Convergence in trace

Question

I am having difficulties with this problem.
Given a non-negative compact self-adjoint operator $\gamma$ i.e. $\langle \gamma u, u\rangle \geq 0$ for all $u$. Denote the cut-off function $\chi \in C^\infty_c\left(\mathbb{R}^d\right)$ satisfying $\chi$ is non-increasing, $\chi\left(\left|x\right|\right) = 1$ if $\left|x\right| \leq 1$ and $\chi\left(\left|x\right|\right) = 0$ if $\left|x\right| \geq 2$. Define $\chi_R = \chi\left(\frac{x}{R}\right)$ for all $R > 0$. Can we prove that $\text{Trace}\left(\chi_R \gamma \chi_R \right) \to \text{Trace}\left(\gamma\right)$ when $R \to \infty$?
I tried to look for the relation between eigenvalues and eigenfunctions of $\chi_R \gamma \chi_R$ and $\gamma$ but I failed. Thank you very much for your help.

Answer 1

Two basic facts:

1. An operator $T$ on $L^2(\mathbb R^d)$ is Hilbert-Schmidt if and only if there exists $k\in L^2(\mathbb R^d\times\mathbb R^d)$ such that $Tf(x)=\int k(x,y)f(y)\,dy$ for all $f\in L^2(\mathbb R^d)$, and in this case, $\mathrm{tr}(T^\ast T)=\lVert k\rVert_2^2$. The adjoint of $T$ is given by $T^\ast f(x)=\int \overline{k(y,x)}f(y)\,dy$. It follows that $T$ is Hilbert-Schmidt if and only if $T^\ast$ is Hilbert-Schmidt and $\mathrm{tr}(T^\ast T)=\mathrm{tr}(TT^\ast)$.
2. If $(T_n)$ is a sequence of non-negative bounded operators and $T_n\nearrow T$ in strong operator topology, then $\mathrm{tr}(T_n)\nearrow \mathrm{tr}(T)$. This is true since $\langle e_i,T_n e_i\rangle\nearrow \langle e_i,T e_i\rangle$ for every member $e_i$ of an ONB and the convergence of the traces follows from the monotone convergence theorem.

As $$ \langle f,\chi_R^2 f\rangle=\int\chi_R(x)^2\lvert f(x)\rvert^2\,dx\nearrow \int\lVert f(x)\rvert^2\,dx=\langle f,If\rangle $$ for all $f\in L^2(\mathbb R^d)$, we can use the facts 1 and 2 above to get $$ \mathrm{tr}(\chi_R \gamma\chi_R)=\mathrm{tr}(\gamma^{1/2}\chi_R)^\ast(\gamma^{1/2}\chi_R))=\mathrm{tr}(\gamma^{1/2}\chi_R^2\gamma^{1/2})\nearrow \mathrm{tr}(\gamma^{1/2}I\gamma^{1/2})=\mathrm{tr}(\gamma). $$ Remark: As noted in the comments, I assume that $(\chi_R(x))$ is increasing in $R$.