{0}^{1} k(x,x)\,dx= \sum_{j}(Ke_{j},e_{k})(e_{j},e_{k})=\sum_{j}(Ke_{j},e_{j})= \mbox{tr}(K). $$

Answer 2

Can we accept as fact the following representation of the delta function? \begin{align} \delta(x-y) = \sum_{j\in \mathbb{Z}} e_{j}(x)\overline{e_{j}(y)} \end{align} If so, then \begin{align} \mathrm{tr}(K) &= \sum_{j\in \mathbb{Z}} \int_{0}^{1}Ke_{j}(x)\overline{e_{j}(x)} \, \mathrm{d}x \\ &= \sum_{j\in \mathbb{Z}} \int_{0}^{1}\int_{0}^{1}k(x,y)e_{j}(y)\overline{e_{j}(x)} \, \mathrm{d}y \, \mathrm{d}x \\ &= \int_{0}^{1}\int_{0}^{1}k(x,y)\sum_{j\in \mathbb{Z}} e_{j}(y)\overline{e_{j}(x)} \, \mathrm{d}y \, \mathrm{d}x \qquad \text{(due to absolute convergence)}\\ &= \int_{0}^{1}\int_{0}^{1}k(x,y)\delta(x-y) \, \mathrm{d}y \, \mathrm{d}x\\ &= \int_{0}^{1}k(x,x) \, \mathrm{d}x \end{align} If you can't accept the resolution of identity, you'll probably need to choose an explicit representation of each $e_{j}$ then use the Dirichlet kernel to do essentially what is done here, but with more pain.