Let $A,B$ on $L^2(\mathbb R^d,\mathbb C^d)$ such that $0\leq A\leq B$ AND $A,B$ are trace-class. Then their densities are given by $$\rho_A(x):=\sum_jA\varphi_j(x)\overline{\varphi_j(x)},\qquad \rho_A(x):=\sum_jB\varphi_j(x)\overline{\varphi_j(x)},$$where $(\varphi_j)$ is an orthonormal basis on $L^2(\mathbb R^d,\mathbb C^d)$. Equivalently, $\rho_A(x)=K_A(x,x)$ where $K_A=K_A(x,y)$ is the integral kernel of $A$ (and similarly for $B$). I am trying to prove the relation $0\leq \rho_A\leq \rho_B$.
Not sure where to start with this one.
Attempted sketch of proof, please check carefully.
Consider $A, B$ trace-class. They can be represented as an integral Hilbert-Schmidt operator $$ A f : x \mapsto \int K_A(x,y) f(y) dy \qquad , \qquad B f : x \mapsto \int K_B(x,y) f(y) dy $$ with $K_A$ and $K_B$ unique up to a.s. equality. For every measurable set $C\subset \mathbb{R}^d$, writing $[C]$ for the indicator function of $C$, using $\cdot$ for the pointwise multiplication of function, $$ A_C : f \mapsto [C] \cdot A (f \cdot [C]) = \left( x \mapsto \int [C](x) K_A(x,y) [C](y) f(y) dy \right) $$ is trace-class with kernel $(x,y) \mapsto [C](x) [C](y) K_A(x,y)$. Note that for every $f\in L^2(C, \mathbb C^d)$, $A_C f = [C] \cdot A f$.
We check that $\mathrm{Tr}(A_C) \leq \mathrm{Tr}(A)$: to see this, define an orthonormal basis $(e_i)_{i\in I}$ of $L^2(C, \mathbb C^d)$ and complete it with $(f_j)_{j\in J}$ to an orthonormal basis of $L^2(\mathbb R^d, \mathbb C^d)$, then write $$ \mathrm{Tr}(A_C) = \sum_{i\in I} \langle e_i, A_C e_i \rangle = \sum_{i\in I} \langle e_i, A e_i \rangle \leq \sum_{i\in I} \langle e_i, A e_i \rangle + \sum_{j\in J} \langle f_j, A f_j \rangle = \mathrm{Tr}(A) . $$ Similarly with $B$, and using that $0\leq \langle e_i, A e_i\rangle \leq \langle e_i, B e_i \rangle$, we get $0 \leq \mathrm{Tr}(A_C) \leq\mathrm{Tr}(B_C)$. We can also express the trace from the kernels: $$ 0 \leq \mathrm{Tr}(A_C) = \int [C](x) K_A(x,x) [C](x) dx = \int_C K_A(x,x) dx \leq \int_C K_B(x,x) dx . $$ This holds for every $C$ measurable. Take $C = \{ x : K_A(x,x) > K_B(x,x) \}$: necessarily $C$ has Lebesgue measure zero, meaning $K_A(x,x) \leq K_B(x,x) $ for Lebesgue-almost all $x\in \mathbb R^d$.